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t birth represents 28.6% of his ﬁnal height” and that, in prerevolutionary China, the average length of a Chinese boy at birth was 18.9 inches. From this, Seligman deduces that the mean height of mature Chinese men is .9 8 1 66.08 inches, or 5 feet 6.08 inches 8 6 .2 CASE STUDY ❍ 253 He then assumes that the distribution of the heights of men in China follows a normal distribution (“as it does in the U.S.”) with a mean of 66 inches and a standard deviation equal to 2.7 inches, “a ﬁgure that looks about right for that mean.” 1. Using Seligman’s assumptions, calculate the probability that a single adult Chinese man, chosen at random, will be less than or equal to 5 feet tall, or equivalently, 60 inches tall. 2. Do the results in part 1 agree with Seligman’s odds? 3. Comment on the validity of Seligman’s assumptions. Are there any basic ﬂaws in his reasoning? 4. Based on the results of parts 1 and 3, do you think that Deng Xiaoping took height into account in selecting his successor? Sampling Distributions GENERAL OBJECTIVES In the past several chapters, we studied populations and the parameters that describe them. These populations were either discrete or continuous, and we used probability as a tool for determining how likely certain sample outcomes might be. In this chapter, our focus changes as we begin to study samples and the statistics that describe them. These sample statistics are used to make inferences about the corresponding population parameters. This chapter involves sampling and sampling distributions, which describe the behavior of sample statistics in repeated sampling. CHAPTER INDEX ● The Central Limit Theorem (7.4) ● Random samples (7.2) ● The sampling distribution of the sample mean, x (7.5) ● The sampling distribution of the sample proportion, pˆ (7.6) ● Sampling plans and experimental designs (7.2) ● Statistical process control: x and p charts (7.7) ● Statistics and sampling distributions (7.3) How Do I Calculate Probabilities for the Sample Mean x ? How Do I Calculate Probabilities for the Sample Proportion ˆp? 7 © PictureNet/CORBIS Sampling the Roulette at Monte Carlo How would you like to try your hand at gambling without the risk of losing? You could do it by simulating the gambling process, making imaginary bets, and observing the results. This technique, called a Monte Carlo procedure, is the topic of the case study at the end of this chapter. 254 7.2 SAMPLING PLANS AND EXPERIMENTAL DESIGNS ❍ 255 INTRODUCTION 7.1 Parameter ⇔ Population Statistic ⇔ Sample In the previous three chapters, you have learned a lot about probability distributions, such as the binomial and normal distributions. The shape of the normal distribution is determined by its mean m and its standard deviation s, whereas the shape of the binomial distribution is determined by p. These numerical descriptive measures— called parameters—are needed to calculate the probability of observing sample results. In practical situations, you may be able to decide which type of probability distribution to use as a model, but the values of the parameters that specify its exact form are unknown. Here are two examples: • A pollster is sure that the responses to his “agree/disagree” questions will follow a binomial distribution, but p, the proportion of those who “agree” in the population, is unknown. • An agronomist believes that the yield per acre of a variety of wheat is approximately normally distributed, but the mean m and standard deviation s of the yields are unknown. In these cases, you must rely on the sample to learn about these parameters. The proportion of those who “agree” in the pollster’s sample provides information about the actual value of p. The mean and standard deviation of the agronomist’s sample approximate the actual values of m and s. If you want the sample to provide reliable information about the population, however, you must select your sample in a certain way! SAMPLING PLANS AND EXPERIMENTAL DESIGNS 7.2 The way a sample is selected is called the sampling plan or experimental design and determines the quantity of information in the sample. Knowing the sampling plan used in a particular situation will often allow you to measure the reliability or goodness of your inference. Simple random sampling is a commonly used sampling plan in which every sample of size n has the same chance of being selected. For example, suppose you want to select a sample of size n 2 from a population containing N 4 objects. If the four objects are identiﬁed by the symbols x1, x2, x3, and x4, there are six distinct pairs that could be selected, as listed in Table 7.1. If the sample of n 2 observations is selected so that each of these six samples has the same chance of selection, given by 1/6, then the resulting sample is called a simple random sample, or just a random sample. TABLE 7.1 ● Ways of Selecting a Sample of Size 2 from 4 Objects Sample Observations in Sample 1 2 3 4 5 6 x1, x2 x1, x3 x1, x4 x2, x3 x2, x4 x3, x4 256 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS EXAMPLE 7.1 TABLE 7.2 Definition If a sample of n elements is selected from a population of N elements using a sampling plan in which each of the possible samples has the same chance of selection, then the sampling is said to be random and the resulting sample is a simple random sample. Perfect random sampling is difficult to achieve in practice. If the size of the population N is small, you might write each of N numbers on a poker chip, mix the chips, and select a sample of n chips. The numbers that you select correspond to the n measurements that appear in the sample. Since this method is not always very practical, a simpler and more reliable method uses random numbers—digits generated so that the values 0 to 9 occur randomly and with equal frequency. These numbers can be generated by computer or may even be available on your scientiﬁc calculator. Alternatively, Table 10 in Appendix I is a table of random numbers that you can use to select a random sample. A computer database at a downtown law ﬁrm contains ﬁles for N 1000 clients. The ﬁrm wants to select n 5 ﬁles for review. Select a simple random sample of 5 ﬁles from this database. Solution You must ﬁrst label each ﬁle with a number from 1 to 1000. Perhaps the ﬁles are stored alphabetically, and the computer has already assigned a number to each. Then generate a sequence of 10 three-digit random numbers. If you are using Table 10 of Appendix I, select a random starting point and use a portion of the table similar to the one shown in Table 7.2. The random starting point ensures that you will not use the same sequence over and over again. The ﬁrst three digits of Table 7.2 indicate the number of the ﬁrst ﬁle to be reviewed. The random number 001 corresponds to ﬁle #1, and the last ﬁle, #1000, corresponds to the random number 000. Using Table 7.2, you would choose the ﬁve ﬁles numbered 155, 450, 32, 882, and 350 for review. Alternately, you might choose to read across the lines, and choose ﬁles 155, 350, 989, 450 and 369 for review. ● Portion of a Table of Random Numbers 98924 28630 50856 21121 35026 36933 78812 26053 15574 45045 03225 88292 The situation described in Example 7.1 is called an observational study because the data already existed before you decided to observe or describe their characteristics. Most sample surveys, in which information is gathered with a questionnaire, fall into this category. Computer databases make it possible to assign identiﬁcation numbers to each element even when the population is large and to select a simple random sample. You must be careful when conducting a sample survey, however, to watch for these frequently occurring problems: • Nonresponse: You have carefully selected your random sample and sent out your questionnaires, but only 50% of those surveyed return their questionnaires. Are the responses you received still representative of the entire population, or are they biased because only those people who were particularly opinionated about the subject chose to respond? EXAMPLE 7.2 7.2 SAMPLING PLANS AND EXPERIMENTAL DESIGNS ❍ 257 • Undercoverage: You have selected your random sample using telephone records as a database. Does the database you used systematically exclude certain segments of the population—perhaps those who do not have telephones? • Wording bias: Your questionnaire may have questions that are too complicated or tend to confuse the reader. Possibly the questions are sensitive in nature—for example, “Have you ever used drugs?” or “Have you ever cheated on your income tax?”—and the respondents will not answer truthfully. Methods have been devised to solve some of these problems, but only if you know that they exist. If your survey is biased by any of these problems, then your conclusions will not be very reliable, even though you did select a random sample! Some research involves experimentation, in which an experimental condition or treatment is imposed on the experimental units. Selecting a simple random sample is more difficult in this situation. A research chemist is testing a new method for measuring the amount of titanium (Ti) in ore samples. She chooses 10 ore samples of the same weight for her experiment. Five of the samples will be measured using a standard method, and the other 5 using the new method. Use random numbers to assign the 10 ore samples to the new and standard groups. Do these data represent a simple random sample from the population? Solution There are really two populations in this experiment. They consist of titanium measurements, using either the new or standard method, for all possible ore samples of this weight. These populations do not exist in fact; they are hypothetical populations, envisioned in the mind of the researcher. Thus, it is impossible to select a simple random sample using the methods of Example 7.1. Instead, the researcher selects what she believes are 10 representative ore samples and hopes that these samples will behave as if t
hey had been randomly selected from the two populations. The researcher can, however, randomly select the ﬁve samples to be measured with each method. Number the samples from 1 to 10. The ﬁve samples selected for the new method may correspond to 5 one-digit random numbers. Use this sequence of random digits generated on a scientiﬁc calculator: 948247817184610 Since you cannot select the same ore sample twice, you must skip any digit that has already been chosen. Ore samples 9, 4, 8, 2, and 7 will be measured using the new method. The other samples—1, 3, 5, 6, and 10—will be measured using the standard method. In addition to simple random sampling, there are other sampling plans that involve randomization and therefore provide a probabilistic basis for inference making. Three such plans are based on stratiﬁed, cluster, and systematic sampling. When the population consists of two or more subpopulations, called strata, a sampling plan that ensures that each subpopulation is represented in the sample is called a stratiﬁed random sample. Definition Stratiﬁed random sampling involves selecting a simple random sample from each of a given number of subpopulations, or strata. 258 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS Citizens’ opinions about the construction of a performing arts center could be collected using a stratiﬁed random sample with city voting wards as strata. National polls usually involve some form of stratiﬁed random sampling with states as strata. Another form of random sampling is used when the available sampling units are groups of elements, called clusters. For example, a household is a cluster of individuals living together. A city block or a neighborhood might be a convenient sampling unit and might be considered a cluster for a given sampling plan. Definition A cluster sample is a simple random sample of clusters from the available clusters in the population. When a particular cluster is included in the sample, a census of every element in the cluster is taken. Sometimes the population to be sampled is ordered, such as an alphabetized list of people with driver’s licenses, a list of utility users arranged by service addresses, or a list of customers by account numbers. In these and other situations, one element is chosen at random from the ﬁrst k elements, and then every kth element thereafter is included in the sample. Definition A 1-in-k systematic random sample involves the random selection of one of the ﬁrst k elements in an ordered population, and then the systematic selection of every kth element thereafter. Not all sampling plans, however, involve random selection. You have probably heard of the nonrandom telephone polls in which those people who wish to express support for a question call one “900 number” and those opposed call a second “900 number.” Each person must pay for his or her call. It is obvious that those people who call do not represent the population at large. This type of sampling plan is one form of a convenience sample—a sample that can be easily and simply obtained without random selection. Advertising for subjects who will be paid a fee for participating in an experiment produces a convenience sample. Judgment sampling allows the sampler to decide who will or will not be included in the sample. Quota sampling, in which the makeup of the sample must reﬂect the makeup of the population on some preselected characteristic, often has a nonrandom component in the selection process. Remember that nonrandom samples can be described but cannot be used for making inferences! All sampling plans used for making inferences must involve randomization! 7.2 EXERCISES BASIC TECHNIQUES 7.1 A population consists of N 500 experimental units. Use a random number table to select a random sample of n 20 experimental units. (HINT: Since you need to use three-digit numbers, you can assign 2 three-digit numbers to each of the sampling units in the manner shown in the table.) What is the probability that each experimental unit is selected for inclusion in the sample? Experimental Units Random Numbers 1 2 3 4 . . . 499 500 001, 501 002, 502 003, 503 004, 504 . . . 499, 999 500, 000 7.2 A political analyst wishes to select a sample of n 20 people from a population of 2000. Use the random number table to identify the people to be included in the sample. 7.3 A population contains 50,000 voters. Use the random number table to identify the voters to be included in a random sample of n 15. 7.4 A small city contains 20,000 voters. Use the random number table to identify the voters to be included in a random sample of n 15. 7.5 Every 10th Person A random sample of public opinion in a small town was obtained by selecting every 10th person who passed by the busiest corner in the downtown area. Will this sample have the characteristics of a random sample selected from the town’s citizens? Explain. 7.6 Parks and Recreation A questionnaire was mailed to 1000 registered municipal voters selected at random. Only 500 questionnaires were returned, and of the 500 returned, 360 respondents were strongly opposed to a surcharge proposed to support the city Parks and Recreation Department. Are you willing to accept the 72% ﬁgure as a valid estimate of the percentage in the city who are opposed to the surcharge? Why or why not? 7.7 DMV Lists In many states, lists of possible jurors are assembled from voter registration lists and Department of Motor Vehicles records of licensed drivers and car owners. In what ways might this list not cover certain sectors of the population adequately? 7.8 Sex and Violence One question on a survey questionnaire is phrased as follows: “Don’t you agree that there is too much sex and violence during prime TV viewing hours?” Comment on possible problems with the responses to this question. Suggest a better way to pose the question. APPLICATIONS 7.9 Cancer in Rats The Press Enterprise identiﬁed a byproduct of chlorination called MX that has been linked to cancer in rats.1 A scientist wants to conduct a validation study using 25 rats in the experimental group, each to receive a ﬁxed dose of MX, and 25 rats in a control group that will receive no MX. Determine a randomization scheme to assign the 50 individual rats to the two groups. 7.10 Racial Bias? Does the race of an interviewer matter? This question was investigated by Chris Gilberg 7.2 SAMPLING PLANS AND EXPERIMENTAL DESIGNS ❍ 259 and colleagues and reported in an issue of Chance magazine.2 The interviewer asked, “Do you feel that affirmative action should be used as an occupation selection criteria?” with possible answers of yes or no. a. What problems might you expect with responses to this question when asked by interviewers of different ethnic origins? b. When people were interviewed by an African- American, the response was about 70% in favor of affirmative action, approximately 35% when interviewed by an Asian, and approximately 25% when interviewed by a Caucasian. Do these results support your answer in part a? 7.11 Native American Youth In the American Journal of Human Biology, Chery Smith and Stefanie Fila reported on a study of urban Native American youth.3 The study objective was to determine the appropriateness of an assessment tool to identify dietary measurements for use in this population. The subjects were Native American youth attending an after-school program in Minneapolis, MN. All 61 children between the ages of 9 and 13 who satisﬁed the requirements of the study objectives were included in the experiment. a. Describe the sampling plan used to select study par- ticipants. b. What chance mechanism was used to select this sample of 61 Native American 9- to 13-year-old individuals? c. Can valid inferences be made using the results of this study? Why or why not? d. If you had to devise an alternative sampling plan, what would you change? 7.12 Blood Thinner A study of an experimental blood thinner was conducted to determine whether it works better than the simple aspirin tablet in warding off heart attacks and strokes.4 The study, reported in the Press Enterprise, involved 19,185 people who had suffered heart attacks, strokes, or pain from clogged arteries. Each person was randomly assigned to take either aspirin or the experimental drug for 1 to 3 years. Assume that each person was equally likely to be assigned one of the two medications. a. Devise a randomization plan to assign the medica- tions to the patients. b. Will there be an equal number of patients in each treatment group? Explain. 7.13 Going to the Moon Two different Gallup Polls were conducted for CNN/USA Today, both of which involved people’s feelings about the U.S. space 260 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS program.5 Here is a question from each poll, along with the responses of the sampled Americans: Space Exploration CNN/USA Today/Gallup Poll. Dec. 5–7, 2003. Nationwide: "Would you favor or oppose a new U.S. space program that would send astronauts to the moon?" Form A (N 510, MoE 5) Favor % 53 Oppose % 45 No Opinion % 2 12/03 "Would you favor or oppose the U.S. government spending billions of dollars to send astronauts to the moon?" Form B (N 494, MoE 5) Favor % 31 Oppose % 67 No Opinion % 2 12/03 a. Read the two poll questions. Which of the two wordings is more unbiased? Explain. b. Look at the responses for the two different polls. How would you explain the large differences in the percentages either favoring or opposing the new program? 7.14 Ask America A nationwide policy survey titled “Ask America” was sent by the National Republican Congressional Committee to voters in the Fortyfourth Congressional District, asking for opinions on a variety of political issues.6 Here are some questions from the survey: • • In recent years has the federal government grown more or less intrusive in your personal and business affairs? Is President Bush right in trying to rein in the size and scope of the federal government against the wishes of the big government Democrats? •
Do you believe the death penalty is a deterrrent to crime? • Do you agree that the obstructionist Democrats should not be allowed to gain control of the U.S. Congress in the upcoming elections? Comment on the effect of wording bias on the responses gathered using this survey. STATISTICS AND SAMPLING DISTRIBUTIONS 7.3 When you select a random sample from a population, the numerical descriptive measures you calculate from the sample are called statistics. These statistics vary or change for each different random sample you select; that is, they are random variables. The probability distributions for statistics are called sampling distributions because, in repeated sampling, they provide this information: • What values of the statistic can occur • How often each value occurs Definition The sampling distribution of a statistic is the probability distribution for the possible values of the statistic that results when random samples of size n are repeatedly drawn from the population. There are three ways to ﬁnd the sampling distribution of a statistic: 1. Derive the distribution mathematically using the laws of probability. 2. Use a simulation to approximate the distribution. That is, draw a large number of samples of size n, calculating the value of the statistic for each sample, and tabulate the results in a relative frequency histogram. When the number of EXAMPLE 7.3 7.3 STATISTICS AND SAMPLING DISTRIBUTIONS ❍ 261 samples is large, the histogram will be very close to the theoretical sampling distribution. 3. Use statistical theorems to derive exact or approximate sampling distributions. The next example demonstrates how to derive the sampling distributions of two statistics for a very small population. A population consists of N 5 numbers: 3, 6, 9, 12, 15. If a random sample of size n 3 is selected without replacement, ﬁnd the sampling distributions for the sample mean x and the sample median m. Solution You are sampling from the population shown in Figure 7.1. It contains ﬁve distinct numbers and each is equally likely, with probability p(x) 1/5. You can easily ﬁnd the population mean and median as m 3 6 9 12 15 5 9 and M 9 FI GUR E 7. 1 Probability histogram for the N 5 population values in Example 7.3 ● p(x) .4 .2 3 6 9 12 15 x Sampling distributions can be either discrete or continuous. There are 10 possible random samples of size n 3 and each is equally likely, with probability 1/10. These samples, along with the calculated values of x and m for each, are listed in Table 7.3. You will notice that some values of x are more likely than others because they occur in more than one sample. For example, 3 2 .3 .2 and P(m 6) P(x 8) 0 1 1 0 The values in Table 7.3 are tabulated, and the sampling distributions for x and m are shown in Table 7.4 and Figure 7.2. Since the population of N 5 values is symmetric about the value x 9, both the population mean and the median equal 9. It would seem reasonable, therefore, to consider using either x or m as a possible estimator of M m 9. Which estimator would you choose? From Table 7.3, you see that, in using m as an estimator, you would be in error by 9 6 3 with probability .3 or by 9 12 3 with probability .3. That is, the error in estimation using m would be 3 with probability .6. In using x, however, an error of 3 would occur with probability only .2. On these grounds alone, you may wish to use x as an estimator in preference to m. 262 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS Values of x and m for Simple Random Sampling TABLE 7.3 TABLE 7.4 ● when n 3 and N 5 x Sample Values Sample 1 2 3 4 5 6 7 8 9 10 3, 6, 9 3, 6, 12 3, 6, 15 3, 9, 12 3, 9, 15 3, 12, 15 6, 9, 12 6, 9, 15 6, 12, 15 9, 12, 15 6 7 8 8 9 10 9 10 11 12 m 6 6 6 9 9 12 9 9 12 12 Sampling Distributions for (a) the Sample Mean ● and (b) the Sample Median m (a) (b) p(x) x 6 9 12 6 7 8 9 10 11 12 .1 .1 .2 .2 .2 .1 .1 F IG URE 7. 2 Probability histograms for the sampling distributions of the sample mean, x, and the sample median, m, in Example 7.3 ● p(x) .4 .3 .2 .1 p (m) .3 .4 .3 p(m) .4 .3 .2 .1 6 7 8 9 10 11 12 x 6 7 8 9 10 11 12 m Almost every statistic has a mean and a standard deviation (or standard error) describing its center and spread. It was not too difficult to derive these sampling distributions in Example 7.3 because the number of elements in the population was very small. When this is not the case, you may need to use one of these methods: • Use a simulation to approximate the sampling distribution empirically. • Rely on statistical theorems and theoretical results. One important statistical theorem that describes the sampling distribution of statistics that are sums or averages is presented in the next section. 7.4 THE CENTRAL LIMIT THEOREM ❍ 263 THE CENTRAL LIMIT THEOREM 7.4 The Central Limit Theorem states that, under rather general conditions, sums and means of random samples of measurements drawn from a population tend to have an approximately normal distribution. Suppose you toss a balanced die n 1 time. The random variable x is the number observed on the upper face. This familiar random variable can take six values, each with probability 1/6, and its probability distribution is shown in Figure 7.3. The shape of the distribution is ﬂat or uniform and symmetric about the mean m 3.5, with a standard deviation s 1.71. (See Section 4.8 and Exercise 4.84.) ● FI GUR E 7. 3 Probability distribution for x, the number appearing on a single toss of a die 1/ Now, take a sample of size n 2 from this population; that is, toss two dice and record the sum of the numbers on the two upper faces, Sxi x1 x2. Table 7.5 shows the 36 possible outcomes, each with probability 1/36. The sums are tabulated, and each of the possible sums is divided by n 2 to obtain an average. The result is the sampling distribution of x Sxi/n, shown in Figure 7.4. You should notice the dramatic difference in the shape of the sampling distribution. It is now roughly moundshaped but still symmetric about the mean m 3.5. TABLE 7.5 ● Sums of the Upper Faces of Two Dice Second Die First Die 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 264 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS F IG URE 7. 4 Sampling distribution of x for n 2 dice ● .15 .10 ) r a b x ( p .05 0 1 2 3 4 5 x-bar 6 Average of Two Dice Using MINITAB, we generated the sampling distributions of x when n 3 and n 4. For n 3, the sampling distribution in Figure 7.5 clearly shows the mound shape of the normal probability distribution, still centered at m 3.5. Notice also that the spread of the distribution is slowly decreasing as the sample size n increases. Figure 7.6 dramatically shows that the distribution of x is approximately normally distributed based on a sample as small as n 4. This phenomenon is the result of an important statistical theorem called the Central Limit Theorem (CLT). F IG URE 7. 5 MINITAB sampling distribution of x for n 3 dice ● FIGURE 7.6 MINITAB sampling distribution of x for n 4 dice ● .15 .10 .05 0 .15 .10 .05 ) -bar 6 Average of Three Dice 0 1 2 3 4 5 Average of Four Dice x-bar 6 7.4 THE CENTRAL LIMIT THEOREM ❍ 265 Central Limit Theorem If random samples of n observations are drawn from a nonnormal population with ﬁnite mean m and standard deviation s, then, when n is large, the sampling distribution of the sample mean x is approximately normally distributed, with mean m and standard deviation s n The approximation becomes more accurate as n becomes large. Regardless of its shape, the sampling distribution of x always has a mean identical to the mean of the sampled population and a standard deviation equal to the population standard deviation s divided by n. Consequently, the spread of the distribution of sample means is considerably less than the spread of the sampled population. The Central Limit Theorem can be restated to apply to the sum of the sample measurements Sxi , which, as n becomes large, also has an approximately normal distribution with mean nm and standard deviation sn. The Java applet called The Central Limit Theorem can be used to perform a simulation for the sampling distributions of the average of one, two, three or four dice. Figure 7.7 shows the applet after the pair of dice (n 2) has been tossed 2500 times. This is not as hard as it seems, since you need only press the button 25 times. The simulation shows the possible values for x Sxi/10 and also shows the mean and standard deviation for these 2500 measurements. The mean, 3.5, is exactly equal to m 3.5. What is the standard deviation for these 2500 measurements? Is it close to the theoretical value, s/n 1.21? You will use this applet again for the MyApplet Exercises at the end of the chapter. 1.71 2 FI GUR E 7. 7 Central Limit Theorem applet ● 266 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS The sampling distribution of x always has a mean m and standard deviation s/n. The CLT helps describe its shape. The important contribution of the Central Limit Theorem is in statistical inference. Many estimators that are used to make inferences about population parameters are sums or averages of the sample measurements. When the sample size is sufficiently large, you can expect these estimators to have sampling distributions that are approximately normal. You can then use the normal distribution to describe the behavior of these estimators in repeated sampling and evaluate the probability of observing certain sample results. As in Chapter 6, these probabilities are calculated using the standard normal random variable an e M or t a im t s E z d n io iat d r da n a t S ev As you reread the Central Limit Theorem, you may notice that the approximation is valid as long as the sample size n is “large”—but how large is “large”? Unfortunately, there is no clear answer to this question. The appropriate value of n depends on the shape of the population from which you sample as well as on how you want to use the approximation. However, these guidelines will help: HOW DO I DECIDE WHEN THE SAMPLE SIZE IS LARGE ENOUGH? • If the sampled popula
tion is normal, then the sampling distribution of x will also be normal, no matter what sample size you choose. This result can be proven theoretically, but it should not be too difficult for you to accept without proof. • When the sampled population is approximately symmetric, the sampling distribution of x becomes approximately normal for relatively small values of n. Remember how rapidly the “ﬂat” distribution in the dice example became mound-shaped (n 3). • When the sampled population is skewed, the sample size n must be larger, with n at least 30 before the sampling distribution of x becomes approximately normal. These guidelines suggest that, for many populations, the sampling distribution of x will be approximately normal for moderate sample sizes; an exception to this rule occurs in sampling a binomial population when either p or q (1 p) is very small. As speciﬁc applications of the Central Limit Theorem arise, we will give you the appropriate sample size n. 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN If the population mean m is unknown, you might choose several statistics as an estimator; the sample mean x and the sample median m are two that readily come to mind. Which should you use? Consider these criteria in choosing the estimator for m: 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 267 Is it easy or hard to calculate? • • Does it produce estimates that are consistently too high or too low? • Is it more or less variable than other possible estimators? The sampling distributions for x and m with n 3 for the small population in Example 7.3 showed that, in terms of these criteria, the sample mean performed better than the sample median as an estimator of m. In many situations, the sample mean x has desirable properties as an estimator that are not shared by other competing estimators; therefore, it is more widely used. THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN, –x • • • If a random sample of n measurements is selected from a population with mean m and standard deviation s, the sampling distribution of the sample mean x will have mean m and standard deviation† s n If the population has a normal distribution, the sampling distribution of x will be exactly normally distributed, regardless of the sample size, n. If the population distribution is nonnormal, the sampling distribution of x will be approximately normally distributed for large samples (by the Central Limit Theorem). Standard Error Definition The standard deviation of a statistic used as an estimator of a population parameter is also called the standard error of the estimator (abbreviated SE) because it refers to the precision of the estimator. Therefore, the standard deviation of x—given by s/n—is referred to as the standard error of the mean (abbreviated as SE(x) or just SE). †When repeated samples of size n are randomly selected from a ﬁnite population with N elements whose mean is m and whose variance is s 2, the standard deviation of x is s N n n 1 N where s 2 is the population variance. When N is large relative to the sample size n, (N n)(N 1) is approximately equal to 1, and the standard deviation of x is s n 268 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS How Do I Calculate Probabilities for the Sample Mean –x ? If you know that the sampling distribution of x is normal or approximately normal, you can describe the behavior of the sample mean x by calculating the probability of observing certain values of x in repeated sampling. 1. Find m and calculate SE (x) s/n. 2. Write down the event of interest in terms of x, and locate the appropriate area on the normal curve. 3. Convert the necessary values of x to z-values using x z s/ m n 4. Use Table 3 in Appendix I to calculate the probability. Exercise Reps (Fill in the Blanks) A. You take a random sample of size n 36 from a distribution with mean m 75 and s 12. The sampling distribution of x will be approximately with a mean of and a standard deviation (or standard error) of . B. To ﬁnd the probability that the sample mean exceeds 80, write down the event of interest. When x 80, x z s/ m n Find the probability: P(x ) P(z ) 1 C. To ﬁnd the probability that the sample mean is between 70 and 72, write down the event of interest. When x 70 and x 72, x z s/ m n Find the probability: x and z s/ m n P( x ) P( z ) Progress Report • Still having trouble? Try again using the Exercise Reps at the end of this section. • No problems? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 269 EXAMPLE 7.4 The duration of Alzheimer’s disease from the onset of symptoms until death ranges from 3 to 20 years; the average is 8 years with a standard deviation of 4 years. The administrator of a large medical center randomly selects the medical records of 30 deceased Alzheimer’s patients from the medical center’s database, and records the average duration. Find the approximate probabilities for these events: If x is normal, x is normal for any n. If x is not normal, x is approximately normal for large n. 1. The average duration is less than 7 years. 2. The average duration exceeds 7 years. 3. The average duration lies within 1 year of the population mean m 8. Solution Since the administrator has selected a random sample from the database at this medical center, he can draw conclusions about only past, present, or future patients with Alzheimer’s disease at this medical center. If, on the other hand, this medical center can be considered representative of other medical centers in the country, it may be possible to draw more far-reaching conclusions. What can you say about the shape of the sampled population? It is not symmetric, because the mean m 8 does not lie halfway between the maximum and minimum values. Since the mean is closer to the minimum value, the distribution is skewed to the right, with a few patients living a long time after the onset of the disease. Regardless of the shape of the population distribution, however, the sampling distribution of x has a mean m 8 and standard deviation s/n 4/30 .73. In addition, because the sample size is n 30, the Central Limit Theorem ensures the approximate normality of the sampling distribution of x. 1. The probability that x is less than 7 is given by the shaded area in Figure 7.8. To ﬁnd this area, you need to calculate the value of z corresponding to x 7: m x 8 1.37 7 z s n / 73 . From Table 3 in Appendix I, you can ﬁnd the cumulative area corresponding to z 1.37 and P(x 7) P(z 1.37) .0853 FI GUR E 7. 8 The probability that x is less than 7 for Example 7.4 ● f(x) P(x < 7) 7 –1.37 µ = 8 0 x z [NOTE: You must use s/n (not s) in the formula for z because you are ﬁnding an area under the sampling distribution for x, not under the probability distribution for x.] 270 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS Remember that for continuous random variables, there is no probability assigned to a single point. Therefore, P(x 7) P(x 7). 2. The event that x exceeds 7 is the complement of the event that x is less than 7. Thus, the probability that x exceeds 7 is P(x 7) 1 P(x 7) 1 .0853 .9147 3. The probability that x lies within 1 year of m 8 is the shaded area in Fig- ure 7.9. The z-value corresponding to x 7 is z 1.37, from part 1, and the z-value for x 9 is m x 8 1.37 9 z n s / 73 . The probability of interest is P(7 x 9) P(1.37 z 1.37) .9147 .0853 .8294 F IG URE 7. 9 The probability that x lies within 1 year of m 8 for Example 7.4 ● f(x) P(7 < x < 9) 7 –1.37 µ = 8 0 9 1.37 x z Example 7.4 can be solved using the Normal Probabilities for Means applet. If you enter the values for x, s, m, and n (press “Enter” to record each change) and adjust the dropdown list at the bottom of the applet, you can calculate the area to the right or left of z0, the area in two tails, or the area between z0 and z0. Conversely, if you need to ﬁnd the value of x that cuts off a certain area under the curve, enter the area in the box marked “prob:” at the bottom of the applet, and the applet will provide the value of x. The applet in Figure 7.10 is set to calculate P(7 x 9) .829, correct to three decimal places. You will use this applet for the MyApplet Exercises at the end of the chapter. FI GUR E 7. 10 Normal Probabilities for Means applet ● 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 271 EXAMPLE 7.5 To avoid difficulties with the Federal Trade Commission or state and local consumer protection agencies, a beverage bottler must make reasonably certain that 12-ounce bottles actually contain 12 ounces of beverage. To determine whether a bottling machine is working satisfactorily, one bottler randomly samples 10 bottles per hour and measures the amount of beverage in each bottle. The mean x of the 10 ﬁll measurements is used to decide whether to readjust the amount of beverage delivered per bottle by the ﬁlling machine. If records show that the amount of ﬁll per bottle is normally distributed, with a standard deviation of .2 ounce, and if the bottling machine is set to produce a mean ﬁll per bottle of 12.1 ounces, what is the approximate probability that the sample mean x of the 10 test bottles is less than 12 ounces? Solution The mean of the sampling distribution of the sample mean x is identical to the mean of the population of bottle ﬁlls—namely, m 12.1 ounces—and the standard error of x is s .2 SE .063 n 10 (NOTE: s is the standard deviation of the population of bottle ﬁlls, and n is the number of bottles in the sample.) Since the amount of ﬁll is normally distributed, x is also normally distributed, as shown in Figure 7.11. To ﬁnd the probability that x is less than 12 ounces, express the value x 12 in units of standard deviations: m x 2.1 1.59 1 12 z n s / 3 06 . Then P(x 12) P(z 1.59) .0559 .056 272 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS ● f(x) F IG URE 7. 11 Probability distribution of x, the mean of the n 10 bottle ﬁlls, for Example 7.5 12 –1.59 µ = 12.1 0 x z Thus,
if the machine is set to deliver an average ﬁll of 12.1 ounces, the mean ﬁll x of a sample of 10 bottles will be less than 12 ounces with a probability equal to .056. When this danger signal occurs (x is less than 12), the bottler takes a larger sample to recheck the setting of the ﬁlling machine. 7.5 EXERCISES EXERCISE REPS (FILL IN THE BLANKS) These exercises refer back to the MyPersonal Trainer section on page 268. 7.15 You take a random sample of size n 49 from a distribution with mean m 53 and s 21. The sampling distribution of x will be approximately mean of and a standard deviation (or standard error) of with a . 7.16 Refer to Exercise 7.15. To ﬁnd the probability that the sample mean is greater than 55, write down the event of interest. When x 55, x z s m n / Find the probability: P(x ) P(z ) 1 7.17 You take a random sample of size n 40 from a distribution with mean m 100 and s 20. The sampling distribution of x will be approximately with a mean of and a standard deviation (or standard error) of . 7.18 Refer to Exercise 7.17. To ﬁnd the probability that the sample mean is between 105 and 110, write down the event of interest. When x 105 and x 110, x z s m n / x and z s m n / Find the probability: P( x ) P( z ) BASIC TECHNIQUES 7.19 Random samples of size n were selected from populations with the means and variances given here. Find the mean and standard deviation of the sampling distribution of the sample mean in each case: a. n 36, m 10, s 2 9 b. n 100, m 5, s 2 4 c. n 8, m 120, s 2 1 7.20 Refer to Exercise 7.19. a. If the sampled populations are normal, what is the sampling distribution of x for parts a, b, and c? b. According to the Central Limit Theorem, if the sampled populations are not normal, what can be said about the sampling distribution of x for parts a, b, and c? 7.21 A population consists of N = 5 numbers: 1, 3, 5, 6, and 7. It can be shown that the mean EX0721 and standard deviation for this population are m = 4.4 and s = 2.15, respectively. a. Construct a probability histogram for this population. b. Use the random number table, Table 10 in Appendix I, to select a random sample of size n = 10 with replacement from the population. Calculate the sample mean, x. Repeat this procedure, calculating the sample mean x for your second sample. (HINT: Assign the random digits 0 and 1 to the measurement x = 1; assign digits 2 and 3 to the measurement x = 3, and so on.) c. To simulate the sampling distribution of x, we have selected 50 more samples of size n = 10 with replacement, and have calculated the corresponding sample means. Construct a relative frequency histogram for these 50 values of x. What is the shape of this distribution? 4.8 3.0 4.6 5.0 4.4 4.2 5.9 4.1 4.6 4.2 4.2 5.7 3.4 4.1 4.2 4.5 4.2 4.9 5.1 5.2 4.3 4.4 4.1 3.4 5.4 4.3 4.8 4.0 5.9 4.8 5.0 5.0 3.7 5.0 3.6 4.0 5.1 4.3 4.3 5.0 3.3 4.8 4.3 4.5 4.5 4.7 4.2 4.5 3.9 4.9 7.22 Refer to Exercise 7.21. a. Use the data entry method in your calculator to ﬁnd the mean and standard deviation of the 50 values of x given in Exercise 7.21, part c. b. Compare the values calculated in part a to the theoretical mean m and the theoretical standard deviation s/n for the sampling distribution of x. How 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 273 close do the values calculated from the 50 measurements come to the theoretical values? c. n 4 f. n 25 b. n 2 e. n 16 7.23 A random sample of n observations is selected from a population with standard deviation s 1. Calculate the standard error of the mean (SE) for these values of n: a. n 1 d. n 9 g. n 100 7.24 Refer to Exercise 7.23. Plot the standard error of the mean (SE) versus the sample size n and connect the points with a smooth curve. What is the effect of increasing the sample size on the standard error? 7.25 Suppose a random sample of n 25 observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 12. a. Give the mean and the standard deviation of the sampling distribution of the sample mean x. b. Find the probability that x exceeds 110. c. Find the probability that the sample mean deviates from the population mean m 106 by no more than 4. APPLICATIONS 7.26 Faculty Salaries Suppose that college faculty with the rank of professor at two-year institutions earn an average of $64,571 per year7 with a standard deviation of $4,000. In an attempt to verify this salary level, a random sample of 60 professors was selected from a personnel database for all two-year institutions in the United States. a. Describe the sampling distribution of the sample mean x. b. Within what limits would you expect the sample average to lie, with probability .95? c. Calculate the probability that the sample mean x is greater than $66,000? d. If your random sample actually produced a sample mean of $66,000, would you consider this unusual? What conclusion might you draw? 7.27 Measurement Error When research chemists perform experiments, they may obtain slightly different results on different replications, even when the experiment is performed identically each time. These 274 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS differences are due to a phenomenon called “measurement error.” a. List some variables in a chemical experiment that might cause some small changes in the ﬁnal response measurement. b. If you want to make sure that your measurement error is small, you can replicate the experiment and take the sample average of all the measurements. To decrease the amount of variability in your average measurement, should you use a large or a small number of replications? Explain. 7.28 Tomatoes Explain why the weight of a package of one dozen tomatoes should be approximately normally distributed if the dozen tomatoes represent a random sample. 7.29 Bacteria in Water Use the Central Limit Theorem to explain why a Poisson random variable— say, the number of a particular type of bacteria in a cubic foot of water—has a distribution that can be approximated by a normal distribution when the mean m is large. (HINT: One cubic foot of water contains 1728 cubic inches of water.) 7.30 Paper Strength A manufacturer of paper used for packaging requires a minimum strength of 20 pounds per square inch. To check on the quality of the paper, a random sample of 10 pieces of paper is selected each hour from the previous hour’s production and a strength measurement is recorded for each. The standard deviation s of the strength measurements, computed by pooling the sum of squares of deviations of many samples, is known to equal 2 pounds per square inch, and the strength measurements are normally distributed. a. What is the approximate sampling distribution of the sample mean of n 10 test pieces of paper? b. If the mean of the population of strength measurements is 21 pounds per square inch, what is the approximate probability that, for a random sample of n 10 test pieces of paper, x 20? c. What value would you select for the mean paper strength m in order that P(x 20) be equal to .001? 7.31 Potassium Levels The normal daily human potassium requirement is in the range of 2000 to 6000 milligrams (mg), with larger amounts required during hot summer weather. The amount of potassium in food varies, depending on the food. For example, there are approximately 7 mg in a cola drink, 46 mg in a beer, 630 mg in a banana, 300 mg in a carrot, and 440 mg in a glass of orange juice. Suppose the distribution of potassium in a banana is normally distributed, with mean equal to 630 mg and standard deviation equal to 40 mg per banana. You eat n 3 bananas per day, and T is the total number of milligrams of potassium you receive from them. a. Find the mean and standard deviation of T. b. Find the probability that your total daily intake of potassium from the three bananas will exceed 2000 mg. (HINT: Note that T is the sum of three random variables, x1, x2, and x3, where x1 is the amount of potassium in banana number 1, etc.) 7.32 Deli Sales The total daily sales, x, in the deli section of a local market is the sum of the sales generated by a ﬁxed number of customers who make purchases on a given day. a. What kind of probability distribution do you expect the total daily sales to have? Explain. b. For this particular market, the average sale per customer in the deli section is $8.50 with s $2.50. If 30 customers make deli purchases on a given day, give the mean and standard deviation of the probability distribution of the total daily sales, x. 7.33 Normal Temperatures In Exercise 1.67, Allen Shoemaker derived a distribution of human body temperatures with a distinct mound shape.8 Suppose we assume that the temperatures of healthy humans is approximately normal with a mean of 98.6 degrees and a standard deviation of 0.8 degrees. a. If 130 healthy people are selected at random, what is the probability that the average temperature for these people is 98.25 degrees or lower? b. Would you consider an average temperature of 98.25 degrees to be an unlikely occurrence, given that the true average temperature of healthy people is 98.6 degrees? Explain. 7.34 Sports and Achilles Tendon Injuries Some sports that involve a signiﬁcant amount of running, jumping, or hopping put participants at risk for Achilles tendinopathy (AT), an inﬂammation and thickening of the Achilles tendon. A study in The American Journal of Sports Medicine looked at the diameter (in mm) of the affected and nonaffected tendons for patients who participated in these types of sports activities.9 Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters (mm) with a standard deviation of 1.95 mm. a. What is the probability that a randomly selected sample of 31 patients would produce an average diameter of 6.5 mm or less for the nonaffected tendon? b. When the diameters of the affected tendon were measured for a sample of 31 patients, the average diameter was 9.80. If the average tendon 7.6 THE SAMPLING DIST
RIBUTION OF THE SAMPLE PROPORTION ❍ 275 diameter in the population of patients with AT is no different than the average diameter of the nonaffected tendons (5.97 mm), what is the probability of observing an average diameter of 9.80 or higher? c. What conclusions might you draw from the results of part b? THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION 7.6 Q: How do you know if it’s binomial or not? A: Look to see if the measurement taken on a single experimental unit in the sample is a “success/failure” type. If so, it’s probably binomial. ● FI GUR E 7. 12 Sampling distribution of the binomial random variable x and the sample proportion ˆp There are many practical examples of the binomial random variable x. One common application involves consumer preference or opinion polls, in which we use a random sample of n people to estimate the proportion p of people in the population who have a speciﬁed characteristic. If x of the sampled people have this characteristic, then the sample proportion x ˆp n can be used to estimate the population proportion p (Figure 7.12).† The binomial random variable x has a probability distribution p(x), described in Chapter 5, with mean np and standard deviation npq. Since ˆp is simply the value of x, the sampling distribution of ˆp is identical to x, expressed as a proportion ˆp the probability distribution of x, except that it has a new scale along the horizontal axis. n 0.3 0.2 0.1 0.0 0 0 1 2 3 4 1/5 2/5 3/5 4/5 x ˆp 5 1 Because of this change of scale, the mean and standard deviation of ˆp are also rescaled, so that the mean of the sampling distribution of ˆp is p, and its standard error is SE( ˆp) p q where q 1 p n Finally, just as we can approximate the probability distribution of x with a normal distribution when the sample size n is large, we can do the same with the sampling distribution of ˆp. †A “hat” placed over the symbol of a population parameter denotes a statistic used to estimate the population parameter. For example, the symbol ˆp denotes the sample proportion. 276 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS PROPERTIES OF THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION, ˆp • If a random sample of n observations is selected from a binomial population with parameter p, then the sampling distribution of the sample proportion x ˆp n will have a mean p and a standard deviation SE( ˆp) p q where q 1 p n EXAMPLE 7.6 • When the sample size n is large, the sampling distribution of ˆp can be approximated by a normal distribution. The approximation will be adequate if np 5 and nq 5. In a survey, 500 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 60% agreed that the genders are equal and should have equal opportunities to participate in sports. Describe the sampling distribution of the sample proportion ˆp of parents who agree that the genders are equal and should have equal opportunities. Solution You can assume that the 500 parents represent a random sample of the parents of all boys and girls in the United States and that the true proportion in the population is equal to some unknown value that you can call p. The sampling distribution of ˆp can be approximated by a normal distribution,† with mean equal to p (see Figure 7.13) and standard error q SE( ˆp) p n ● ˆ f(p) F IG URE 7. 13 The sampling distribution for ˆp based on a sample of n 500 parents for Example 7.6 p 2SE .044 ˆp †Checking the conditions that allow the normal approximation to the distribution of ˆp, you can see that n 500 is adequate for values of p near .60 because nˆp 300 and n ˆq 200 are both greater than 5. 7.6 THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION ❍ 277 You can see from Figure 7.13 that the sampling distribution of ˆp is centered over its mean p. Even though you do not know the exact value of p (the sample proportion ˆp .60 may be larger or smaller than p), an approximate value for the standard deviation of the sampling distribution can be found using the sample proportion ˆp .60 to approximate the unknown value of p. Thus, SE p qˆ q ˆp n n (.60 .40) .022 ( ) 0 0 5 Therefore, approximately 95% of the time, ˆp will fall within 2SE .044 of the (unknown) value of p. How Do I Calculate Probabilities for the Sample Proportion ˆp? 1. Find the necessary values of n and p. 2. Check whether the normal approximation to the binomial distribution is appro- priate (np 5 and nq 5). 3. Write down the event of interest in terms of ˆp, and locate the appropriate area on the normal curve. 4. Convert the necessary values of ˆp to z-values using z ˆp p q p n 5. Use Table 3 in Appendix I to calculate the probability. Exercise Reps (Fill in the Blanks) A. You take a random sample of size n 36 from a binomial distribution with mean p .4. The sampling distribution of ˆp will be approximately with a mean of . B. To ﬁnd the probability that the sample proportion exceeds .5, write down the and a standard deviation (or standard error) of event of interest. When ˆp .5, z ˆp p q p n Find the probability: P( ˆp ) P(z ) 1 (continued) 278 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS C. To ﬁnd the probability that the sample proportion is between .5 and .6, write down the event of interest. When ˆp .5 and ˆp .6, z ˆp p q p n and z ˆp p q p n Find the probability: P( ˆp ) P( z ) Progress Report • Still having trouble? Try again using the Exercise Reps at the end of this section. • No problems? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. EXAMPLE 7.7 Refer to Example 7.6. Suppose the proportion p of parents in the population is actually equal to .55. What is the probability of observing a sample proportion as large as or larger than the observed value ˆp .60? Solution Figure 7.14 shows the sampling distribution of ˆp when p .55, with the observed value ˆp .60 located on the horizontal axis. The probability of observing a sample proportion ˆp equal to or larger than .60 is approximated by the shaded area in the upper tail of this normal distribution with p .55 and SE p q (.55 .45) .0222 ( ) 0 0 5 n F IG URE 7. 14 The sampling distribution of ˆp for n 500 and p .55 for Example 7.7 ● f(ˆp) P(ˆp ≥ .60) p = .55 .60 ˆp To ﬁnd this shaded area, ﬁrst calculate the z-value corresponding to ˆp .60: p ˆp .55 2.25 .60 z n 222 .0 p q/ 7.6 THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION ❍ 279 Using Table 3 in Appendix I, you ﬁnd P( ˆp .60) P(z 2.25) 1 .9878 .0122 That is, if you were to select a random sample of n 500 observations from a population with proportion p equal to .55, the probability that the sample proportion ˆp would be as large as or larger than .60 is only .0122. When the normal distribution was used in Chapter 6 to approximate the binomial probabilities associated with x, a correction of .5 was applied to improve the approximation. The equivalent correction here is (.5/n). For example, for ˆp .60 the value of z with the correction is z1 (.60 .001) .55 .45) (.55 ( ) 0 0 5 2.20 with P( ˆp .60) .0139. To two-decimal-place accuracy, this value agrees with the earlier result. When n is large, the effect of using the correction is generally negligible. You should solve problems in this and the remaining chapters without the correction factor unless you are speciﬁcally instructed to use it. 7.6 EXERCISES EXERCISE REPS (FILL IN THE BLANKS) These exercises refer back to the MyPersonal Trainer section on page 277. 7.35 You take a random sample of size n 50 from a binomial distribution with a mean of p .7. The sampling distribution of ˆp will be approximately mean of and a standard deviation (or standard error) of . with a 7.36 To ﬁnd the probability that the sample proportion is less than .8, write down the event of interest. When ˆp .8, ˆp p q p z n Find the probability: P( ˆp ) P(z ) BASIC TECHNIQUES 7.37 Random samples of size n were selected from binomial populations with population parameters p given here. Find the mean and the standard deviation of the sampling distribution of the sample proportion ˆp in each case: a. n 100, p .3 b. n 400, p .1 c. n 250, p .6 7.38 Is it appropriate to use the normal distribution to approximate the sampling distribution of ˆp in the following circumstances? a. n 50, p .05 b. n 75, p .1 c. n 250, p .99 280 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS 7.39 Random samples of size n 75 were selected from a binomial population with p .4. Use the normal distribution to approximate the following probabilities: a. P( ˆp .43) b. P(.35 ˆp .43) 7.40 Random samples of size n 500 were selected from a binomial population with p .1. a. Is it appropriate to use the normal distribution to approximate the sampling distribution of ˆp? Check to make sure the necessary conditions are met. Using the results of part a, ﬁnd these probabilities: b. ˆp .12 c. ˆp .10 d. ˆp lies within .02 of p 7.41 Calculate SE( ˆp) for n 100 and these values of p: a. p .01 d. p .50 g. p .99 h. Plot SE( ˆp) versus p on graph paper and sketch a b. p .10 e. p .70 c. p .30 f. p .90 smooth curve through the points. For what value of p is the standard deviation of the sampling distribution of ˆp a maximum? What happens to the standard error when p is near 0 or near 1.0? 7.42 a. Is the normal approximation to the sampling distribution of ˆp appropriate when n 400 and p .8? b. Use the results of part a to ﬁnd the probability that ˆp is greater than .83. c. Use the results of part a to ﬁnd the probability that ˆp lies between .76 and .84. APPLICATIONS 7.43 Losing Weight News reports tell us that the average American is overweight. Many of us have tried to trim down to our weight when we ﬁnished high school or college. And, in fact, only 19% of adults say they do not suffer from weight-loss woes.10 Suppose that the 19% ﬁgure is correct, and that a random sample of n 100 adults is selected. a. Does the distribution of ˆp, the sample proportion of adults who do not suffer from excess weight, have an approximate normal distrib
ution? If so, what is its mean and standard deviation? b. What is the probability that the sample proportion ˆp exceeds .25? c. What is the probability that ˆp lies within the interval .25 to .30? d. What might you conclude about p if the sample proportion exceeded .30? 7.44 Prescription Costs The cost of brand-name prescriptions are set to provide support for research and development of these drugs, which may cover as many as 20 years. Nonetheless, a majority of Americans say that costs of prescription drugs (66%), hospital costs (64%), and doctors visits (55%) are unreasonably high.11 Suppose that you take a random sample of n 1000 adults. Let ˆp be the proportion of adults who say that prescription drugs are unreasonably high. a. What is the exact distribution of ˆp? How can you approximate the distribution of ˆp? b. What is the probability that ˆp exceeds .68? c. What is the probability that ˆp lies between .64 and .68? d. Would a sample percentage of 70% contradict the reported value of 66%? 7.45 M&M’S According to the M&M’S website, the average percentage of brown M&M’S candies in a package of milk chocolate M&M’S is 13%.12 (This percentage varies, however, among the different types of packaged M&M’S.) Suppose you randomly select a package of milk chocolate M&M’S that contains 55 candies and determine the proportion of brown candies in the package. a. What is the approximate distribution of the sample proportion of brown candies in a package that contains 55 candies? b. What is the probability that the sample percentage of brown candies is less than 20%? c. What is the probability that the sample percentage exceeds 35%? d. Within what range would you expect the sample proportion to lie about 95% of the time? 7.46 The “Cheeseburger Bill” In the spring of 2004, the U.S. Congress considered a bill that would prevent Americans from suing fast-food giants like McDonald’s for making them overweight. Although the fast-food industry may not be to blame, a study by Children’s Hospital in Boston reports that about twothirds of adult Americans and about 15% of children and adolescents are overweight.13 A random sample of 100 children is selected. a. What is the probability that the sample percentage of overweight children exceeds 25%? 7.7 A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) ❍ 281 b. What is the probability that the sample percentage of overweight children is less than 12%? c. Would it be unusual to ﬁnd that 30% of the sam- pled children were overweight? Explain. 7.47 Oh, Nuts! Are you a chocolate “purist,” or do you like other ingredients in your chocolate? American Demographics reports that almost 75% of consumers like traditional ingredients such as nuts or caramel in their chocolate. They are less enthusiastic about the taste of mint or coffee that provide more distinctive ﬂavors.14 A random sample of 200 consumers is selected and the number who like nuts or caramel in their chocolate is recorded. a. What is the approximate sampling distribution for the sample proportion ˆp? What are the mean and standard deviation for this distribution? b. What is the probability that the sample percentage is greater than 80%? c. Within what limits would you expect the sample proportion to lie about 95% of the time? A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) 7.7 Statistical process control (SPC) methodology was developed to monitor, control, and improve products and services. Steel bearings must conform to size and hardness speciﬁcations, industrial chemicals must have a low prespeciﬁed level of impurities, and accounting ﬁrms must minimize and ultimately eliminate incorrect bookkeeping entries. It is often said that statistical process control consists of 10% statistics, 90% engineering and common sense. We can statistically monitor a process mean and tell when the mean falls outside preassigned limits, but we cannot tell why it is out of control. Answering this last question requires knowledge of the process and problemsolving ability—the other 90%! Product quality is usually monitored using statistical control charts. Measurements on a process variable to be monitored change over time. The cause of a change in the variable is said to be assignable if it can be found and corrected. Other variation— small haphazard changes due to alteration in the production environment—that is not controllable is regarded as random variation. If the variation in a process variable is solely random, the process is said to be in control. The ﬁrst objective in statistical process control is to eliminate assignable causes of variation in the process variable and then get the process in control. The next step is to reduce variation and get the measurements on the process variable within speciﬁcation limits, the limits within which the measurements on usable items or services must fall. Once a process is in control and is producing a satisfactory product, the process variables are monitored with control charts. Samples of n items are drawn from the process at speciﬁed intervals of time, and a sample statistic is computed. These statistics are plotted on the control chart, so that the process can be checked for shifts in the process variable that might indicate control problems. A Control Chart for the Process Mean: The ¯x Chart Assume that n items are randomly selected from the production process at equal intervals and that measurements are recorded on the process variable. If the process is in control, the sample means should vary about the population mean m in a random manner. Moreover, according to the Central Limit Theorem, the sampling distribution 282 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS of x should be approximately normal, so that almost all of the values of x fall into the interval (m 3 SE) m 3(s/n). Although the exact values of m and s are unknown, you can obtain accurate estimates by using the sample measurements. Every control chart has a centerline and control limits. The centerline for the x chart is the estimate of m, the grand average of all the sample statistics calculated from the measurements on the process variable. The upper and lower control limits are placed three standard deviations above and below the centerline. If you monitor the process mean based on k samples of size n taken at regular intervals, the centerline is x, the average of the sample means, and the control limits are at x 3(s/n), with s estimated by s, the standard deviation of the nk measurements. A statistical process control monitoring system samples the inside diameters of n 4 bearings each hour. Table 7.6 provides the data for k 25 hourly samples. Construct an x chart for monitoring the process mean. Solution The sample mean was calculated for each of the k 25 samples. For example, the mean for sample 1 is x .992 1.007 1.016 .991 4 1.0015 EXAMPLE 7.8 TABLE 7.6 ● n 4 Bearings per Sample 25 Hourly Samples of Bearing Diameters, Sample Sample Measurements Sample Mean 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 .992 1.015 .988 .996 1.015 1.000 .989 .994 1.018 .997 1.020 1.007 1.016 .982 1.001 .992 1.020 .993 .978 .984 .990 1.015 .983 1.011 .987 1.007 .984 .993 1.020 1.006 .982 1.009 1.010 1.016 1.005 .986 .986 1.002 .995 1.000 1.008 .988 .987 1.006 1.009 1.012 .983 .990 1.012 .987 1.016 .976 1.011 1.004 1.002 1.005 1.019 1.009 .990 .989 1.002 .981 1.010 1.011 .983 1.001 1.015 1.006 1.002 .983 1.010 1.003 .997 .991 1.007 .991 1.000 .981 .999 1.001 .989 .994 .990 1.011 1.001 .989 .995 .999 .987 1.002 .996 .986 1.001 .982 .986 1.007 .989 1.002 1.008 .995 1.00150 .99375 .99325 1.00475 1.00600 .99400 1.00275 1.00075 1.00875 .99800 .99925 .99225 1.00675 .99375 .99650 .99925 1.00225 .99675 .99200 .99050 1.00475 .99750 .99300 1.00550 .99400 The sample means are shown in the last column of Table 7.6. The centerline is located at the average of the sample means, or 675 .9987 9 x 24. 5 2 7.7 A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) ❍ 283 The calculated value of s, the sample standard deviation of all nk 4(25) 100 observations, is s .011458, and the estimated standard error of the mean of n 4 observations is s 58 14 .01 .005729 4 n The upper and lower control limits are found as s .9987 3(.005729) 1.015887 UCL x 3 n and s .9987 3(.005729) .981513 LCL x 3 n Figure 7.15 shows a MINITAB printout of the x chart constructed from the data. If you assume that the samples used to construct the x chart were collected when the process was in control, the chart can now be used to detect changes in the process mean. Sample means are plotted periodically, and if a sample mean falls outside the control limits, a warning should be conveyed. The process should be checked to locate the cause of the unusually large or small mean. FI GUR E 7. 15 MINITAB x chart for Example 7..02 1.01 1.00 0.99 0.98 Xbar Chart of Diameter UCL 1.01589 X 0.9987 LCL 0.98151 1 3 5 7 9 11 13 15 17 19 21 23 25 Sample A Control Chart for the Proportion Defective: The p Chart Sometimes the observation made on an item is simply whether or not it meets speciﬁcations; thus, it is judged to be defective or nondefective. If the fraction defective produced by the process is p, then x, the number of defectives in a sample of n items, has a binomial distribution. To monitor a process for defective items, samples of size n are selected at periodic intervals and the sample proportion ˆp is calculated. When the process is in control, ˆp should fall into the interval p 3SE, where p is the proportion of defectives in the population (or the process fraction defective) with standard error SE p q p(1 p) n n 284 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS The process fraction defective is unknown but can be estimated by the average of the k sample proportions: S ˆpi p k and the standard error is estimated by SE p(1 p) n The centerline for the p chart is located at p, and the upper and lower control limits are UCL p 3p(1 p) n and LCL p 3p(1 p) n EXAMPLE 7.9 TABLE 7.7 A manu
facturer of ballpoint pens randomly samples 400 pens per day and tests each to see whether the ink ﬂow is acceptable. The proportions of pens judged defective each day over a 40-day period are listed in Table 7.7. Construct a control chart for the proportion ˆp defective in samples of n 400 pens selected from the process. ● Proportions of Defectives in Samples of n 400 Pens Proportion Proportion Proportion Proportion Day Day Day Day 1 2 3 4 5 6 7 8 9 10 .0200 .0125 .0225 .0100 .0150 .0200 .0275 .0175 .0200 .0250 11 12 13 14 15 16 17 18 19 20 .0100 .0175 .0250 .0175 .0275 .0200 .0225 .0100 .0175 .0200 21 22 23 24 25 26 27 28 29 30 .0300 .0200 .0125 .0175 .0225 .0150 .0200 .0250 .0150 .0175 31 32 33 34 35 36 37 38 39 40 .0225 .0175 .0225 .0100 .0125 .0300 .0200 .0150 .0150 .0225 Solution The estimate of the process proportion defective is the average of the k 40 sample proportions in Table 7.7. Therefore, the centerline of the control chart is located at S ˆpi p k .0200 .0125 .0225 40 00 .019 6 .7 0 4 An estimate of SE, the standard error of the sample proportions, is p) (.019 (0 p(1 ) 0 4 n .981) .00683 and 3SE (3)(.00683) .0205. Therefore, the upper and lower control limits for the p chart are located at UCL p 3SE .0190 .0205 .0395 and LCL p 3SE .0190 .0205 .0015 7.7 A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) ❍ 285 Or, since p cannot be negative, LCL 0. The p control chart is shown in Figure 7.16. Note that all 40 sample proportions fall within the control limits. If a sample proportion collected at some time in the future falls outside the control limits, the manufacturer should be concerned about an increase in the defective rate. He should take steps to look for the possible causes of this increase. FI GUR E 7. 16 MINITAB p chart for Example 7..04 0.03 0.02 0.01 0.00 P Chart of Defects UCL 0.03948 p 0.019 LCL 0 1 5 9 13 17 25 29 33 37 21 Day Other commonly used control charts are the R chart, which is used to monitor variation in the process variable by using the sample range, and the c chart, which is used to monitor the number of defects per item. 7.7 EXERCISES BASIC TECHNIQUES 7.48 The sample means were calculated for 30 samples of size n 10 for a process that was judged to be in control. The means of the 30 x-values and the standard deviation of the combined 300 measurements were x 20.74 and s .87, respectively. a. Use the data to determine the upper and lower con- trol limits for an x chart. b. What is the purpose of an x chart? c. Construct an x chart for the process and explain how it can be used. 7.49 The sample means were calculated for 40 samples of size n 5 for a process that was judged to be in control. The means of the 40 values and the standard deviation of the combined 200 measurements were x 155.9 and s 4.3, respectively. a. Use the data to determine the upper and lower con- trol limits for an x chart. b. Construct an x chart for the process and explain how it can be used. 7.50 Explain the difference between an x chart and a p chart. 7.51 Samples of n 100 items were selected hourly over a 100-hour period, and the sample proportion of defectives was calculated each hour. The mean of the 100 sample proportions was .035. a. Use the data to ﬁnd the upper and lower control limits for a p chart. b. Construct a p chart for the process and explain how it can be used. 7.52 Samples of n 200 items were selected hourly over a 100-hour period, and the sample proportion of defectives was calculated each hour. The mean of the 100 sample proportions was .041. a. Use the data to ﬁnd the upper and lower control limits for a p chart. 286 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS b. Construct a p chart for the process and explain how it can be used. APPLICATIONS 7.53 Black Jack A gambling casino records and plots the mean daily gain or loss from ﬁve blackjack tables on an x chart. The overall mean of the sample means and the standard deviation of the combined data over 40 weeks were x $10,752 and s $1605, respectively. a. Construct an x chart for the mean daily gain per blackjack table. b. How can this x chart be of value to the manager of the casino? 7.54 Brass Rivets A producer of brass rivets randomly samples 400 rivets each hour and calculates the proportion of defectives in the sample. The mean sample proportion calculated from 200 samples was equal to .021. Construct a control chart for the proportion of defectives in samples of 400 rivets. Explain how the control chart can be of value to a manager. EX0755 7.55 Lumber Specs The manager of a building-supplies company randomly samples incoming lumber to see whether it meets quality speciﬁcations. From each shipment, 100 pieces of 2 4 lumber are inspected and judged according to whether they are ﬁrst (acceptable) or second (defective) grade. The proportions of second-grade 2 4s recorded for 30 shipments were as follows: .14 .21 .14 .21 .15 .20 .19 .23 .18 .22 .22 .19 .19 .26 .23 .20 .22 .13 .18 .12 .22 .23 .19 .21 .25 .15 .20 .17 .21 .26 Construct a control chart for the proportion of secondgrade 2 4s in samples of 100 pieces of lumber. Explain how the control chart can be of use to the manager of the building-supplies company. 7.56 Coal Burning Power Plant A coal-burning power plant tests and measures three specimens of coal each day to monitor the percentage of ash in the coal. The overall mean of 30 daily sample means and the combined standard deviation of all the data were x 7.24 and s .07, respectively. Construct an x chart for the process and explain how it can be of value to the manager of the power plant. EX0757 7.57 Nuclear Power Plant The data in the table are measures of the radiation in air particulates at a nuclear power plant. Four measurements were recorded at weekly intervals over a 26-week period. Use the data to construct an x chart and plot the 26 values of x. Explain how the chart can be used. Week Radiation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 .031 .025 .029 .035 .022 .030 .019 .027 .034 .017 .022 .016 .015 .029 .031 .014 .019 .024 .029 .032 .041 .034 .021 .029 .016 .020 .032 .026 .029 .037 .024 .029 .019 .028 .032 .016 .020 .018 .017 .028 .029 .016 .019 .024 .027 .030 .042 .036 .022 .029 .017 .021 .030 .025 .031 .034 .022 .030 .018 .028 .033 .018 .020 .017 .018 .029 .030 .016 .021 .024 .028 .031 .038 .036 .024 .030 .017 .020 .031 .025 .030 .035 .023 .030 .019 .028 .033 .018 .021 .017 .017 .029 .031 .017 .020 .025 .028 .030 .039 .035 .022 .029 .016 .022 7.58 Baseball Bats A hardwoods manufacturing plant has several different production lines to make baseball bats of different weights. One such production line is designed to produce bats weighing 32 ounces. During a period of time when the production process was known to be in statistical control, the average bat weight was found to be 31.7 ounces. The observed data were gathered from 50 samples, each consisting of 5 measurements. The standard deviation of all samples was found to be s .2064 ounces. Construct an x-chart to monitor the 32-ounce bat production process. 7.59 More Baseball Bats Refer to Exercise 7.58 and suppose that during a day when the state of the 32-ounce bat production process was unknown, the following measurements were obtained at hourly intervals. x Hour Hour x 1 2 3 31.6 32.5 33.4 4 5 6 33.1 31.6 31.8 Each measurement represents a statistic computed from a sample of ﬁve bat weights selected from the production process during a certain hour. Use the control chart constructed in Exercise 7.58 to monitor the process. CHAPTER REVIEW ❍ 287 CHAPTER REVIEW Key Concepts and Formulas I. Sampling Plans and Experimental Designs 1. Simple random sampling a. Each possible sample of size n is equally likely to occur. b. Use a computer or a table of random num- bers. 3. Probabilities involving the sample mean can be calculated by standardizing the value of x using z: m x z n s / IV. Sampling Distribution of the Sample c. Problems are nonresponse, undercoverage, Proportion and wording bias. 2. Other sampling plans involving randomization a. Stratiﬁed random sampling b. Cluster sampling c. Systematic 1-in-k sampling 3. Nonrandom sampling a. Convenience sampling b. Judgment sampling c. Quota sampling II. Statistics and Sampling Distributions 1. Sampling distributions describe the possible values of a statistic and how often they occur in repeated sampling. 2. Sampling distributions can be derived mathematically, approximated empirically, or found using statistical theorems. 3. The Central Limit Theorem states that sums and averages of measurements from a nonnormal population with ﬁnite mean m and standard deviation s have approximately normal distributions for large samples of size n. III. Sampling Distribution of the Sample Mean 1. When samples of size n are randomly drawn from a normal population with mean m and variance s 2, the sample mean x has a normal distribution with mean m and standard deviation s/n. 2. When samples of size n are randomly drawn from a nonnormal population with mean m and variance s 2, the Central Limit Theorem ensures that the sample mean x will have an approximately normal distribution with mean m and standard deviation s/n when n is large (n 30). 1. When samples of size n are drawn from a binomial population with parameter p, the sample proportion ˆp will have an approximately normal distribution with mean p and standard deviation pq/n as long as np 5 and nq 5. 2. Probabilities involving the sample proportion can be calculated by standardizing the value ˆp using z: z ˆp p q p n V. Statistical Process Control 1. To monitor a quantitative process, use an x chart. Select k samples of size n and calculate the overall mean x and the standard deviation s of all nk measurements. Create upper and lower control limits as s x 3 n If a sample mean exceeds these limits, the process is out of control. 2. To monitor a binomial process, use a p chart. Select k samples of size n and calculate the average of the sample proport
ions as S ˆpi p k Create upper and lower control limits as p 3p(1 p) n If a sample proportion exceeds these limits, the process is out of control. 288 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS The Central Limit Theorem at Work MINITAB provides a perfect tool for exploring the way the Central Limit Theorem works in practice. Remember that, according to the Central Limit Theorem, if random samples of size n are drawn from a nonnormal population with mean m and standard deviation s, then when n is large, the sampling distribution of the sample mean x will be approximately normal with the same mean m and with standard error s/n. Let’s try sampling from a nonnormal population with the help of MINITAB. In a new MINITAB worksheet, generate 100 samples of size n 30 from a nonnormal distribution called the exponential distribution. Use Calc Random Data Exponential. Type 100 for the number of rows of data, and store the results in C1–C30 (see Figure 7.17). Leave the mean at the default of 1.0, the threshold at 0.0, and click OK. The data are generated and stored in the worksheet. Use Graph Histogram Simple to look at the distribution of some of the data—say, C1 (as in Figure 7.18). Notice that the distribution is not mound-shaped; it is highly skewed to the right. F IG URE 7. 17 ● For the exponential distribution that we have used, the mean and standard deviation are m 1 and s 1, respectively. Check the descriptive statistics for one of the columns (use Stat Basic Statistics Display Descriptive Statistics), and you will ﬁnd that the 100 observations have a sample mean and standard deviation that are both close to but not exactly equal to 1. Now, generate 100 values of x based on samples of size n 30 by creating a column of means for the 100 rows. Use Calc Row Statistics, and select Mean. To average the entries in all 30 columns, select or type C1–C30 in the Input variables box, and store the results in C31 (see Figure 7.19). You can now look at the distribution of the sample means using Graph Histogram Simple, selecting C31 and clicking OK. The distribution of the 100 sample means generated for our example is shown in Figure 7.20. FI GUR E 7. 18 ● MY MINITAB ❍ 289 F IGU RE 7 .1 9 ● 290 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS F IG URE 7. 20 ● Notice the distinct mound shape of the distribution in Figure 7.20 compared to the original distribution in Figure 7.18. Also, if you check the descriptive statistics for C31, you will ﬁnd that the mean and standard deviation of our 100 sample means are not too different from the theoretical values, m 1 and s/n 1/30 .18. (For our data, the sample mean is .9645 and the standard deviation is .1875.) Since we had only 100 samples, our results are not exactly equal to the theoretical values. If we had generated an inﬁnite number of samples, we would have gotten an exact match. This is the Central Limit Theorem at work! Supplementary Exercises 7.60 A ﬁnite population consists of four elements: 6, 1, 3, 2. a. How many different samples of size n 2 can be selected from this population if you sample without replacement? (Sampling is said to be without replacement if an element cannot be selected twice for the same sample.) b. List the possible samples of size n 2. c. Compute the sample mean for each of the samples given in part b. d. Find the sampling distribution of x. Use a probability histogram to graph the sampling distribution of x. e. If all four population values are equally likely, calculate the value of the population mean m. Do any of the samples listed in part b produce a value of x exactly equal to m? 7.61 Refer to Exercise 7.60. Find the sampling distribution for x if random samples of size n 3 are selected without replacement. Graph the sampling distribution of x. 7.62 Lead Pipes Studies indicate that drinking water supplied by some old lead-lined city piping systems may contain harmful levels of lead. An important study of the Boston water supply system showed that the distribution of lead content readings for individual water specimens had a mean and standard deviation of approximately .033 milligrams per liter (mg/l) and .10 mg/l, respectively.15 a. Explain why you believe this distribution is or is not normally distributed. b. Because the researchers were concerned about the shape of the distribution in part a, they calculated the average daily lead levels at 40 different locations on each of 23 randomly selected days. What can you say about the shape of the distribution of the average daily lead levels from which the sample of 23 days was taken? c. What are the mean and standard deviation of the distribution of average lead levels in part b? 7.63 Biomass The total amount of vegetation held by the earth’s forests is important to both ecologists and politicians because green plants absorb carbon dioxide. An underestimate of the earth’s vegetative mass, or biomass, means that much of the carbon dioxide emitted by human activities (primarily fossilburning fuels) will not be absorbed, and a climatealtering buildup of carbon dioxide will occur. Studies16 indicate that the biomass for tropical woodlands, thought to be about 35 kilograms per square meter (kg/m2), may in fact be too high and that tropical biomass values vary regionally—from about 5 to 55 kg/m2. Suppose you measure the tropical biomass in 400 randomly selected square-meter plots. a. Approximate s, the standard deviation of the biomass measurements. b. What is the probability that your sample average is within two units of the true average tropical biomass? c. If your sample average is x 31.75, what would you conclude about the overestimation that concerns the scientists? 7.64 Hard Hats The safety requirements for hard hats worn by construction workers and others, established by the American National Standards Institute (ANSI), specify that each of three hats pass the following test. A hat is mounted on an aluminum head form. An 8-pound steel ball is dropped on the hat from a height of 5 feet, and the resulting force is measured at the bottom of the head form. The force exerted on the head form by each of the three hats must be less than 1000 pounds, and the average of the three must be less than 850 pounds. (The relationship between this test and actual human head damage is unknown.) Suppose SUPPLEMENTARY EXERCISES ❍ 291 the exerted force is normally distributed, and hence a sample mean of three force measurements is normally distributed. If a random sample of three hats is selected from a shipment with a mean equal to 900 and s 100, what is the probability that the sample mean will satisfy the ANSI standard? 7.65 Imagery and Memory A research psychologist is planning an experiment to determine whether the use of imagery—picturing a word in your mind— affects people’s ability to memorize. He wants to use two groups of subjects: a group that memorizes a set of 20 words using the imagery technique, and a control group that does not use imagery. a. Use a randomization technique to divide a group of 20 subjects into two groups of equal size. b. How can the researcher randomly select the group of 20 subjects? c. Suppose the researcher offers to pay subjects $50 each to participate in the experiment and uses the ﬁrst 20 students who apply. Would this group behave as if it were a simple random sample of size n 20? 7.66 Legal Abortions The results of a Newsweek poll concerning views on abortion given in the table that follows show that there is no consensus on this issue among Americans.17 Newsweek Poll conducted by Princeton Survey Research Associates International. Oct. 26–27, 2006. N 1002 adults nationwide. MoE 3 (for all adults). “Which side of the political debate on the abortion issue do you sympathize with more: the right-to-life movement that believes abortion is the taking of human life and should be outlawed; OR, the pro-choice movement that believes a woman has the right to choose what happens to her body, including deciding to have an abortion?” (Options rotated) Right-to-Life % Pro-Choice % Neither % Unsure % ALL adults Republicans Democrats Independents 39 62 25 35 53 31 69 57 3 4 2 4 5 3 4 4 a. Is this an observational study or a planned experi- ment? b. Is there the possibility of problems in responses arising because of the somewhat sensitive nature of the subject? What kinds of biases might occur? 7.67 Sprouting Radishes A biology experiment was designed to determine whether sprouting radish seeds inhibit the germination of lettuce seeds.18 Three 292 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS 10-centimeter petri dishes were used. The ﬁrst contained 26 lettuce seeds, the second contained 26 radish seeds, and the third contained 13 lettuce seeds and 13 radish seeds. a. Assume that the experimenter had a package of 50 radish seeds and another of 50 lettuce seeds. Devise a plan for randomly assigning the radish and lettuce seeds to the three treatment groups. b. What assumptions must the experimenter make about the packages of 50 seeds in order to assure randomness in the experiment? 7.68 9/11 A study of about n 1000 individuals in the United States during September 21–22, 2001, revealed that 43% of the respondents indicated that they were less willing to ﬂy following the events of September 11, 2001.19 a. Is this an observational study or a designed experi- ment? b. What problems might or could have occurred because of the sensitive nature of the subject? What kinds of biases might have occurred? 7.69 Telephone Service Suppose a telephone company executive wishes to select a random sample of n 20 (a small number is used to simplify the exercise) out of 7000 customers for a survey of customer attitudes concerning service. If the customers are numbered for identiﬁcation purposes, indicate the customers whom you will include in your sample. Use the random number table and explain how you selected your sample. 7.70 Rh Positive The proportion of individuals with an Rh-positive blood type is 85%. You have a random sample of n 500 in
dividuals. a. What are the mean and standard deviation of ˆp, the sample proportion with Rh-positive blood type? b. Is the distribution of ˆp approximately normal? Justify your answer. c. What is the probability that the sample proportion ˆp exceeds 82%? d. What is the probability that the sample proportion lies between 83% and 88%? e. 99% of the time, the sample proportion would lie between what two limits? 7.71 What survey design is used in each of these situations? a. A random sample of n 50 city blocks is se- lected, and a census is done for each single-family dwelling on each block. b. The highway patrol stops every 10th vehicle on a given city artery between 9:00 A.M. and 3:00 P.M. to perform a routine traffic safety check. c. One hundred households in each of four city wards are surveyed concerning a pending city tax relief referendum. d. Every 10th tree in a managed slash pine plantation is checked for pine needle borer infestation. e. A random sample of n 1000 taxpayers from the city of San Bernardino is selected by the Internal Revenue Service and their tax returns are audited. 7.72 Elevator Loads The maximum load (with a generous safety factor) for the elevator in an office building is 2000 pounds. The relative frequency distribution of the weights of all men and women using the elevator is mound-shaped (slightly skewed to the heavy weights), with mean m equal to 150 pounds and standard deviation s equal to 35 pounds. What is the largest number of people you can allow on the elevator if you want their total weight to exceed the maximum weight with a small probability (say, near .01)? (HINT: If x1, x2, . . . , xn are independent observations made on a random variable x, and if x has mean m and variance s 2, then the mean and variance of Sxi are nm and ns 2, respectively. This result was given in Section 7.4.) 7.73 Wiring Packages The number of wiring packages that can be assembled by a company’s employees has a normal distribution, with a mean equal to 16.4 per hour and a standard deviation of 1.3 per hour. a. What are the mean and standard deviation of the number x of packages produced per worker in an 8-hour day? b. Do you expect the probability distribution for x to be mound-shaped and approximately normal? Explain. c. What is the probability that a worker will produce at least 135 packages per 8-hour day? 7.74 Wiring Packages, continued Refer to Exercise 7.73. Suppose the company employs 10 assemblers of wiring packages. a. Find the mean and standard deviation of the company’s daily (8-hour day) production of wiring packages. b. What is the probability that the company’s daily production is less than 1280 wiring packages per day? 7.75 Defective Lightbulbs The table lists the number of defective 60-watt lightbulbs EX0775 found in samples of 100 bulbs selected over 25 days from a manufacturing process. Assume that during these 25 days the manufacturing process was not producing an excessively large fraction of defectives. Day Defectives Day Defectives Day Defectives 1 4 11 2 21 2 2 2 12 4 22 2 3 5 13 3 23 3 4 8 14 4 24 5 5 3 15 0 25 3 6 4 7 4 8 5 9 6 16 2 17 3 18 19 1 4 10 1 20 0 a. Construct a p chart to monitor the manufacturing process, and plot the data. b. How large must the fraction of defective items be in a sample selected from the manufacturing process before the process is assumed to be out of control? c. During a given day, suppose a sample of 100 items is selected from the manufacturing process and 15 defective bulbs are found. If a decision is made to shut down the manufacturing process in an attempt to locate the source of the implied controllable variation, explain how this decision might lead to erroneous conclusions. 7.76 Lightbulbs, continued A hardware store chain purchases large shipments of lightbulbs from the manufacturer described in Exercise 7.75 and speciﬁes that each shipment must contain no more than 4% defectives. When the manufacturing process is in control, what is the probability that the hardware store’s speciﬁcations are met? 7.77 Lightbulbs, again Refer to Exercise 7.75. During a given week the number of defective bulbs in each of ﬁve samples of 100 were found to be 2, 4, 9, 7, and 11. Is there reason to believe that the production process has been producing an excessive proportion of defectives at any time during the week? 7.78 Canned Tomatoes During long production runs of canned tomatoes, the average EX0778 weights (in ounces) of samples of ﬁve cans of standard-grade tomatoes in pureed form were taken at 30 control points during an 11-day period. These results are shown in the table.20 When the machine is performing normally, the average weight per can is 21 ounces with a standard deviation of 1.20 ounces. SUPPLEMENTARY EXERCISES ❍ 293 a. Compute the upper and lower control limits and the centerline for the x chart. b. Plot the sample data on the x chart and determine whether the performance of the machine is in control. Sample Average Number Weight Sample Average Number Weight 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 23.1 21.3 22.0 21.4 21.8 20.6 20.1 21.4 21.5 20.2 20.3 20.1 21.7 21.0 21.6 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 21.4 20.4 22.8 21.1 20.7 21.6 22.4 21.3 21.1 20.1 21.2 19.9 21.1 21.6 21.3 Source: Adapted from J. Hackl, Journal of Quality Technology, April 1991. Used with permission. 7.79 Pepsi or Coke? The battle for consumer preference continues between Pepsi and Coke. How can you make your preferences known? There is a web page where you can vote for one of these colas if you click on the link that says PAY CASH for your opinion. Explain why the respondents do not represent a random sample of the opinions of purchasers or drinkers of these drinks. Explain the types of distortions that could creep into an Internet opinion poll. 7.80 Strawberries An experimenter wants to ﬁnd an appropriate temperature at which to store fresh strawberries to minimize the loss of ascorbic acid. There are 20 storage containers, each with controllable temperature, in which strawberries can be stored. If two storage temperatures are to be used, how would the experimenter assign the 20 containers to one of the two storage temperatures? 7.81 Filling Soda Cans A bottler of soft drinks packages cans in six-packs. Suppose that the ﬁll per can has an approximate normal distribution with a mean of 12 ﬂuid ounces and a standard deviation of 0.2 ﬂuid ounces. a. What is the distribution of the total ﬁll for a case of 24 cans? b. What is the probability that the total ﬁll for a case is less than 286 ﬂuid ounces? 294 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS c. If a six-pack of soda can be considered a random sample of size n 6 from the population, what is the probability that the average ﬁll per can for a six-pack of soda is less than 11.8 ﬂuid ounces? 7.82 Total Packing Weight Packages of food whose average weight is 16 ounces with a standard deviation of 0.6 ounces are shipped in boxes of 24 packages. If the package weights are approximately normally distributed, what is the probability that a box of 24 packages will weigh more than 392 ounces (24.5 pounds)? 7.83 Electronic Components A manufacturing process is designed to produce an electronic compo- Exercises 7.84 Dice Refer to the die-tossing experiment with n 1 in Section 7.4 in which x is the number on the upper face of a single balanced die. a. Use the formulas in Section 4.8 to verify that m 3.5 and s 1.71 for this population. b. Use the Central Limit Theorem applet to toss a single die at least 2000 times. (Your simulation can be done quickly by using the button.) What are the mean and standard deviation of these 2000 observations? What is the shape of the histogram? c. Compare the results of part b to the actual probability distribution shown in Figure 7.3 and the actual mean and standard deviation in part a. They should be similar! 7.85 Dice Two balanced dice are thrown, and the average number on the two upper faces is recorded. a. Use the values m 3.5 and s 1.71 from Exercise 7.84. What are the theoretical mean and standard deviation of the sampling distribution for x? b. Use the Central Limit Theorem applet to toss a single die at least 2000 times. (Your simulation can be done quickly by using the What are the mean and standard deviation of these 2000 observations? What is the shape of the histogram? button.) c. Compare the results of part b to the actual probability distribution shown in Figure 7.4 and the actual mean and standard deviation in part a. nent for use in small portable television sets. The components are all of standard size and need not conform to any measurable characteristic, but are sometimes inoperable when emerging from the manufacturing process. Fifteen samples were selected from the process at times when the process was known to be in statistical control. Fifty components were observed within each sample, and the number of inoperable components was recorded. 6, 7, 3, 5, 6, 8, 4, 5, 7, 3, 1, 6, 5, 4, 5 Construct a p chart to monitor the manufacturing process. 7.86 Repeat the instructions in Exercise 7.85 when three dice are tossed. 7.87 Repeat the instructions in Exercise 7.85 when four dice are tossed. 7.88 Suppose a random sample of n 5 observations is selected from a population that is normally distributed, with mean equal to 1 and standard deviation equal to .36. a. Give the mean and the standard deviation of the sampling distribution of x. b. Find the probability that x exceeds 1.3, using the Normal Probabilities for Means applet. c. Find the probability that the sample mean x is less than .5. d. Find the probability that the sample mean deviates from the population mean m 1 by more than .4. 7.89 Batteries A certain type of automobile battery is known to last an average of 1110 days with a standard deviation of 80 days. If 400 of these batteries are selected, use the Normal Probabilities for Means applet to ﬁnd the following probabilities for the average length of life of the selected batteries:
a. The average is between 1100 and 1110. b. The average is greater than 1120. c. The average is less than 900. CASE STUDY CASE STUDY ❍ 295 Sampling the Roulette at Monte Carlo The technique of simulating a process that contains random elements and repeating the process over and over to see how it behaves is called a Monte Carlo procedure. It is widely used in business and other ﬁelds to investigate the properties of an operation that is subject to random effects, such as weather, human behavior, and so on. For example, you could model the behavior of a manufacturing company’s inventory by creating, on paper, daily arrivals and departures of manufactured products from the company’s warehouse. Each day a random number of items produced by the company would be received into inventory. Similarly, each day a random number of orders of varying random sizes would be shipped. Based on the input and output of items, you could calculate the inventory—that is, the number of items on hand at the end of each day. The values of the random variables, the number of items produced, the number of orders, and the number of items per order needed for each day’s simulation would be obtained from theoretical distributions of observations that closely model the corresponding distributions of the variables that have been observed over time in the manufacturing operation. By repeating the simulation of the supply, the shipping, and the calculation of daily inventory for a large number of days (a sampling of what might really happen), you can observe the behavior of the plant’s daily inventory. The Monte Carlo procedure is particularly valuable because it enables the manufacturer to see how the daily inventory would behave when certain changes are made in the supply pattern or in some other aspect of the operation that could be controlled. In an article entitled “The Road to Monte Carlo,” Daniel Seligman comments on the Monte Carlo method, noting that although the technique is widely used in business schools to study capital budgeting, inventory planning, and cash ﬂow management, no one seems to have used the procedure to study how well we might do if we were to gamble at Monte Carlo.21 To follow up on this thought, Seligman programmed his personal computer to simulate the game of roulette. Roulette involves a wheel with its rim divided into 38 pockets. Thirty-six of the pockets are numbered 1 to 36 and are alternately colored red and black. The two remaining pockets are colored green and are marked 0 and 00. To play the game, you bet a certain amount of money on one or more pockets. The wheel is spun and turns until it stops. A ball falls into a slot on the wheel to indicate the winning number. If you have money on that number, you win a speciﬁed amount. For example, if you were to play the number 20, the payoff is 35 to 1. If the wheel does not stop at that number, you lose your bet. Seligman decided to see how his nightly gains (or losses) would fare if he were to bet $5 on each turn of the wheel and repeat the process 200 times each night. He did this 365 times, thereby simulating the outcomes of 365 nights at the casino. Not surprisingly, the mean “gain” per $1000 evening for the 365 nights was a loss of $55, the average of the winnings retained by the gambling house. The surprise, according to Seligman, was the extreme variability of the nightly “winnings.” Seven times out of the 365 evenings, the ﬁctitious gambler lost the $1000 stake, and only once did he win a maximum of $1160. On 141 nights, the loss exceeded $250. 1. To evaluate the results of Seligman’s Monte Carlo experiment, ﬁrst ﬁnd the prob- ability distribution of the gain x on a single $5 bet. 2. Find the expected value and variance of the gain x from part 1. 296 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS 3. Find the expected value and variance for the evening’s gain, the sum of the gains or losses for the 200 bets of $5 each. 4. Use the results of part 2 to evaluate the probability of 7 out of 365 evenings resulting in a loss of the total $1000 stake. 5. Use the results of part 3 to evaluate the probability that the largest evening’s win- nings were as great as $1160. Large-Sample Estimation 8 © Associated Press GENERAL OBJECTIVE In previous chapters, you learned about the probability distributions of random variables and the sampling distributions of several statistics that, for large sample sizes, can be approximated by a normal distribution according to the Central Limit Theorem. This chapter presents a method for estimating population parameters and illustrates the concept with practical examples. The Central Limit Theorem and the sampling distributions presented in Chapter 7 play a key role in evaluating the reliability of the estimates. CHAPTER INDEX ● Choosing the sample size (8.9) ● Estimating the difference between two binomial propor- tions (8.6) ● Estimating the difference between two population means (8.6) ● Interval estimation (8.5) ● Large-sample conﬁdence intervals for a population mean or proportion (8.5) ● One-sided conﬁdence bounds (8.8) ● Picking the best point estimator (8.4) ● Point estimation for a population mean or proportion (8.4) ● Types of estimators (8.3) How Reliable Is That Poll? Do the national polls conducted by the Gallup and Harris organizations, the news media, and others provide accurate estimates of the percentages of people in the United States who have a variety of eating habits? The case study at the end of this chapter examines the reliability of a poll conducted by CBS News using the theory of large-sample estimation. How Do I Estimate a Population Mean or Proportion? How Do I Choose the Sample Size? 297 298 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION WHERE WE’VE BEEN 8.1 The ﬁrst seven chapters of this book have given you the building blocks you will need to understand statistical inference and how it can be applied in practical situations. The ﬁrst three chapters were concerned with using descriptive statistics, both graphical and numerical, to describe and interpret sets of measurements. In the next three chapters, you learned about probability and probability distributions—the basic tools used to describe populations of measurements. The binomial and the normal distributions were emphasized as important for practical applications. The seventh chapter provided the link between probability and statistical inference. Many statistics are either sums or averages calculated from sample measurements. The Central Limit Theorem states that, even if the sampled populations are not normal, the sampling distributions of these statistics will be approximately normal when the sample size n is large. These statistics are the tools you use for inferential statistics—making inferences about a population using information contained in a sample. WHERE WE’RE GOING— STATISTICAL INFERENCE 8.2 Parameter ⇔ Population Statistic ⇔ Sample Inference—speciﬁcally, decision making and prediction—is centuries old and plays a very important role in most peoples’ lives. Here are some applications: • The government needs to predict short- and long-term interest rates. • A broker wants to forecast the behavior of the stock market. • A metallurgist wants to decide whether a new type of steel is more resistant to high temperatures than the current type. • A consumer wants to estimate the selling price of her house before putting it on the market. There are many ways to make these decisions or predictions, some subjective and some more objective in nature. How good will your predictions or decisions be? Although you may feel that your own built-in decision-making ability is quite good, experience suggests that this may not be the case. It is the job of the mathematical statistician to provide methods of statistical inference making that are better and more reliable than just subjective guesses. Statistical inference is concerned with making decisions or predictions about parameters—the numerical descriptive measures that characterize a population. Three parameters you encountered in earlier chapters are the population mean m, the population standard deviation s, and the binomial proportion p. In statistical inference, a practical problem is restated in the framework of a population with a speciﬁc parameter of interest. For example, the metallurgist could measure the average coefficients of expansion for both types of steel and then compare their values. Methods for making inferences about population parameters fall into one of two categories: • Estimation: Estimating or predicting the value of the parameter • Hypothesis testing: Making a decision about the value of a parameter based on some preconceived idea about what its value might be EXAMPLE 8.1 EXAMPLE 8.2 8.3 TYPES OF ESTIMATORS ❍ 299 The circuits in computers and other electronics equipment consist of one or more printed circuit boards (PCB), and computers are often repaired by simply replacing one or more defective PCBs. In an attempt to ﬁnd the proper setting of a plating process applied to one side of a PCB, a production supervisor might estimate the average thickness of copper plating on PCBs using samples from several days of operation. Since he has no knowledge of the average thickness m before observing the production process, his is an estimation problem. The supervisor in Example 8.1 is told by the plant owner that the thickness of the copper plating must not be less than .001 inch in order for the process to be in control. To decide whether or not the process is in control, the supervisor might formulate a test. He could hypothesize that the process is in control—that is, assume that the average thickness of the copper plating is .001 or greater—and use samples from several days of operation to decide whether or not his hypothesis is correct. The supervisor’s decision-making approach is called a test of hypothesis. Which method of inference should be used? That is, should the parameter be estimated, or shou
ld you test a hypothesis concerning its value? The answer is dictated by the practical question posed and is often determined by personal preference. Since both estimation and tests of hypotheses are used frequently in scientiﬁc literature, we include both methods in this and the next chapter. A statistical problem, which involves planning, analysis, and inference making, is incomplete without a measure of the goodness of the inference. That is, how accurate or reliable is the method you have used? If a stockbroker predicts that the price of a stock will be $80 next Monday, will you be willing to take action to buy or sell your stock without knowing how reliable her prediction is? Will the prediction be within $1, $2, or $10 of the actual price next Monday? Statistical procedures are important because they provide two types of information: • Methods for making the inference • A numerical measure of the goodness or reliability of the inference TYPES OF ESTIMATORS 8.3 To estimate the value of a population parameter, you can use information from the sample in the form of an estimator. Estimators are calculated using information from the sample observations, and hence, by deﬁnition they are also statistics. Definition An estimator is a rule, usually expressed as a formula, that tells us how to calculate an estimate based on information in the sample. Estimators are used in two different ways: • Point estimation: Based on sample data, a single number is calculated to estimate the population parameter. The rule or formula that describes this calculation is called the point estimator, and the resulting number is called a point estimate. 300 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION EXAMPLE 8.3 • Interval estimation: Based on sample data, two numbers are calculated to form an interval within which the parameter is expected to lie. The rule or formula that describes this calculation is called the interval estimator, and the resulting pair of numbers is called an interval estimate or conﬁdence interval. A veterinarian wants to estimate the average weight gain per month of 4-month-old golden retriever pups that have been placed on a lamb and rice diet. The population consists of the weight gains per month of all 4-month-old golden retriever pups that are given this particular diet. The veterinarian wants to estimate the unknown parameter m, the average monthly weight gain for this hypothetical population. One possible estimator based on sample data is the sample mean, x Sxi/n. It could be used in the form of a single number or point estimate—for instance, 3.8 pounds—or you could use an interval estimate and estimate that the average weight gain will be between 2.7 and 4.9 pounds. Both point and interval estimation procedures use information provided by the sampling distribution of the speciﬁc estimator you have chosen to use. We will begin by discussing point estimation and its use in estimating population means and proportions. POINT ESTIMATION 8.4 In a practical situation, there may be several statistics that could be used as point estimators for a population parameter. To decide which of several choices is best, you need to know how the estimator behaves in repeated sampling, described by its sampling distribution. By way of analogy, think of ﬁring a revolver at a target. The parameter of interest is the bull’s-eye, at which you are ﬁring bullets. Each bullet represents a single sample estimate, ﬁred by the revolver, which represents the estimator. Suppose your friend ﬁres a single shot and hits the bull’s-eye. Can you conclude that he is an excellent shot? Would you stand next to the target while he ﬁres a second shot? Probably not, because you have no measure of how well he performs in repeated trials. Does he always hit the bull’s-eye, or is he consistently too high or too low? Do his shots cluster closely around the target, or do they consistently miss the target by a wide margin? Figure 8.1 shows several target conﬁgurations. Which target would you pick as belonging to the best shot? parameter target’s bull’s-eye estimator bullet or arrow F IG URE 8. 1 Which marksman is best? ● Consistently below bulls-eye Consistently above bull’s-eye Off bull’s-eye by a wide margin Best marksmanship 8.4 POINT ESTIMATION ❍ 301 Sampling distributions provide information that can be used to select the best estimator. What characteristics would be valuable? First, the sampling distribution of the point estimator should be centered over the true value of the parameter to be estimated. That is, the estimator should not consistently underestimate or overestimate the parameter of interest. Such an estimator is said to be unbiased. Definition An estimator of a parameter is said to be unbiased if the mean of its distribution is equal to the true value of the parameter. Otherwise, the estimator is said to be biased. The sampling distributions for an unbiased estimator and a biased estimator are shown in Figure 8.2. The sampling distribution for the biased estimator is shifted to the right of the true value of the parameter. This biased estimator is more likely than an unbiased one to overestimate the value of the parameter. FIGURE 8.2 Distributions for biased and unbiased estimators ● Unbiased estimator Biased estimator True value of parameter The second desirable characteristic of an estimator is that the spread (as measured by the variance) of the sampling distribution should be as small as possible. This ensures that, with a high probability, an individual estimate will fall close to the true value of the parameter. The sampling distributions for two unbiased estimators, one with a small variance† and the other with a larger variance, are shown in Figure 8.3. FI GUR E 8. 3 Comparison of estimator variability ● Estimator with smaller variance Estimator with larger variance True value of parameter †Statisticians usually use the term variance of an estimator when in fact they mean the variance of the sampling distribution of the estimator. This contractive expression is used almost universally. 302 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Naturally, you would prefer the estimator with the smaller variance because the estimates tend to lie closer to the true value of the parameter than in the distribution with the larger variance. In real-life sampling situations, you may know that the sampling distribution of an estimator centers about the parameter that you are attempting to estimate, but all you have is the estimate computed from the n measurements contained in the sample. How far from the true value of the parameter will your estimate lie? How close is the marksman’s bullet to the bull’s-eye? The distance between the estimate and the true value of the parameter is called the error of estimation. Definition The distance between an estimate and the estimated parameter is called the error of estimation. In this chapter, you may assume that the sample sizes are always large and, therefore, that the unbiased estimators you will study have sampling distributions that can be approximated by a normal distribution (because of the Central Limit Theorem). Remember that, for any point estimator with a normal distribution, the Empirical Rule states that approximately 95% of all the point estimates will lie within two (or more exactly, 1.96) standard deviations of the mean of that distribution. For unbiased estimators, this implies that the difference between the point estimator and the true value of the parameter will be less than 1.96 standard deviations or 1.96 standard errors (SE). This quantity, called the 95% margin of error (or simply the “margin of error”), provides a practical upper bound for the error of estimation (see Figure 8.4). It is possible that the error of estimation will exceed this margin of error, but that is very unlikely. F IG URE 8. 4 Sampling distribution of an unbiased estimator ● 95% 1.96SE 1.96SE True value Sample estimator Margin of error Margin of error A particular estimate 95% Margin of error 1.96 Standard error POINT ESTIMATION OF A POPULATION PARAMETER • Point estimator: a statistic calculated using sample measurements 95% Margin of error: 1.96 Standard error of the estimator • The sampling distributions for two unbiased point estimators were discussed in Chapter 7. It can be shown that both of these point estimators have the minimum variability of all unbiased estimators and are thus the best estimators you can ﬁnd in each situation. 8.4 POINT ESTIMATION ❍ 303 The variability of the estimator is measured using its standard error. However, you might have noticed that the standard error usually depends on unknown parameters such as s or p. These parameters must be estimated using sample statistics such as s and ˆp. Although not exactly correct, experimenters generally refer to the estimated standard error as the standard error. How Do I Estimate a Population Mean or Proportion? • To estimate the population mean m for a quantitative population, the point estimator x is unbiased with standard error estimated as s SE † n The 95% margin of error when n 30 is estimated as s 1.96 n • To estimate the population proportion p for a binomial population, the point estimator ˆp x/n is unbiased, with standard error estimated as qˆ SE ˆp n The 95% margin of error is estimated as qˆ 1.96ˆp n Assumptions: npˆ 5 and n qˆ 5. EXAMPLE 8.4 An environmentalist is conducting a study of the polar bear, a species found in and around the Arctic Ocean. Their range is limited by the availability of sea ice, which they use as a platform to hunt seals, the mainstay of their diet. The destruction of its habitat on the Arctic ice, which has been attributed to global warming, threatens the bear’s survival as a species; it may become extinct within the century.1 A random sample of n 50 polar bears produced an average weight of 980 pounds with a standard deviation of 105 pounds. Use this information to estimate the average weight of all Arctic polar bears. Sol
ution The random variable measured is weight, a quantitative random variable best described by its mean m. The point estimate of m, the average weight of all Arctic polar bears, is x 980 pounds. The margin of error is estimated as s 5 10 29.10 29 pounds 1.96 0 n 5 1.96 SE 1.96 †When you sample from a normal distribution, the statistic (x m)/(s/ n) has a t distribution, which will be discussed in Chapter 10. When the sample is large, this statistic is approximately normally distributed whether the sampled population is normal or nonnormal. 304 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION You can be fairly conﬁdent that the sample estimate of 980 pounds is within 29 pounds of the population mean. In reporting research results, investigators often attach either the sample standard deviation s (sometimes called SD) or the standard error s/n (usually called SE or SEM) to the estimates of population means. You should always look for an explanation somewhere in the text of the report that tells you whether the investigator is reporting x SD or x SE. In addition, the sample means and standard deviations or standard errors are often presented as “error bars” using the graphical format shown in Figure 8.5. F IG URE 8. 5 Plot of treatment means and their standard errors ● 15 e 10 s n o p s e R 5 SE SE A B Treatments EXAMPLE 8.5 In addition to the average weight of the Arctic polar bear, the environmentalist from Example 8.4 is also interested in the opinions of adults on the subject of global warming. In particular, he wants to estimate the proportion of adults who think that global warming is a very serious problem. In a random sample of n 100 adults, 73% of the sample indicated that, from what they have heard or read, global warming is a very serious problem. Estimate the true population proportion of adults who believe that global warming is a very serious problem, and ﬁnd the margin of error for the estimate. Solution The parameter of interest is now p, the proportion of individuals in the population who believe that global warming is a very serious problem. The best estimator of p is the sample proportion pˆ, which for this sample is pˆ .73. In order to ﬁnd the margin of error, you can approximate the value of p with its estimate pˆ .73: 1.96 SE 1.96 pˆ qˆ 1.96 .73 7) .09 ( .2 0 0 1 n With this margin of error, you can be fairly conﬁdent that the estimate of .73 is within .09 of the true value of p. Hence, you can conclude that the true value of p could be as small as .64 or as large as .82. This margin of error is quite large when compared to the estimate itself and reﬂects the fact that large samples are required to achieve a small margin of error when estimating p. 8.4 POINT ESTIMATION ❍ 305 TABLE 8.1 ● Some Calculated Values of pq p .1 .2 .3 .4 .5 pq .09 .16 .21 .24 .25 pq .30 .40 .46 .49 .50 p .6 .7 .8 .9 pq .24 .21 .16 .09 pq .49 .46 .40 .30 Table 8.1 shows how the numerator of the standard error of pˆ changes for various values of p. Notice that, for most values of p—especially when p is between .3 and .7—there is very little change in pq, the numerator of SE, reaching its maximum value when p .5. This means that the margin of error using the estimator pˆ will also be a maximum when p .5. Some pollsters routinely use the maximum margin of error—often called the sampling error—when estimating p, in which case they calculate 1.96 SE 1.96.5( .5) or sometimes 2 SE 2.5( .5) n n Gallup, Harris, and Roper polls generally use sample sizes of approximately 1000, so their margin of error is 1.96. ) .031 5 (. 5 0 0 0 1 or approximately 3% In this case, the estimate is said to be within 3 percentage points of the true population proportion. 8.4 EXERCISES BASIC TECHNIQUES 8.1 Explain what is meant by “margin of error” in point estimation. 8.2 What are two characteristics of the best point estimator for a population parameter? 8.3 Calculate the margin of error in estimating a population mean m for these values: a. n 30, s 2 .2 b. n 30, s 2 .9 c. n 30, s 2 1.5 8.4 Refer to Exercise 8.3. What effect does a larger population variance have on the margin of error? 8.5 Calculate the margin of error in estimating a population mean m for these values: a. n 50, s2 4 b. n 500, s2 4 c. n 5000, s2 4 8.6 Refer to Exercise 8.5. What effect does an increased sample size have on the margin of error? 8.7 Calculate the margin of error in estimating a binomial proportion for each of the following values of n. Use p .5 to calculate the standard error of the estimator. a. n 30 c. n 400 b. n 100 d. n 1000 8.8 Refer to Exercise 8.7. What effect does increasing the sample size have on the margin of error? 8.9 Calculate the margin of error in estimating a binomial proportion p using samples of size n 100 and the following values for p: b. p .3 a. p .1 d. p .7 e. p .9 f. Which of the values of p produces the largest mar- c. p .5 gin of error? 306 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.10 Suppose you are writing a questionnaire for a sample survey involving n 100 individuals. The questionnaire will generate estimates for several different binomial proportions. If you want to report a single margin of error for the survey, which margin of error from Exercise 8.9 is the correct one to use? 8.11 A random sample of n 900 observations from a binomial population produced x 655 successes. Estimate the binomial proportion p and calculate the margin of error. 8.12 A random sample of n 50 observations from a quantitative population produced x 56.4 and s2 2.6. Give the best point estimate for the population mean m, and calculate the margin of error. APPLICATIONS 8.13 The San Andreas Fault Geologists are interested in shifts and movements of the earth’s surface indicated by fractures (cracks) in the earth’s crust. One of the most famous large fractures is the San Andreas fault in California. A geologist attempting to study the movement of the relative shifts in the earth’s crust at a particular location found many fractures in the local rock structure. In an attempt to determine the mean angle of the breaks, she sampled n 50 fractures and found the sample mean and standard deviation to be 39.8° and 17.2°, respectively. Estimate the mean angular direction of the fractures and ﬁnd the margin of error for your estimate. 8.14 Biomass Estimates of the earth’s biomass, the total amount of vegetation held by the earth’s forests, are important in determining the amount of unabsorbed carbon dioxide that is expected to remain in the earth’s atmosphere.2 Suppose a sample of 75 one-square-meter plots, randomly chosen in North America’s boreal (northern) forests, produced a mean biomass of 4.2 kilograms per square meter (kg/m2), with a standard deviation of 1.5 kg/m2. Estimate the average biomass for the boreal forests of North America and ﬁnd the margin of error for your estimate. Source: Reprinted with permission from Science News, the weekly newsmagazine of Science, copyright 1989 by Science Services, Inc. 8.15 Consumer Conﬁdence An increase in the rate of consumer savings is frequently tied to a lack of conﬁdence in the economy and is said to be an indicator of a recessional tendency in the economy. A random sampling of n 200 savings accounts in a local community showed a mean increase in savings account values of 7.2% over the past 12 months, with a stan- dard deviation of 5.6%. Estimate the mean percent increase in savings account values over the past 12 months for depositors in the community. Find the margin of error for your estimate. 8.16 Multimedia Kids Do our children spend as much time enjoying the outdoors and playing with family and friends as previous generations did? Or are our children spending more and more time glued to the television, computer, and other multimedia equipment? A random sample of 250 children between the ages of 8 and 18 showed that 170 children had a TV in their bedroom and that 120 of them had a video game player in their bedroom. a. Estimate the proportion of all 8- to 18-year-olds who have a TV in their bedroom, and calculate the margin of error for your estimate. b. Estimate the proportion of all 8- to 18-year-olds who have a video game player in their bedroom, and calculate the margin of error for your estimate. 8.17 Legal Immigration At a time in U.S. history when there appears to be genuine concern about the number of illegal aliens living in the United States, there also appears to be concern over the number of legal immigrants allowed to move to the United States. In a recent poll that included questions about both legal and illegal immigrants to the United States, 51% of the n 900 registered voters interviewed indicated that the U.S. should decrease the number of legal immigrants entering the United States.3 a. What is a point estimate for the proportion of U.S. registered voters who feel that the United States should decrease the number of legal immigrants entering the United States? Calculate the margin of error. b. The poll reports a margin of error of 3%. How was the reported margin of error calculated so that it can be applied to all of the questions in the survey? 8.18 Summer Vacations One of the major costs involved in planning a summer vacation is the cost of lodging. Even within a particular chain of hotels, costs can vary substantially depending on the type of room and the amenities offered.4 Suppose that we randomly select 50 billing statements from each of the computer databases of the Marriott, Radisson, and Wyndham hotel chains, and record the nightly room rates. Marriott Radisson Wyndham Sample average Sample standard deviation $170 17.5 $145 10 $150 16.5 a. Describe the sampled population(s). b. Find a point estimate for the average room rate for the Marriott hotel chain. Calculate the margin of error. c. Find a point estimate for the average room rate for the Radisson hotel chain. Calculate the margin of error. d. Find a point estimate for the average room rate for the Wyndham hotel chain. Calculate the margin of error.
e. Display the results of parts b, c, and d graphically, using the form shown in Figure 8.5. Use this display to compare the average room rates for the three hotel chains. 8.19 “900” Numbers Radio and television stations often air controversial issues during broadcast time and ask viewers to indicate their agreement or disagreement with a given stand on the issue. A poll is conducted by asking those viewers who agree to call a certain 900 telephone number and those who disagree to call a second 900 telephone number. All respondents pay a fee for their calls. a. Does this polling technique result in a random sample? b. What can be said about the validity of the results of such a survey? Do you need to worry about a margin of error in this case? 8.5 INTERVAL ESTIMATION ❍ 307 8.20 Men On Mars? The Mars twin rovers, Spirit and Opportunity, which roamed the surface of Mars several years ago, found evidence that there was once water on Mars, raising the possibility that there was once life on the planet. Do you think that the United States should pursue a program to send humans to Mars? An opinion poll conducted by the Associated Press indicated that 49% of the 1034 adults surveyed think that we should pursue such a program.5 a. Estimate the true proportion of Americans who think that the United States should pursue a program to send humans to Mars. Calculate the margin of error. b. The question posed in part a was only one of many questions concerning our space program that were asked in the opinion poll. If the Associated Press wanted to report one sampling error that would be valid for the entire poll, what value should they report? 8.21 Hungry Rats In an experiment to assess the strength of the hunger drive in rats, 30 previously trained animals were deprived of food for 24 hours. At the end of the 24-hour period, each animal was put into a cage where food was dispensed if the animal pressed a lever. The length of time the animal continued pressing the bar (although receiving no food) was recorded for each animal. If the data yielded a sample mean of 19.3 minutes with a standard deviation of 5.2 minutes, estimate the true mean time and calculate the margin of error. INTERVAL ESTIMATION 8.5 like lariat roping: parameter fence post interval estimate lariat An interval estimator is a rule for calculating two numbers—say, a and b—to create an interval that you are fairly certain contains the parameter of interest. The concept of “fairly certain” means “with high probability.” We measure this probability using the conﬁdence coefficient, designated by 1 a. Definition The probability that a conﬁdence interval will contain the estimated parameter is called the conﬁdence coefficient. For example, experimenters often construct 95% conﬁdence intervals. This means that the conﬁdence coefficient, or the probability that the interval will contain the estimated parameter, is .95. You can increase or decrease your amount of certainty by changing the conﬁdence coefficient. Some values typically used by experimenters are .90, .95, .98, and .99. Consider an analogy—this time, throwing a lariat at a fence post. The fence post represents the parameter that you wish to estimate, and the loop formed by the lariat represents the conﬁdence interval. Each time you throw your lariat, you hope to rope the fence post; however, sometimes your lariat misses. In the same way, each time 308 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION you draw a sample and construct a conﬁdence interval for a parameter, you hope to include the parameter in your interval, but, just like the lariat, sometimes you miss. Your “success rate”—the proportion of intervals that “rope the post” in repeated sampling—is the conﬁdence coefficient. Constructing a Confidence Interval When the sampling distribution of a point estimator is approximately normal, an interval estimator or conﬁdence interval can be constructed using the following reasoning. For simplicity, assume that the conﬁdence coefficient is .95 and refer to Figure 8.6. F IG URE 8. 6 Parameter 1.96 SE ● 95% Parameter Parameter 1.96 SE Estimator • We know that, of all possible values of the estimator that we might select, 95% of them will be in the interval Parameter 1.96 SE shown in Figure 8.6. • Since the value of the parameter is unknown, consider constructing the interval estimator 1.96 SE which has the same width as the ﬁrst interval, but has a variable center. • How often will this interval work properly and enclose the parameter of interest? Refer to Figure 8.7. F IG URE 8. 7 Some 95% conﬁdence intervals ● 95% Interval 1 Interval 2 Interval 3 like a game of ring toss: parameter peg interval estimate ring 8.5 INTERVAL ESTIMATION ❍ 309 The ﬁrst two intervals work properly—the parameter (marked with a dotted line) is contained within both intervals. The third interval does not work, since it fails to enclose the parameter. This happened because the value of the estimator at the center of the interval was too far away from the parameter. Fortunately, values of the estimator only fall this far away 5% of the time—our procedure will work properly 95% of the time! You may want to change the conﬁdence coefficient from (1 a) .95 to another conﬁdence level (1 a). To accomplish this, you need to change the value z 1.96, which locates an area .95 in the center of the standard normal curve, to a value of z that locates the area (1 a) in the center of the curve, as shown in Figure 8.8. Since the total area under the curve is 1, the remaining area in the two tails is a, and each tail contains area a/2. The value of z that has “tail area” a/2 to its right is called za/2, and the area between za/2 and za/2 is the conﬁdence coefficient (1 a). Values of za/2 that are typically used by experimenters will become familiar to you as you begin to construct conﬁdence intervals for different practical situations. Some of these values are given in Table 8.2. FI GUR E 8. 8 Location of za/2 ● f(z) (1 – α) α/2 α/2 –z α/2 0 z α/2 z A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL (Point estimator) za/2 (Standard error of the estimator) where za/2 is the z-value with an area a/2 in the right tail of a standard normal distribution. This formula generates two values; the lower conﬁdence limit (LCL) and the upper conﬁdence limit (UCL). TABLE 8.2 ● Values of z Commonly Used for Confidence Intervals Conﬁdence coefficient, (1 a) .90 .95 .98 .99 a .10 .05 .02 .01 a/2 .05 .025 .01 .005 za/2 1.645 1.96 2.33 2.58 310 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Large-Sample Confidence Interval for a Population Mean m Practical problems very often lead to the estimation of m, the mean of a population of quantitative measurements. Here are some examples: • The average achievement of college students at a particular university • The average strength of a new type of steel • The average number of deaths per age category • The average demand for a new cosmetics product When the sample size n is large, the sample mean x is the best point estimator for the population mean m. Since its sampling distribution is approximately normal, it can be used to construct a conﬁdence interval according to the general approach given earlier. A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR A POPULATION MEAN m s x za/2 n where za/2 is the z-value corresponding to an area a/2 in the upper tail of a standard normal z distribution, and n Sample size s Standard deviation of the sampled population If s is unknown, it can be approximated by the sample standard deviation s when the sample size is large (n 30) and the approximate conﬁdence interval is s x za/2 n Another way to ﬁnd the large-sample conﬁdence interval for a population mean m is to begin with the statistic m n / x z s which has a standard normal distribution. If you write za/2 as the value of z with area a/2 to its right, then you can write m x za/2 1 a Pza/2 n s / You can rewrite this inequality as s s za/2 x m za/2 n n s s m x za/2 x za/2 n n EXAMPLE 8.6 A 95% conﬁdence interval tells you that, if you were to construct many of these intervals (all of which would have slightly different endpoints), 95% of them would enclose the population mean. ● FI GUR E 8. 9 Twenty conﬁdence intervals for the mean for Example 8.6 8.5 INTERVAL ESTIMATION ❍ 311 so that Px za/2 s s 1 a m x za/2 n n Both x za/2(s/n) and x za/2(s/n), the lower and upper conﬁdence limits, are actually random quantities that depend on the sample mean x. Therefore, in repeated sampling, the random interval, x za/2(s/n), will contain the population mean m with probability (1 a). A scientist interested in monitoring chemical contaminants in food, and thereby the accumulation of contaminants in human diets, selected a random sample of n 50 male adults. It was found that the average daily intake of dairy products was x 756 grams per day with a standard deviation of s 35 grams per day. Use this sample information to construct a 95% conﬁdence interval for the mean daily intake of dairy products for men. Solution Since the sample size of n 50 is large, the distribution of the sample mean x is approximately normally distributed with mean m and standard error estimated by s/n. The approximate 95% conﬁdence interval is x 1.96 s n 3 5 756 1.96 0 5 756 9.70 Hence, the 95% conﬁdence interval for m is from 746.30 to 765.70 grams per day. Interpreting the Confidence Interval What does it mean to say you are “95% conﬁdent” that the true value of the population mean m is within a given interval? If you were to construct 20 such intervals, each using different sample information, your intervals might look like those shown in Figure 8.9. Of the 20 intervals, you might expect that 95% of them, or 19 out of 20, will perform as planned and contain m within their upper and lower bounds. 20 16 12 312 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Remember that you cannot be absolutely sure that any one particular interval contains the mean m. You will never know whether your particular interval is one of
the 19 that “worked,” or whether it is the one interval that “missed.” Your conﬁdence in the estimated interval follows from the fact that when repeated intervals are calculated, 95% of these intervals will contain m. You can try this experiment on your own using the Java applet called Interpreting Conﬁdence Intervals. The applet shown in Figure 8.10(a) shows the calculation of a 95% conﬁdence interval for m when n 50 and s 35. For this particular conﬁdence interval, we used the One Sample button. You can see the value of m shown as a vertical green line on your monitor (gray in Figure 8.10). Notice that this conﬁdence interval worked properly and enclosed the vertical line between its upper and lower limits. Figure 8.10(b) shows the calculation of 100 such intervals, using the 100 Samples button. The intervals that fail to work properly are shown in red on your monitor (black in Figure 8.10). How many intervals fail to work? Is it close to the 95% conﬁdence that we claim to have? You will use this applet again for the MyApplet Exercises section at the end of the chapter. F IG URE 8. 10 Interpeting Conﬁdence Intervals applet ● (a) (b) A good conﬁdence interval has two desirable characteristics: • • It is as narrow as possible. The narrower the interval, the more exactly you have located the estimated parameter. It has a large conﬁdence coefficient, near 1. The larger the conﬁdence coefficient, the more likely it is that the interval will contain the estimated parameter. EXAMPLE 8.7 Construct a 99% conﬁdence interval for the mean daily intake of dairy products for adult men in Example 8.6. 8.5 INTERVAL ESTIMATION ❍ 313 Solution To change the conﬁdence level to .99, you must ﬁnd the appropriate value of the standard normal z that puts area (1 a) .99 in the center of the curve. This value, with tail area a/2 .005 to its right, is found from Table 8.2 to be z 2.58 (see Figure 8.11). The 99% conﬁdence interval is then x 2.58 s n 756 2.58(4.95) 756 12.77 or 743.23 to 768.77 grams per day. This conﬁdence interval is wider than the 95% conﬁdence interval in Example 8.6. FI GUR E 8. 11 Standard normal values for a 99% conﬁdence interval ● f(z) Right Tail Area .05 .025 .01 .005 z-Value 1.645 1.96 2.33 2.58 .005 –2.58 .99 0 α/2 = .005 2.58 z The increased width is necessary to increase the conﬁdence, just as you might want a wider loop on your lariat to ensure roping the fence post! The only way to increase the conﬁdence without increasing the width of the interval is to increase the sample size, n. The standard error of x, s SE n measures the variability or spread of the values of x. The more variable the population data, measured by s, the more variable will be x, and the standard error will be larger. On the other hand, if you increase the sample size n, more information is available for estimating m. The estimates should fall closer to m and the standard error will be smaller. You can use the Exploring Conﬁdence Intervals applet, shown in Figure 8.12, to see the effect of changing the sample size n, the standard deviation s, and the conﬁdence coefficient 1 a on the width of the conﬁdence interval. The conﬁdence intervals of Examples 8.6 and 8.7 are approximate because you substituted s as an approximation for s. That is, instead of the conﬁdence coefficient being .95, the value speciﬁed in the example, the true value of the coefficient may be .92, .94, or .97. But this discrepancy is of little concern from a practical point of view; as far as your “conﬁdence” is concerned, there is little difference among these conﬁdence coefficients. Most interval estimators used in statistics yield approximate conﬁdence intervals because the assumptions upon which they are based are not satisﬁed exactly. Having made this point, we will not continue to refer to conﬁdence intervals as “approximate.” It is of little practical concern as long as the actual conﬁdence coefficient is near the value speciﬁed. 314 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION F IG URE 8. 12 Exploring Conﬁdence Intervals applet ● Large-Sample Confidence Interval for a Population Proportion p Many research experiments or sample surveys have as their objective the estimation of the proportion of people or objects in a large group that possess a certain characteristic. Here are some examples: • The proportion of sales that can be expected in a large number of customer contacts • The proportion of seeds that germinate • The proportion of “likely” voters who plan to vote for a particular political candidate Each is a practical example of the binomial experiment, and the parameter to be estimated is the binomial proportion p. When the sample size is large, the sample proportion, x pˆ n Total number of successes Total number of trials is the best point estimator for the population proportion p. Since its sampling distribution is approximately normal, with mean p and standard error SE pq/n, pˆ can be used to construct a conﬁdence interval according to the general approach given in this section. A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR A POPULATION PROPORTION p q pˆ za/2p n where za/2 is the z-value corresponding to an area a/2 in the right tail of a standard normal z distribution. Since p and q are unknown, they are estimated using 8.5 INTERVAL ESTIMATION ❍ 315 the best point estimators: pˆ and qˆ. The sample size is considered large when the normal approximation to the binomial distribution is adequate—namely, when npˆ 5 and nqˆ 5. EXAMPLE 8.8 A random sample of 985 “likely” voters—those who are likely to vote in the upcoming election—were polled during a phone-athon conducted by the Republican Party. Of those surveyed, 592 indicated that they intended to vote for the Republican candidate in the upcoming election. Construct a 90% conﬁdence interval for p, the proportion of likely voters in the population who intend to vote for the Republican candidate. Based on this information, can you conclude that the candidate will win the election? Solution The point estimate for p is x 5 9 2 .601 pˆ n 5 8 9 and the standard error is (5 qˆ (.601 pˆ ) 8 9 n .399) .016 The z-value for a 90% conﬁdence interval is the value that has area a/2 .05 in the upper tail of the z distribution, or z.05 1.645 from Table 8.2. The 90% conﬁdence interval for p is thus qˆ pˆ 1.645pˆ n .601 .026 or .575 p .627. You estimate that the percentage of likely voters who intend to vote for the Republican candidate is between 57.5% and 62.7%. Will the candidate win the election? Assuming that she needs more than 50% of the vote to win, and since both the upper and lower conﬁdence limits exceed this minimum value, you can say with 90% conﬁdence that the candidate will win. There are some problems, however, with this type of sample survey. What if the voters who consider themselves “likely to vote” do not actually go to the polls? What if a voter changes his or her mind between now and election day? What if a surveyed voter does not respond truthfully when questioned by the campaign worker? The 90% conﬁdence interval you have constructed gives you 90% conﬁdence only if you have selected a random sample from the population of interest. You can no longer be assured of “90% conﬁdence” if your sample is biased, or if the population of voter responses changes before the day of the election! You may have noticed that the point estimator with its 95% margin of error looks very similar to a 95% conﬁdence interval for the same parameter. This close relationship exists for most of the parameters estimated in this book, but it is not true in general. Sometimes the best point estimator for a parameter does not fall in the middle of the best conﬁdence interval; the best conﬁdence interval may not even be a function of the best point estimator. Although this is a theoretical distinction, you should remember that there is a difference between point and interval estimation, and that the choice between the two depends on the preference of the experimenter. 316 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.5 EXERCISES BASIC TECHNIQUES 8.22 Find and interpret a 95% conﬁdence interval for a population mean m for these values: a. n 36, x 13.1, s2 3.42 b. n 64, x 2.73, s2 .1047 8.23 Find a 90% conﬁdence interval for a population mean m for these values: a. n 125, x .84, s2 .086 b. n 50, x 21.9, s2 3.44 c. Interpret the intervals found in parts a and b. 8.24 Find a (1 a)100% conﬁdence interval for a population mean m for these values: a. a .01, n 38, x 34, s2 12 b. a .10, n 65, x 1049, s2 51 c. a .05, n 89, x 66.3, s2 2.48 8.25 A random sample of n 300 observations from a binomial population produced x 263 successes. Find a 90% conﬁdence interval for p and interpret the interval. 8.26 Suppose the number of successes observed in n 500 trials of a binomial experiment is 27. Find a 95% conﬁdence interval for p. Why is the conﬁdence interval narrower than the conﬁdence interval in Exercise 8.25? 8.27 A random sample of n measurements is selected from a population with unknown mean m and known standard deviation s 10. Calculate the width of a 95% conﬁdence interval for m for these values of n: c. n 400 a. n 100 8.28 Compare the conﬁdence intervals in Exercise 8.27. What effect does each of these actions have on the width of a conﬁdence interval? a. Double the sample size b. Quadruple the sample size b. n 200 8.29 Refer to Exercise 8.28. a. Calculate the width of a 90% conﬁdence interval for m when n 100. b. Calculate the width of a 99% conﬁdence interval for m when n 100. c. Compare the widths of 90%, 95%, and 99% conﬁdence intervals for m. What effect does increasing the conﬁdence coefficient have on the width of the conﬁdence interval? APPLICATIONS 8.30 A Chemistry Experiment Due to a variation in laboratory techniques, impurities in materials, and other unknown factors, the results of an experiment in a chemistry laboratory will not always yield the same numerical answer. In an electrolysis experiment, a class measured the amount of coppe
r precipitated from a saturated solution of copper sulfate over a 30-minute period. The n 30 students calculated a sample mean and standard deviation equal to .145 and .0051 mole, respectively. Find a 90% conﬁdence interval for the mean amount of copper precipitated from the solution over a 30-minute period. 8.31 Acid Rain Acid rain, caused by the reaction of certain air pollutants with rainwater, appears to be a growing problem in the northeastern United States. (Acid rain affects the soil and causes corrosion on exposed metal surfaces.) Pure rain falling through clean air registers a pH value of 5.7 (pH is a measure of acidity: 0 is acid; 14 is alkaline). Suppose water samples from 40 rainfalls are analyzed for pH, and x and s are equal to 3.7 and .5, respectively. Find a 99% conﬁdence interval for the mean pH in rainfall and interpret the interval. What assumption must be made for the conﬁdence interval to be valid? 8.32 MP3 Players Do you own an iPod Nano or a Sony Walkman Bean? These and other brands of MP3 players are becoming more and more popular among younger Americans. An iPod survey reported that 54% of 12- to 17-year-olds, 30% of 18- to 34-year-olds, and 13% of 35- to 54-year-olds own MP3 players.6 Suppose that these three estimates are based on random samples of size 400, 350, and 362, respectively. a. Construct a 95% conﬁdence interval estimate for the proportion of 12- to 17-year-olds who own an MP3 player. b. Construct a 95% conﬁdence interval estimate for the proportion of 18- to 34-year-olds who own an MP3 player. 8.33 Hamburger Meat The meat department of a local supermarket chain packages ground beef using meat trays of two sizes: one designed to hold approximately 1 pound of meat, and one that holds approximately 3 pounds. A random sample of 35 packages in the smaller meat trays produced weight measurements with an average of 1.01 pounds and a standard deviation of .18 pound. a. Construct a 99% conﬁdence interval for the average weight of all packages sold in the smaller meat trays by this supermarket chain. b. What does the phrase “99% conﬁdent” mean? c. Suppose that the quality control department of this supermarket chain intends that the amount of ground beef in the smaller trays should be 1 pound on average. Should the conﬁdence interval in part a concern the quality control department? Explain. 8.34 Legal Abortions The results of a Newsweek poll concerning views on abortion given in Exercise 7.66 showed that of n 1002 adults, 39% favored the “right-to-life” stand, while 53% were “pro-choice.”7 The poll reported a margin of error of plus or minus 3%. a. Construct a 90% conﬁdence interval for the proportion of adult Americans who favor the “right-tolife” position. b. Construct a 90% conﬁdence interval for the proportion of adult Americans who favor the “pro-choice” position. 8.35 SUVs A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A random sample of 500 registrations are selected from a Department of Motor Vehicles database, and 68 are classiﬁed as sports utility vehicles. a. Use a 95% conﬁdence interval to estimate the proportion of sports utility vehicles in California. b. How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers.) 8.36 e-Shopping In a report of why e-shoppers abandon their online sales transactions, Alison Stein Wellner8 found that “pages took too long to load” and “site was so confusing that I couldn’t ﬁnd the product” were the two complaints heard most often. Based on 8.5 INTERVAL ESTIMATION ❍ 317 customers’ responses, the average time to complete an online order form will take 4.5 minutes. Suppose that n 50 customers responded and that the standard deviation of the time to complete an online order is 2.7 minutes. a. Do you think that x, the time to complete the online order form, has a mound-shaped distribution? If not, what shape would you expect? b. If the distribution of the completion times is not normal, you can still use the standard normal distribution to construct a conﬁdence interval for m, the mean completion time for online shoppers. Why? c. Construct a 95% conﬁdence interval for m, the mean completion time for online orders. 8.37 What’s Normal? What is normal, when it comes to people’s body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker9 in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. a. Construct a 99% conﬁdence interval for the average body temperature of healthy people. b. Does the conﬁdence interval constructed in part a contain the value 98.6 degrees, the usual average temperature cited by physicians and others? If not, what conclusions can you draw? 8.38 Rocking the Vote How likely are you to vote in the next presidential election? A random sample of 300 adults was taken, and 192 of them said that they always vote in presidential elections. a. Construct a 95% conﬁdence interval for the proportion of adult Americans who say they always vote in presidential elections. b. An article in American Demographics reports this percentage of 67%.10 Based on the interval constructed in part a, would you disagree with their reported percentage? Explain. c. Can we use the interval estimate from part a to estimate the actual proportion of adult Americans who vote in the 2008 presidential election? Why or why not? 318 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS A problem equally as important as the estimation of a single population mean m for a quantitative population is the comparison of two population means. You may want to make comparisons like these: • The average scores on the Medical College Admission Test (MCAT) for students whose major was biochemistry and those whose major was biology • The average yields in a chemical plant using raw materials furnished by two different suppliers • The average stem diameters of plants grown on two different types of nutrients For each of these examples, there are two populations: the ﬁrst with mean and vari1, and the second with mean and variance m2 and s 2 ance m1 and s 2 2. A random sample of n1 measurements is drawn from population 1, and a second random sample of size n2 is independently drawn from population 2. Finally, the estimates of the population parameters are calculated from the sample data using the estimators x1, s2 1, x2, and s 2 2 as shown in Table 8.3. TABLE 8.3 ● Samples From Two Quantitative Populations Mean Variance Population 1 m1 s 2 1 Population 2 m2 s 2 2 Sample 1 Sample 2 Mean Variance Sample Size x1 s2 1 n1 x2 s2 2 n2 Intuitively, the difference between two sample means would provide the maximum information about the actual difference between two population means, and this is in fact the case. The best point estimator of the difference (m1 m2) between the population means is (x1 x2). The sampling distribution of this estimator is not difficult to derive, but we state it here without proof. PROPERTIES OF THE SAMPLING DISTRIBUTION 1 x— OF (x— SAMPLE MEANS 2), THE DIFFERENCE BETWEEN TWO When independent random samples of n1 and n2 observations have been selected from populations with means m1 and m2 and variances s 2 the sampling distribution of the difference (x1 x2) has the following properties: 2, respectively, 1 and s 2 1. The mean of (x1 x2) is m1 m2 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 319 and the standard error is SE 2 1 s s 2 2 n1 n2 which can be estimated as SE ns2 s2 1 n 1 2 2 when the sample sizes are large. 2. If the sampled populations are normally distributed, then the sampling distribution of (x1 x2) is exactly normally distributed, regardless of the sample size. 3. If the sampled populations are not normally distributed, then the sampling distribution of (x1 x2) is approximately normally distributed when n1 and n2 are both 30 or more, due to the Central Limit Theorem. Since (m1 m2) is the mean of the sampling distribution, it follows that (x1 x2) is an unbiased estimator of (m1 m2) with an approximately normal distribution when n1 and n2 are large. That is, the statistic z (x1 x2) (m1 m2) ns2 s2 1 n 1 2 2 has an approximately standard normal z distribution, and the general procedures of Section 8.5 can be used to construct point and interval estimates. Although the choice between point and interval estimation depends on your personal preference, most experimenters choose to construct conﬁdence intervals for two-sample problems. The appropriate formulas for both methods are given next. LARGE-SAMPLE POINT ESTIMATION OF (m1 m2) Point estimator: (x1 x2) 95% Margin of error: 1.96 SE 1.96 ns2 s2 1 n 1 2 2 A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR (m1 m2) (x1 x2) za/2 2 ns2 s2 1 n 1 2 The wearing qualities of two types of automobile tires were compared by road-testing samples of n1 n2 100 tires for each type. The number of miles until wearout was deﬁned as a speciﬁc amount of tire wear. The test results are given in Table 8.4. Estimate (m1 m2), the difference in mean miles to wearout, using a 99% conﬁdence interval. Is there a difference in the average wearing quality for the two types of tires? Right Tail Area .05 .025 .01 .005 z-Value 1.645 1.96 2.33 2.58 EXAMPLE 8.9 320 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION TABLE 8.4 ● Sample Data Summary for Two Types of Tires Tire 1 x1 26,400 miles 1 1,440,000 s2 Tire 2 x2 25,100 miles 2 1,960,000 s2 Solution The point estimate of (m1 m2) is (x1 x2) 26,400 25,100 1300 miles and the standard error of (x1 x2) is estimated as 1,44 SE 0, 0000 1,96 ns2 s2 000 184.4 miles The 99% conﬁdence interval is calculated as (x1 x2) 2.58 ns2 s2 1 n 1 2 2 1300 2.58(184.4) 1300 475.8 or 824.2 (m1 m2) 1775.8. The difference in the average miles to wearout for the two types of tires is estimated to lie between LCL 824.2 and UCL 17
75.8 miles of wear. Based on this conﬁdence interval, can you conclude that there is a difference in the average miles to wearout for the two types of tires? If there were no difference in the two population means, then m1 and m2 would be equal and (m1 m2) 0. If you look at the conﬁdence interval you constructed, you will see that 0 is not one of the possible values for (m1 m2). Therefore, it is not likely that the means are the same; you can conclude that there is a difference in the average miles to wearout for the two types of tires. The conﬁdence interval has allowed you to make a decision about the equality of the two population means. The scientist in Example 8.6 wondered whether there was a difference in the average daily intakes of dairy products between men and women. He took a sample of n 50 adult women and recorded their daily intakes of dairy products in grams per day. He did the same for adult men. A summary of his sample results is listed in Table 8.5. Construct a 95% conﬁdence interval for the difference in the average daily intakes of dairy products for men and women. Can you conclude that there is a difference in the average daily intakes for men and women? If 0 is not in the interval, you can conclude that there is a difference in the population means. EXAMPLE 8.10 TABLE 8.5 ● Sample Values for Daily Intakes of Dairy Products Sample Size Sample Mean Sample Standard Deviation Men Women 50 756 35 50 762 30 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 321 Solution The conﬁdence interval is constructed using a value of z with tail area a/2 .025 to its right; that is, z.025 1.96. Using the sample standard deviations to approximate the unknown population standard deviations, the 95% conﬁdence interval is (x1 x2) 1.96 (756 762) 1.963 s2 1 n 1 ns2 12.78 or 18.78 (m1 m2) 6.78. Look at the possible values for (m1 m2) in the conﬁdence interval. It is possible that the difference (m1 m2) could be negative (indicating that the average for women exceeds the average for men), it could be positive (indicating that men have the higher average), or it could be 0 (indicating no difference between the averages). Based on this information, you should not be willing to conclude that there is a difference in the average daily intakes of dairy products for men and women. Examples 8.9 and 8.10 deserve further comment with regard to using sample esti- mates in place of unknown parameters. The sampling distribution of (x1 x2) (m1 m2) 2 1 2 s s 2 n1 n2 has a standard normal distribution for all sample sizes when both sampled populations are normal and an approximate standard normal distribution when the sampled 1 and s 2 populations are not normal but the sample sizes are large ( 30). When s 2 2 are not known and are estimated by the sample estimates s2 2, the resulting statistic will still have an approximate standard normal distribution when the sample sizes are large. The behavior of this statistic when the population variances are unknown and the sample sizes are small will be discussed in Chapter 10. 1 and s2 8.6 EXERCISES BASIC TECHNIQUES 8.39 Independent random samples were selected from populations 1 and 2. The sample sizes, means, and variances are as follows: Population 1 35 12.7 1.38 2 49 7.4 4.14 Sample Size Sample Mean Sample Variance a. Find a 95% conﬁdence interval for estimating the difference in the population means (m1 m2). Sample Size Sample Mean Sample Variance Population 1 64 2.9 0.83 2 64 5.1 1.67 b. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the means for the two populations? Explain. 8.40 Independent random samples were selected from populations 1 and 2. The sample sizes, means, and variances are as follows: 322 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION a. Find a 90% conﬁdence interval for the difference in the population means. What does the phrase “90% conﬁdent” mean? b. Find a 99% conﬁdence interval for the difference in the population means. Can you conclude that there is a difference in the two population means? Explain. APPLICATIONS 8.41 Selenium A small amount of the trace element selenium, 50–200 micrograms (mg) per day, is considered essential to good health. Suppose that random samples of n1 n2 30 adults were selected from two regions of the United States and that a day’s intake of selenium, from both liquids and solids, was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 30 adults from region 1 were x1 167.1 and s1 24.3 mg, respectively. The corresponding statistics for the 30 adults from region 2 were x2 140.9 and s2 17.6. Find a 95% conﬁdence interval for the difference in the mean selenium intakes for the two regions. Interpret this interval. 8.42 9-1-1 A study was conducted to compare the mean numbers of police emergency calls per 8-hour shift in two districts of a large city. Samples of 100 8-hour shifts were randomly selected from the police records for each of the two regions, and the number of emergency calls was recorded for each shift. The sample statistics are listed here: Region 1 100 2.4 1.44 2 100 3.1 2.64 Sample Size Sample Mean Sample Variance Find a 90% conﬁdence interval for the difference in the mean numbers of police emergency calls per shift between the two districts of the city. Interpret the interval. 8.43 Teaching Biology In developing a standard for assessing the teaching of precollege sciences in the United States, an experiment was conducted to evaluate a teacher-developed curriculum, “Biology: A Community Context” (BACC) that was standards-based, activity-oriented, and inquiry-centered. This approach was compared to the historical presentation through lecture, vocabulary, and memorized facts. Students were tested on biology concepts that featured biological knowledge and process skills in the traditional sense. The perhaps not-so-startling results from a test on biology concepts, published in The American Biology Teacher, are shown in the following table.11 Pretest: All BACC Classes Pretest: All Traditional Posttest: All BACC Classes Posttest: All Traditional Mean 13.38 14.06 18.5 16.5 Sample Size Standard Deviation 372 368 365 298 5.59 5.45 8.03 6.96 a. Find a 95% conﬁdence interval for the mean score for the posttest for all BACC classes. b. Find a 95% conﬁdence interval for the mean score for the posttest for all traditional classes. c. Find a 95% conﬁdence interval for the difference in mean scores for the posttest BACC classes and the posttest traditional classes. d. Does the conﬁdence interval in c provide evidence that there is a real difference in the posttest BACC and traditional class scores? Explain. Source: From “Performance Assessment of a Standards-Based High School Biology Curriculum,” by W. Leonard, B. Speziale, and J. Pernick in The American Biology Teacher, 2001, 63(5), 310–316. Reprinted by permission of National Association of Biology Teachers. 8.44 Are You Dieting? An experiment was conducted to compare two diets A and B designed for weight reduction. Two groups of 30 overweight dieters each were randomly selected. One group was placed on diet A and the other on diet B, and their weight losses were recorded over a 30-day period. The means and standard deviations of the weight-loss measurements for the two groups are shown in the table. Find a 95% conﬁdence interval for the difference in mean weight loss for the two diets. Interpret your conﬁdence interval. Diet A xA 21.3 sA 2.6 Diet B xB 13.4 sB 1.9 8.45 Starting Salaries College graduates are getting more for their degrees as starting salaries rise. To compare the starting salaries of college graduates majoring in chemical engineering and computer science, random samples of 50 recent college graduates in each major were selected and the following information obtained. Major Chemical engineering Computer science Mean $53,659 51,042 SD 2225 2375 a. Find a point estimate for the difference in starting salaries of college students majoring in chemical engineering and computer science. What is the margin of error for your estimate? 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 323 b. Based upon the results in part a, do you think that there is a signiﬁcant difference in starting salaries for chemical engineers and computer scientists? Explain. c. Do the intervals in parts a and b contain the value (m1 m2) 0? Why is this of interest to the researcher? 8.46 Biology Skills Refer to Exercise 8.43. In addition to tests involving biology concepts, students were also tested on process skills. The results of pretest and posttest scores, published in The American Biology Teacher, are given below.11 Pretest: All BACC Classes Pretest: All Traditional Posttest: All BACC Classes Posstest: All Traditional Mean 10.52 11.97 14.06 12.96 Sample Size Standard Deviation 395 379 376 308 4.79 5.39 5.65 5.93 a. Find a 95% conﬁdence interval for the mean score on process skills for the posttest for all BACC classes. b. Find a 95% conﬁdence interval for the mean score on process skills for the posttest for all traditional classes. c. Find a 95% conﬁdence interval for the difference in mean scores on process skills for the posttest BACC classes and the posttest traditional classes. d. Does the conﬁdence interval in c provide evidence that there is a real difference in the mean process skills scores between posttest BACC and traditional class scores? Explain. Source: From “Peformance Assessment of a Standards-Based High School Biology Curriculum,” by W. Leonard, B. Speziale, and J. Pernick in The American Biology Teacher, 2001, 63(5), 310–316. Reprinted by permission of National Association of Biology Teachers. 8.47 Hotel Costs Refer to Exercise 8.18. The means and standard deviations for 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table:4 Marriott Radisson Wyndham Sample Average Sample Standard Deviation $170 17.5 $145 10 $160 16.5 a.
Find a 95% conﬁdence interval for the difference in the average room rates for the Marriott and the Wyndham hotel chains. b. Find a 99% conﬁdence interval for the difference in the average room rates for the Radisson and the Wyndham hotel chains. d. Do the data indicate a difference in the average room rates between the Marriott and the Wyndham chains? Between the Radisson and the Wyndham chains? 8.48 Noise and Stress To compare the effect of stress in the form of noise on the ability to perform a simple task, 70 subjects were divided into two groups. The ﬁrst group of 30 subjects acted as a control, while the second group of 40 were the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to ﬁnish the task was recorded for each subject and the following summary was obtained: Control Experimental n x s 30 15 minutes 4 minutes 40 23 minutes 10 minutes a. Find a 99% conﬁdence interval for the difference in mean completion times for these two groups. b. Based on the conﬁdence interval in part a, is there sufficient evidence to indicate a difference in the average time to completion for the two groups? Explain. 8.49 What’s Normal, continued Of the 130 people in Exercise 8.37, 65 were female and 65 were male.9 The means and standard deviation of their temperatures are shown below. Sample Mean Standard Deviation Men 98.11 0.70 Women 98.39 0.74 Find a 95% conﬁdence interval for the difference in the average body temperatures for males versus females. Based on this interval, can you conclude that there is a difference in the average temperatures for males versus females? Explain. 324 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION ESTIMATING THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS 8.7 A simple extension of the estimation of a binomial proportion p is the estimation of the difference between two binomial proportions. You may wish to make comparisons like these: • The proportion of defective items manufactured in two production lines • The proportion of female voters and the proportion of male voters who favor an equal rights amendment • The germination rates of untreated seeds and seeds treated with a fungicide These comparisons can be made using the difference ( p1 p2) between two binomial proportions, p1 and p2. Independent random samples consisting of n1 and n2 trials are drawn from populations 1 and 2, respectively, and the sample estimates pˆ1 and pˆ2 are calculated. The unbiased estimator of the difference ( p1 p2) is the sample difference ( pˆ1 pˆ2). PROPERTIES OF THE SAMPLING DISTRIBUTION OF THE DIFFERENCE (pˆ 1 pˆ 2) BETWEEN TWO SAMPLE PROPORTIONS Assume that independent random samples of n1 and n2 observations have been selected from binomial populations with parameters p1 and p2, respectively. The sampling distribution of the difference between sample proportions ( pˆ1 pˆ2) x2 x1 n n 1 2 has these properties: 1. The mean of ( pˆ1 pˆ2) is p1 p2 and the standard error is 2 1 p SE p 2q 1q n n 2 1 which is estimated as ˆ2 ˆ1 pˆ SE pˆ 2q 1q n n 1 2 2. The sampling distribution of ( pˆ1 pˆ2) can be approximated by a normal dis- tribution when n1 and n2 are large, due to the Central Limit Theorem. Although the range of a single proportion is from 0 to 1, the difference between two proportions ranges from 1 to 1. To use a normal distribution to approximate the distribution of ( pˆ1 pˆ2), both pˆ1 and pˆ2 should be approximately normal; that is, n1pˆ1 5, n1qˆ1 5, and n2 pˆ2 5, n2qˆ2 5. The appropriate formulas for point and interval estimation are given next. 8.7 ESTIMATING THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍ 325 LARGE-SAMPLE POINT ESTIMATION OF (p1 p2) Point estimator: ( pˆ1 pˆ2) ˆ2 ˆ1 pˆ 95% Margin of error: 1.96 SE 1.96pˆ 2q 1q n n 1 2 A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR ( p1 p2) ˆ2 ˆ1 pˆ ( pˆ1 pˆ2) za/2pˆ 2q 1q n n 1 2 Assumption: n1 and n2 must be sufficiently large so that the sampling distribution of ( pˆ1 pˆ2) can be approximated by a normal distribution—namely, if n1pˆ1, n1qˆ1, n2pˆ2, and n2qˆ2 are all greater than 5. EXAMPLE 8.11 A bond proposal for school construction will be submitted to the voters at the next municipal election. A major portion of the money derived from this bond issue will be used to build schools in a rapidly developing section of the city, and the remainder will be used to renovate and update school buildings in the rest of the city. To assess the viability of the bond proposal, a random sample of n1 50 residents in the developing section and n2 100 residents from the other parts of the city were asked whether they plan to vote for the proposal. The results are tabulated in Table 8.6. TABLE 8.6 ● Sample Values for Opinion on Bond Proposal Developing Section Rest of the City Sample Size Number Favoring Proposal Proportion Favoring Proposal 50 38 .76 100 65 .65 1. Estimate the difference in the true proportions favoring the bond proposal with 2. a 99% conﬁdence interval. If both samples were pooled into one sample of size n 150, with 103 in favor of the proposal, provide a point estimate of the proportion of city residents who will vote for the bond proposal. What is the margin of error? Solution 1. The best point estimate of the difference ( p1 p2) is given by ( pˆ1 pˆ2) .76 .65 .11 and the standard error of ( pˆ1 pˆ2) is estimated as ˆ2 (.76 ˆ1 pˆ pˆ 2q 1q n n (0 .24) (.65 For a 99% conﬁdence interval, z.005 2.58, and the approximate 99% conﬁdence interval is found as .35) .0770 ) 0 1 )( 0 5 1 2 326 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION ˆ2 ˆ1 pˆ ( pˆ1 pˆ2) z.005pˆ 2q 1q n n 1 2 .11 (2.58)(.0770) .11 .199 or (.089, .309). Since this interval contains the value ( p1 p2) 0, it is possible that p1 p2, which implies that there may be no difference in the proportions favoring the bond issue in the two sections of the city. If there is no difference in the two proportions, then the two samples are not really different and might well be combined to obtain an overall estimate of the proportion of the city residents who will vote for the bond issue. If both samples are pooled, then n 150 and 2. 0 3 pˆ 1 .69 0 5 1 Therefore, the point estimate of the overall value of p is .69, with a margin of error given by 1.96(.69 .31) 1.96(.0378) .074 ( ) 0 5 1 Notice that .69 .074 produces the interval .62 to .76, which includes only proportions greater than .5. Therefore, if voter attitudes do not change adversely prior to the election, the bond proposal should pass by a reasonable majority. 8.7 EXERCISES BASIC TECHNIQUES 8.50 Independent random samples of n1 500 and n2 500 observations were selected from binomial populations 1 and 2, and x1 120 and x2 147 successes were observed. a. What is the best point estimator for the difference ( p1 p2) in the two binomial proportions? b. Calculate the approximate standard error for the statistic used in part a. c. What is the margin of error for this point estimate? 8.51 Independent random samples of n1 800 and n2 640 observations were selected from binomial populations 1 and 2, and x1 337 and x2 374 successes were observed. a. Find a 90% conﬁdence interval for the difference ( p1 p2) in the two population proportions. Interpret the interval. b. What assumptions must you make for the conﬁ- dence interval to be valid? Are these assumptions met? 8.52 Independent random samples of n1 1265 and n2 1688 observations were selected from binomial populations 1 and 2, and x1 849 and x2 910 successes were observed. a. Find a 99% conﬁdence interval for the difference ( p1 p2) in the two population proportions. What does “99% conﬁdence” mean? b. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the two binomial proportions? Explain. APPLICATIONS 8.53 M&M’S Does Mars, Incorporated use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plain M&M’S contained 12 red candies, and another random sample of 32 peanut M&M’S contained 8 red candies. a. Construct a 95% conﬁdence interval for the difference in the proportions of red candies for the plain and peanut varieties. 8.7 ESTIMATING THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍ 327 b. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the proportions of red candies for the plain and peanut varieties? Explain. 8.54 Different Realities The voters in the midterm 2006 elections showed that Democrat and Republican differences extend beyond matters of opinion, and actually include the way they see the world.12 The three most important issues mentioned by Democrats and Republicans are listed below. Republicans (n 995) Democrats (n 1094) 1st Terrorism 42% Iraq war 60% 2nd Economy 41% Economy 44% 3rd Iraq war 37% Health care 44% Use a large-sample estimation procedure to compare the proportions of Republicans and Democrats who mentioned the economy as an important issue in the elections. Explain your conclusions. 8.55 Baseball Fans The ﬁrst day of baseball comes in late March, ending in October with the World Series. Does fan support grow as the season goes on? Two CNN/USA Today/Gallup polls, one conducted in March and one in November, both involved random samples of 1001 adults aged 18 and older. In the March sample, 45% of the adults claimed to be fans of professional baseball, while 51% of the adults in the November sample claimed to be fans.13 a. Construct a 99% conﬁdence interval for the difference in the proportion of adults who claim to be fans in March versus November. b. Does the data indicate that the proportion of adults who claim to be fans increases in November, around the time of the World Series? Explain. 8.56 Baseball and Steroids Refer to Exercise 8.55. In the March opinion poll, suppose that 451 adults identiﬁed themselves as baseball fans, while the other 550 were not fans. The following question was posed: Should major league baseball pla
yers be tested for steroids or other performance enhancing drugs, or not? Suppose that 410 of the baseball fans and 505 of those who are not fans answered yes to this question. a. Construct a 95% conﬁdence interval for the difference in the proportion of adults (fans versus nonfans) who favor mandatory drug testing for professional baseball players. b. Does the data indicate that there is a difference in the proportion of fans versus nonfans who favor mandatory drug testing for professional baseball players? Explain. 8.57 Catching a Cold Do well-rounded people get fewer colds? A study on the Chronicle of Higher Education was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities.14 Suppose that of the 276 healthy men and women tested, n1 96 had only a few social outlets and n2 105 were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: Few Social Outlets Many Social Outlets Sample Size Percent with Colds 96 62% 105 35% a. Construct a 99% conﬁdence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected ﬁnding? 8.58 Union, Yes! A sampling of political candidates—200 randomly chosen from the West and 200 from the East—was classiﬁed according to whether the candidate received backing by a national labor union and whether the candidate won. In the West, 120 winners had union backing, and in the East, 142 winners were backed by a national union. Find a 95% conﬁdence interval for the difference between the proportions of union-backed winners in the West versus the East. Interpret this interval. 8.59 Birth Order and College Success In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were ﬁrstborn or only children. In a sample of 100 nongraduates of comparable age and socioeconomic background, the number of ﬁrstborn or only children was 54. Estimate the difference between the proportions of ﬁrstborn or only children in the two populations from which these samples were drawn. Use a 90% conﬁdence interval and interpret your results. 8.60 Generation Next Born between 1980 and 1990, Generation Next have lived in a post–Cold War world and a time of relative economic prosperity in America, but they have also experienced 328 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION September 11th and the fear of another attack, two Gulf Wars, the tragedy at Columbine High School, Hurricane Katrina, and the increasing polarization of public discourse. More than any who came before, Generation Next is engaged with technology, and the vast majority is dependent upon it.15 Suppose that a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Construct a 98% conﬁdence interval for the difference in the proportions of female and male students who decided to attend college in order to make more money. b. What does it mean to say that you are “98% conﬁdent”? c. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the proportions of female and male students who decided to attend college in order to make more money? 8.61 Excedrin or Tylenol? In a study to compare the effects of two pain relievers it was found that of n1 200 randomly selectd individuals instructed to use the ﬁrst pain reliever, 93% indicated that it relieved their pain. Of n2 450 randomly selected individuals instructed to use the second pain reliever, 96% indicated that it relieved their pain. a. Find a 99% conﬁdence interval for the difference in the proportions experiencing relief from pain for these two pain relievers. b. Based on the conﬁdence interval in part a, is there sufficient evidence to indicate a difference in the proportions experiencing relief for the two pain relievers? Explain. 8.62 Auto Accidents Last year’s records of auto accidents occurring on a given section of highway were classiﬁed according to whether the resulting damage was $1000 or more and to whether a physical injury resulted from the accident. The data follows: Under $1000 $1000 or More Number of Accidents Number Involving Injuries 32 10 41 23 a. Estimate the true proportion of accidents involving injuries when the damage was $1000 or more for similar sections of highway and ﬁnd the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under $1000 and those with damage of $1000 or more. Use a 95% conﬁdence interval. ONE-SIDED CONFIDENCE BOUNDS 8.8 The conﬁdence intervals discussed in Sections 8.5–8.7 are sometimes called two-sided conﬁdence intervals because they produce both an upper (UCL) and a lower (LCL) bound for the parameter of interest. Sometimes, however, an experimenter is interested in only one of these limits; that is, he needs only an upper bound (or possibly a lower bound) for the parameter of interest. In this case, you can construct a one-sided conﬁdence bound for the parameter of interest, such as m, p, m1 m2, or p1 p2. When the sampling distribution of a point estimator is approximately normal, an argument similar to the one in Section 8.5 can be used to show that one-sided conﬁdence bounds, constructed using the following equations when the sample size is large, will contain the true value of the parameter of interest (1 a)100% of the time in repeated sampling. A (1 a)100% LOWER CONFIDENCE BOUND (LCB) (Point estimator) za (Standard error of the estimator) A (1 a)100% UPPER CONFIDENCE BOUND (UCB) (Point estimator) za (Standard error of the estimator) The z-value used for a (1 a)100% one-sided conﬁdence bound, za, locates an area a in a single tail of the normal distribution as shown in Figure 8.13. 8.9 CHOOSING THE SAMPLE SIZE ❍ 329 FI GUR E 8. 13 z-value for a one-sided conﬁdence bound ● f(z) EXAMPLE 8.12 α 0 zα z A corporation plans to issue some short-term notes and is hoping that the interest it will have to pay will not exceed 11.5%. To obtain some information about this problem, the corporation marketed 40 notes, one through each of 40 brokerage ﬁrms. The mean and standard deviation for the 40 interest rates were 10.3% and .31%, respectively. Since the corporation is interested in only an upper limit on the interest rates, ﬁnd a 95% upper conﬁdence bound for the mean interest rate that the corporation will have to pay for the notes. Solution Since the parameter of interest is m, the point estimator is x with stans dard error SE . The conﬁdence coefficient is .95, so that a .05 and n z.05 1.645. Therefore, the 95% upper conﬁdence bound is UCB x 1.645 s 1 .3 10.3 .0806 10.3806 10.3 1.645 0 n 4 Thus, you can estimate that the mean interest rate that the corporation will have to pay on its notes will be less than 10.3806%. The corporation should not be concerned about its interest rates exceeding 11.5%. How conﬁdent are you of this conclusion? Fairly conﬁdent, because intervals constructed in this manner contain m 95% of the time. CHOOSING THE SAMPLE SIZE 8.9 Designing an experiment is essentially a plan for buying a certain amount of information. Just as the price you pay for a video game varies depending on where and when you buy it, the price of statistical information varies depending on how and where the information is collected. As when you buy any product, you should buy as much statistical information as you can for the minimum possible cost. The total amount of relevant information in a sample is controlled by two factors: • The sampling plan or experimental design: the procedure for collecting the information • The sample size n: the amount of information you collect 330 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION You can increase the amount of information you collect by increasing the sample size, or perhaps by changing the type of sampling plan or experimental design you are using. We will discuss the simplest sampling plan—random sampling from a relatively large population—and focus on ways to choose the sample size n needed to purchase a given amount of information. A researcher makes little progress in planning an experiment before encountering the problem of sample size. How many measurements should be included in the sample? How much information does the researcher want to buy? The total amount of information in the sample will affect the reliability or goodness of the inferences made by the researcher, and it is this reliability that the researcher must specify. In a statistical estimation problem, the accuracy of the estimate is measured by the margin of error or the width of the conﬁdence interval. Since both of these measures are a function of the sample size, specifying the accuracy determines the necessary sample size. For instance, suppose you want to estimate the average daily yield m of a chemical process and you need the margin of error to be less than 4 tons. This means that, approximately 95% of the time in repeated sampling, the distance between the sample mean x and the population mean m will be less than 1.96 SE. You want this quantity to be less than 4. That is, s 4 n or 1.96 1.96 SE 4 Solving for n, you obtain 962 n 1. s 2 or 4 n .24s 2 If you know s, the population standard deviation, you can substitute its value into the formula and solve for n. If s is unknown—which is usually the case—you can use the best approximation available: • An estimate s obtained from a previous sample • A range estimate based on knowledge of the largest a
nd smallest possible measurements: s Range/4 For this example, suppose that a prior study of the chemical process produced a sample standard deviation of s 21 tons. Then n .24s 2 .24(21)2 105.8 Using a sample of size n 106 or larger, you could be reasonably certain (with probability approximately equal to .95) that your estimate of the average yield will be within 4 tons of the actual average yield. The solution n 106 is only approximate because you had to use an approximate value for s to calculate the standard error of the mean. Although this may bother you, it is the best method available for selecting the sample size, and it is certainly better than guessing! Sometimes researchers request a different conﬁdence level than the 95% conﬁdence speciﬁed by the margin of error. In this case, the half-width of the conﬁdence interval provides the accuracy measure for your estimate; that is, the bound B on the error of your estimate is s B n za/2 This method for choosing the sample size can be used for all four estimation procedures presented in this chapter. The general procedure is described next. 8.9 CHOOSING THE SAMPLE SIZE ❍ 331 How Do I Choose the Sample Size? Determine the parameter to be estimated and the standard error of its point estimator. Then proceed as follows: 1. Choose B, the bound on the error of your estimate, and a conﬁdence coefficient (1 a). 2. For a one-sample problem, solve this equation for the sample size n: za/2 (Standard error of the estimator) B where za/2 is the value of z having area a/2 to its right. 3. For a two-sample problem, set n1 n2 n and solve the equation in step 2. [NOTE: For most estimators (all presented in this textbook), the standard error is a function of the sample size n.] Exercise Reps Fill in the blanks in the table below and ﬁnd the necessary sample sizes. The ﬁrst problem has been done for you. p or s p .4 Bound, Solve This Inequality B q .1 1.96p n .1 Sample Size n ———93 Type of Data One or Two Samples Margin of Error Binomial 1 Quantitative 2 q 1.96p n s 1.96 n s 6 1 s1 s2 6 2 2 p1 p2 .4 .05 1 p 1.96p 2q 1q n n 2 1 1 2 n n1 n2 .05 n1 n2 Progress Report • Still having trouble with sample sizes? Try again using the Exercise Reps at the end of this section. • Mastered the sample size problem? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. EXAMPLE 8.13 Producers of polyvinyl plastic pipe want to have a supply of pipes sufficient to meet marketing needs. They wish to survey wholesalers who buy polyvinyl pipe in order to estimate the proportion who plan to increase their purchases next year. What sample size is required if they want their estimate to be within .04 of the actual proportion with probability equal to .90? 332 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Solution For this particular example, the bound B on the error of the estimate is .04. Since the conﬁdence coefficient is (1 a) .90, a must equal .10 and a/2 is .05. The z-value corresponding to an area equal to .05 in the upper tail of the z distribution is z.05 1.645. You then require 1.645 SE 1.645p q .04 n In order to solve this equation for n, you must substitute an approximate value of p into the equation. If you want to be certain that the sample is large enough, you should use p .5 (substituting p .5 will yield the largest possible solution for n because the maximum value of pq occurs when p q .5). Then 1.645(.5) (.5) .04 n or )(.5) 20.56 5 4 n (1.6 0 4 . n (20.56)2 422.7 Therefore, the producers must include at least 423 wholesalers in its survey if it wants to estimate the proportion p correct to within .04. EXAMPLE 8.14 A personnel director wishes to compare the effectiveness of two methods of training industrial employees to perform a certain assembly operation. A number of employees are to be divided into two equal groups: the ﬁrst receiving training method 1 and the second training method 2. Each will perform the assembly operation, and the length of assembly time will be recorded. It is expected that the measurements for both groups will have a range of approximately 8 minutes. For the estimate of the difference in mean times to assemble to be correct to within 1 minute with a probability equal to .95, how many workers must be included in each training group? Solution Letting B 1 minute, you get 2 1 1.96 1 s s 2 2 n1 n2 Since you wish n1 to equal n2, you can let n1 n2 n and obtain the equation 1.96 1 2 1 s s 2 2 n n As noted above, the variability (range) of each method of assembly is approximately 2 s 2. Since the range, equal to 8 minutes, is approx1 s 2 the same, and hence s 2 imately equal to 4s, you have 4s 8 or s 2 8.9 CHOOSING THE SAMPLE SIZE ❍ 333 Substituting this value for s1 and s2 in the earlier equation, you get 1 (2 1.96(2 )2 )2 n n 1.968 1 n n 1.968 Solving, you have n 31. Thus, each group should contain at least n 31 workers. Table 8.7 provides a summary of the formulas used to determine the sample sizes required for estimation with a given bound on the error of the estimate or conﬁdence interval width W (W 2B). Notice that to estimate p, the sample size formula uses s 2 pq, whereas to estimate ( p1 p2), the sample size formula uses s 2 1 p1q1 and s 2 2 p2q2. TABLE 8.7 ● Sample Size Formulas Parameter Estimator Sample Size Assumptions m x m1 m2 x1 x2 p pˆ p1 p2 pˆ1 pˆ2 s 2 n z 2 a /2 2 B a/2(s s 2 n z 2 2 2) n1 n2 n 1 2 B n z 2 pq a /2 2 B 2 )z n (.25 a/2 2 B p .5 p2q2) n1 n2 n n z 2 a/2(p1q 1 B 2 or or z 2 5) n 2(.2 a/2 2 B n1 n2 n p1 p2 .5 and 8.9 EXERCISES EXERCISE REPS These exercises refer back to the MyPersonal Trainer section on page 331. 8.63 Fill in the blanks in the table below and ﬁnd the necessary sample sizes. Type of Data One or Two Samples Margin of Error Binomial 1 s 1.96 n p or s p .5 s 10 Bound, Solve This Inequality B Sample Size .05 2 .05 n 2 n 334 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.64 Fill in the blanks in the table below and ﬁnd the necessary sample sizes. Type of Data One or Two Samples Margin of Error Quantitative 2 Bound, Solve This p or s Inequality B s1 s2 10 4 4 Sample Size n1 n2 2 p1 p2 .5 1 p 1.96p 2q 1q n n 1 2 .10 .10 n1 n2 BASIC TECHNIQUES 8.65 Find a 90% one-sided upper conﬁdence bound for the population mean m for these values: a. n 40, s2 65, x 75 b. n 100, s 2.3, x 1.6 8.66 Find a 99% lower conﬁdence bound for the binomial proportion p when a random sample of n 400 trials produced x 196 successes. 8.67 Independent random samples of size 50 are drawn from two quantitative populations, producing the sample information in the table. Find a 95% upper conﬁdence bound for the difference in the two population means. Sample 1 Sample 2 Sample Size Sample Mean Sample Standard Deviation 50 12 5 50 10 7 8.68 Suppose you wish to estimate a population mean based on a random sample of n observations, and prior experience suggests that s 12.7. If you wish to estimate m correct to within 1.6, with probability equal to .95, how many observations should be included in your sample? 8.69 Suppose you wish to estimate a binomial parameter p correct to within .04, with probability equal to .95. If you suspect that p is equal to some value between .1 and .3 and you want to be certain that your sample is large enough, how large should n be? (HINT: When calculating the standard error, use the value of p in the interval .1 p .3 that will give the largest sample size.) 8.70 Independent random samples of n1 n2 n observations are to be selected from each of two populations 1 and 2. If you wish to estimate the difference between the two population means correct to within 1 s 2 .17, with probability equal to .90, how large should n1 and n2 be? Assume that you know s 2 2 27.8. 8.71 Independent random samples of n1 n2 n observations are to be selected from each of two binomial populations 1 and 2. If you wish to estimate the difference in the two population proportions correct to within .05, with probability equal to .98, how large should n be? Assume that you have no prior information on the values of p1 and p2, but you want to make certain that you have an adequate number of observations in the samples. APPLICATIONS 8.72 Operating Expenses A random sampling of a company’s monthly operating expenses for n 36 months produced a sample mean of $5474 and a standard deviation of $764. Find a 90% upper conﬁdence bound for the company’s mean monthly expenses. 8.73 Legal Immigration Exercise 8.17 discussed a research poll conducted for Fox News by Opinion Dynamics concerning opinions about the number of legal immigrants entering the United States.3 Suppose you were designing a poll of this type. a. Explain how you would select your sample. What problems might you encounter in this process? b. If you wanted to estimate the percentage of the population who agree with a particular statement in your survey questionnaire correct to within 1% with probability .95, approximately how many people would have to be polled? 8.74 Political Corruption A questionnaire is designed to investigate attitudes about political corruption in government. The experimenter would like to survey two different groups—Republicans and Democrats—and compare the responses to various “yes/no” questions for the two groups. The experimenter requires that the sampling error for the difference in the proportion of yes responses for the two groups is no more than 3 percentage points. If the two samples are the same size, how large should the samples be? 8.75 Less Red Meat! Americans are becoming more conscious of the importance of good nutrition, and some researchers believe that we may be altering our diets to include less red meat and more fruits and vegetables. To test this theory, a researcher decides to select hospital nutritional records for subjects surveyed 10 years ago and to compare the average amount of beef consumed per year to the amounts consumed by an equal number of subjects she will interview this year. She knows that the amoun
t of beef consumed annually by Americans ranges from 0 to approximately 104 pounds. How many subjects should the researcher select for each group if she wishes to estimate the difference in the average annual per-capita beef consumption correct to within 5 pounds with 99% conﬁdence? 8.76 Red Meat, continued Refer to Exercise 8.75. The researcher selects two groups of 400 subjects each and collects the following sample information on the annual beef consumption now and 10 years ago: Ten Years Ago This Year Sample Mean Sample Standard Deviation 73 25 63 28 a. The researcher would like to show that per-capita beef consumption has decreased in the last 10 years, so she needs to show that the difference in the averages is greater than 0. Find a 99% lower conﬁdence bound for the difference in the average per-capita beef consumptions for the two groups. b. What conclusions can the researcher draw using the conﬁdence bound from part a? 8.77 Hunting Season If a wildlife service wishes to estimate the mean number of days of hunting per hunter for all hunters licensed in the state during a given season, with a bound on the error of estimation equal to 2 hunting days, how many hunters must be 8.9 CHOOSING THE SAMPLE SIZE ❍ 335 included in the survey? Assume that data collected in earlier surveys have shown s to be approximately equal to 10. 8.78 Polluted Rain Suppose you wish to estimate the mean pH of rainfalls in an area that suffers heavy pollution due to the discharge of smoke from a power plant. You know that s is in the neighborhood of .5 pH, and you wish your estimate to lie within .1 of m, with a probability near .95. Approximately how many rainfalls must be included in your sample (one pH reading per rainfall)? Would it be valid to select all of your water specimens from a single rainfall? Explain. 8.79 pH in Rainfall Refer to Exercise 8.78. Suppose you wish to estimate the difference between the mean acidity for rainfalls at two different locations, one in a relatively unpolluted area along the ocean and the other in an area subject to heavy air pollution. If you wish your estimate to be correct to the nearest .1 pH, with probability near .90, approximately how many rainfalls (pH values) would have to be included in each sample? (Assume that the variance of the pH measurements is approximately .25 at both locations and that the samples will be of equal size.) 8.80 GPAs You want to estimate the difference in grade point averages between two groups of college students accurate to within .2 grade point, with probability approximately equal to .95. If the standard deviation of the grade point measurements is approximately equal to .6, how many students must be included in each group? (Assume that the groups will be of equal size.) 8.81 Selenium, again Refer to the comparison of the daily adult intake of selenium in two different regions of the United States in Exercise 8.41. Suppose you wish to estimate the difference in the mean daily intakes between the two regions correct to within 5 micrograms, with probability equal to .90. If you plan to select an equal number of adults from the two regions (i.e., n1 n2), how large should n1 and n2 be? 336 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION CHAPTER REVIEW Key Concepts and Formulas I. Types of Estimators 1. Point estimator: a single number is calculated to estimate the population parameter. 2. Interval estimator: two numbers are calculated to form an interval that, with a certain amount of conﬁdence, contains the parameter. II. Properties of Good Estimators 1. Unbiased: the average value of the estimator equals the parameter to be estimated. 2. Minimum variance: of all the unbiased estimators, the best estimator has a sampling distribution with the smallest standard error. 3. The margin of error measures the maximum distance between the estimator and the true value of the parameter. III. Large-Sample Point Estimators To estimate one of four population parameters when the sample sizes are large, use the following point estimators with the appropriate margins of error. Parameter Point Estimator m p x x pˆ n m1 m2 x1 x2 p1 p 2 (pˆ1 pˆ 2) x2 x1 n n 1 2 95% Margin of Error s 1.96 n 1.96pˆ qˆ n 2 s 1.96s ˆ2 ˆ1 pˆ 1.96pˆ 2 1 n n1 2 2 1q n 1 2q n 2 IV. Large-Sample Interval Estimators To estimate one of four population parameters when the sample sizes are large, use the following interval estimators. Parameter m p m1 m2 p1 p2 (1 a)100% Conﬁdence Interval s x za/2 n pˆ za/2pˆ qˆ n (x1 x2) za/2s (pˆ 1 pˆ 2) za/2pˆ 2 s ˆ2 ˆ1 pˆ 2 1 n1 n 2 2 2q n 2 1q n 1 1. All values in the interval are possible values for the unknown population parameter. 2. Any values outside the interval are unlikely to be the value of the unknown parameter. 3. To compare two population means or propor- tions, look for the value 0 in the conﬁdence interval. If 0 is in the interval, it is possible that the two population means or proportions are equal, and you should not declare a difference. If 0 is not in the interval, it is unlikely that the two means or proportions are equal, and you can conﬁdently declare a difference. V. One-Sided Confidence Bounds Use either the upper () or lower () twosided bound, with the critical value of z changed from za/2 to za. VI. Choosing the Sample Size 1. Determine the size of the margin of error, B, that you are willing to tolerate. 2. Choose the sample size by solving for n or n n1 n2 in the inequality: za/2 B, where SE is a function of the sample size n. 3. For quantitative populations, estimate the population standard deviation using a previously calculated value of s or the range approximation s Range/4. 4. For binomial populations, use the conservative approach and approximate p using the value p .5. Supplementary Exercises 8.82 State the Central Limit Theorem. Of what value is the Central Limit Theorem in large-sample statistical estimation? 8.83 A random sample of n 64 observations has a mean x 29.1 and a standard deviation s 3.9. a. Give the point estimate of the population mean m and ﬁnd the margin of error for your estimate. b. Find a 90% conﬁdence interval for m. What does “90% conﬁdent” mean? c. Find a 90% lower conﬁdence bound for the population mean m. Why is this bound different from the lower conﬁdence limit in part b? d. How many observations do you need to estimate m to within .5, with probability equal to .95? 8.84 Independent random samples of n1 50 and n2 60 observations were selected from populations 1 and 2, respectively. The sample sizes and computed sample statistics are given in the table: Sample Size Sample Mean Sample Standard Deviation Population 1 5 100.4 0.8 2 60 96.2 1.3 Find a 90% conﬁdence interval for the difference in population means and interpret the interval. 8.85 Refer to Exercise 8.84. Suppose you wish to estimate (m1 m2) correct to within .2, with probability equal to .95. If you plan to use equal sample sizes, how large should n1 and n2 be? 8.86 A random sample of n 500 observations from a binomial population produced x 240 successes. a. Find a point estimate for p, and ﬁnd the margin of error for your estimator. b. Find a 90% conﬁdence interval for p. Interpret this interval. 8.87 Refer to Exercise 8.86. How large a sample is required if you wish to estimate p correct to within .025, with probability equal to .90? 8.88 Independent random samples of n1 40 and n2 80 observations were selected from binomial populations 1 and 2, respectively. The number of successes in the two samples were x1 17 and x2 23. Find a 99% conﬁdence interval for the difference between the two binomial population proportions. Interpret this interval. 8.89 Refer to Exercise 8.88. Suppose you wish to estimate (p1 p2) correct to within .06, with probability equal to .99, and you plan to use equal sample sizes—that is, n1 n2. How large should n1 and n2 be? 8.90 Ethnic Cuisine Ethnic groups in America buy differing amounts of various food products because of their ethnic cuisine. Asians buy fewer canned vegetables than do other groups, and Hispanics purchase more cooking oil. A researcher interested in market segmentation for these two groups would like to estimate the proportion of households that select certain brands for various products. If the researcher wishes these estimates to be within .03 with probability .95, SUPPLEMENTARY EXERCISES ❍ 337 how many households should she include in the samples? Assume that the sample sizes are equal. 8.91 Women on Wall Street Women on Wall Street can earn large salaries, but may need to make sacriﬁces in their personal lives. In fact, many women in the securities industry have to make signiﬁcant personal sacriﬁces. A survey of 482 women and 356 men found that only half of the women have children, compared to three-quarters of the men surveyed.16 a. What are the values of pˆ1 and pˆ2 for the women and men in this survey? b. Find a 95% conﬁdence interval for the difference in the proportion of women and men on Wall Street who have children. c. What conclusions can you draw regarding the groups compared in part b? 8.92 Smoking and Blood Pressure An experiment was conducted to estimate the effect of smoking on the blood pressure of a group of 35 cigarette smokers. The difference for each participant was obtained by taking the difference in the blood pressure readings at the beginning of the experiment and again ﬁve years later. The sample mean increase, measured in millimeters of mercury, was x 9.7. The sample standard deviation was s 5.8. Estimate the mean increase in blood pressure that one would expect for cigarette smokers over the time span indicated by the experiment. Find the margin of error. Describe the population associated with the mean that you have estimated. 8.93 Blood Pressure, continued Using a conﬁdence coefficient equal to .90, place a conﬁdence interval on the mean increase in blood pressure for Exercise 8.92. 8.94 Iodine Concentration Based on repeated measurements of the iodine concentration in a solution, a che
mist reports the concentration as 4.614, with an “error margin of .006.” a. How would you interpret the chemist’s “error margin”? b. If the reported concentration is based on a random sample of n 30 measurements, with a sample standard deviation s .017, would you agree that the chemist’s “error margin” is .006? 8.95 Heights If it is assumed that the heights of men are normally distributed with a standard deviation of 2.5 inches, how large a sample should be taken to be fairly sure (probability .95) that the sample mean does 338 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION not differ from the true mean (population mean) by more than .50 in absolute value? 8.96 Chicken Feed An experimenter fed different rations, A and B, to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration A, 13 died, and of the chicks fed ration B, 6 died. a. Construct a 98% conﬁdence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations? 8.97 Antibiotics You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results x 34 ounces per hour and s 3. Estimate the mean hourly yield for the process using a 95% conﬁdence interval. 8.98 Cheese and Soda The average American has become accustomed to eating away from home, especially at fast-food restaurants. Partly as a result of this fast-food habit, the per-capita consumption of cheese (the main ingredient in pizza) and nondiet soft drinks has risen dramatically from a decade ago. A study in American Demographics reports that the average American consumes 25.7 pounds of cheese and drinks 40 gallons (or approximately 645 8-ounce servings) of nondiet soft drinks per year.17 To test the accuracy of these reported averages, a random sample of 40 consumers is selected, and these summary statistics are recorded: Cheese (lb/yr) Soft Drinks (gal/yr) Sample Mean Sample Standard Deviation 28.1 3.8 39.2 4.5 Use your knowledge of statistical estimation to estimate the average per-capita annual consumption for these two products. Does this sample cause you to support or to question the accuracy of the reported averages? Explain. 8.99 Healthy Eating Don’t Americans know that eating pizza and french fries leads to being overweight? In the same American Demographics article referenced in Exercise 8.98, a survey of women who are the main meal preparers in their households reported these results: • 90% know that obesity causes health problems. • 80% know that high fat intake may lead to health problems. • 86% know that cholesterol is a health problem. • 88% know that sodium may have negative effects on health. a. Suppose that this survey was based on a random sample of 750 women. How accurate do you expect the percentages given above to be in estimating the actual population percentages? (HINT: If these are the only four percentages for which you need a margin of error, a conservative estimate for p is p .80.) b. If you want to decrease your sampling error to 1%, how large a sample should you take? 8.100 Sunﬂowers In an article in the Annals of Botany, a researcher reported the basal stem diameters of two groups of dicot sunﬂowers: those that were left to sway freely in the wind and those that were artiﬁcially supported.18 A similar experiment was conducted for monocot maize plants. Although the authors measured other variables in a more complicated experimental design, assume that each group consisted of 64 plants (a total of 128 sunﬂower and 128 maize plants). The values shown in the table are the sample means plus or minus the standard error. Free-Standing Supported Sunﬂower Maize 35.3 .72 32.1 .72 16.2 .41 14.6 .40 Use your knowledge of statistical estimation to compare the free-standing and supported basal diameters for the two plants. Write a paragraph describing your conclusions, making sure to include a measure of the accuracy of your inference. 8.101 A Female President? For a number of years, nearly all Americans say that they would vote for a woman for president IF she were qualiﬁed, and IF she were from their own political party. But is America ready for a female president? A CBS/New York Times poll asked this question of a random sample of 1229 adults, with the following results: 19 % Responding “Yes” Now 1999 55% 60 51 48 61 55 48% 46 49 47 44 54 Total Men Women Republicans Democrats Independents a. Construct a 95% conﬁdence interval for the proportion of all Americans who now believe that America is ready for a female president. b. If there were n1 610 men and n2 619 women in the sample, construct a 95% conﬁdence interval for the difference in the proportion of men and women who now believe that America is ready for a female president. Can you conclude that the proportion of men who now believe that America is ready for a female president is larger than the proportion of women? Explain. c. Look at the percentages of “yes” responses for Republicans, Democrats and Independents now compared to the percentages in 1999. Can you think of a reason why the percentage of Democrats might have changed so dramatically? 8.102 College Costs A dean of men wishes to estimate the average cost of the freshman year at a particular college correct to within $500, with a probability of .95. If a random sample of freshmen is to be selected and each asked to keep ﬁnancial data, how many must be included in the sample? Assume that the dean knows only that the range of expenditures will vary from approximately $4800 to $13,000. 8.103 Quality Control A quality-control engineer wants to estimate the fraction of defectives in a large lot of ﬁlm cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around .05. How large a sample should he take if he wants to estimate the true fraction to within .01, using a 95% conﬁdence interval? 8.104 Circuit Boards Samples of 400 printed circuit boards were selected from each of two production lines A and B. Line A produced 40 defectives, and line B produced 80 defectives. Estimate the difference in the actual fractions of defectives for the two lines with a conﬁdence coefficient of .90. 8.105 Circuit Boards II Refer to Exercise 8.104. Suppose 10 samples of n 400 printed circuit boards were tested and a conﬁdence interval was constructed for p for each of the ten samples. What is the probability that exactly one of the intervals will not contain the true value of p? That at least one interval will not contain the true value of p? 8.106 Ice Hockey The ability to accelerate rapidly is an important attribute for an ice hockey player. G. Wayne Marino investigated some of the variables related to the acceleration and speed of a hockey player from a stopped position.20 Sixty-nine hockey players, varsity and intramural, from the University of Illinois were included in the experiment. Each player SUPPLEMENTARY EXERCISES ❍ 339 was required to move as rapidly as possible from a stopped position to cover a distance of 6 meters. The means and standard deviations of some of the variables recorded for each of the 69 skaters are shown in the table: Mean SD Weight (kilograms) Stide Length (meters) Stride Rate (strides/second) Average Acceleration (meters/second2) Instantaneous Velocity (meters/second) Time to Skate (seconds) 75.270 1.110 3.310 2.962 5.753 1.953 9.470 .205 .390 .529 .892 .131 a. Give the formula that you would use to construct a 95% conﬁdence interval for one of the population means (e.g., mean time to skate the 6-meter distance). b. Construct a 95% conﬁdence interval for the mean time to skate. Interpret this interval. 8.107 Ice Hockey, continued Exercise 8.106 presented statistics from a study of fast starts by ice hockey skaters. The mean and standard deviation of the 69 individual average acceleration measurements over the 6-meter distance were 2.962 and .529 meters per second, respectively. a. Find a 95% conﬁdence interval for this population mean. Interpret the interval. b. Suppose you were dissatisﬁed with the width of this conﬁdence interval and wanted to cut the interval in half by increasing the sample size. How many skaters (total) would have to be included in the study? 8.108 Ice Hockey, continued The mean and standard deviation of the speeds of the sample of 69 skaters at the end of the 6-meter distance in Exercise 8.106 were 5.753 and .892 meters per second, respectively. a. Find a 95% conﬁdence interval for the mean velocity at the 6-meter mark. Interpret the interval. b. Suppose you wanted to repeat the experiment and you wanted to estimate this mean velocity correct to within .1 second, with probability .99. How many skaters would have to be included in your sample? 8.109 School Workers In addition to teachers and administrative staff, schools also have many other employees, including bus drivers, custodians, and cafeteria workers. The average hourly wage is $14.18 for bus drivers, $12.61 for custodians, and $10.33 for cafeteria workers.21 Suppose that a school district 340 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION employs n 36 bus drivers who earn an average of $11.45 per hour with a standard deviation of s $2.84. Find a 95% conﬁdence interval for the average hourly wage of bus drivers in school districts similar to this one. Does your conﬁdence interval enclose the stated average of $14.18? What can you conclude about the hourly wages for bus drivers in this school district? 8.110 Recidivism An experimental rehabilitation technique was used on released convicts. It was shown that 79 of 121 men subjected to the technique pursued useful and crime-free lives for a three-year period following prison release. Find a 95% conﬁdence interval for p, the probability that a convict subjected to the rehabilitation technique will follow a crime-free existence for
at least three years after prison release. 8.111 Speciﬁc Gravity If 36 measurements of the speciﬁc gravity of aluminum had a mean of 2.705 and a standard deviation of .028, construct a 98% conﬁdence interval for the actual speciﬁc gravity of aluminum. 8.112 Audiology Research In a study to establish the absolute threshold of hearing, 70 male college freshmen were asked to participate. Each subject was seated in a soundproof room and a 150 H tone was presented at a large number of stimulus levels in a Exercises 8.116 Refer to the Interpreting Conﬁdence Intervals applet. a. Suppose that you have a random sample of size n 50 from a population with unknown mean m and known standard deviation s 35. Calculate the half width of a 95% conﬁdence interval for m. What would the width of this interval be? b. Use the button to create a single conﬁdence interval for m. What is the width of this interval? Compare your results to the calculation you did in part a. 8.117 Refer to the Interpreting Conﬁdence Intervals applet. a. Use the button to create 10 conﬁdence intervals for m. b. What do you notice about the widths of these intervals? randomized order. The subject was instructed to press a button if he detected the tone; the experimenter recorded the lowest stimulus level at which the tone was detected. The mean for the group was 21.6 db with s 2.1. Estimate the mean absolute threshold for all college freshmen and calculate the margin of error. 8.113 Right- or Left-Handed A researcher classiﬁed his subjects as innately right-handed or lefthanded by comparing thumbnail widths. He took a sample of 400 men and found that 80 men could be classiﬁed as left-handed according to his criterion. Estimate the proportion of all males in the population who would test to be left-handed using a 95% conﬁdence interval. 8.114 The Citrus Red Mite An entomologist wishes to estimate the average development time of the citrus red mite correct to within .5 day. From previous experiments it is known that s is in the neighborhood of 4 days. How large a sample should the entomologist take to be 95% conﬁdent of her estimate? 8.115 The Citrus Red Mite, continued A grower believes that one in ﬁve of his citrus trees are infected with the citrus red mite mentioned in Exercise 8.114. How large a sample should be taken if the grower wishes to estimate the proportion of his trees that are infected with citrus red mite to within .08? c. How many of the intervals work properly and en- close the true value of m? d. Try this simulation again by clicking the button a few more times and counting the number of intervals that work correctly. Is it close to our 95% conﬁdence level? 8.118 Refer to the Interpreting Conﬁdence Intervals applet. a. Use the button to create one hundred conﬁdence intervals for m. b. What do you notice about the widths of these intervals? c. How many of the intervals work properly and en- close the true value of m? d. Try this simulation again by clicking the button a few more times and counting the number of intervals that work correctly. Is it close to our 95% conﬁdence level? 8.119 Suppose that a random sample of size n is selected from a population with mean m 750 and standard deviation s. The Exploring Conﬁdence Intervals applet shows the sampling distribution of x and a representative conﬁdence interval, calculated as s x za/2 n a. The applet loads with n 50, s 35, and x 756. Calculate the half-width of a 95% conﬁdence interval for m. b. Calculate the upper and lower conﬁdence limits and compare these limits to the endpoints of the interval shown in the applet. c. Does the conﬁdence interval work properly? That is, does it enclose the true value of m 750? 8.120 Refer to the Exploring Conﬁdence Intervals applet. a. Use the applet to ﬁnd the values of za/2 for a 99% conﬁdence interval. For a 95% conﬁdence interval? For a 90% conﬁdence interval? b. What effect does reducing the conﬁdence level have on the width of the conﬁdence interval? CASE STUDY ❍ 341 c. A narrower interval indicates a more precise estimate of m, consisting of a smaller range of values. To obtain a more precise estimate by using a smaller z-value, what has been sacriﬁced? 8.121 Refer to the Exploring Conﬁdence Intervals applet. a. Move the slider marked “n” from bottom to top. b. What is the effect of increasing the sample size on the standard error of x? On the width of the conﬁdence interval? c. Can you think of a practical explanation for the phenomena you observe in part b? 8.122 Refer to the Exploring Conﬁdence Intervals applet. a. Move the slider marked “sigma” from bottom to top. b. What is the effect of increasing the variability on the standard error of x? On the width of the conﬁdence interval? c. Can you think of a practical explanation for the phenomena you observe in part b? CASE STUDY How Reliable Is That Poll? CBS News: How and Where America Eats When Americans eat out at restaurants, most choose American food; however, tastes for Mexican, Chinese, and Italian food vary from region to region of the U.S. In a CBS telephone survey22 conducted October 30 through November 1, 2005, it was found that 39% of families ate together 7 nights a week, slightly less than the 46% of families who reported eating together 7 nights a week in a 1990 survey by CBS. Most Americans, both men and women, do some of the cooking when meals are cooked at home, as reported in the following table where we compare the number of evening meals personally cooked per week by men and women. Number of Meals Cooked 3 or less 4 or more Men Women 76 33 24 67 How often Americans eat out at restaurants is largely a function of income. “While most households earning over $50,000 got restaurant food for dinner at least once in the last week, 75% of those earning under $15,000 did not do so at all.” Income All Under $15,000 $15–$30,000 $30–$50,000 Over $50,000 None 1–3 Nights 4 or More Nights 47 75 58 59 31 49 19 39 38 64 4 6 3 3 5 342 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION In spite of all the negative publicity about obesity and high calories associated with burgers and fries, many Americans continue to eat fast food to save time within busy schedules. Fast Food Nights With Kids Without Kids Fast Food Nights Men Women 0 47 59 0 46 63 1 30 20 1 28 20 2–3 19 16 2–3 20 15 4 4 5 4 6 2 Fifty-three percent of families with kids ate fast food at least once last week, compared with 41% of families without kids. Furthermore, 54% of men ate fast food at least once last week, compared with only 37% of women. The description of the survey methods that gave rise to this data was stated as follows: “This poll was conducted among a nationwide random sample of 936 adults, interviewed by telephone October 30 through November 1, 2005. The error due to sampling for results based on the entire sample could be plus or minus three percentage points.” 1. Verify the margin of error of 3 percentage points as stated for the sample of n 936 adults. Suppose that the sample contained an equal number of men and women or 468 men and 468 women. What is the margin of error for men and for women? 2. Do the numbers in the tables indicate the number of people/families in the cate- gories? If not, what do those numbers represent? 3. a. Construct a 95% conﬁdence interval for the proportion of Americans who ate together seven nights a week. b. Construct a 95% conﬁdence interval for the difference in the proportion of women and men who personally cook at least 4 meals per week. c. Construct a 95% conﬁdence interval for the proportion of Americans who eat out at restaurants at least once a week. 4. If these questions were asked today, would you expect the responses to be similar to those reported here or would you expect them to differ signiﬁcantly? 9 Large-Sample Tests of Hypotheses © Scott Olson/Getty Images An Aspirin a Day . . . ? Will an aspirin a day reduce the risk of heart attack? A very large study of U.S. physicians showed that a single aspirin taken every other day reduced the risk of heart attack in men by one-half. However, three days later, a British study reported a completely opposite conclusion. How could this be? The case study at the end of this chapter looks at how the studies were conducted, and you will analyze the data using large-sample techniques. GENERAL OBJECTIVE In this chapter, the concept of a statistical test of hypothesis is formally introduced. The sampling distributions of statistics presented in Chapters 7 and 8 are used to construct largesample tests concerning the values of population parameters of interest to the experimenter. CHAPTER INDEX ● Large-sample test about (m1 m2) (9.4) ● Large-sample test about a population mean m (9.3) ● A statistical test of hypothesis (9.2) ● Testing a hypothesis about (p1 p2) (9.6) ● Testing a hypothesis about a population proportion p (9.5) Rejection Regions, p-Values, and Conclusions How Do I Calculate b? 343 344 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES TESTING HYPOTHESES ABOUT POPULATION PARAMETERS 9.1 In practical situations, statistical inference can involve either estimating a population parameter or making decisions about the value of the parameter. For example, if a pharmaceutical company is fermenting a vat of antibiotic, samples from the vat can be used to estimate the mean potency m for all of the antibiotic in the vat. In contrast, suppose that the company is not concerned about the exact mean potency of the antibiotic, but is concerned only that it meet the minimum government potency standards. Then the company can use samples from the vat to decide between these two possibilities: • The mean potency m does not exceed the minimum allowable potency. • The mean potency m exceeds the minimum allowable potency. The pharmaceutical company’s problem illustrates a statistical test of hypothesis. The reasoning used in a statistical test of hypothesis is similar to the process in a court trial. In trying a person for theft, the court must decide between inn
ocence and guilt. As the trial begins, the accused person is assumed to be innocent. The prosecution collects and presents all available evidence in an attempt to contradict the innocent hypothesis and hence obtain a conviction. If there is enough evidence against innocence, the court will reject the innocence hypothesis and declare the defendant guilty. If the prosecution does not present enough evidence to prove the defendant guilty, the court will ﬁnd him not guilty. Notice that this does not prove that the defendant is innocent, but merely that there was not enough evidence to conclude that the defendant was guilty. We use this same type of reasoning to explain the basic concepts of hypothesis testing. These concepts are used to test the four population parameters discussed in Chapter 8: a single population mean or proportion (m or p) and the difference between two population means or proportions (m1 m2 or p1 p2). When the sample sizes are large, the point estimators for each of these four parameters have normal sampling distributions, so that all four large-sample statistical tests follow the same general pattern. A STATISTICAL TEST OF HYPOTHESIS 9.2 A statistical test of hypothesis consists of ﬁve parts: 1. The null hypothesis, denoted by H0 2. The alternative hypothesis, denoted by Ha 3. The test statistic and its p-value 4. The rejection region 5. The conclusion When you specify these ﬁve elements, you deﬁne a particular test; changing one or more of the parts creates a new test. Let’s look at each part of the statistical test of hypothesis in more detail. 1–2 Definition The two competing hypotheses are the alternative hypothesis Ha, generally the hypothesis that the researcher wishes to support, and the null hypothesis H0, a contradiction of the alternative hypothesis. 9.2 A STATISTICAL TEST OF HYPOTHESIS ❍ 345 As you will soon see, it is easier to show support for the alternative hypothesis by proving that the null hypothesis is false. Hence, the statistical researcher always begins by assuming that the null hypothesis H0 is true. The researcher then uses the sample data to decide whether the evidence favors Ha rather than H0 and draws one of these two conclusions: • Reject H0 and conclude that Ha is true. • Accept (do not reject) H0 as true. EXAMPLE 9.1 You wish to show that the average hourly wage of carpenters in the state of California is different from $14, which is the national average. This is the alternative hypothesis, written as 2 1 Ha : m 14 The null hypothesis is H0 : m 14 You would like to reject the null hypothesis, thus concluding that the California mean is not equal to $14. EXAMPLE 9.2 A milling process currently produces an average of 3% defectives. You are interested in showing that a simple adjustment on a machine will decrease p, the proportion of defectives produced in the milling process. Thus, the alternative hypothesis is 2 1 Ha : p .03 and the null hypothesis is H0 : p .03 If you can reject H0, you can conclude that the adjusted process produces fewer than 3% defectives. There is a difference in the forms of the alternative hypotheses given in Examples 9.1 and 9.2. In Example 9.1, no directional difference is suggested for the value of m; that is, m might be either larger or smaller than $14 if Ha is true. This type of test is called a two-tailed test of hypothesis. In Example 9.2, however, you are speciﬁcally interested in detecting a directional difference in the value of p; that is, if Ha is true, the value of p is less than .03. This type of test is called a one-tailed test of hypothesis. The decision to reject or accept the null hypothesis is based on information contained in a sample drawn from the population of interest. This information takes these forms: Two-tailed ⇔ Look for a sign in Ha. One-tailed ⇔ Look for a or sign in Ha. 3 • Test statistic: a single number calculated from the sample data • p-value: a probability calculated using the test statistic Either or both of these measures act as decision makers for the researcher in deciding whether to reject or accept H0. 346 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.3 For the test of hypothesis in Example 9.1, the average hourly wage x for a random sample of 100 California carpenters might provide a good test statistic for testing H0 : m 14 versus Ha : m 14 If the null hypothesis H0 is true, then the sample mean should not be too far from the population mean m 14. Suppose that this sample produces a sample mean x 15 with standard deviation s 2. Is this sample evidence likely or unlikely to occur, if in fact H0 is true? You can use two measures to ﬁnd out. Since the sample size is large, the sampling distribution of x is approximately normal with mean m 14 and standard error s/n, estimated as 3 4 s 2 SE .2 00 n 1 • The test statistic x 15 lies m 15 x 14 5 z n s / .2 standard deviations from the population mean m. • The p-value is the probability of observing a test statistic as extreme as or more extreme than the observed value, if in fact H0 is true. For this example, we deﬁne “extreme” as far below or far above what we would have expected. That is, p-value P(z 5) P(z 5) 0 The large value of the test statistic and the small p-value mean that you have observed a very unlikely event, if indeed H0 is true and m 14. How do you decide whether to reject or accept H0? The entire set of values that the test statistic may assume is divided into two sets, or regions. One set, consisting of values that support the alternative hypothesis and lead to rejecting H0, is called the rejection region. The other, consisting of values that support the null hypothesis, is called the acceptance region. For example, in Example 9.1, you would be inclined to believe that California’s average hourly wage was different from $14 if the sample mean is either much less than $14 or much greater than $14. The two-tailed rejection region consists of very small and very large values of x, as shown in Figure 9.1. In Example 9.2, since you want to prove that the percentage of defectives has decreased, you would be inclined to reject H0 for values of pˆ that are much smaller than .03. Only small values of pˆ belong in the left-tailed rejection region shown in Figure 9.2. When the rejection region is in the left tail of the distribution, the test is called a left-tailed test. A test with its rejection region in the right tail is called a right-tailed test. F IG URE 9. 1 Rejection and acceptance regions for Example 9.1 ● Rejection region Acceptance region Rejection region x Critical value $14 Critical value 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 347 FI GUR E 9 .2 Rejection and acceptance regions for Example 9.2 ● Rejection region Acceptance region Critical value .03 p 5 If the test statistic falls into the rejection region, then the null hypothesis is rejected. If the test statistic falls into the acceptance region, then either the null hypothesis is accepted or the test is judged to be inconclusive. We will clarify the different types of conclusions that are appropriate as we consider several practical examples of hypothesis tests. Finally, how do you decide on the critical values that separate the acceptance and rejection regions? That is, how do you decide how much statistical evidence you need before you can reject H0? This depends on the amount of conﬁdence that you, the researcher, want to attach to the test conclusions and the signiﬁcance level a, the risk you are willing to take of making an incorrect decision. Definition A Type I error for a statistical test is the error of rejecting the null hypothesis when it is true. The level of signiﬁcance (signiﬁcance level) for a statistical test of hypothesis is a P(Type I error) P(falsely rejecting H0) P(rejecting H0 when it is true) This value a represents the maximum tolerable risk of incorrectly rejecting H0. Once this signiﬁcance level is ﬁxed, the rejection region can be set to allow the researcher to reject H0 with a ﬁxed degree of conﬁdence in the decision. In the next section, we will show you how to use a test of hypothesis to test the value of a population mean m. As we continue, we will clarify some of the computational details and add some additional concepts to complete your understanding of hypothesis testing. A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN 9.3 Consider a random sample of n measurements drawn from a population that has mean m and standard deviation s. You want to test a hypothesis of the form† 1 2 The null hypothesis will always have an “equals” sign attached. H0 : m m0 where m0 is some hypothesized value for m, versus a one-tailed alternative hypothesis: Ha : m m0 The subscript zero indicates the value of the parameter speciﬁed by H0. Notice that H0 provides an exact value for the parameter to be tested, whereas Ha gives a range of possible values for m. †Note that if the test rejects the null hypothesis m m0 in favor of the alternative hypothesis m m0, then it will certainly reject a null hypothesis that includes m m0, since this is even more contradictory to the alternative hypothesis. For this reason, in this text we state the null hypothesis for a one-tailed test as m m0 rather than m m0. 348 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES The Essentials of the Test The sample mean x is the best estimate of the actual value of m, which is presently in question. What values of x would lead you to believe that H0 is false and m is, in fact, greater than the hypothesized value? The values of x that are extremely large would imply that m is larger than hypothesized. Hence, you should reject H0 if x is too large. The next problem is to deﬁne what is meant by “too large.” Values of x that lie too many standard deviations to the right of the mean are not very likely to occur. Those values have very little area to their right. Hence, you can deﬁne “too large” as being too many standard deviations away from m0. But what is “too many”? This question can be answ
ered using the signiﬁcance level a, the probability of rejecting H0 when H0 is true. Remember that the standard error of x is estimated as s SE n Since the sampling distribution of the sample mean x is approximately normal when n is large, the number of standard deviations that x lies from m0 can be measured using the standardized test statistic, m x 0 z n / s which has an approximate standard normal distribution when H0 is true and m m0. The signiﬁcance level a is equal to the area under the normal curve lying above the rejection region. Thus, if you want a .01, you will reject H0 when x is more than 2.33 standard deviations to the right of m0. Equivalently, you will reject H0 if the standardized test statistic z is greater than 2.33 (see Figure 9.3). 3 4 F IG URE 9. 3 The rejection region for a right-tailed test with a .01 ● f(z) α = .01 0 2.33 z Acceptance region Rejection region EXAMPLE 9.4 The average weekly earnings for female social workers is $670. Do men in the same positions have average weekly earnings that are higher than those for women? A random sample of n 40 male social workers showed x $725 and s $102. Test the appropriate hypothesis using a .01. 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 349 For one-tailed tests, look for directional words like “greater,” “less than,” “higher,” “lower,” etc. Solution You would like to show that the average weekly earnings for men are higher than $670, the women’s average. Hence, if m is the average weekly earnings for male social workers, you can set out the formal test of hypothesis in steps: 1–2 Null and alternative hypotheses: H0: m 670 versus Ha: m 670 3 4 5 If the test is two-tailed, you will not see any directional words. The experimenter is only looking for a “difference” from the hypothesized value. Test statistic: Using the sample information, with s as an estimate of the population standard deviation, calculate x 0 6 7 6 70 z 3.41 Rejection region: For this one-tailed test, values of x much larger than 670 would lead you to reject H0; or, equivalently, values of the standardized test statistic z in the right tail of the standard normal distribution. To control the risk of making an incorrect decision as a .01, you must set the critical value separating the rejection and acceptance regions so that the area in the right tail is exactly a .01. This value is found in Table 3 of Appendix I to be z 2.33, as shown in Figure 9.3. The null hypothesis will be rejected if the observed value of the test statistic, z, is greater than 2.33. Conclusion: Compare the observed value of the test statistic, z 3.41, with the critical value necessary for rejection, z 2.33. Since the observed value of the test statistic falls in the rejection region, you can reject H0 and conclude that the average weekly earnings for male social workers are higher than the average for female social workers. The probability that you have made an incorrect decision is a .01. If you wish to detect departures either greater or less than m0, then the alternative hypothesis is two-tailed, written as Ha : m m0 which implies either m m0 or m m0. Values of x that are either “too large” or “too small” in terms of their distance from m0 are placed in the rejection region. If you choose a .01, the area in the rejection region is equally divided between the two tails of the normal distribution, as shown in Figure 9.4. Using the standardized test statistic z, you can reject H0 if z 2.58 or z 2.58. For different values of a, the critical values of z that separate the rejection and acceptance regions will change accordingly. FI GUR E 9 .4 The rejection region for a two-tailed test with a .01 ● f(z) α/2 = .005 α/2 = .005 –2.58 0 Rejection region 2.58 z Rejection region 350 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.5 The daily yield for a local chemical plant has averaged 880 tons for the last several years. The quality control manager would like to know whether this average has changed in recent months. She randomly selects 50 days from the computer database and computes the average and standard deviation of the n 50 yields as x 871 tons and s 21 tons, respectively. Test the appropriate hypothesis using a .05. Solution 1–2 Null and alternative hypotheses: H0: m 880 versus Ha: m 880 3 4 5 Test statistic: The point estimate for m is x. Therefore, the test statistic is m x 80 8 0 3.03 z 0 n 5 / 1/ 2 87 1 s Rejection region: For this two-tailed test, you use values of z in both the right and left tails of the standard normal distribution. Using a .05, the critical values separating the rejection and acceptance regions cut off areas of a/2 .025 in the right and left tails. These values are z 1.96 and the null hypothesis will be rejected if z 1.96 or z 1.96. Conclusion: Since z 3.03 and the calculated value of z falls in the rejection region, the manager can reject the null hypothesis that m 880 tons and conclude that it has changed. The probability of rejecting H0 when H0 is true and a .05, a fairly small probability. Hence, she is reasonably conﬁdent that the decision is correct. LARGE-SAMPLE STATISTICAL TEST FOR m 1. Null hypothesis: H0 : m m0 2. Alternative hypothesis: Two-Tailed Test Ha : m m0 One-Tailed Test Ha : m m0 (or, Ha : m m0) m m x 0 0 estimated as z 3. Test statistic. Rejection region: Reject H0 when One-Tailed Test z za (or z za when the alternative hypothesis is Ha : m m0) Two-Tailed Test z za/2 or z za/2 α α/2 0 zα –zα/2 0 α/2 zα/2 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 351 Assumptions: The n observations in the sample are randomly selected from the population and n is large—say, n 30. Calculating the p-Value In the previous examples, the decision to reject or accept H0 was made by comparing the calculated value of the test statistic with a critical value of z based on the significance level a of the test. However, different signiﬁcance levels may lead to different conclusions. For example, if in a right-tailed test, the test statistic is z 2.03, you can reject H0 at the 5% level of signiﬁcance because the test statistic exceeds z 1.645. However, you cannot reject H0 at the 1% level of signiﬁcance, because the test statistic is less than z 2.33 (see Figure 9.5). To avoid any ambiguity in their conclusions, some experimenters prefer to use a variable level of signiﬁcance called the p-value for the test. Definition The p-value or observed signiﬁcance level of a statistical test is the smallest value of a for which H0 can be rejected. It is the actual risk of committing a Type I error, if H0 is rejected based on the observed value of the test statistic. The p-value measures the strength of the evidence against H0. In the right-tailed test with observed test statistic z 2.03, the smallest critical value you can use and still reject H0 is z 2.03. For this critical value, the risk of an incorrect decision is P(z 2.03) 1 .9788 .0212 This probability is the p-value for the test. Notice that it is actually the area to the right of the calculated value of the test statistic. FI GUR E 9 .5 Variable rejection regions ● f(z) .0500 .0212 .0100 0 1.645 2.03 2.33 z p-value tail area (one or two tails) “beyond” the observed value of the test statistic A small p-value indicates that the observed value of the test statistic lies far away from the hypothesized value of m. This presents strong evidence that H0 is false and should be rejected. Large p-values indicate that the observed test statistic is not far from the hypothesized mean and does not support rejection of H0. How small does the p-value need to be before H0 can be rejected? 352 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.6 Definition If the p-value is less than or equal to a preassigned signiﬁcance level a, then the null hypothesis can be rejected, and you can report that the results are statistically signiﬁcant at level a. In the previous instance, if you choose a .05 as your signiﬁcance level, H0 can be rejected because the p-value is less than .05. However, if you choose a .01 as your signiﬁcance level, the p-value (.0212) is not small enough to allow rejection of H0. The results are signiﬁcant at the 5% level, but not at the 1% level. You might see these results reported in professional journals as signiﬁcant (p .05).† Refer to Example 9.5. The quality control manager wants to know whether the daily yield at a local chemical plant—which has averaged 880 tons for the last several years—has changed in recent months. A random sample of 50 days gives an average yield of 871 tons with a standard deviation of 21 tons. Calculate the p-value for this two-tailed test of hypothesis. Use the p-value to draw conclusions regarding the statistical test. Solution The rejection region for this two-tailed test of hypothesis is found in both tails of the normal probability distribution. Since the observed value of the test statistic is z 3.03, the smallest rejection region that you can use and still reject H0 is z 3.03. For this rejection region, the value of a is the p-value: p-value P(z 3.03) P(z 3.03) (1 .9988) .0012 .0024 Notice that the two-tailed p-value is actually twice the tail area corresponding to the calculated value of the test statistic. If this p-value .0024 is less than or equal to the preassigned level of signiﬁcance a, H0 can be rejected. For this test, you can reject H0 at either the 1% or the 5% level of signiﬁcance. If you are reading a research report, how small should the p-value be before you decide to reject H0? Many researchers use a “sliding scale” to classify their results. • • • • If the p-value is less than .01, H0 is rejected. The results are highly signiﬁcant. If the p-value is between .01 and .05, H0 is rejected. The results are statistically signiﬁcant. If the p-value is between .05 and .10, H0 is usually not rejected. The results are only tending toward statistical signiﬁcance. If the p-value is greater than .10, H0 is not rejected. The results are not statistically signiﬁcant. EXAMPLE 9.
7 Standards set by government agencies indicate that Americans should not exceed an average daily sodium intake of 3300 milligrams (mg). To ﬁnd out whether Americans are exceeding this limit, a sample of 100 Americans is selected, and the mean and standard deviation of daily sodium intake are found to be 3400 mg and 1100 mg, respectively. Use a .05 to conduct a test of hypothesis. †In reporting statistical signiﬁcance, many researchers write (p .05) or (P .05) to mean that the p-value of the test was smaller than .05, making the results signiﬁcant at the 5% level. The symbol p or P in the expression has no connection with our notation for probability or with the binomial parameter p. 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 353 Solution The hypotheses to be tested are versus Ha : m 3300 H0 : m 3300 and the test statistic is 91 small p-value ⇔ large z-value small p-value ⇒ reject H0 How small? p-value a The two approaches developed in this section yield the same conclusions. • The critical value approach: Since the signiﬁcance level is a .05 and the test is one-tailed, the rejection region is determined by a critical value with tail area equal to a .05; that is, H0 can be rejected if z 1.645. Since z .91 is not greater than the critical value, H0 is not rejected (see Figure 9.6). • The p-value approach: Calculate the p-value, the probability that z is greater than or equal to z .91: p-value P(z .91) 1 .8186 .1814 The null hypothesis can be rejected only if the p-value is less than or equal to the speciﬁed 5% signiﬁcance level. Therefore, H0 is not rejected and the results are not statistically signiﬁcant (see Figure 9.6). There is not enough evidence to indicate that the average daily sodium intake exceeds 3300 mg. F IG URE 9. 6 Rejection region and p-value for Example 9.7 ● f(z) p-value = .1814 α = .05 0 .91 1.645 z Reject H0 (z > 1.645) You can use the Large-Sample Test of a Population Mean applet to visualize the p-values for either one- or two-tailed tests of the population mean m (Figure 9.7). Remember, however, that these large-sample z-tests are restricted to samples of size n 30. The applet does not prohibit you from entering a value of n 30; you’ll have to be careful to check the sample size before you start! The procedure follows the same pattern as with previous applets. You enter the values of x, n, and s—remember to press “Enter” after each entry to record the changes. The applet will calculate z (using full accuracy) and give you the option of choosing one- or two-tailed p-values (Area to Left, Area to Right, or Two Tails), as well as a Middle area that you will not need. 354 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES F IG URE 9. 7 Large-Sample Test of a Population Mean applet ● For the data of Example 9.7, the p-value is the one-tailed area to the right of z .909. Do the results shown in the applet conﬁrm our conclusions in Example 9.7? Remember that the applet uses full accuracy for the calculation of z and its corresponding probability. This means that the probability we calculate using Table 3 in Appendix I may be slightly different from the probability shown in the applet. Notice that these two approaches are actually the same, as shown in Figure 9.6. As soon as the calculated value of the test statistic z becomes larger than the critical value, za, the p-value becomes smaller than the signiﬁcance level a. You can use the most convenient of the two methods; the conclusions you reach will always be the same! The p-value approach does have two advantages, however: • Statistical output from packages such as MINITAB usually reports the p-value of the test. • Based on the p-value, your test results can be evaluated using any signiﬁcance level you wish to use. Many researchers report the smallest possible signiﬁcance level for which their results are statistically signiﬁcant. Sometimes it is easy to confuse the signiﬁcance level a with the p-value (or observed signiﬁcance level). They are both probabilities calculated as areas in the tails of the sampling distribution of the test statistic. However, the signiﬁcance level a is preset by the experimenter before collecting the data. The p-value is linked directly to the data and actually describes how likely or unlikely the sample results are, assuming that H0 is true. The smaller the p-value, the more unlikely it is that H0 is true! 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 355 Rejection Regions, p-Values, and Conclusions The signiﬁcance level, a, lets you set the risk that you are willing to take of making an incorrect decision in a test of hypothesis. • To set a rejection region, choose a critical value of z so that the area in the tail(s) of the z-distribution is (are) either a for a one-tailed test or a/2 for a two-tailed test. Use the right tail for an upper-tailed test and the left tail for a lower-tailed test. Reject H0 when the test statistic exceeds the critical value and falls in the rejection region. • To ﬁnd a p-value, ﬁnd the area in the tail “beyond” the test statistic. If the test is one-tailed, this is the p-value. If the test is two-tailed, this is only half the p-value and must be doubled. Reject H0 when the p-value is less than a. Exercise Reps A. Critical value approach: Fill in the blanks in the table below. The ﬁrst prob- lem has been done for you. Signiﬁcance One or Test Level Statistic a .05 z 1.4 z 2.46 a .01 z 0.74 a .05 z 6.12 a .01 Two-tailed Two-tailed Two-Tailed Test? One-tailed (upper) One-tailed (upper) Rejection Critical Value Region Conclusion 1.645 z 1.645 Do not reject H0 B. p-value approach: Fill in the blanks in the table below. The ﬁrst problem has been done for you. Test Statistic z 1.4 z 2.46 z 0.74 z 6.12 Signiﬁcance Level a .05 a .01 a .05 a .01 One or Two-Tailed Test? p-Value One-tailed (upper) .0808 One-tailed (upper) Two-tailed Two-tailed p-Value b a? No Conclusion Do not reject H0 Progress Report • Still having trouble with p-values and rejection regions? Try again using the Excercise Reps at the end of this section. • Mastered p-values and rejection regions? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. 356 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES Two Types of Errors You might wonder why, when H0 was not rejected in the previous example, we did not say that H0 was deﬁnitely true and m 3300. This is because, if we choose to accept H0, we must have a measure of the probability of error associated with this decision. Since there are two choices in a statistical test, there are also two types of errors that can be made. In the courtroom trial, a defendant could be judged not guilty when he’s really guilty, or vice versa—the same is true in a statistical test. In fact, the null hypothesis may be either true or false, regardless of the decision the experimenter makes. These two possibilities, along with the two decisions that can be made by the researcher, are shown in Table 9.1. TABLE 9.1 ● Decision Table Null Hypothesis Decision True False Reject H0 Accept H0 Type I error Correct decision Correct decision Type II error a P (reject H0 when H0 true) b P (accept H0 when H0 false) In addition to the Type I error with probability a deﬁned earlier in this section, it is possible to commit a second error, called a Type II error, which has probability b. Definition A Type I error for a statistical test is the error of rejecting the null hypothesis when it is true. The probability of making a Type I error is denoted by the symbol a. A Type II error for a statistical test is the error of accepting the null hypothesis when it is false and some alternative hypothesis is true. The probability of making a Type II error is denoted by the symbol b. Notice that the probability of a Type I error is exactly the same as the level of signiﬁcance a and is therefore controlled by the researcher. When H0 is rejected, you have an accurate measure of the reliability of your inference; the probability of an incorrect decision is a. However, the probability b of a Type II error is not always controlled by the experimenter. In fact, when H0 is false and Ha is true, you may not be able to specify an exact value for m, but only a range of values. This makes it difficult, if not impossible, to calculate b. Without a measure of reliability, it is not wise to conclude that H0 is true. Rather than risk an incorrect decision, you should withhold judgment, concluding that you do not have enough evidence to reject H0. Instead of accepting H0, you should not reject or fail to reject H0. Keep in mind that “accepting” a particular hypothesis means deciding in its favor. Regardless of the outcome of a test, you are never certain that the hypothesis you “accept” is true. There is always a risk of being wrong (measured by a or b). Consequently, you never “accept” H0 if b is unknown or its value is unacceptable to you. When this situation occurs, you should withhold judgment and collect more data. The Power of a Statistical Test The goodness of a statistical test is measured by the size of the two error rates: a, the probability of rejecting H0 when it is true, and b, the probability of accepting H0 when H0 is false and Ha is true. A “good” test is one for which both of these error rates are 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 357 small. The experimenter begins by selecting a, the probability of a Type I error. If he or she also decides to control the value of b, the probability of accepting H0 when Ha is true, then an appropriate sample size is chosen. Another way of evaluating a test is to look at the complement of a Type II error— that is, rejecting H0 when Ha is true—which has probability 1 b P(reject H0 when Ha is true) The quantity (1 b) is called the power of the test because it measures the probability of taking the action that we wish to have occur—that is, rejecting the null hypothesis when it is false and Ha is true. Defin
ition The power of a statistical test, given as 1 b P(reject H0 when Ha is true) measures the ability of the test to perform as required. A graph of (1 b), the probability of rejecting H0 when in fact H0 is false, as a function of the true value of the parameter of interest is called the power curve for the statistical test. Ideally, you would like a to be small and the power (1 b) to be large. EXAMPLE 9.8 Refer to Example 9.5. Calculate b and the power of the test (1 b) when m is actually equal to 870 tons. Solution The acceptance region for the test of Example 9.5 is located in the interval [m0 1.96(s/n)]. Substituting numerical values, you get 2 1 or 874.18 to 885.82 880 1.96 0 5 The probability of accepting H0, given m 870, is equal to the area under the sampling distribution for the test statistic x in the interval from 874.18 to 885.82. Since x is normally distributed with a mean of 870 and SE 21/50 2.97, b is equal to the area under the normal curve with m 870 located between 874.18 and 885.82 (see Figure 9.8). Calculating the z-values corresponding to 874.18 and 885.82, you get F IGU RE 9 .8 Calculating b in Example 9.8 ● f(x) α /2 = .025 Ha true: µ = 870 Ho true: µ = 880 β α /2 = .025 870 874.18 Rejection region µ 0 = 880 Acceptance region 885.82 x Rejection region 358 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES x 874 870 m 8 .1 1.41 z1 50 n / s/ 1 m 2 .8 5.33 z2 50 n / s/ 1 870 885 2 2 x Then b P(accept H0 when m 870) P(874.18 x 885.82 when m 870) P(1.41 z 5.33) You can see from Figure 9.8 that the area under the normal curve with m 870 above x 885.82 (or z 5.33) is negligible. Therefore, b P(z 1.41) From Table 3 in Appendix I you can ﬁnd b 1 .9207 .0793 Hence, the power of the test is 1 b 1 .0793 .9207 The probability of correctly rejecting H0, given that m is really equal to 870, is .9207, or approximately 92 chances in 100. You can use the Power of a z-Test applet to calculate the power for the hypothesis test in Example 9.8, and also for the same test when the sample size is changed. Refer to Figure 9.9. The applet in Figure 9.9 shows a sample size of n 50. The slider at the bottom of the applet allows you to change the true value of m; the power is recalculated as the mean changes. What is the true value of m and the power of the test shown in the applet? Compare this to the value found in Table 9.2. The slider on the left side of the applet allows you to change a, and the slider on the right allows you to change the sample size n. Remember that n must be 30 for the z-test to be appropriate. You will use these applets to explore power using the MyApplet Exercises at the end of the chapter. F IGU RE 9 .9 Power of a z-Test applet ● 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 359 Values of (1 b) can be calculated for various values of ma different from m0 880 to measure the power of the test. For example, if ma 885, b P(874.18 x 885.82 when m 885) P(3.64 z .28) .6103 0 .6103 and the power is (1 b) .3897. Table 9.2 shows the power of the test for various values of ma, and a power curve is graphed in Figure 9.10. Note that the power of the test increases as the distance between ma and m0 increases. The result is a U-shaped curve for this two-tailed test. TABLE 9.2 ● Value of (1 b) for Various Values of ma for Example 9.8 ma 865 870 872 875 877 880 (1 b) .9990 .9207 .7673 .3897 .1726 .0500 ma 883 885 888 890 895 (1 b) .1726 .3897 .7673 .9207 .9990 FI GUR E 9 .1 0 Power curve for Example 9.8 ● Power, 1 – β 1.0 .8 .6 .4 .2 865 870 875 880 885 890 895 µ There are many important links among the two error rates, a and b, the power, (1 b), and the sample size, n. Look at the two curves shown in Figure 9.8. • If a (the sum of the two tail areas in the curve on the right) is increased, the shaded area corresponding to b decreases, and vice versa. • The only way to decrease b for a ﬁxed a is to “buy” more information—that is, increase the sample size n. What would happen to the area b as the curve on the left is moved closer to the curve on the right (m 880)? With the rejection region in the right curve ﬁxed, the value of b will increase. What effect does this have on the power of the test? Look at Figure 9.10. 360 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES You may also want to use the Power of a z-Test applet to help you visualize the following statements: • As the distance between the true (ma) and hypothesized (m0) values of the mean increases, the power (1 b) increases. The test is better at detecting differences when the distance is large. • The closer the true value (ma) gets to the hypothesized value (m0), the less power (1 b) the test has to detect the difference. • The only way to increase the power (1 b) for a ﬁxed a is to “buy” more information—that is, increase the sample size, n. The experimenter must decide on the values of a and b—measuring the risks of the possible errors he or she can tolerate. He or she also must decide how much power is needed to detect differences that are practically important in the experiment. Once these decisions are made, the sample size can be chosen by consulting the power curves corresponding to various sample sizes for the chosen test. How Do I Calculate b? 1. Find the critical value or values of x used to separate the acceptance and rejec- tion regions. 2. Using one or more values for m consistent with the alternative hypothesis Ha, calculate the probability that the sample mean x falls in the acceptance region. This produces the value b P(accept Ha when m ma). 3. Remember that the power of the test is (1 b). 2.2 EXERCISES 9.3 EXERCISE REPS These exercises refer back to the MyPersonal Trainer section on page 355. 9.1 Critical Value Approach Fill in the blanks in the table below. Test Statistic z 0.88 z 2.67 z 5.05 z 1.22 Signiﬁcance Level a .05 a .05 a .01 a .01 One or Two-Tailed Test? Two-tailed One-tailed (lower) Two-tailed One-tailed (lower) Critical Value Rejection Region Conclusion 9.2 p-value Approach Fill in the blanks in the table below. Test Statistic z 3.01 z 2.47 z 1.30 z 2.88 Signiﬁcance Level a .05 a .05 a .01 a .01 One or Two-Tailed Test? Two-tailed One-tailed (upper) Two-tailed One-tailed (lower) p-Value p-Value b a? Conclusion BASIC TECHNIQUES 9.3 Find the appropriate rejection regions for the large-sample test statistic z in these cases: a. A right-tailed test with a .01 b. A two-tailed test at the 5% signiﬁcance level c. A left-tailed test at the 1% signiﬁcance level d. A two-tailed test with a .01 9.4 Find the p-value for the following large-sample z tests: a. A right-tailed test with observed z 1.15 b. A two-tailed test with observed z 2.78 c. A left-tailed test with observed z 1.81 9.5 For the three tests given in Exercise 9.4, use the p-value to determine the signiﬁcance of the results. Explain what “statistically signiﬁcant” means in terms of rejecting or accepting H0 and Ha. 9.6 A random sample of n 35 observations from a quantitative population produced a mean x 2.4 and a standard deviation s .29. Suppose your research objective is to show that the population mean m exceeds 2.3. a. Give the null and alternative hypotheses for the test. b. Locate the rejection region for the test using a 5% signiﬁcance level. c. Find the standard error of the mean. d. Before you conduct the test, use your intuition to decide whether the sample mean x 2.4 is likely or unlikely, assuming that m 2.3. Now conduct the test. Do the data provide sufficient evidence to indicate that m 2.3? 9.7 Refer to Exercise 9.6. a. Calculate the p-value for the test statistic in part d. b. Use the p-value to draw a conclusion at the 5% sig- niﬁcance level. c. Compare the conclusion in part b with the conclusion reached in part d of Exercise 9.6. Are they the same? 9.8 Refer to Exercise 9.6. You want to test H0 : m 2.3 against Ha : m 2.3. a. Find the critical value of x used for rejecting H0. b. Calculate b P(accept H0 when m 2.4). c. Repeat the calculation of b for m 2.3, 2.5, and 2.6. d. Use the values of b from parts b and c to graph the power curve for the test. 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 361 9.9 A random sample of 100 observations from a quantitative population produced a sample mean of 26.8 and a sample standard deviation of 6.5. Use the p-value approach to determine whether the population mean is different from 28. Explain your conclusions. APPLICATIONS 9.10 Airline Occupancy Rates High airline occupancy rates on scheduled ﬂights are essential to corporate proﬁtability. Suppose a scheduled ﬂight must average at least 60% occupancy in order to be profitable, and an examination of the occupancy rate for 120 10:00 A.M. ﬂights from Atlanta to Dallas showed a mean occupancy per ﬂight of 58% and a standard deviation of 11%. a. If m is the mean occupancy per ﬂight and if the company wishes to determine whether or not this scheduled ﬂight is unproﬁtable, give the alternative and the null hypotheses for the test. b. Does the alternative hypothesis in part a imply a one- or two-tailed test? Explain. c. Do the occupancy data for the 120 ﬂights suggest that this scheduled ﬂight is unproﬁtable? Test using a .05. 9.11 Hamburger Meat Exercise 8.33 involved the meat department of a local supermarket chain that packages ground beef in trays of two sizes. The smaller tray is intended to hold 1 pound of meat. A random sample of 35 packages in the smaller meat tray produced weight measurements with an average of 1.01 pounds and a standard deviation of .18 pound. a. If you were the quality control manager and wanted to make sure that the average amount of ground beef was indeed 1 pound, what hypotheses would you test? b. Find the p-value for the test and use it to perform the test in part a. c. How would you, as the quality control manager, report the results of your study to a consumer interest group? 9.12 Invasive Species In a study of the pernicious weed giant hogweed, one of the tallest herbaceous species in Europe, Jan Pergl1 and associates compared the d
ensity of these plants in both managed and unmanaged sites within the Caucasus region of Russia. In its native area, the average density was found to be 5 plants/m2. In an invaded area in the Czech Republic, 362 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES a sample of n 50 plants produced an average density of 11.17 plants/m2 with a standard deviation of 3.9 plants/m2. a. Does the invaded area in the Czech Republic have an average density of giant hogweed that is different from m 5 at the a .05 level of signiﬁcance? b. What is the p-value associated with the test in part a? Can you reject H0 at the 5% level of signiﬁcance using the p-value? 9.13 Potency of an Antibiotic A drug manufacturer claimed that the mean potency of one of its antibiotics was 80%. A random sample of n 100 capsules were tested and produced a sample mean of x 79.7% with a standard deviation of s .8%. Do the data present sufficient evidence to refute the manufacturer’s claim? Let a .05. a. State the null hypothesis to be tested. b. State the alternative hypothesis. c. Conduct a statistical test of the null hypothesis and state your conclusion. 9.14 Flextime Many companies are becoming involved in ﬂextime, in which a worker schedules his or her own work hours or compresses work weeks. A company that was contemplating the installation of a ﬂextime schedule estimated that it needed a minimum mean of 7 hours per day per assembly worker in order to operate effectively. Each of a random sample of 80 of the company’s assemblers was asked to submit a tentative ﬂextime schedule. If the mean number of hours per day for Monday was 6.7 hours and the standard deviation was 2.7 hours, do the data provide sufﬁcient evidence to indicate that the mean number of hours worked per day on Mondays, for all of the company’s assemblers, will be less than 7 hours? Test using a .05. 9.15 Does College Pay Off? An article in Time describing various aspects of American life indicated that higher educational achievement paid off! College grads work 7.4 hours per day, fewer than those with less than a college education.2 Suppose that the average work day for a random sample of n 100 individuals who had less than a four-year college education was calculated to be x 7.9 hours with a standard deviation of s 1.9 hours. a. Use the p-value approach to test the hypothesis that the average number of hours worked by individuals having less than a college degree is greater than individuals having a college degree. At what level can you reject H0? b. If you were a college graduate, how would you state your conclusion to put yourself in the best possible light? c. If you were not a college graduate, how might you state your conclusion? 9.16 What’s Normal? What is normal, when it comes to people’s body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker3 in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. Does the data indicate that the average body temperature for healthy humans is different from 98.6 degrees, the usual average temperature cited by physicians and others? Test using both methods given in this section. a. Use the p-value approach with a .05. b. Use the critical value approach with a .05. c. Compare the conclusions from parts a and b. Are they the same? d. The 98.6 standard was derived by a German doctor in 1868, who claimed to have recorded 1 million temperatures in the course of his research.4 What conclusions can you draw about his research in light of your conclusions in parts a and b? 9.17 Sports and Achilles Tendon Injuries Some sports that involve a signiﬁcant amount of running, jumping, or hopping put participants at risk for Achilles tendinopathy (AT), an inﬂammation and thickening of the Achilles tendon. A study in The American Journal of Sports Medicine looked at the diameter (in mm) of the affected tendons for patients who participated in these types of sports activities.5 Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters (mm). When the diameters of the affected tendon were measured for a random sample of 31 patients, the average diameter was 9.80 with a standard deviation of 1.95 mm. Is there sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.97 mm? Test at the 5% level of signiﬁcance. 9.4 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 363 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS 9.4 In many situations, the statistical question to be answered involves a comparison of two population means. For example, the U.S. Postal Service is interested in reducing its massive 350 million gallons/year gasoline bill by replacing gasoline-powered trucks with electric-powered trucks. To determine whether signiﬁcant savings in operating costs are achieved by changing to electric-powered trucks, a pilot study should be undertaken using, say, 100 conventional gasoline-powered mail trucks and 100 electricpowered mail trucks operated under similar conditions. The statistic that summarizes the sample information regarding the difference in population means (m1 m2) is the difference in sample means (x1 x2). Therefore, in testing whether the difference in sample means indicates that the true difference in population means differs from a speciﬁed value, (m1 m2) D0, you can use the standard error of (x1 x2), 1 2 estimated by SE s s 2 2 2 ns2 s2 1 n 1 2 n1 n2 in the form of a z-statistic to measure how many standard deviations the difference (x1 x2) lies from the hypothesized difference D0. The formal testing procedure is described next. LARGE-SAMPLE STATISTICAL TEST FOR (m1 m2) 1. Null hypothesis: H0 : (m1 m2) D0, where D0 is some speciﬁed difference that you wish to test. For many tests, you will hypothesize that there is no difference between m1 and m2; that is, D0 0. 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : (m1 m2) D0 Ha : (m1 m2) D0 [or Ha : (m1 m2) D0] 3. Test statistic: z (x1 ) D0 x2 S E (x1 x2) D0 n s2 1 n 1 s2 2 2 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test z za/2 or z za/2 z za [or z za when the alternative hypothesis is Ha : (m1 m2) D0] or when p-value a 364 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.9 α α/2 0 zα –zα/2 0 α/2 zα/2 Assumptions: The samples are randomly and independently selected from the two populations and n1 30 and n2 30. To determine whether car ownership affects a student’s academic achievement, two random samples of 100 male students were each drawn from the student body. The grade point average for the n1 100 non-owners of cars had an average and variance equal to x1 2.70 and s2 2 .40 for the n2 100 car 1 .36, while x2 2.54 and s2 owners. Do the data present sufficient evidence to indicate a difference in the mean achievements between car owners and nonowners of cars? Test using a .05. Solution To detect a difference, if it exists, between the mean academic achievements for non-owners of cars m1 and car owners m2, you will test the null hypothesis that there is no difference between the means against the alternative hypothesis that (m1 m2 ) 0; that is, H0 : (m1 m2) D0 0 versus Ha : (m1 m2) 0 Substituting into the formula for the test statistic, you get z (x1 x2) D0 ns2 s2 1 n 1 2 2 2.70 2.54 0 .4 0 0 1 6 .3 0 0 1 1.84 test statistic critical value ⇔ reject H0 • The critical value approach: Using a two-tailed test with signiﬁcance level a .05, you place a/2 .025 in each tail of the z distribution and reject H0 if z 1.96 or z 1.96. Since z 1.84 does not exceed 1.96 and is not less than 1.96, H0 cannot be rejected (see Figure 9.11). That is, there is insufficient evidence to declare a difference in the average academic achievements for the two groups. Remember that you should not be willing to accept H0—declare the two means to be the same—until b is evaluated for some meaningful values of (m1 m2). F IG URE 9. 11 Rejection region and p-value for Example 9.9 ● f(z) 1 2 p-value 1 2 p-value –1.84 0 1.84 z Reject Ho (z < –1.96) Reject Ho (z > 1.96) 9.4 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 365 • The p-value approach: Calculate the p-value, the probability that z is greater than z 1.84 plus the probability that z is less than z 1.84, as shown in Figure 9.11: p-value P(z 1.84) P(z 1.84) (1 .9671) .0329 .0658 The p-value lies between .10 and .05, so you can reject H0 at the .10 level but not at the .05 level of signiﬁcance. Since the p-value of .0658 exceeds the speciﬁed signiﬁcance level a .05, H0 cannot be rejected. Again, you should not be willing to accept H0 until b is evaluated for some meaningful values of (m1 m2). Hypothesis Testing and Confidence Intervals Whether you use the critical value or the p-value approach for testing hypotheses about (m1 m2), you will always reach the same conclusion because the calculated value of the test statistic and the critical value are related exactly in the same way that the p-value and the signiﬁcance level a are related. You might remember that the conﬁdence intervals constructed in Chapter 8 could also be used to answer questions about the difference between two population means. In fact, for a two-tailed test, the (1 a)100% conﬁdence interval for the parameter of interest can be used to test its value, just as you did informally in Chapter 8. The value of a indicated by the conﬁdence coefficient in the conﬁdence interval is equivalent to the signiﬁcance level a in the statistical test. For a one-tailed test, the equivalent conﬁdence interval approach would use the one-sided conﬁdence bounds in Section 8.8 with conﬁdence coefficient a. In addition, by using the conﬁdence interval approach, you gain a range of possible values for the parameter of interest, regardless of the outcome of the test of hypothesis. • • If the conﬁdence inte
rval you construct contains the value of the parameter speciﬁed by H0, then that value is one of the likely or possible values of the parameter and H0 should not be rejected. If the hypothesized value lies outside of the conﬁdence limits, the null hypothesis is rejected at the a level of signiﬁcance. Construct a 95% conﬁdence interval for the difference in average academic achievements between car owners and non-owners. Using the conﬁdence interval, can you conclude that there is a difference in the population means for the two groups of students? Solution For the large-sample statistics discussed in Chapter 8, the 95% conﬁdence interval is given as Point estimator 1.96 (Standard error of the estimator) For the difference in two population means, the conﬁdence interval is approximated as EXAMPLE 9.10 (x1 x2) 1.96 (2.70 2.54) 1.96 2 ns2 s2 1 n 1 6 .3 0 0 1 2 0 .4 0 0 1 .16 .17 366 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES or .01 (m1 m2) .33. This interval gives you a range of possible values for the difference in the population means. Since the hypothesized difference, (m1 m2) 0, is contained in the conﬁdence interval, you should not reject H0. Look at the signs of the possible values in the conﬁdence interval. You cannot tell from the interval whether the difference in the means is negative (), positive (), or zero (0)—the latter of the three would indicate that the two means are the same. Hence, you can really reach no conclusion in terms of the question posed. There is not enough evidence to indicate that there is a difference in the average achievements for car owners versus non-owners. The conclusion is the same one reached in Example 9.9. 9.4 EXERCISES BASIC TECHNIQUES 9.18 Independent random samples of 80 measurements were drawn from two quantitative populations, 1 and 2. Here is a summary of the sample data: Sample 1 Sample 2 80 9.7 38.4 80 11.6 27.9 Sample Size Sample Mean Sample Variance a. If your research objective is to show that m1 is larger than m2, state the alternative and the null hypotheses that you would choose for a statistical test. b. Is the test in part a one- or two-tailed? c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that H0 is true and the two population means are the same? d. p-value approach: Find the p-value for the test. Test for a signiﬁcant difference in the population means at the 1% signiﬁcance level. e. Critical value approach: Find the rejection region when a .01. Do the data provide sufficient evidence to indicate a difference in the population means? 9.19 Independent random samples of 36 and 45 observations are drawn from two quantitative populations, 1 and 2, respectively. The sample data summary is shown here: Sample 1 Sample 2 Sample Size Sample Mean Sample Variance 36 1.24 .0560 45 1.31 .0540 Do the data present sufficient evidence to indicate that the mean for population 1 is smaller than the mean for population 2? Use one of the two methods of testing presented in this section, and explain your conclusions. 9.20 Suppose you wish to detect a difference between m1 and m2 (either m1 m2 or m1 m2) and, instead of running a two-tailed test using a .05, you use the following test procedure. You wait until you have collected the sample data and have calculated x1 and x2. If x1 is larger than x2, you choose the alternative hypothesis Ha : m1 m2 and run a one-tailed test placing a1 .05 in the upper tail of the z distribution. If, on the other hand, x2 is larger than x1, you reverse the procedure and run a one-tailed test, placing a2 .05 in the lower tail of the z distribution. If you use this procedure and if m1 actually equals m2, what is the probability a that you will conclude that m1 is not equal to m2 (i.e., what is the probability a that you will incorrectly reject H0 when H0 is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data. APPLICATIONS 9.21 Cure for the Common Cold? An experiment was planned to compare the mean time (in days) required to recover from a common cold for persons given a daily dose of 4 milligrams (mg) of vitamin C versus those who were not given a vitamin supplement. Suppose that 35 adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the two groups were as follows: 9.4 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 367 No Vitamin Supplement 4 mg Vitamin C Sample Size Sample Mean Sample Standard Deviation 35 6.9 2.9 35 5.8 1.2 a. Suppose your research objective is to show that the use of vitamin C reduces the mean time required to recover from a common cold and its complications. Give the null and alternative hypotheses for the test. Is this a one- or a two-tailed test? b. Conduct the statistical test of the null hypothesis in part a and state your conclusion. Test using a .05. 9.22 Healthy Eating Americans are becoming more conscious about the importance of good nutrition, and some researchers believe we may be altering our diets to include less red meat and more fruits and vegetables. To test the theory that the consumption of red meat has decreased over the last 10 years, a researcher decides to select hospital nutrition records for 400 subjects surveyed 10 years ago and to compare their average amount of beef consumed per year to amounts consumed by an equal number of subjects interviewed this year. The data are given in the table. Ten Years Ago This Year Sample Mean Sample Standard Deviation 73 25 63 28 a. Do the data present sufficient evidence to indicate that per-capita beef consumption has decreased in the last 10 years? Test at the 1% level of signiﬁcance. b. Find a 99% lower conﬁdence bound for the difference in the average per-capita beef consumptions for the two groups. (This calculation was done as part of Exercise 8.76.) Does your conﬁdence bound conﬁrm your conclusions in part a? Explain. What additional information does the conﬁdence bound give you? 9.23 Lead Levels in Drinking Water Analyses of drinking water samples for 100 homes in each of two different sections of a city gave the following means and standard deviations of lead levels (in parts per million): Section 1 Section 2 Sample Size Mean Standard Deviation 100 34.1 5.9 100 36.0 6.0 a. Calculate the test statistic and its p-value (observed signiﬁcance level) to test for a difference in the two population means. Use the p-value to evaluate the statistical signiﬁcance of the results at the 5% level. b. Use a 95% conﬁdence interval to estimate the difference in the mean lead levels for the two sections of the city. c. Suppose that the city environmental engineers will be concerned only if they detect a difference of more than 5 parts per million in the two sections of the city. Based on your conﬁdence interval in part b, is the statistical signiﬁcance in part a of practical signiﬁcance to the city engineers? Explain. 9.24 Starting Salaries, again In an attempt to compare the starting salaries for college graduates who majored in chemical engineering and computer science (see Exercise 8.45), random samples of 50 recent college graduates in each major were selected and the following information obtained. Mean Major SD Chemical Engineering Computer Science $53,659 51,042 2225 2375 a. Do the data provide sufficient evidence to indicate a difference in average starting salaries for college graduates who majored in chemical engineering and computer science? Test using a .05. b. Compare your conclusions in part a with the results of part b in Exercise 8.45. Are they the same? Explain. 9.25 Hotel Costs In Exercise 8.18, we explored the average cost of lodging at three different hotel chains.6 We randomly select 50 billing statements from the computer databases of the Marriott, Radisson, and Wyndham hotel chains, and record the nightly room rates. A portion of the sample data is shown in the table. Marriott Radisson Sample Average Sample Standard Deviation $170 17.5 $145 10 a. Before looking at the data, would you have any preconceived idea about the direction of the difference between the average room rates for these two hotels? If not, what null and alternative hypotheses should you test? b. Use the critical value approach to determine if there is a signiﬁcant difference in the average room rates for the Marriott and the Radisson hotel chains. Use a .01. c. Find the p-value for this test. Does this p-value con- ﬁrm the results of part b? 368 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES 9.26 Hotel Costs II Refer to Exercise 9.25. The table below shows the sample data collected to compare the average room rates at the Wyndham and Radisson hotel chains.6 Wyndham Radisson Sample Average Sample Standard Deviation $150 16.5 $145 10 a. Do the data provide sufficient evidence to indicate a difference in the average room rates for the Wyndham and the Radisson hotel chains? Use a .05. b. Construct a 95% conﬁdence interval for the difference in the average room rates for the two chains. Does this interval conﬁrm your conclusions in part a? 9.27 MMT in Gasoline The addition of MMT, a compound containing manganese (Mn), to gasoline as an octane enhancer has caused concern about human exposure to Mn because high intakes have been linked to serious health effects. In a study of ambient air concentrations of ﬁne Mn, Wallace and Slonecker (Journal of the Air and Waste Management Association) presented the accompanying summary information about the amounts of ﬁne Mn (in nanograms per cubic meter) in mostly rural national park sites and in mostly urban California sites.7 National Parks California Mean Standard Deviation Number of Sites .94 1.2 36 2.8 2.8 26 a. Is there sufficient evidence to indicate that the mean concentrations differ for these two types of sites at the a .05 level of signiﬁcance? Use the
largesample z-test. What is the p-value of this test? b. Construct a 95% conﬁdence interval for (m1 m2). Does this interval conﬁrm your conclusions in part a? 9.28 Noise and Stress In Exercise 8.48, you compared the effect of stress in the form of noise on the ability to perform a simple task. Seventy subjects were divided into two groups; the ﬁrst group of 30 subjects acted as a control, while the second group of 40 was the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to ﬁnish the task was recorded for each subject and the following summary was obtained: Control Experimental n x s 30 15 minutes 4 minutes 40 23 minutes 10 minutes a. Is there sufficient evidence to indicate that the average time to complete the task was longer for the experimental “rock music” group? Test at the 1% level of signiﬁcance. b. Construct a 99% one-sided upper bound for the difference (control experimental) in average times for the two groups. Does this interval conﬁrm your conclusions in part a? 9.29 What’s Normal II Of the 130 people in Exercise 9.16, 65 were female and 65 were male.3 The means and standard deviations of their temperatures are shown below. Men Women Sample Mean Standard Deviation 98.11 0.70 98.39 0.74 a. Use the p-value approach to test for a signiﬁcant difference in the average temperatures for males versus females. b. Are the results signiﬁcant at the 5% level? At the 1% level? A LARGE-SAMPLE TEST OF HYPOTHESIS FOR A BINOMIAL PROPORTION 9.5 When a random sample of n identical trials is drawn from a binomial population, the sample proportion pˆ has an approximately normal distribution when n is large, with mean p and standard error q SE p n 9.5 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR A BINOMIAL PROPORTION ❍ 369 When you test a hypothesis about p, the proportion in the population possessing a certain attribute, the test follows the same general form as the large-sample tests in Sections 9.3 and 9.4. To test a hypothesis of the form H0 : p p0 versus a one- or two-tailed alternative Ha : p p0 or Ha : p p0 or Ha : p p0 the test statistic is constructed using pˆ, the best estimator of the true population proportion p. The sample proportion pˆ is standardized, using the hypothesized mean and standard error, to form a test statistic z, which has a standard normal distribution if H0 is true. This large-sample test is summarized next. LARGE-SAMPLE STATISTICAL TEST FOR p 1. Null hypothesis: H0 : p p0 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : p p0 Ha : p p0 (or, Ha : p p0) 3. Test statistic: z pˆ p0 SE pˆ p0 q0 p0 n x with pˆ n where x is the number of successes in n binomial trials.† 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test z za/2 or z za/2 z za (or z za when the alternative hypothesis is Ha : p p0) or when p-value a α α/2 0 zα –zα/2 0 α/2 zα/2 Assumption: The sampling satisﬁes the assumptions of a binomial experiment (see Section 5.2), and n is large enough so that the sampling distribution of pˆ can be approximated by a normal distribution (np0 5 and nq0 5). †An equivalent test statistic can be found by multiplying the numerator and denominator by z by n to obtain p n x 0 z 0q np 0 370 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.11 Regardless of age, about 20% of American adults participate in ﬁtness activities at least twice a week. However, these ﬁtness activities change as the people get older, and occasionally participants become nonparticipants as they age. In a local survey of n 100 adults over 40 years old, a total of 15 people indicated that they participated in a ﬁtness activity at least twice a week. Do these data indicate that the participation rate for adults over 40 years of age is signiﬁcantly less than the 20% ﬁgure? Calculate the p-value and use it to draw the appropriate conclusions. Solution Assuming that the sampling procedure satisﬁes the requirements of a binomial experiment, you can answer the question posed using a one-tailed test of hypothesis: H0 : p .2 versus Ha : p .2 Begin by assuming that H0 is true—that is, the true value of p is p0 .2. Then pˆ x/n will have an approximate normal distribution with mean p0 and standard error p0q0/n. (NOTE: This is different from the estimation procedure in which the unknown standard error is estimated by pˆqˆ/n.) The observed value of pˆ is 15/100 .15 and the test statistic is pˆ p0 q0 p0 .15 .20 .80) (.20 ( ) 0 0 1 1.25 z n p-value a ⇔ reject H0 p-value a ⇔ do not reject H0 The p-value associated with this test is found as the area under the standard normal curve to the left of z 1.25 as shown in Figure 9.12. Therefore, p-value P(z 1.25) .1056 F IG URE 9. 12 p-value for Example 9.11 ● f(z) p-value = .1056 –1.25 0 z If you use the guidelines for evaluating p-values, then .1056 is greater than .10, and you would not reject H0. There is insufficient evidence to conclude that the percentage of adults over age 40 who participate in ﬁtness activities twice a week is less than 20%. Statistical Significance and Practical Importance It is important to understand the difference between results that are “signiﬁcant” and results that are practically “important.” In statistical language, the word signiﬁcant does not necessarily mean “important,” but only that the results could not have occurred by chance. For example, suppose that in Example 9.11, the researcher had used 9.5 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR A BINOMIAL PROPORTION ❍ 371 n 400 adults in her experiment and had observed the same sample proportion. The test statistic is now pˆ p0 q0 p0 .15 .20 .80) (.20 ( ) 0 0 4 2.50 z n with p-value P(z 2.50) .0062 Now the results are highly signiﬁcant: H0 is rejected, and there is sufficient evidence to indicate that the percentage of adults over age 40 who participate in physical ﬁtness activities is less than 20%. However, is this drop in activity really important? Suppose that physicians would be concerned only about a drop in physical activity of more than 10%. If there had been a drop of more than 10% in physical activity, this would imply that the true value of p was less than .10. What is the largest possible value of p? Using a 95% upper one-sided conﬁdence bound, you have qˆ pˆ 1.645pˆ n .15 1.645(.15 .85) ( ) 0 0 4 .15 .029 or p .179. The physical activity for adults aged 40 and older has dropped from 20%, but you cannot say that it has dropped below 10%. So, the results, although statistically signiﬁcant, are not practically important. In this book, you will learn how to determine whether results are statistically signiﬁcant. When you use these procedures in a practical situation, however, you must also make sure the results are practically important. 9.5 EXERCISES BASIC TECHNIQUES 9.30 A random sample of n 1000 observations from a binomial population produced x 279. a. If your research hypothesis is that p is less than .3, what should you choose for your alternative hypothesis? Your null hypothesis? b. What is the critical value that determines the rejec- tion region for your test with a .05? c. Do the data provide sufficient evidence to indicate that p is less than .3? Use a 5% signiﬁcance level. 9.31 A random sample of n 1400 observations from a binomial population produced x 529. a. If your research hypothesis is that p differs from .4, what hypotheses should you test? b. Calculate the test statistic and its p-value. Use the p-value to evaluate the statistical signiﬁcance of the results at the 1% level. c. Do the data provide sufficient evidence to indicate that p is different from .4? 9.32 A random sample of 120 observations was selected from a binomial population, and 72 successes were observed. Do the data provide sufficient evidence to indicate that p is greater than .5? Use one of the two methods of testing presented in this section, and explain your conclusions. APPLICATIONS 9.33 Childhood Obesity According to PARADE magazine’s “What America Eats” survey involving 372 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES n 1015 adults, almost half of parents say their children’s weight is ﬁne.8 Only 9% of parents describe their children as overweight. However, the American Obesity Association says the number of overweight children and teens is at least 15%. Suppose that the number of parents in the sample is n 750 and the number of parents who describe their children as overweight is x 68. a. How would you proceed to test the hypothesis that the proportion of parents who describe their children as overweight is less than the actual proportion reported by the American Obesity Association? b. What conclusion are you able to draw from these data at the a .05 level of signiﬁcance? c. What is the p-value associated with this test? 9.34 Plant Genetics A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that 75% of the offspring resulting from this cross will have red ﬂowers. To test this claim, 100 seeds from this cross were collected and germinated, and 58 plants had red petals. a. What hypothesis should you use to test the geneti- cist’s claim? b. Calculate the test statistic and its p-value. Use the p-value to evaluate the statistical signiﬁcance of the results at the 1% level. 9.35 Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a community public health department instituted a screening program to provide for the early detection of breast cancer and to increase the survival rate p of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were periodically screened by the program and who were diagnosed to have the disease. Let x represent the number of those in the sample who survive the disease. a. If you wish to detect whether the community screening program
has been effective, state the null hypothesis that should be tested. b. State the alternative hypothesis. c. If 164 women in the sample of 200 survive the disease, can you conclude that the community screening program was effective? Test using a .05 and explain the practical conclusions from your test. d. Find the p-value for the test and interpret it. 9.36 Sweet Potato Whiteﬂy Suppose that 10% of the ﬁelds in a given agricultural area are infested with the sweet potato whiteﬂy. One hundred ﬁelds in this area are randomly selected, and 25 are found to be infested with whiteﬂy. a. Assuming that the experiment satisﬁes the conditions of the binomial experiment, do the data indicate that the proportion of infested ﬁelds is greater than expected? Use the p-value approach, and test using a 5% signiﬁcance level. b. If the proportion of infested ﬁelds is found to be signiﬁcantly greater than .10, why is this of practical signiﬁcance to the agronomist? What practical conclusions might she draw from the results? 9.37 Brown or Blue? An article in the Washington Post stated that nearly 45% of the U.S. population is born with brown eyes, although they don’t necessarily stay that way.9 To test the newspaper’s claim, a random sample of 80 people was selected, and 32 had brown eyes. Is there sufficient evidence to dispute the newspaper’s claim regarding the proportion of browneyed people in the United States? Use a .01. 9.38 Colored Contacts Refer to Exercise 9.37. Contact lenses, worn by about 26 million Americans, come in many styles and colors. Most Americans wear soft lenses, with the most popular colors being the blue varieties (25%), followed by greens (24%), and then hazel or brown. A random sample of 80 tinted contact lens wearers was checked for the color of their lenses. Of these people, 22 wore blue lenses and only 15 wore green lenses.9 a. Do the sample data provide sufficient evidence to indicate that the proportion of tinted contact lens wearers who wear blue lenses is different from 25%? Use a .05. b. Do the sample data provide sufficient evidence to indicate that the proportion of tinted contact lens wearers who wear green lenses is different from 24%? Use a .05. c. Is there any reason to conduct a one-tailed test for either part a or b? Explain. 9.39 A Cure for Insomnia An experimenter has prepared a drug-dose level that he claims will induce sleep for at least 80% of people suffering from insomnia. After examining the dosage we feel that his claims regarding the effectiveness of his dosage are inﬂated. In an attempt to disprove his claim, we administer his prescribed dosage to 50 insomniacs and observe that 9.6 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍ 373 37 of them have had sleep induced by the drug dose. Is there enough evidence to refute his claim at the 5% level of signiﬁcance? percentage of adults who say that they always vote is different from the percentage reported in Time? Test using a .01. 9.40 Who Votes? About three-fourths of voting age Americans are registered to vote, but many do not bother to vote on Election Day. Only 64% voted in 1992, and 60% in 2000, but turnout in off-year elections is even lower. An article in Time stated that 35% of adult Americans are registered voters who always vote.10 To test this claim, a random sample of n 300 adult Americans was selected and x 123 were registered regular voters who always voted. Does this sample provide sufficient evidence to indicate that the 9.41 Man’s Best Friend The Humane Society reports that there are approximately 65 million dogs owned in the United States and that approximately 40% of all U.S. households own at least one dog.11 In a random sample of 300 households, 114 households said that they owned at least one dog. Does this data provide sufficient evidence to indicate that the proportion of households with at least one dog is different from that reported by the Humane Society? Test using a .05. A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS 9.6 When random and independent samples are selected from two binomial populations, the focus of the experiment may be the difference ( p1 p2) in the proportions of individuals or items possessing a speciﬁed characteristic in the two populations. In this situation, you can use the difference in the sample proportions ( pˆ1 pˆ2) along with its standard error, 2 1 p SE p 2q 1q n n 2 1 in the form of a z-statistic to test for a signiﬁcant difference in the two population proportions. The null hypothesis to be tested is usually of the form H0 : p1 p2 or H0 : ( p1 p2) 0 Remember: Each trial results in one of two outcomes (S or F). versus either a one- or two-tailed alternative hypothesis. The formal test of hypothesis is summarized in the next display. In estimating the standard error for the z-statistic, you should use the fact that when H0 is true, the two population proportions are equal to some common value—say, p. To obtain the best estimate of this common value, the sample data are “pooled” and the estimate of p is pˆ Total number of successes Total number of trials x x1 2 n n 2 1 Remember that, in order for the difference in the sample proportions to have an approximately normal distribution, the sample sizes must be large and the proportions should not be too close to 0 or 1. 374 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES LARGE-SAMPLE STATISTICAL TEST FOR (p1 p2) 1. Null hypothesis: H0 : ( p1 p2) 0 or equivalently H0 : p1 p2 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : ( p1 p2) 0 Ha : ( p1 p2) 0 [or Ha : ( p1 p2) 0] 2) 0 3. Test statistic: z ( pˆ1 pˆ S E pˆ1 pˆ 2 2 1 p p 2q 1q n n 2 1 where pˆ1 x1/n1 and pˆ2 x2/n2. Since the common value of p1 p2 p (used in the standard error) is unknown, it is estimated by pˆ1 pˆ x1 pˆ 2 n n 2 1 z and the test statistic is ( pˆ1 pˆ2) 0 ˆ ˆ p p ˆq ˆq n n 1 2 or z pˆ1 pˆ2 pˆqˆ 1 1 n n 1 2 4. Rejection region: Reject H0 when Two-Tailed Test z za/2 or z za/2 One-Tailed Test z za [or z za when the alternative hypothesis is Ha : ( p1 p2) 0] or when p-value a α α/2 0 zα –zα/2 0 α/2 zα/2 Assumptions: Samples are selected in a random and independent manner from two binomial populations, and n1 and n2 are large enough so that the sampling distribution of ( pˆ1 pˆ2) can be approximated by a normal distribution. That is, n1pˆ1, n1qˆ1, n2 pˆ2, and n2qˆ2 should all be greater than 5. EXAMPLE 9.12 The records of a hospital show that 52 men in a sample of 1000 men versus 23 women in a sample of 1000 women were admitted because of heart disease. Do these data present sufficient evidence to indicate a higher rate of heart disease among men admitted to the hospital? Use a .05. Solution Assume that the number of patients admitted for heart disease has an approximate binomial probability distribution for both men and women with parameters 9.6 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍ 375 p1 and p2, respectively. Then, since you wish to determine whether p1 p2, you will test the null hypothesis p1 p2—that is, H0 : ( p1 p2) 0—against the alternative hypothesis Ha : p1 p2 or, equivalently, Ha : ( p1 p2) 0. To conduct this test, use the z-test statistic and approximate the standard error using the pooled estimate of p. Since Ha implies a one-tailed test, you can reject H0 only for large values of z. Thus, for a .05, you can reject H0 if z 1.645 (see Figure 9.13). The pooled estimate of p required for the standard error is x1 .0375 pˆ n 3 000 2 0 2 1 5 0 10 x 2 n 2 1 FI GUR E 9 .1 3 Location of the rejection region in Example 9.12 ● f(z) α = .05 0 1.645 z Rejection region and the test statistic is pˆ1 pˆ2 pˆqˆ 1 1 n n z 1 2 .052 .023 (.0375)(.9625) 00 1 00 10 1 10 3.41 Since the computed value of z falls in the rejection region, you can reject the hypothesis that p1 p2. The data present sufficient evidence to indicate that the percentage of men entering the hospital because of heart disease is higher than that of women. (NOTE: This does not imply that the incidence of heart disease is higher in men. Perhaps fewer women enter the hospital when afflicted with the disease!) How much higher is the proportion of men than women entering the hospital with heart disease? A 95% lower one-sided conﬁdence bound will help you ﬁnd the lowest likely value for the difference. ˆ1 pˆ ˆ2 ( pˆ1 pˆ 2) 1.645pˆ 2q 1q n n 90 (.052 .023) 1.645.05 .029 .014 (. 2 0 0 1 2 1 09 48) .02 (. 3 0 1 77) 0 or ( p1 p2) .015. The proportion of men is roughly 1.5% higher than women. Is this of practical importance? This is a question for the researcher to answer. In some situations, you may need to test for a difference D0 (other than 0) between two binomial proportions. If this is the case, the test statistic is modiﬁed for testing 376 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES H0 : ( p1 p2) D0, and a pooled estimate for a common p is no longer used in the standard error. The modiﬁed test statistic is z ( pˆ1 pˆ2) D0 ˆ2 ˆ1 pˆ pˆ 2q 1q n n 1 2 Although this test statistic is not used often, the procedure is no different from other large-sample tests you have already mastered! 9.6 EXERCISES BASIC TECHNIQUES 9.42 Independent random samples of n1 140 and n2 140 observations were randomly selected from binomial populations 1 and 2, respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to which parameter, p1 or p2, is the larger, but you want to detect only a difference between the two parameters if one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, ( pˆ1 pˆ2). Make sure to use the pooled estimate for the common value of p. c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely
or unlikely observation, assuming that H0 is true and the two population proportions are the same? d. p-value approach: Find the p-value for the test. Test for a signiﬁcant difference in the population proportions at the 1% signiﬁcance level. e. Critical value approach: Find the rejection region when a .01. Do the data provide sufficient evidence to indicate a difference in the population proportions? 9.43 Refer to Exercise 9.42. Suppose, for practical reasons, you know that p1 cannot be larger than p2. a. Given this knowledge, what should you choose as the alternative hypothesis for your statistical test? The null hypothesis? b. Does your alternative hypothesis in part a imply a one- or two-tailed test? Explain. c. Conduct the test and state your conclusions. Test using a .05. 9.44 Independent random samples of 280 and 350 observations were selected from binomial populations 1 and 2, respectively. Sample 1 had 132 successes, and sample 2 had 178 successes. Do the data present sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population 2? Use one of the two methods of testing presented in this section, and explain your conclusions. APPLICATIONS 9.45 Treatment versus Control An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50. The ﬁrst group, the control group, received no treatment for the infection. The second group received the drug. After a 30-day period, the proportions of survivors, pˆ1 and pˆ2, in the two groups were found to be .36 and .60, respectively. a. Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use a .05. b. Use a 95% conﬁdence interval to estimate the actual difference in the cure rates for the treated versus the control groups. 9.46 Movie Marketing Marketing to targeted age groups has become a standard method of advertising, even in movie theater advertising. Advertisers use computer software to track the demographics of moviegoers and then decide on the type of products to advertise before a particular movie.12 One statistic that might be of interest is how frequently adults with children under 18 attend movies as compared to those without children. Suppose that a theater database is used to randomly select 1000 adult ticket purchasers. These adults are then surveyed and asked whether they 9.6 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍ 377 were frequent moviegoers—that is, do they attend movies 12 or more times a year? The results are shown in the table: With Children Without Children under 18 Sample Size Number Who Attend 12 Times per Year 440 123 560 145 a. Is there a signiﬁcant difference in the population proportions of frequent moviegoers in these two demographic groups? Use a .01. b. Why would a statistically signiﬁcant difference in these population proportions be of practical importance to the advertiser? 9.47 M&M’S In Exercise 8.53, you investigated whether Mars, Inc., uses the same proportion of red M&M’S in its plain and peanut varieties. Random samples of plain and peanut M&M’S provide the following sample data for the experiment: Plain Peanut Sample Size Number of Red M&M’S 56 12 32 8 Use a test of hypothesis to determine whether there is a signiﬁcant difference in the proportions of red candies for the two types of M&M’S. Let a .05 and compare your results with those of Exercise 8.53. 9.48 Hormone Therapy and Alzheimer’s Disease In the last few years, many research studies have shown that the purported beneﬁts of hormone replacement therapy (HRT) do not exist, and in fact, that hormone replacement therapy actually increases the risk of several serious diseases. A four-year experiment involving 4532 women, reported in The Press Enterprise, was conducted at 39 medical centers. Half of the women took placebos and half took Prempro, a widely prescribed type of hormone replacement therapy. There were 40 cases of dementia in the hormone group and 21 in the placebo group.13 Is there sufficient evidence to indicate that the risk of dementia is higher for patients using Prempro? Test at the 1% level of signiﬁcance. 9.50 Clopidogrel and Aspirin A large study was conducted to test the effectiveness of clopidogrel in combination with aspirin in warding off heart attacks and strokes.14 The trial involved more than 15,500 people 45 years of age or older from 32 countries, including the United States, who had been diagnosed with cardiovascular disease or had multiple risk factors. The subjects were randomly assigned to one of two groups. After two years, there was no difference in the risk of heart attack, stroke, or dying from heart disease between those who took clopidogrel and low-dose aspirin daily and those who took low-dose aspirin plus a dummy pill. The two-drug combination actually increased the risk of dying (5.4% versus 3.8%) or dying speciﬁcally from cardiovascular disease (3.9% versus 2.2%). a. The subjects were randomly assigned to one of the two groups. Explain how you could use the random number table to make these assignments. b. No sample sizes were given in the article: however, let us assume that the sample sizes for each group were n1 7720 and n2 7780. Determine whether the risk of dying was signiﬁcantly different for the two groups. c. What do the results of the study mean in terms of practical signiﬁcance? 9.51 Baby’s Sleeping Position Does a baby’s sleeping position affect the development of motor skills? In one study, published in the Archives of Pediatric Adolescent Medicine, 343 full-term infants were examined at their 4-month checkups for various developmental milestones, such as rolling over, grasping a rattle, reaching for an object, and so on.15 The baby’s predominant sleep position—either prone (on the stomach) or supine (on the back) or side—was determined by a telephone interview with the parent. The sample results for 320 of the 343 infants for whom information was received are shown here: Prone Supine or Side Number of Infants Number That Roll Over 121 93 199 119 9.49 HRT, continued Refer to Exercise 9.48. Calculate a 99% lower one-sided conﬁdence bound for the difference in the risk of dementia for women using hormone replacement therapy versus those who do not. Would this difference be of practical importance to a woman considering HRT? Explain. The researcher reported that infants who slept in the side or supine position were less likely to roll over at the 4-month checkup than infants who slept primarily in the prone position (P .001). Use a large-sample test of hypothesis to conﬁrm or refute the researcher’s conclusion. 378 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES SOME COMMENTS ON TESTING HYPOTHESES 9.7 A statistical test of hypothesis is a fairly clear-cut procedure that enables an experimenter to either reject or accept the null hypothesis H0, with measured risks a and b. The experimenter can control the risk of falsely rejecting H0 by selecting an appropriate value of a. On the other hand, the value of b depends on the sample size and the values of the parameter under test that are of practical importance to the experimenter. When this information is not available, an experimenter may decide to select an affordable sample size, in the hope that the sample will contain sufficient information to reject the null hypothesis. The chance that this decision is in error is given by a, whose value has been set in advance. If the sample does not provide sufficient evidence to reject H0, the experimenter may wish to state the results of the test as “The data do not support the rejection of H0” rather than accepting H0 without knowing the chance of error b. Some experimenters prefer to use the observed p-value of the test to evaluate the strength of the sample information in deciding to reject H0. These values can usually be generated by computer and are often used in reports of statistical results: • • • • If the p-value is greater than .05, the results are reported as NS—not signiﬁcant at the 5% level. If the p-value lies between .05 and .01, the results are reported as P .05— signiﬁcant at the 5% level. If the p-value lies between .01 and .001, the results are reported as P .01— “highly signiﬁcant” or signiﬁcant at the 1% level. If the p-value is less than .001, the results are reported as P .001—“very highly signiﬁcant” or signiﬁcant at the .1% level. Still other researchers prefer to construct a conﬁdence interval for a parameter and perform a test informally. If the value of the parameter speciﬁed by H0 is included within the upper and lower limits of the conﬁdence interval, then “H0 is not rejected.” If the value of the parameter speciﬁed by H0 is not contained within the interval, then “H0 is rejected.” These results will agree with a two-tailed test; one-sided conﬁdence bounds are used for one-tailed alternatives. Finally, consider the choice between a one- and two-tailed test. In general, experimenters wish to know whether a treatment causes what could be a beneﬁcial increase in a parameter or what might be a harmful decrease in a parameter. Therefore, most tests are two-tailed unless a one-tailed test is strongly dictated by practical considerations. For example, assume you will sustain a large ﬁnancial loss if the mean m is greater than m0 but not if it is less. You will then want to detect values larger than m0 with a high probability and thereby use a right-tailed test. In the same vein, if pollution levels higher than m0 cause critical health risks, then you will certainly wish to detect levels higher than m0 with a right-tailed test of hypothesis. In any case, the choice of a one- or two-tailed test should be dictated by the practical consequences that result from a decision to reject or not reject H0 in favor of the alternative. CHAPTER REVIEW Key Concepts
and Formulas I. Parts of a Statistical Test 1. Null hypothesis: a contradiction of the alterna- tive hypothesis 2. Alternative hypothesis: the hypothesis the researcher wants to support 3. Test statistic and its p-value: sample evidence calculated from the sample data 4. Rejection region—critical values and signiﬁcance levels: values that separate rejection and nonrejection of the null hypothesis 5. Conclusion: Reject or do not reject the null hypothesis, stating the practical signiﬁcance of your conclusion II. Errors and Statistical Significance 1. The signiﬁcance level a is the probability of rejecting H0 when it is in fact true. 2. The p-value is the probability of observing a test statistic as extreme as or more extreme than the one observed; also, the smallest value of a for which H0 can be rejected. 3. When the p-value is less than the signiﬁcance level a, the null hypothesis is rejected. This happens when the test statistic exceeds the critical value. CHAPTER REVIEW ❍ 379 4. In a Type II error, b is the probability of accepting H0 when it is in fact false. The power of the test is (1 b), the probability of rejecting H0 when it is false. III. Large-Sample Test Statistics Using the z Distribution To test one of the four population parameters when the sample sizes are large, use the following test statistics: Parameter Test Statistic pˆ p0 q0 p0 n m1 m2 z (x1 x2) D0 2 n s 1 s n1 2 2 2 p1 p2 z pˆ1 pˆ 2 pˆqˆ 1 1 n n 1 2 or z (pˆ1 pˆ 2) D0 ˆ2 ˆ1 pˆ pˆ 2q n 1q n 1 2 Supplementary Exercises Starred () exercises are optional. 9.52 a. Deﬁne a and b for a statistical test of hypoth- esis. b. For a ﬁxed sample size n, if the value of a is de- creased, what is the effect on b? c. In order to decrease both a and b for a particular alternative value of m, how must the sample size change? 9.53 What is the p-value for a test of hypothesis? How is it calculated for a large-sample test? 9.54 What conditions must be met so that the z test can be used to test a hypothesis concerning a population mean m? 9.55 Deﬁne the power of a statistical test. As the alternative value of m gets farther from m0, how is the power affected? 9.56 Acidity in Rainfall Refer to Exercise 8.31 and the collection of water samples to estimate the mean acidity (in pH) of rainfalls in the northeastern United States. As noted, the pH for pure rain falling through clean air is approximately 5.7. The sample of n 40 rainfalls produced pH readings with x 3.7 and s .5. Do the data provide sufficient evidence to indicate that the mean pH for rainfalls is more acidic (Ha : m 5.7 pH) than pure rainwater? Test using a .05. Note that this inference is appropriate only for the area in which the rainwater specimens were collected. 380 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES 9.57 Washing Machine Colors A manufacturer of automatic washers provides a particular model in one of three colors. Of the ﬁrst 1000 washers sold, it is noted that 400 were of the ﬁrst color. Can you conclude that more than one-third of all customers have a preference for the ﬁrst color? a. Find the p-value for the test. b. If you plan to conduct your test using a .05, what will be your test conclusions? 9.58 Generation Next Born between 1980 and 1990, Generation Next was the topic of Exercise 8.60.16 In a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Is there a signiﬁcant difference in the population proportions of female and male students who decided to attend college in order to make more money? Use a .01. b. Can you think of any reason why a statistically signiﬁcant difference in these population proportions might be of practical importance? To whom might this difference be important? 9.59 Bass Fishing The pH factor is a measure of the acidity or alkalinity of water. A reading of 7.0 is neutral; values in excess of 7.0 indicate alkalinity; those below 7.0 imply acidity. Loren Hill states that the best chance of catching bass occurs when the pH of the water is in the range 7.5 to 7.9.17 Suppose you suspect that acid rain is lowering the pH of your favorite ﬁshing spot and you wish to determine whether the pH is less than 7.5. a. State the alternative and null hypotheses that you would choose for a statistical test. b. Does the alternative hypothesis in part a imply a one- or a two-tailed test? Explain. c. Suppose that a random sample of 30 water specimens gave pH readings with x 7.3 and s .2. Just glancing at the data, do you think that the difference x 7.5 .2 is large enough to indicate that the mean pH of the water samples is less than 7.5? (Do not conduct the test.) 9.60 Pennsylvania Lottery A central Pennsylvania attorney reported that the Northumberland County district attorney’s (DA) office trial record showed only 6 convictions in 27 trials from January to mid-July 1997. Four central Pennsylvania county DAs responded, “Don’t judge us by statistics!”18 a. If the attorney’s information is correct, would you reject a claim by the DA of a 50% or greater conviction rate? b. The actual records show that there have been 455 guilty pleas and 48 cases that have gone to trial. Even assuming that the 455 guilty pleas are the only convictions of the 503 cases reported, what is the 95% conﬁdence interval for p, the true proportion of convictions by this district attorney? c. Using the results of part b, are you willing to reject a ﬁgure of 50% or greater for the true conviction rate? Explain. 9.61 White-Tailed Deer In an article entitled “A Strategy for Big Bucks,” Charles Dickey discusses studies of the habits of white-tailed deer that indicate that they live and feed within very limited ranges— approximately 150 to 205 acres.19 To determine whether there was a difference between the ranges of deer located in two different geographic areas, 40 deer were caught, tagged, and ﬁtted with small radio transmitters. Several months later, the deer were tracked and identiﬁed, and the distance x from the release point was recorded. The mean and standard deviation of the distances from the release point were as follows: Location 1 Location 2 Sample Size Sample Mean Sample Standard Deviation 40 2980 ft 1140 ft 40 3205 ft 963 ft a. If you have no preconceived reason for believing one population mean is larger than another, what would you choose for your alternative hypothesis? Your null hypothesis? b. Does your alternative hypothesis in part a imply a one- or a two-tailed test? Explain. c. Do the data provide sufficient evidence to indicate that the mean distances differ for the two geographic locations? Test using a .05. d. Now conduct a statistical test of the hypotheses in part a and state your conclusions. Test using a .05. Compare your statistically based decision with your intuitive decision in part c. 9.62 Female Models In a study to assess various effects of using a female model in automobile advertising, 100 men were shown photographs of two automobiles matched for price, color, and size, but of different makes. One of the automobiles was shown with a female model to 50 of the men (group A), and both automobiles were shown without the model to the other 50 men (group B). In group A, the automobile shown with the model was judged as more expensive by 37 men; in group B, the same automobile was judged as the more expensive by 23 men. Do these results indicate that using a female model inﬂuences the perceived cost of an automobile? Use a one-tailed test with a .05. 9.63 Bolts Random samples of 200 bolts manufactured by a type A machine and 200 bolts manufactured by a type B machine showed 16 and 8 defective bolts, respectively. Do these data present sufficient evidence to suggest a difference in the performance of the machine types? Use a .05. 9.64 Biomass Exercise 7.63 reported that the biomass for tropical woodlands, thought to be about 35 kilograms per square meter (kg/m2), may in fact be too high and that tropical biomass values vary regionally—from about 5 to 55 kg/m2.20 Suppose you measure the tropical biomass in 400 randomly selected square-meter plots and obtain x 31.75 and s 10.5. Do the data present sufficient evidence to indicate that scientists are overestimating the mean biomass for tropical woodlands and that the mean is in fact lower than estimated? a. State the null and alternative hypotheses to be tested. b. Locate the rejection region for the test with a .01. c. Conduct the test and state your conclusions. 9.65 Adolescents and Social Stress In a study to compare ethnic differences in adolescents’ social stress, researchers recruited subjects from three middle schools in Houston, Texas.21 Social stress among four ethnic groups was measured using the Social Attitudinal Familial and Environment Scale for Children (SAFE-C). In addition, demographic information about the 316 students was collected using self-administered questionnaires. A tabulation of student responses to a question regarding their socioeconomic status (SES) compared with other families in which the students chose one of ﬁve responses (much worse off, somewhat worse off, about the same, better off, or much better off ) resulted in the tabulation that follows. European African American American American American Hispanic Asian Sample Size About the Same 144 68 66 42 77 48 19 8 SUPPLEMENTARY EXERCISES ❍ 381 a. Do these data support the hypothesis that the proportion of adolescent African Americans who state that their SES is “about the same” exceeds that for adolescent Hispanic Americans? b. Find the p-value for the test. c. If you plan to test using a .05, what is your conclusion? 9.66 Adolescents and Social Stress, continued Refer to Exercise 9.65. Some thought should have been given to designing a test for which b is tolerably low when p1 exceeds p2 by an important amount. For example, ﬁnd a common sample size n for a test with a .05 and b .20 when in fact p1 exceeds p2 by 0.1. (H
INT: The maximum value of p(1 p) .25.) 9.67 Losing Weight In a comparison of the mean 1-month weight losses for women aged 20–30 years, these sample data were obtained for each of two diets: Diet I Diet II Sample Size n Sample Mean x Sample Variance s 2 40 10 lb 4.3 40 8 lb 5.7 Do the data provide sufficient evidence to indicate that diet I produces a greater mean weight loss than diet II? Use a .05. 9.68 Increased Yield An agronomist has shown experimentally that a new irrigation/fertilization regimen produces an increase of 2 bushels per quadrat (signiﬁcant at the 1% level) when compared with the regimen currently in use. The cost of implementing and using the new regimen will not be a factor if the increase in yield exceeds 3 bushels per quadrat. Is statistical signiﬁcance the same as practical importance in this situation? Explain. 9.69 Breaking Strengths of Cables A test of the breaking strengths of two different types of cables was conducted using samples of n1 n2 100 pieces of each type of cable. Cable I x1 1925 s1 40 Cable II x2 1905 s2 30 Do the data provide sufficient evidence to indicate a difference between the mean breaking strengths of the two cables? Use a .05. 9.70 Put on the Brakes The braking ability was compared for two 2008 automobile models. Random samples of 64 automobiles were tested for each type. The recorded measurement was the distance (in feet) 382 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES required to stop when the brakes were applied at 40 miles per hour. These are the computed sample means and variances: Model I x1 118 1 102 s 2 Model II x2 109 2 87 s 2 Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models? 9.71 Spraying Fruit Trees A fruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage. To test the claim, the grower sprays 200 trees with the new spray and 200 other trees with the standard spray. The following data were recorded: New Spray Standard Spray Mean Yield per Tree x (lb) Variance s 2 240 980 227 820 a. Do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray? Use a .05. b. Construct a 95% conﬁdence interval for the difference between the mean yields for the two sprays. 9.72 Actinomycin D A biologist hypothesizes that high concentrations of actinomycin D inhibit RNA synthesis in cells and hence the production of proteins as well. An experiment conducted to test this theory compared the RNA synthesis in cells treated with two concentrations of actinomycin D: .6 and .7 microgram per milliliter. Cells treated with the lower concentration (.6) of actinomycin D showed that 55 out of 70 developed normally, whereas only 23 out of 70 appeared to develop normally for the higher concentration (.7). Do these data provide sufficient evidence to indicate a difference between the rates of normal RNA synthesis for cells exposed to the two different concentrations of actinomycin D? a. Find the p-value for the test. b. If you plan to conduct your test using a .05, what will be your test conclusions? 9.73 SAT Scores How do California high school students compare to students nationwide in their college readiness, as measured by their SAT scores? The national average scores for the class of 2005 were 508 on the verbal portion and 520 on the math portion.22 Suppose that 100 California students from the class of 2005 were randomly selected and their SAT scores recorded in the following table: Verbal Math Sample Average Sample Standard Deviation 499 98 516 96 a. Do the data provide sufficient evidence to indicate that the average verbal score for all California students in the class of 2005 is different from the national average? Test using a .05. b. Do the data provide sufficient evidence to indicate that the average math score for all California students in the class of 2005 is different from the national average? Test using a .05. c. Could you use this data to determine if there is a difference between the average math and verbal scores for all California students in the class of 2005? Explain your answer. 9.74 A Maze Experiment In a maze running study, a rat is run in a T maze and the result of each run recorded. A reward in the form of food is always placed at the right exit. If learning is taking place, the rat will choose the right exit more often than the left. If no learning is taking place, the rat should randomly choose either exit. Suppose that the rat is given n 100 runs in the maze and that he chooses the right exit x 64 times. Would you conclude that learning is taking place? Use the p-value approach, and make a decision based on this p-value. 9.75 PCBs Polychlorinated biphenyls (PCBs) have been found to be dangerously high in some game birds found along the marshlands of the southeastern coast of the United States. The Federal Drug Administration (FDA) considers a concentration of PCBs higher than 5 parts per million (ppm) in these game birds to be dangerous for human consumption. A sample of 38 game birds produced an average of 7.2 ppm with a standard deviation of 6.2 ppm. Is there sufficient evidence to indicate that the mean ppm of PCBs in the population of game birds exceeds the FDA’s recommended limit of 5 ppm? Use a .01. 9.76* PCBs, continued Refer to Exercise 9.75. a. Calculate b and 1 b if the true mean ppm of PCBs is 6 ppm. b. Calculate b and 1 b if the true mean ppm of PCBs is 7 ppm. c. Find the power, 1 b, when m 8, 9, 10, and 12. Use these values to construct a power curve for the test in Exercise 9.75. d. For what values of m does this test have power greater than or equal to .90? 9.77 9/11 Conspiracy Some Americans believe that the entire 9/11 catastrophe was planned and executed by federal officials in order to provide the United States with a pretext for going to war in the Middle East and as a means of consolidating and extending the power of the then-current administration. This group of Americans is larger than you think. A Scripps-Howard poll of n 1010 adults in August of 2006 found that 36% of American consider such a scenario very or somewhat likely!23 In a follow-up poll, a random sample of n 100 adult Americans found that 26 of those sampled agreed that the conspiracy theory was either likely or somewhat likely. Does this sample contradict the reported 36% ﬁgure? Test at the a .05 level of signiﬁcance. 9.78 Heights and Gender It is a well-accepted fact that males are taller on the average than females. But how much taller? The genders of 105 biomedical students (Exercise 1.54) were also recorded and the data are summarized below: Males Females Sample Size Sample Mean Sample Standard Deviation 48 69.58 2.62 77 64.43 2.58 a. Perform a test of hypothesis to either conﬁrm or refute our initial claim that males are taller on the average than females? Use a .01. b. If the results of part a show that our claim was correct, construct a 99% conﬁdence one-sided lower conﬁdence bound for the average difference in heights between male and female college students. How much taller are males than females? 9.79 English as a Second Language The state of California is working very hard to make sure that all elementary-aged students whose native language is not English become proﬁcient in English by the sixth Exercises MYAPPLET EXERCISES ❍ 383 grade. Their progress is monitored each year using the California English Language Development Test.24 The results for two school districts in southern California for a recent school year are shown below. District Riverside Palm Springs Number of Students Tested Percentage Fluent 6124 40 5512 37 Does this data provide sufficient statistical evidence to indicate that the percentage of students who are ﬂuent in English differs for these two districts? Test using a .01. 9.80 Breaststroke Swimmers How much training time does it take to become a world-class breaststroke swimmer? A survey published in The American Journal of Sports Medicine reported the number of meters per week swum by two groups of swimmers—those who competed only in breaststroke and those who competed in the individual medley (which includes breaststroke). The number of meters per week practicing the breaststroke swim was recorded and the summary statistics are shown below.25 Breaststroke Individual Medley Sample Size Sample Mean Sample Standard Deviation 130 9017 7162 80 5853 1961 Is there sufficient evidence to indicate a difference in the average number of meters swum by these two groups of swimmers? Test using a .01. 9.81 Breaststroke, continued Refer to Exercise 9.80. a. Construct a 99% conﬁdence interval for the difference in the average number of meters swum by breaststroke versus individual medley swimmers. b. How much longer do pure breaststroke swimmers practice that stroke than individual medley swimmers? What is the practical reason for this difference? 9.82 School Workers In Exercise 8.109, the average hourly wage for public school cafeteria workers was given as $10.33.26 If n 40 randomly selected public school cafeteria workers within one school district are found to have an average hourly wage of x $9.75 with a standard deviation of s $1.65, would this information contradict the reported average of $10.33? a. What are the null and alternative hypotheses to be tested? b. Use the Large-Sample Test of a Population Mean applet to ﬁnd the observed value of the test statistic. 384 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES c. Use the Large-Sample Test of a Population Mean applet to ﬁnd the p-value of this test. d. Based on your results from part c, what conclu- sions can you draw about the average hourly wage of $10.33? 9.83 Daily Wages The daily wages in a particular industry are normally distributed with a mean of $94 and a standard deviation of $11.88. Suppose a company in this industry employs 40 workers and pays them $91.50 per
week on the average. Can these workers be viewed as a random sample from among all workers in the industry? a. What are the null and alternative hypotheses to be tested? b. Use the Large-Sample Test of a Population Mean applet to ﬁnd the observed value of the test statistic. c. Use the Large-Sample Test of a Population Mean applet to ﬁnd the p-value for this test. d. If you planned to conduct your test using a .01, what would be your test conclusions? e. Was it necessary to know that the daily wages are normally distributed? Explain your answer. 9.84 Refer to Example 9.8. Use the Power of a z-Test applet to verify the power of the test of H0: m 880 versus Ha: m 880 for values of m equal to 870, 875, 880, 885 and 890. Check your answers against the values shown in Table 9.2. 9.85 Refer to Example 9.8. a. Use the method given in Example 9.8 to calculate the power of the test of H0: m 880 versus Ha: m 880 when n 30 and the true value of m is 870 tons. b. Repeat part a using n 70 and m 870 tons. c. Use the Power of a z-Test applet to verify your hand-calculated results in parts a and b. d. What is the effect of increasing the sample size on the power of the test? 9.86 Use the appropriate slider on the Power of a z-Test applet to answer the following questions. Write a sentence for each part, describing what you see using the applet. a. What effect does increasing the sample size have on the power of the test? b. What effect does increasing the distance between the true value of m and the hypothesized value, m 880, have on the power of the test? c. What effect does decreasing the signiﬁcance level a have on the power of the test? CASE STUDY An Aspirin a Day . . . ? On Wednesday, January 27, 1988, the front page of the New York Times read, “Heart attack risk found to be cut by taking aspirin: Lifesaving effects seen.” A very large study of U.S. physicians showed that a single aspirin tablet taken every other day reduced by one-half the risk of heart attack in men.27 Three days later, a headline in the Times read, “Value of daily aspirin disputed in British study of heart attacks.” How could two seemingly similar studies, both involving doctors as participants, reach such opposite conclusions? The U.S. physicians’ health study consisted of two randomized clinical trials in one. The ﬁrst tested the hypothesis that 325 milligrams (mg) of aspirin taken every other day reduces mortality from cardiovascular disease. The second tested whether 50 mg of b-carotene taken on alternate days decreases the incidence of cancer. From names on an American Medical Association computer tape, 261,248 male physicians between the ages of 40 and 84 were invited to participate in the trial. Of those who responded, 59,285 were willing to participate. After the exclusion of those physicians who had a history of medical disorders, or who were currently taking aspirin or had negative reactions to aspirin, 22,071 physicians were randomized into one of four treatment groups: (1) buffered aspirin and b-carotene, (2) buffered aspirin and a CASE STUDY ❍ 385 b-carotene placebo, (3) aspirin placebo and b-carotene, and (4) aspirin placebo and b-carotene placebo. Thus, half were assigned to receive aspirin and half to receive b-carotene. The study was conducted as a double-blind study, in which neither the participants nor the investigators responsible for following the participants knew to which group a participant belonged. The results of the American study concerning myocardial infarctions (the technical name for heart attacks) are given in the following table: American Study Aspirin (n 11,037) Placebo (n 11,034) Myocardial Infarction Fatal Nonfatal Total 5 99 104 18 171 189 The objective of the British study was to determine whether 500 mg of aspirin taken daily would reduce the incidence of and mortality from cardiovascular disease. In 1978 all male physicians in the United Kingdom were invited to participate. After the usual exclusions, 5139 doctors were randomly allocated to take aspirin, unless some problem developed, and one-third were randomly allocated to avoid aspirin. Placebo tablets were not used, so the study was not blind! The results of the British study are given here: British Study Aspirin (n 3429) Control (n 1710) Myocardial Infarction Fatal Nonfatal Total 89 (47.3) 80 (42.5) 169 (89.8) 47 (49.6) 41 (43.3) 88 (92.9) To account for unequal sample sizes, the British study reported rates per 10,000 subject-years alive (given in parentheses). 1. Test whether the American study does in fact indicate that the rate of heart attacks for physicians taking 325 mg of aspirin every other day is signiﬁcantly different from the rate for those on the placebo. Is the American claim justiﬁed? 2. Repeat the analysis using the data from the British study in which one group took 500 mg of aspirin every day and the control group took none. Based on their data, is the British claim justiﬁed? 3. Can you think of some possible reasons the results of these two studies, which were alike in some respects, produced such different conclusions? Inference from Small Samples 10 © CORBIS SYGMA GENERAL OBJECTIVE The basic concepts of large-sample statistical estimation and hypothesis testing for practical situations involving population means and proportions were introduced in Chapters 8 and 9. Because all of these techniques rely on the Central Limit Theorem to justify the normality of the estimators and test statistics, they apply only when the samples are large. This chapter supplements the large-sample techniques by presenting small-sample tests and conﬁdence intervals for population means and variances. Unlike their large-sample counterparts, these small-sample techniques require the sampled populations to be normal, or approximately so. CHAPTER INDEX ● Comparing two population variances (10.7) ● Inferences concerning a population variance (10.6) ● Paired-difference test: Dependent samples (10.5) ● Small-sample assumptions (10.8) ● Small-sample inferences concerning the difference in two means: Independent random samples (10.4) ● Small-sample inferences concerning a population mean (10.3) ● Student’s t distribution (10.2) How Do I Decide Which Test to Use? Would You Like a Four-Day Workweek? Will a ﬂexible workweek schedule result in positive beneﬁts for both employer and employee? Four obvious beneﬁts are (1) less time traveling from ﬁeld positions to the office, (2) fewer employees parked in the parking lot, (3) reduced travel expenses, and (4) allowance for employees to have another day off. But does the ﬂexible workweek make employees more efficient and cause them to take fewer sick and personal days? The answers to some of these questions are posed in the case study at the end of this chapter. 386 10.2 STUDENT’S t DISTRIBUTION ❍ 387 INTRODUCTION 10.1 Suppose you need to run an experiment to estimate a population mean or the difference between two means. The process of collecting the data may be very expensive or very time-consuming. If you cannot collect a large sample, the estimation and test procedures of Chapters 8 and 9 are of no use to you. This chapter introduces some equivalent statistical procedures that can be used when the sample size is small. The estimation and testing procedures involve these familiar parameters: • A single population mean, m • The difference between two population means, (m1 m2) • A single population variance, s 2 • The comparison of two population variances, s 2 1 and s 2 2 Small-sample tests and conﬁdence intervals for binomial proportions will be omitted from our discussion.† STUDENT’S t DISTRIBUTION 10.2 In conducting an experiment to evaluate a new but very costly process for producing synthetic diamonds, you are able to study only six diamonds generated by the process. How can you use these six measurements to make inferences about the average weight m of diamonds from this process? In discussing the sampling distribution of x in Chapter 7, we made these points: • When the original sampled population is normal, x and z (x m)/(s/n) both have normal distributions, for any sample size. • When the original sampled population is not normal, x, z (x m)/(s/n), and z ( x m)/(s/n) all have approximately normal distributions, if the sample size is large. Unfortunately, when the sample size n is small, the statistic (x m)/(s/n) does not have a normal distribution. Therefore, all the critical values of z that you used in Chapters 8 and 9 are no longer correct. For example, you cannot say that x will lie within 1.96 standard errors of m 95% of the time. This problem is not new; it was studied by statisticians and experimenters in the early 1900s. To ﬁnd the sampling distribution of this statistic, there are two ways to proceed: • Use an empirical approach. Draw repeated samples and compute ( x m)/(s/n) for each sample. The relative frequency distribution that you construct using these values will approximate the shape and location of the sampling distribution. • Use a mathematical approach to derive the actual density function or curve that describes the sampling distribution. †A small-sample test for the binomial parameter p will be presented in Chapter 15. When n 30, the Central Limit Theorem will not guarantee that m x n s/ is approximately normal. 388 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES This second approach was used by an Englishman named W.S. Gosset in 1908. He derived a complicated formula for the density function of m x t n s/ for random samples of size n from a normal population, and he published his results under the pen name “Student.” Ever since, the statistic has been known as Student’s t. It has the following characteristics: • • It is mound-shaped and symmetric about t 0, just like z. It is more variable than z, with “heavier tails”; that is, the t curve does not approach the horizontal axis as quickly as z does. This is because the t statistic involves two random quantities, x and s, whereas the z statistic inv
olves only the sample mean, x. You can see this phenomenon in Figure 10.1. • The shape of the t distribution depends on the sample size n. As n increases, the variability of t decreases because the estimate s of s is based on more and more information. Eventually, when n is inﬁnitely large, the t and z distributions are identical! Normal distribution t distribution 0 The divisor (n 1) in the formula for the sample variance s2 is called the number of degrees of freedom (df ) associated with s2. It determines the shape of the t distribution. The origin of the term degrees of freedom is theoretical and refers to the number of independent squared deviations in s2 that are available for estimating s 2. These degrees of freedom may change for different applications and, since they specify the correct t distribution to use, you need to remember to calculate the correct degrees of freedom for each application. The table of probabilities for the standard normal z distribution is no longer useful in calculating critical values or p-values for the t statistic. Instead, you will use Table 4 in Appendix I, which is partially reproduced in Table 10.1. When you index a particular number of degrees of freedom, the table records ta, a value of t that has tail area a to its right, as shown in Figure 10.2. ● F IG URE 10 .1 Standard normal z and the t distribution with 5 degrees of freedom For a one-sample t, df n 1. F IG URE 10 .2 Tabulated values of Student’s t ● f(t) 0 a ta t TABLE 10.1 ● Format of the Student’s t Table from Table 4 in Appendix I 10.2 STUDENT’S t DISTRIBUTION ❍ 389 df 1 2 3 4 5 6 7 8 9 . . . 26 27 28 29 inf. t.100 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 . . . 1.315 1.314 1.313 1.311 1.282 t.050 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 . . . 1.706 1.703 1.701 1.699 1.645 t.025 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 . . . 2.056 2.052 2.048 2.045 1.960 t.010 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 . . . 2.479 2.473 2.467 2.462 2.326 t.005 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 . . . 2.779 2.771 2.763 2.756 2.576 df 1 2 3 4 5 6 7 8 9 . . . 26 27 28 29 inf. EXAMPLE 10.1 For a t distribution with 5 degrees of freedom, the value of t that has area .05 to its right is found in row 5 in the column marked t.050. For this particular t distribution, the area to the right of t 2.015 is .05; only 5% of all values of the t statistic will exceed this value. You can use the Student’s t Probabilities applet to ﬁnd the t-value described in Example 10.1. The ﬁrst applet, shown in Figure 10.3, provides t-values and their two-tailed probabilities, while the second applet provides t-values and one-tailed probabilities. Use the slider on the right side of the applet to select the proper degrees of freedom. For Example 10.1, you should choose df 5 and type .10 in the box marked “prob:” at the bottom of the ﬁrst applet. The applet will provide the value of t that puts .05 in one tail of the t distribution. The second applet will show the identical t for a one-tailed area of .05. The applet in Figure 10.3 shows t 2.02 which is correct to two decimal places. We will use this applet for the MyApplet Exercises at the end of the chapter. FI GUR E 10. 3 Student’s t Probabilities applet ● 390 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES EXAMPLE 10.2 Suppose you have a sample of size n 10 from a normal distribution. Find a value of t such that only 1% of all values of t will be smaller. Solution The degrees of freedom that specify the correct t distribution are df n 1 9, and the necessary t-value must be in the lower portion of the distribution, with area .01 to its left, as shown in Figure 10.4. Since the t distribution is symmetric about 0, this value is simply the negative of the value on the right-hand side with area .01 to its right, or t.01 2.821. F IG URE 10 .4 t Distribution for Example 10.2 ● f(t) .01 –2.821 0 t Comparing the t and z Distributions Look at one of the columns in Table 10.1. As the degrees of freedom increase, the critical value of t decreases until, when df inf., the critical t-value is the same as the critical z-value for the same tail area. You can use the Comparing t and z applet to visualize this concept. Look at the three applets in Figure 10.5, which show the critical values for t.025 compared with z.025 for df 8, 29 and 100. (The slider on the right side of the applet allows you to change the df.) The red curve (black in Figure 10.5) is the standard normal distribution, with z.025 1.96. F IG URE 10 .5 Comparing t and z applet ● Assumptions for one-sample t: • • Random sample Normal distribution 10.3 SMALL-SAMPLE INFERENCES CONCERNING A POPULATION MEAN ❍ 391 The blue curve is the t distribution. With 8 df, you can clearly see a difference in the t and z curves, especially in the critical values that cut off an area of .025 in the tails. As the degrees of freedom increase, the difference in the shapes of t and z becomes very similar, as do their critical values, until at df 100, there is almost no difference. This helps to explain why we use n 30 as the somewhat arbitrary dividing line between large and small samples. When n 30 (df 29), the critical values of t are quite close to their normal counterparts. Rather than produce a t table with rows for many more degrees of freedom, the critical values of z are sufficient when the sample size reaches n 30. Assumptions behind Student’s t Distribution The critical values of t allow you to make reliable inferences only if you follow all the rules; that is, your sample must meet these requirements speciﬁed by the t distribution: • The sample must be randomly selected. • The population from which you are sampling must be normally distributed. These requirements may seem quite restrictive. How can you possibly know the shape of the probability distribution for the entire population if you have only a sample? If this were a serious problem, however, the t statistic could be used in only very limited situations. Fortunately, the shape of the t distribution is not affected very much as long as the sampled population has an approximately mound-shaped distribution. Statisticians say that the t statistic is robust, meaning that the distribution of the statistic does not change signiﬁcantly when the normality assumption is violated. How can you tell whether your sample is from a normal population? Although there are statistical procedures designed for this purpose, the easiest and quickest way to check for normality is to use the graphical techniques of Chapter 2: Draw a dotplot or construct a stem and leaf plot. As long as your plot tends to “mound up” in the center, you can be fairly safe in using the t statistic for making inferences. The random sampling requirement, on the other hand, is quite critical if you want to produce reliable inferences. If the sample is not random, or if it does not at least behave as a random sample, then your sample results may be affected by some unknown factor and your conclusions may be incorrect. When you design an experiment or read about experiments conducted by others, look critically at the way the data have been collected! SMALL-SAMPLE INFERENCES CONCERNING A POPULATION MEAN 10.3 As with large-sample inference, small-sample inference can involve either estimation or hypothesis testing, depending on the preference of the experimenter. We explained the basics of these two types of inference in the earlier chapters, and we use them again now, with a different sample statistic, t ( x m)/(s/n), and a different sampling distribution, the Student’s t, with (n 1) degrees of freedom. 392 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES SMALL-SAMPLE HYPOTHESIS TEST FOR m 1. Null hypothesis: H0 : m m0 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : m m0 Ha : m m0 (or, Ha : m m0) x m 0 3. Test statistic: t n / s 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test t ta/2 or t ta/2 t ta (or t ta when the alternative hypothesis is Ha : m m0) or when p-value a α α/2 0 αt –t α/2 0 α/2 t α/2 The critical values of t, ta, and ta/2 are based on (n 1) degrees of freedom. These tabulated values can be found using Table 4 of Appendix I or the Student’s t Probabilities applet. Assumption: The sample is randomly selected from a normally distributed population. SMALL-SAMPLE (1 a)100% CONFIDENCE INTERVAL FOR m s x ta/2 n where s/n is the estimated standard error of x, often referred to as the standard error of the mean. A new process for producing synthetic diamonds can be operated at a proﬁtable level only if the average weight of the diamonds is greater than .5 karat. To evaluate the proﬁtability of the process, six diamonds are generated, with recorded weights .46, .61, .52, .48, .57, and .54 karat. Do the six measurements present sufficient evidence to indicate that the average weight of the diamonds produced by the process is in excess of .5 karat? Solution The population of diamond weights produced by this new process has mean m, and you can set out the formal test of hypothesis in steps, as you did in Chapter 9: EXAMPLE 10.3 10.3 SMALL-SAMPLE INFERENCES CONCERNING A POPULATION MEAN ❍ 393 1–2 Null and alternative hypotheses: H0: m .5 versus Ha: m .5 3 4 5 Test statistic: You can use your calculator to verify that the mean and standard deviation for the six diamond weights are .53 and .0559, respectively. The test statistic is a t statistic, calculated as m x 3 0 t n / 59 .5 1.32 6 / 5 . 0 . 5 s As with the large-sample tests, the test statistic provides evidence for either rejecting or accepting H0 depending on how far from the center of the t distribution it lies. If you choose a 5% level of signiﬁcance (a .05), the rightRejection region: tailed rejection region is found using the critical values of t from Table 4 of Appendix I. With df n 1 5, you can reject H0 if t t.05 2.015, as shown in Figure 10.6. Conclusion: Since the calculated value of the test statistic, 1.
32, does not fall in the rejection region, you cannot reject H0. The data do not present sufficient evidence to indicate that the mean diamond weight exceeds .5 karat. FI GUR E 10. 6 Rejection region for Example 10.3 ● f(t) A 95% conﬁdence interval tells you that, if you were to construct many of these intervals (all of which would have slightly different endpoints), 95% of them would enclose the population mean. .05 0 1.32 2.015 t Reject H0 As in Chapter 9, the conclusion to accept H0 would require the difficult calculation of b, the probability of a Type II error. To avoid this problem, we choose to not reject H0. We can then calculate the lower bound for m using a small-sample lower onesided conﬁdence bound. This bound is similar to the large-sample one-sided conﬁdence bound, except that the critical za is replaced by a critical ta from Table 4. For this example, a 95% lower one-sided conﬁdence bound for m is: s x ta n 55 9 .0 .53 2.015 6 .53 .046 The 95% lower bound for m is m .484. The range of possible values includes mean diamond weights both smaller and greater than .5; this conﬁrms the failure of our test to show that m exceeds .5. 394 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Remember from Chapter 9 that there are two ways to conduct a test of hypothesis: • The critical value approach: Set up a rejection region based on the critical values of the statistic’s sampling distribution. If the test statistic falls in the rejection region, you can reject H0. • The p-value approach: Calculate the p-value based on the observed value of the test statistic. If the p-value is smaller than the signiﬁcance level, a, you can reject H0. If there is no preset signiﬁcance level, use the guidelines in Section 9.3 to judge the statistical signiﬁcance of your sample results. We used the ﬁrst approach in the solution to Example 10.3. We use the second approach to solve Example 10.4. Labels on 1-gallon cans of paint usually indicate the drying time and the area that can be covered in one coat. Most brands of paint indicate that, in one coat, a gallon will cover between 250 and 500 square feet, depending on the texture of the surface to be painted. One manufacturer, however, claims that a gallon of its paint will cover 400 square feet of surface area. To test this claim, a random sample of ten 1-gallon cans of white paint were used to paint 10 identical areas using the same kind of equipment. The actual areas (in square feet) covered by these 10 gallons of paint are given here: 310 376 311 303 412 410 368 365 447 350 Do the data present sufficient evidence to indicate that the average coverage differs from 400 square feet? Find the p-value for the test, and use it to evaluate the statistical signiﬁcance of the results. Solution To test the claim, the hypotheses to be tested are versus Ha : m 400 H0 : m 400 The sample mean and standard deviation for the recorded data are x 365.2 s 48.417 and the test statistic is EXAMPLE 10.4 Remember from Chapter 2 how to calculate x and s using the data entry method on your calculator. m x .2 3 0 t n / 17 8 4 0 4 0 2.27 0 / 1 .4 6 5 s The p-value for this test is the probability of observing a value of the t statistic as contradictory to the null hypothesis as the one observed for this set of data—namely, t 2.27. Since this is a two-tailed test, the p-value is the probability that either t 2.27 or t 2.27. Unlike the z-table, the table for t gives the values of t corresponding to upper-tail areas equal to .100, .050, .025, .010, and .005. Consequently, you can only approximate the upper-tail area that corresponds to the probability that t 2.27. Since the t statistic for this test is based on 9 df, we refer to the row corresponding to df 9 in Table 4. The ﬁve critical values for various tail areas are shown in Figure 10.7, an enlargement of the tail of the t distribution with 9 degrees of freedom. The value t 2.27 falls between t.025 2.262 and t.010 2.821. Therefore, the right-tail area corresponding to the probability that t 2.27 lies between .01 and .025. Since this area represents only half of the p-value, you can write .01 1 (p-value) .025 or 2 .02 p-value .05 10.3 SMALL-SAMPLE INFERENCES CONCERNING A POPULATION MEAN ❍ 395 FI GUR E 10. 7 Calculating the p-value for Example 10.4 (shaded area 1 p-value) 2 ● f(t) .100 .050 .025 .010 .005 2.262 1.383 1.833 2.821 3.250 t 2.27 What does this tell you about the signiﬁcance of the statistical results? For you to reject H0, the p-value must be less than the speciﬁed signiﬁcance level, a. Hence, you could reject H0 at the 5% level, but not at the 2% or 1% level. Therefore, the p-value for this test would typically be reported by the experimenter as p-value .05 (or sometimes P .05) For this test of hypothesis, H0 is rejected at the 5% signiﬁcance level. There is sufficient evidence to indicate that the average coverage differs from 400 square feet. Within what limits does this average coverage really fall? A 95% conﬁdence in- terval gives the upper and lower limits for m as x ta/2 s n 365.2 2.262 7 1 8.4 4 0 1 365.2 34.63 Thus, you can estimate that the average area covered by 1 gallon of this brand of paint lies in the interval 330.6 to 399.8. A more precise interval estimate (a shorter interval) can generally be obtained by increasing the sample size. Notice that the upper limit of this interval is very close to the value of 400 square feet, the coverage claimed on the label. This coincides with the fact that the observed value of t 2.27 is just slightly less than the left-tail critical value of t.025 2.262, making the p-value just slightly less than .05. Most statistical computing packages contain programs that will implement the Student’s t-test or construct a conﬁdence interval for m when the data are properly entered into the computer’s database. Most of these programs will calculate and report the exact p-value of the test, allowing you to quickly and accurately draw conclusions about the statistical signiﬁcance of the results. The results of the MINITAB one-sample t-test and conﬁdence interval procedures are given in Figure 10.8. Besides the observed value of t 2.27 and the conﬁdence interval (330.6, 399.8), the output gives the sample mean, the sample standard deviation, the standard error of the mean (SE Mean s/n), and the exact p-value of the test (P .049). This is consistent with the range for the p-value that we found using Table 4 in Appendix I: .02 p-value .05 396 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES F IG URE 10 .8 MINITAB output for Example 10.4 ● One-Sample T: Area Test of mu = 400 vs not = 400 Variable Area Variable Area N 10 Mean 365.2 StDev 48.4 SE Mean 15.3 95% CI (330.6, 399.8) T -2.27 P 0.049 You can use the Small Sample Test of a Population Mean applet to visualize the p-values for either one- or two-tailed tests of the population mean m. The procedure follows the same pattern as with previous applets. You enter the values of x, n, and s and press “Enter” after each entry; the applet will calculate t and give you the option of choosing one- or two-tailed p-values (Area to Left, Area to Right, or Two Tails), as well as a Middle area that you will not need. F IG URE 10 .9 Small Sample Test of a Population Mean applet ● For the data of Example 10.4, the p-value is the two-tailed area to the right of t 2.273 and to the left of t 2.273. Can you ﬁnd this same p-value in the MINITAB printout shown in Figure 10.9? You can see the value of using the computer output or the Java applet to evaluate statistical results: • The exact p-value eliminates the need for tables and critical values. • All of the numerical calculations are done for you. The most important job—which is left for the experimenter—is to interpret the results in terms of their practical signiﬁcance! 10.3 SMALL-SAMPLE INFERENCES CONCERNING A POPULATION MEAN ❍ 397 10.3 EXERCISES BASIC TECHNIQUES APPLICATIONS b. t.025 for 8 df d. t.025 for 30 df 10.1 Find the following t-values in Table 4 of Appendix I: a. t.05 for 5 df c. t.10 for 18 df 10.2 Find the critical value(s) of t that specify the rejection region in these situations: a. A two-tailed test with a .01 and 12 df b. A right-tailed test with a .05 and 16 df c. A two-tailed test with a .05 and 25 df d. A left-tailed test with a .01 and 7 df 10.3 Use Table 4 in Appendix I to approximate the p-value for the t statistic in each situation: a. A two-tailed test with t 2.43 and 12 df b. A right-tailed test with t 3.21 and 16 df c. A two-tailed test with t 1.19 and 25 df d. A left-tailed test with t 8.77 and 7 df 10.4 Test Scores The test scores on a 100-point test were recorded for 20 students: EX1004 71 73 84 77 93 86 89 68 91 82 67 65 86 76 62 75 75 57 72 84 a. Can you reasonably assume that these test scores have been selected from a normal population? Use a stem and leaf plot to justify your answer. b. Calculate the mean and standard deviation of the scores. c. If these students can be considered a random sample from the population of all students, ﬁnd a 95% conﬁdence interval for the average test score in the population. 10.5 The following n 10 observations are a sample from a normal population: 7.4 7.1 6.5 7.5 7.6 6.3 6.9 7.7 6.5 7.0 a. Find the mean and standard deviation of these data. b. Find a 99% upper one-sided conﬁdence bound for the population mean m. c. Test H0 : m 7.5 versus Ha : m 7.5. Use a .01. d. Do the results of part b support your conclusion in part c? EX1006 10.6 Tuna Fish Is there a difference in the prices of tuna, depending on the method of packaging? Consumer Reports gives the estimated average price for a 6-ounce can or a 7.06-ounce pouch of tuna, based on prices paid nationally in supermarkets.1 These prices are recorded for a variety of different brands of tuna. Light Tuna White Tuna White Tuna in Water in Water in Oil 1.27 1.22 1.19 1.22 .99 1.92 1.23 .85 .65 .69 .60 .53 1.41 1.12 .63 .67 .60 .66 1.49 1.29 1.27 1.35 1.29 1.00 1.27 1.28 Light Tuna in Oil 2.56 1.92 1.30 1.79 1.23 .62 .66
62 .65 .60 .67 Source: Case Study “Pricing of Tuna” Copyright 2001 by Consumers Union of U.S., Inc., Yonkers, NY 10703-1057, a nonproﬁt organization. Reprinted with permission from the June 2001 issue of Consumer Reports® for educational purposes only. No commercial use or reproduction permitted. www.ConsumerReports.org®. Assume that the tuna brands included in this survey represent a random sample of all tuna brands available in the United States. a. Find a 95% conﬁdence interval for the average price for light tuna in water. Interpret this interval. That is, what does the “95%” refer to? b. Find a 95% conﬁdence interval for the average price for white tuna in oil. How does the width of this interval compare to the width of the interval in part a? Can you explain why? c. Find 95% conﬁdence intervals for the other two samples (white tuna in water and light tuna in oil). Plot the four treatment means and their standard errors in a two-dimensional plot similar to Figure 8.5. What kind of broad comparisons can you make about the four treatments? (We will discuss the procedure for comparing more than two population means in Chapter 11.) 10.7 Dissolved O2 Content Industrial wastes and sewage dumped into our rivers and streams absorb oxygen and thereby reduce the amount of dissolved oxygen available for ﬁsh and other forms of aquatic life. One state agency requires a minimum of 5 parts per million (ppm) of dissolved oxygen in order for the oxygen content to be sufficient to support aquatic life. Six water specimens taken from a river at a speciﬁc location during the low-water season (July) gave 398 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES readings of 4.9, 5.1, 4.9, 5.0, 5.0, and 4.7 ppm of dissolved oxygen. Do the data provide sufficient evidence to indicate that the dissolved oxygen content is less than 5 ppm? Test using a .05. 10.8 Lobsters In a study of the infestation of the Thenus orientalis lobster by two types of barnacles, Octolasmis tridens and O. lowei, the carapace lengths (in millimeters) of 10 randomly selected lobsters caught in the seas near Singapore are measured:2 78 50 60 Find a 95% conﬁdence interval for the mean carapace length of the T. orientalis lobsters. 52 58 60 65 66 56 63 EX1009 10.9 Smoking and Lung Capacity It is recognized that cigarette smoking has a deleterious effect on lung function. In a study of the effect of cigarette smoking on the carbon monoxide diffusing capacity (DL) of the lung, researchers found that current smokers had DL readings signiﬁcantly lower than those of either exsmokers or nonsmokers. The carbon monoxide diffusing capacities for a random sample of n 20 current smokers are listed here: 91.052 73.003 103.768 76.014 90.677 92.295 89.222 71.210 100.615 102.754 90.479 73.154 a. Do these data indicate that the mean DL reading 123.086 84.023 82.115 106.755 88.602 61.675 88.017 108.579 for current smokers is signiﬁcantly lower than 100 DL, the average for nonsmokers? Use a .01. b. Find a 99% upper one-sided conﬁdence bound for the mean DL reading for current smokers. Does this bound conﬁrm your conclusions in part a? EX1010 10.10 Brett Favre In Exercise 2.36 (EX0236), the number of passes completed by Brett Favre, quarterback for the Green Bay Packers, was recorded for each of the 16 regular season games in the fall of 2006 (ESPN.com):3 25 15 24 17 22 26 a. A stem and leaf plot of the n 16 observations is 22 5 21 31 28 20 19 24 22 22 shown below: Stem-and-Leaf Display: Favre Stem-and-leaf of Favre N = 16 Leaf Unit = 1.0 LO 01 (4) 2 2222 6 2 445 3 2 6 2 2 8 1 3 1 Based on this plot, is it reasonable to assume that the underlying population is approximately normal, as required for the one-sample t-test? Explain. b. Calculate the mean and standard deviation for Brett Favre’s per game pass completions. c. Construct a 95% conﬁdence interval to estimate the per game pass completions per game for Brett Favre. 10.11 Purifying Organic Compound Organic chemists often purify organic compounds by a method known as fractional crystallization. An experimenter wanted to prepare and purify 4.85 grams (g) of aniline. Ten 4.85-g quantities of aniline were individually prepared and puriﬁed to acetanilide. The following dry yields were recorded: 3.85 3.36 3.80 3.62 3.88 4.01 3.85 3.72 3.90 3.82 Estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline. Use a 95% conﬁdence interval. 10.12 Organic Compounds, continued Refer to Exercise 10.11. Approximately how many 4.85-g specimens of aniline are required if you wish to estimate the mean number of grams of acetanilide correct to within .06 g with probability equal to .95? 10.13 Bulimia Although there are many treatments for bulimia nervosa, some subjects fail to beneﬁt from treatment. In a study to determine which factors predict who will beneﬁt from treatment, an article in the British Journal of Clinical Psychology indicates that self-esteem was one of these important predictors.4 The table gives the mean and standard deviation of self-esteem scores prior to treatment, at posttreatment, and during a follow-up: Pretreatment Posttreatment Follow-up Sample Mean x Standard Deviation s Sample Size n 20.3 5.0 21 26.6 7.4 21 27.7 8.2 20 a. Use a test of hypothesis to determine whether there is sufficient evidence to conclude that the true pretreatment mean is less than 25. b. Construct a 95% conﬁdence interval for the true posttreatment mean. c. In Section 10.4, we will introduce small-sample techniques for making inferences about the difference between two population means. Without the formality of a statistical test, what are you willing to conclude about the differences among the three sampled population means represented by the results in the table? 10.4 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 399 10.14 RBC Counts Here are the red blood cell counts (in 106 cells per microliter) of a EX1014 healthy person measured on each of 15 days: 5.4 5.3 5.3 5.2 5.4 4.9 5.0 5.2 5.4 5.2 5.1 5.2 5.5 5.3 5.2 Find a 95% conﬁdence interval estimate of m, the true mean red blood cell count for this person during the period of testing. 10.15 Hamburger Meat These data are the weights (in pounds) of 27 packages of ground EX1015 beef in a supermarket meat display: 1.08 1.06 .89 .89 .99 1.14 .89 .98 .97 1.38 .96 1.14 1.18 .75 1.12 .92 1.41 .96 1.12 1.18 1.28 1.08 .93 1.17 .83 .87 1.24 a. Interpret the accompanying MINITAB printouts for the one-sample test and estimation procedures. MINITAB output for Exercise 10.15 One-Sample T: Weight Test of mu = 1 vs not = 1 Variable Weight Variable Weight N 27 Mean 1.0522 StDev 0.1657 SE Mean 0.0319 95% CI (0.9867, 1.1178) T 1.64 P 0.113 b. Verify the calculated values of t and the upper and lower conﬁdence limits. 10.16 Cholesterol The serum cholesterol levels of 50 subjects randomly selected from EX1016 the L.A. Heart Data, data from an epidemiological heart disease study on Los Angeles County employees,5 follow. 148 303 262 278 305 304 315 284 227 225 300 174 275 220 306 240 209 229 260 184 368 253 261 221 242 139 169 239 247 282 203 170 254 178 311 249 254 222 204 271 265 212 273 250 276 229 255 299 256 248 a. Construct a histogram for the data. Are the data approximately mound-shaped? b. Use a t-distribution to construct a 95% conﬁdence interval for the average serum cholesterol levels for L.A. County employees. 10.17 Cholesterol, continued Refer to Exercise 10.16. Since n 30, use the methods of Chapter 8 to create a large-sample 95% conﬁdence interval for the average serum cholesterol level for L.A. County employees. Compare the two intervals. (HINT: The two intervals should be quite similar. This is the reason we choose to approximate the sample distribution of m x with a z-distribution when n 30.) n s/ SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS: INDEPENDENT RANDOM SAMPLES 10.4 The physical setting for the problem considered in this section is the same as the one in Section 8.6, except that the sample sizes are no longer large. Independent random samples of n1 and n2 measurements are drawn from two populations, with means and variances m1, s 2 2, and your objective is to make inferences about (m1 m2), the difference between the two population means. 1, m2, and s 2 When the sample sizes are small, you can no longer rely on the Central Limit Theorem to ensure that the sample means will be normal. If the original populations are normal, however, then the sampling distribution of the difference in the sample means, (x1 x2), will be normal (even for small samples) with mean (m1 m2) and standard error 2 s 2 1 s 2 n2 n1 400 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Assumptions for the two-sample (independent) t-test: • • • Random independent samples Normal distributions s1 s2 In Chapters 7 and 8, you used the sample variances, s2 2, to calculate an estimate of the standard error, which was then used to form a large-sample conﬁdence interval or a test of hypothesis based on the large-sample z statistic: 1 and s2 z (x1 x2) (m1 m2) ns2 s2 1 n 1 2 2 Unfortunately, when the sample sizes are small, this statistic does not have an approximately normal distribution—nor does it have a Student’s t distribution. In order to form a statistic with a sampling distribution that can be derived theoretically, you must make one more assumption. Suppose that the variability of the measurements in the two normal populations is the same and can be measured by a common variance s 2. That is, both populations 2 s 2. Then the standard error of the difhave exactly the same shape, and s 2 ference in the two sample means is n1 n2 1 n 1 1 n 2 It can be proven mathematically that, if you use the appropriate sample estimate s2 for the population variance s 2, then the resulting test statistic, t (x1 x2) (m1 m2) s2 1 n 1 n 2 1 has a Student’s t distribution. The only remaining problem is to ﬁnd the sample estimate s2 and the appropriate number of deg
rees of freedom for the t statistic. Remember that the population variance s 2 describes the shape of the normal distributions from which your samples come, so that either s2 2 would give you an estimate of s 2. But why use just one when information is provided by both? A better procedure is to combine the information in both sample variances using a weighted average, in which the weights are determined by the relative amount of information (the number of measurements) in each sample. For example, if the ﬁrst sample contained twice as many measurements as the second, you might consider giving the ﬁrst sample variance twice as much weight. To achieve this result, use this formula: 1 or s2 s2 (n1 1)s2 1 (n2 1)s2 2 n1 n2 2 Remember from Section 10.3 that the degrees of freedom for the one-sample t statis1 has (n1 1) df tic are (n 1), the denominator of the sample estimate s2. Since s2 and s2 2 has (n2 1) df, the total number of degrees of freedom is the sum (n1 1) (n2 1) n1 n2 2 shown in the denominator of the formula for s2. 10.4 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 401 CALCULATION OF s2 • If you have a scientiﬁc calculator, calculate each of the two sample standard deviations s1 and s2 separately, using the data entry procedure for your particular calculator. These values are squared and used in this formula: s2 (n1 1)s2 1 (n2 1)s2 2 n1 n2 2 It can be shown that s2 is an unbiased estimator of the common population variance s 2. If s2 is used to estimate s 2 and if the samples have been randomly and independently drawn from normal populations with a common variance, then the statistic For the two-sample (independent) t-test, df n1 n2 2 t (x1 x2) (m1 m2) s2 1 n 1 1 n 2 has a Student’s t distribution with (n1 n2 2) degrees of freedom. The small-sample estimation and test procedures for the difference between two means are given next. TEST OF HYPOTHESIS CONCERNING THE DIFFERENCE BETWEEN TWO MEANS: INDEPENDENT RANDOM SAMPLES 1. Null hypothesis: H0 : (m1 m2) D0, where D0 is some speciﬁed difference that you wish to test. For many tests, you will hypothesize that there is no difference between m1 and m2; that is, D0 0. 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : (m1 m2) D0 [or Ha : (m1 m2) D0] Ha : (m1 m2) D0 3. Test statistic: t (x1 x2) D0 s2 1 n 1 n 1 2 where s2 (n1 1)s2 1 (n2 1)s2 2 n1 n2 2 4. Rejection region: Reject H0 when One-Tailed Test t ta [or t ta when the alternative hypothesis is Ha : (m1 m2) D0] or when p-value a Two-Tailed Test t ta/2 or t ta/2 (continued) 402 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES TEST OF HYPOTHESIS CONCERNING THE DIFFERENCE BETWEEN TWO MEANS: INDEPENDENT RANDOM SAMPLES (continued) The critical values of t, ta, and ta/2 are based on (n1 n2 2) df. The tabulated values can be found using Table 4 of Appendix I or the Student’s t Probabilities applet. Assumptions: The samples are randomly and independently selected from normally distributed populations. The variances of the populations s 2 equal. 1 and s 2 2 are SMALL-SAMPLE (1 a)100% CONFIDENCE INTERVAL FOR (m1 m2) BASED ON INDEPENDENT RANDOM SAMPLES (x1 x2) ta/2s2 1 n 2 1 n 1 EXAMPLE 10.5 where s2 is the pooled estimate of s 2. A course can be taken for credit either by attending lecture sessions at ﬁxed times and days, or by doing online sessions that can be done at the student’s own pace and at those times the student chooses. The course coordinator wants to determine if these two ways of taking the course resulted in a signiﬁcant difference in achievement as measured by the ﬁnal exam for the course. The following data gives the scores on an examination with 45 possible points for one group of n1 9 students who took the course online, and a second group of n2 9 students who took the course with conventional lectures. Do these data present sufficient evidence to indicate that the grades for students who took the course online are signiﬁcantly higher than those who attended a conventional class? TABLE 10.2 ● Test Scores for Online and Classroom Presentations Online Classroom 32 37 35 28 41 44 35 31 34 35 31 29 25 34 40 27 32 31 Solution Let m1 and m2 be the mean scores for the online group and the classroom group, respectively. Then, since you seek evidence to support the theory that m1 m2, you can test the null hypothesis H0 : m1 m2 [or H0 : (m1 m2) 0] 10.4 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 403 versus the alternative hypothesis Ha : m1 m2 [or Ha : (m1 m2) 0] To conduct the t-test for these two independent samples, you must assume that the sampled populations are both normal and have the same variance s 2. Is this reasonable? Stem and leaf plots of the data in Figure 10.10 show at least a “mounding” pattern, so that the assumption of normality is not unreasonable. FI GUR E 10. 10 Stem and leaf plots for Example 10.5 ● Online Classroom 2 3 3 4 8 124 557 14 2 3 3 4 579 1124 5 0 Furthermore, the standard deviations of the two samples, calculated as s1 4.9441 and s2 4.4752 Stem and leaf plots can help you decide if the normality assumption is reasonable. are not different enough for us to doubt that the two distributions may have the same shape. If you make these two assumptions and calculate (using full accuracy) the pooled estimate of the common variance as s2 1 (n2 1)s2 (n1 1)s2 2 n1 n2 2 8(4.9441)2 8(4.4752)2 9 9 2 22.2361 you can then calculate the test statistic, 9 9 t 1.65 x1 x2 s2 1 1 n n 2 1 35.22 31.56 1 22.23611 The alternative hypothesis Ha : m1 m2 or, equivalently, Ha : (m1 m2) 0 implies that you should use a one-tailed test in the upper tail of the t distribution with (n1 n2 2) 16 degrees of freedom. You can ﬁnd the appropriate critical value for a rejection region with a .05 in Table 4 of Appendix I, and H0 will be rejected if t 1.746. Comparing the observed value of the test statistic t 1.65 with the critical value t.05 1.746, you cannot reject the null hypothesis (see Figure 10.11). There is insufficient evidence to indicate that the online course grades are higher than the conventional course grades at the 5% level of signiﬁcance. If you are using a calculator, don’t round off until the ﬁnal step! FI GUR E 10. 11 Rejection region for Example 10.5 ● f(t) α = .05 0 1.746 t Reject H0 404 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES EXAMPLE 10.6 Find the p-value that would be reported for the statistical test in Example 10.5. Solution The observed value of t for this one-tailed test is t 1.65. Therefore, p-value P(t 1.65) for a t statistic with 16 degrees of freedom. Remember that you cannot obtain this probability directly from Table 4 in Appendix I; you can only bound the p-value using the critical values in the table. Since the observed value, t 1.65, lies between t.100 1.337 and t.050 1.746, the tail area to the right of 1.65 is between .05 and .10. The p-value for this test would be reported as .05 p-value .10 Because the p-value is greater than .05, most researchers would report the results as not signiﬁcant. You can use the Two-Sample t-Test: Independent Samples applet, shown in Figure 10.12, to visualize the p-values for either one- or two-tailed tests of the difference between two population means. The procedure follows the same pattern as with previous applets. You need to enter summary statistics—the values of x1, x2, n1, n2, s1, and s2 and press “Enter” after each entry; the applet will calculate t (assuming equal variances) and give you the option of choosing one- or two-tailed p-values, (Area to Left, Area to Right, or Two Tails), as well as a Middle area that you will not need. ● F IGU RE 1 0. 12 Two-Sample t-Test: Independent Samples applet For the data of Example 10.5, the p-value is the one-tailed area to the right of t 1.65. Does the p-value conﬁrm the conclusions for the test in Example 10.5? 10.4 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 405 EXAMPLE 10.7 Use a lower 95% conﬁdence bound to estimate the difference (m1 m2) in Example 10.5. Does the lower conﬁdence bound indicate that the online average is signiﬁcantly higher than the classroom average? Solution The lower conﬁdence bound takes a familiar form—the point estimator (x1 x2) minus an amount equal to ta times the standard error of the estimator. Substituting into the formula, you can calculate the 95% lower conﬁdence bound: (x1 x2) tas2 (35.22 31.56) 1.74622.23611 1 n 1 n 2 1 1 9 9 Larger s 2/Smaller s 2 3 ⇔ variance assumption is reasonable 3.66 3.88 or (m1 m2) .22. Since the value (m1 m2) 0 is included in the conﬁdence interval, it is possible that the two means are equal. There is insufficient evidence to indicate that the online average is higher than the classroom average. The two-sample procedure that uses a pooled estimate of the common variance s 2 relies on four important assumptions: • The samples must be randomly selected. Samples not randomly selected may introduce bias into the experiment and thus alter the signiﬁcance levels you are reporting. • The samples must be independent. If not, this is not the appropriate statistical procedure. We discuss another procedure for dependent samples in Section 10.5. • The populations from which you sample must be normal. However, moderate departures from normality do not seriously affect the distribution of the test statistic, especially if the sample sizes are nearly the same. • The population variances should be equal or nearly equal to ensure that the procedures are valid. If the population variances are far from equal, there is an alternative procedure for estimation and testing that has an approximate t distribution in repeated sampling. As a rule of thumb, you should use this procedure if the ratio of the two sample variances, 2 s g r e ar L 3 2 s r e ll a m S Since the population variances are not equal, the pooled estimator s2 is no longer appropriate, and each population variance must be estimated by its corresponding sample variance. The resulting test statistic
is (x1 x2) D0 ns2 s2 1 n 1 2 2 406 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES When the sample sizes are small, critical values for this statistic are found using degrees of freedom approximated by the formula 2 2 2 s2 s1 n n 2 1 2 2 2/ 2/ n2) n1 df The degrees of freedom are taken to be the integer part of this result. Computer packages such as MINITAB can be used to implement this procedure, sometimes called Satterthwaite’s approximation, as well as the pooled method described earlier. In fact, some experimenters choose to analyze their data using both methods. As long as both analyses lead to the same conclusions, you need not concern yourself with the equality or inequality of variances. The MINITAB output resulting from the pooled method of analysis for the data of Example 10.5 is shown in Figure 10.13. Notice that the ratio of the two sample variances, (4.94/4.48)2 1.22, is less than 3, which makes the pooled method appropriate. The calculated value of t 1.65 and the exact p-value .059 with 16 degrees of freedom are shown in the last line of the output. The exact p-value makes it quite easy for you to determine the signiﬁcance or nonsigniﬁcance of the sample results. You will ﬁnd instructions for generating this MINITAB output in the section “My MINITAB” at the end of this chapter. F IG URE 10 .1 3 MINITAB output for Example 10.5 ● Two-Sample T-Test and CI: Online, Classroom Two-sample T for Online vs Classroom Online Classroom N 9 9 Mean 35.22 31.56 StDev 4.94 4.48 SE Mean 1.6 1.5 Difference = mu (Online) - mu (Classroom) Estimate for difference: 3.67 95% lower bound for difference: -0.21 T-Test of difference = 0 (vs >): T-Value = 1.65 P-Value = 0.059 DF = 16 Both use Pooled StDev = 4.7155 If there is reason to believe that the normality assumptions have been violated, you can test for a shift in location of two population distributions using the nonparametric Wilcoxon rank sum test of Chapter 15. This test procedure, which requires fewer assumptions concerning the nature of the population probability distributions, is almost as sensitive in detecting a difference in population means when the conditions necessary for the t-test are satisﬁed. It may be more sensitive when the normality assumption is not satisﬁed. 10.4 EXERCISES BASIC TECHNIQUES 10.18 Give the number of degrees of freedom for s2, the pooled estimator of s2, in these cases: a. n1 16, n2 8 b. n1 10, n2 12 c. n1 15, n2 3 10.19 Calculate s2, the pooled estimator for s 2, in these cases: a. n1 10, n2 4, s2 b. n1 12, n2 21, s2 1 3.4, s2 1 18, s2 2 4.9 2 23 10.4 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 407 10.20 Two independent random samples of sizes n1 4 and n2 5 are selected from each of two normal populations: Population 1 Population 2 12 14 3 7 8 7 5 9 6 a. Calculate s2, the pooled estimator of s 2. b. Find a 90% conﬁdence interval for (m1 m2), the difference between the two population means. c. Test H0 : (m1 m2) 0 against Ha : (m1 m2) 0 for a .05. State your conclusions. 10.21 Independent random samples of n1 16 and n2 13 observations were selected from two normal populations with equal variances: Population 1 16 34.6 4.8 2 13 32.2 5.9 Sample Size Sample Mean Sample Variance a. Suppose you wish to detect a difference between the population means. State the null and alternative hypotheses for the test. b. Find the rejection region for the test in part a for a .01. c. Find the value of the test statistic. d. Find the approximate p-value for the test. e. Conduct the test and state your conclusions. 10.22 Refer to Exercise 10.21. Find a 99% conﬁdence interval for (m1 m2). 10.23 The MINITAB printout shows a test for the difference in two population means. MINITAB output for Exercise 10.23 Two-Sample T-Test and CI: Sample 1, Sample 2 Two-sample T for Sample 1 vs Sample 2 Mean 29.00 28.86 N Sample 1 6 Sample 2 7 SE Mean 1.6 1.8 StDev 4.00 4.67 b. What is the observed value of the test statistic? What is the p-value associated with this test? c. What is the pooled estimate s2 of the population variance? d. Use the answers to part b to draw conclusions about the difference in the two population means. e. Find the 95% conﬁdence interval for the difference in the population means. Does this interval conﬁrm your conclusions in part d? APPLICATIONS 10.24 Healthy Teeth Jan Lindhe conducted a study on the effect of an oral antiplaque rinse on plaque buildup on teeth.6 Fourteen people whose teeth were thoroughly cleaned and polished were randomly assigned to two groups of seven subjects each. Both groups were assigned to use oral rinses (no brushing) for a 2-week period. Group 1 used a rinse that contained an antiplaque agent. Group 2, the control group, received a similar rinse except that, unknown to the subjects, the rinse contained no antiplaque agent. A plaque index x, a measure of plaque buildup, was recorded at 4, 7, and 14 days. The mean and standard deviation for the 14-day plaque measurements are shown in the table for the two groups. Control Group Antiplaque Group Sample Size Mean Standard Deviation 7 1.26 .32 7 .78 .32 a. State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse. b. Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using a .05. c. Find the approximate p-value for the test. Difference = mu (Sample 1) - mu (Sample 2) Estimate for difference: 0.14 95% CI for difference: (-5.2, 5.5) T-Test of difference = 0 (vs not =): T-Value = 0.06 P-Value = 0.95 DF = 11 Both use Pooled StDev = 4.38 a. Do the two sample standard deviations indicate that the assumption of a common population variance is reasonable? EX1025 10.25 Tuna, again In Exercise 10.6 we presented data on the estimated average price for a 6-ounce can or a 7.06-ounce pouch of tuna, based on prices paid nationally in supermarkets. A portion of the data is reproduced in the table below. Use the MINITAB printout to answer the questions. 408 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Light Tuna in Water Light Tuna in Oil .99 1.92 1.23 .85 .65 .69 .60 .53 1.41 1.12 .63 .67 .60 .66 2.56 1.92 1.30 1.79 1.23 .62 .66 .62 .65 .60 .67 MINITAB output for Exercise 10.25 Two-Sample T-Test and CI: Water, Oil Two-sample T for Water vs Oil StDev 0.400 0.679 Mean 0.896 1.147 Water Oil N 14 11 SE Mean 0.11 0.20 Difference = mu (Water) - mu (Oil) Estimate for difference: -0.251 95% CI for difference: (-0.700, 0.198) T-Test of difference = 0 (vs not =): T-Value = -1.16 P-Value = 0.260 DF = 23 Both use Pooled StDev = 0.5389 b. Construct a 95% conﬁdence interval estimate of the difference in means for runners and cyclists under the condition of exercising at 80% of maximal oxygen consumption. c. To test for a signiﬁcant difference in compartment pressure at maximal oxygen consumption, should you use the pooled or unpooled t-test? Explain. 10.27 Disinfectants An experiment published in The American Biology Teacher studied the efficacy of using 95% ethanol or 20% bleach as a disinfectant in removing bacterial and fungal contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using eggplant as the plant tissue being cultured.8 Five cuttings per plant were placed on a petri dish for each disinfectant and stored at 25°C for 4 weeks. The observation reported was the number of uncontaminated eggplant cuttings after the 4-week storage. Disinfectant 95% Ethanol 20% Bleach a. Do the data in the table present sufficient evidence to indicate a difference in the average prices of light tuna in water versus oil? Test using a .05. Mean Variance n 3.73 2.78095 15 4.80 .17143 15 Pooled variance 1.47619 b. What is the p-value for the test? c. The MINITAB analysis uses the pooled estimate of s 2. Is the assumption of equal variances reasonable? Why or why not? 10.26 Runners and Cyclists Chronic anterior compartment syndrome is a condition characterized by exercise-induced pain in the lower leg. Swelling and impaired nerve and muscle function also accompany this pain, which is relieved by rest. Susan Beckham and colleagues conducted an experiment involving ten healthy runners and ten healthy cyclists to determine whether there are signiﬁcant differences in pressure measurements within the anterior muscle compartment for runners and cyclists.7 The data summary—compartment pressure in millimeters of mercury (Hg)—is as follows: Condition Rest 80% maximal O2 consumption Maximal O2 consumption Runners Cyclists Standard Standard Mean Deviation Mean Deviation 14.5 12.2 19.1 3.92 11.1 3.49 16.9 11.5 12.2 3.98 4.95 4.47 a. Test for a signiﬁcant difference in compartment pressure between runners and cyclists under the resting condition. Use a .05. a. Are you willing to assume that the underlying vari- ances are equal? b. Using the information from part a, are you willing to conclude that there is a signiﬁcant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested? EX1028 10.28 Titanium A geologist collected 20 different ore samples, all of the same weight, and randomly divided them into two groups. The titanium contents of the samples, found using two different methods, are listed in the table: Method 1 Method 2 .011 .013 .013 .010 .013 .013 .015 .011 .014 .012 .011 .012 .016 .017 .013 .013 .012 .014 .015 .015 a. Use an appropriate method to test for a signiﬁcant difference in the average titanium contents using the two different methods. b. Determine a 95% conﬁdence interval estimate for (m1 m2). Does your interval estimate substantiate your conclusion in part a? Explain. 10.29 Raisins The numbers of raisins in each of 14 miniboxes (1/2-ounce size) were EX1029 counted for a generic brand and for Sunmaid® brand raisins: 10.4 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 409 Generic Brand Sunmaid 25 26 26 26 26 28 27 26 25 28 24 28 27 25 25 28 25 28 29 24 28 24 24 28 30 24 22 2
7 a. Although counts cannot have a normal distribution, do these data have approximately normal distributions? (HINT: Use a histogram or stem and leaf plot.) b. Are you willing to assume that the underlying pop- ulation variances are equal? Why? c. Use the p-value approach to determine whether there is a signiﬁcant difference in the mean numbers of raisins per minibox. What are the implications of your conclusion? 10.30 Dissolved O2 Content, continued Refer to Exercise 10.7, in which we measured the dissolved oxygen content in river water to determine whether a stream had sufficient oxygen to support aquatic life. A pollution control inspector suspected that a river community was releasing amounts of semitreated sewage into a river. To check his theory, he drew ﬁve randomly selected specimens of river water at a location above the town, and another ﬁve below. The dissolved oxygen readings (in parts per million) are as follows: Above Town Below Town 4.8 5.0 5.2 4.7 5.0 4.9 4.9 4.8 5.1 4.9 a. Do the data provide sufficient evidence to indicate that the mean oxygen content below the town is less than the mean oxygen content above? Test using a .05. b. Suppose you prefer estimation as a method of inference. Estimate the difference in the mean dissolved oxygen contents for locations above and below the town. Use a 95% conﬁdence interval. EX1031 10.31 Freestyle Swimmers In an effort to compare the average swimming times for two swimmers, each swimmer was asked to swim freestyle for a distance of 100 yards at randomly selected times. The swimmers were thoroughly rested between laps and did not race against each other, so that each sample of times was an independent random sample. The times for each of 10 trials are shown for the two swimmers. Swimmer 1 Swimmer 2 59.62 59.48 59.65 59.50 60.01 59.74 59.43 59.72 59.63 59.68 59.81 59.32 59.76 59.64 59.86 59.41 59.63 59.50 59.83 59.51 Suppose that swimmer 2 was last year’s winner when the two swimmers raced. Does it appear that the average time for swimmer 2 is still faster than the average time for swimmer 1 in the 100-yard freestyle? Find the approximate p-value for the test and interpret the results. 10.32 Freestyle Swimmers, continued Refer to Exercise 10.31. Construct a lower 95% one-sided conﬁdence bound for the difference in the average times for the two swimmers. Does this interval conﬁrm your conclusions in Exercise 10.31? EX1033 10.33 Comparing NFL Quarterbacks How does Brett Favre, quarterback for the Green Bay Packers, compare to Peyton Manning, quarterback for the Indianapolis Colts? The table below shows the number of completed passes for each athlete during the 2006 NFL football season:3 Brett Favre Peyton Manning 22 20 26 21 15 31 25 22 22 19 17 28 24 5 22 24 25 29 21 22 25 26 14 21 20 25 32 30 27 20 14 21 a. Does the data indicate that there is a difference in the average number of completed passes for the two quarterbacks? Test using a .05. b. Construct a 95% conﬁdence interval for the difference in the average number of completed passes for the two quarterbacks. Does the conﬁdence interval conﬁrm your conclusion in part a? Explain. EX1034 10.34 An Archeological Find An article in Archaeometry involved an analysis of 26 samples of Romano-British pottery, found at four different kiln sites in the United Kingdom.9 The samples were analyzed to determine their chemical composition and the percentage of aluminum oxide in each of 10 samples at two sites is shown below. Island Thorns Ashley Rails 18.3 15.8 18.0 18.0 20.8 17.7 18.3 16.7 14.8 19.1 Does the data provide sufficient information to indicate that there is a difference in the average percentage of aluminum oxide at the two sites? Test at the 5% level of signiﬁcance. 410 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO MEANS: A PAIRED-DIFFERENCE TEST 10.5 To compare the wearing qualities of two types of automobile tires, A and B, a tire of type A and one of type B are randomly assigned and mounted on the rear wheels of each of ﬁve automobiles. The automobiles are then operated for a speciﬁed number of miles, and the amount of wear is recorded for each tire. These measurements appear in Table 10.3. Do the data present sufficient evidence to indicate a difference in the average wear for the two tire types? TABLE 10.3 ● Average Wear for Two Types of Tires Automobile Tire A Tire B 1 2 3 4 5 10.6 9.8 12.3 9.7 8.8 x1 10.24 s1 1.316 10.2 9.4 11.8 9.1 8.3 x2 9.76 s2 1.328 Table 10.3 shows a difference of (x1 x2) (10.24 9.76) .48 between the two sample means, while the standard deviations of both samples are approximately 1.3. Given the variability of the data and the small number of measurements, this is a rather small difference, and you would probably not suspect a difference in the average wear for the two types of tires. Let’s check your suspicions using the methods of Section 10.4. Look at the MINITAB analysis in Figure 10.14. The two-sample pooled t-test is used for testing the difference in the means based on two independent random samples. The calculated value of t used to test the null hypothesis H0 : m1 m2 is t .57 with p-value .582, a value that is not nearly small enough to indicate a signiﬁcant difference in the two population means. The corresponding 95% conﬁdence interval, given as 1.448 (m1 m2) 2.408 is quite wide and also does not indicate a signiﬁcant difference in the population means. ● F IG URE 10 .1 4 MINITAB output using t-test for independent samples for the tire data Two-Sample T-Test and CI: Tire A, Tire B Two-sample T for Tire A vs Tire B Tire A Tire B N 5 5 Mean 10.24 9.76 StDev 1.32 1.33 SE Mean 0.59 0.59 Difference = mu (Tire A) - mu (Tire B) Estimate for difference: 0.480 95% CI for difference: (-1.448, 2.408) T-Test of difference = 0 (vs not =): T-Value = 0.57 P-Value = 0.582 DF = 8 Both use Pooled StDev = 1.3221 10.5 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO MEANS: A PAIRED-DIFFERENCE TEST ❍ 411 Take a second look at the data and you will notice that the wear measurement for type A is greater than the corresponding value for type B for each of the ﬁve automobiles. Wouldn’t this be unlikely, if there’s really no difference between the two tire types? Consider a simple intuitive test, based on the binomial distribution of Chapter 5. If there is no difference in the mean tire wear for the two types of tires, then it is just as likely as not that tire A shows more wear than tire B. The ﬁve automobiles then correspond to ﬁve binomial trials with p P(tire A shows more wear than tire B) .5. Is the observed value of x 5 positive differences shown in Table 10.4 unusual? The probability of observing x 5 or the equally unlikely value x 0 can be found in Table 1 in Appendix I to be 2(.031) .062, which is quite small compared to the likelihood of the more powerful t-test, which had a p-value of .58. Isn’t it peculiar that the t-test, which uses more information (the actual sample measurements) than the binomial test, fails to supply sufficient information for rejecting the null hypothesis? TABLE 10.4 ● Differences in Tire Wear, Using the Data of Table 10.3 d A B Automobile A B 1 2 3 4 5 10.6 9.8 12.3 9.7 8.8 10.2 9.4 11.8 9.1 8.3 .4 .4 .5 .6 .5 d .48 There is an explanation for this inconsistency. The t-test described in Section 10.4 is not the proper statistical test to be used for our example. The statistical test procedure of Section 10.4 requires that the two samples be independent and random. Certainly, the independence requirement is violated by the manner in which the experiment was conducted. The (pair of) measurements, an A and a B tire, for a particular automobile are deﬁnitely related. A glance at the data shows that the readings have approximately the same magnitude for a particular automobile but vary markedly from one automobile to another. This, of course, is exactly what you might expect. Tire wear is largely determined by driver habits, the balance of the wheels, and the road surface. Since each automobile has a different driver, you would expect a large amount of variability in the data from one automobile to another. In designing the tire wear experiment, the experimenter realized that the measurements would vary greatly from automobile to automobile. If the tires (ﬁve of type A and ﬁve of type B) were randomly assigned to the ten wheels, resulting in independent random samples, this variability would result in a large standard error and make it difficult to detect a difference in the means. Instead, he chose to “pair” the measurements, comparing the wear for type A and type B tires on each of the ﬁve automobiles. This experimental design, sometimes called a paired-difference or matched pairs design, allows us to eliminate the car-to-car variability by looking at only the ﬁve difference measurements shown in Table 10.4. These ﬁve differences form a single random sample of size n 5. Notice that in Table 10.4 the sample mean of the differences, d A B, is cal- culated as S di .48 d n 412 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES and is exactly the same as the difference of the sample means: (x1 x2) (10.24 9.76) .48. It should not surprise you that this can be proven to be true in general, and also that the same relationship holds for the population means. That is, the average of the population differences is md (m1 m2) Because of this fact, you can use the sample differences to test for a signiﬁcant difference in the two population means, (m1 m2) md. The test is a single-sample t-test of the difference measurements to test the null hypothesis H0 : md 0 [or H0 : (m1 m2) 0] versus the alternative hypothesis Ha : md 0 [or Ha : (m1 m2) 0] The test procedures take the same form as the procedures used in Section 10.3 and are described next. PAIRED-DIFFERENCE TEST OF HYPOTHESIS FOR (m1 m2) md: DEPENDENT SAMPLES 1. Null hypothesis: H0 : md 0 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : md 0 (or Ha : md 0
) Ha : md 0 d d 0 3. Test statistic: t n n sd/ s d/ where n Number of paired differences d Mean of the sample differences sd Standard deviation of the sample differences 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test t ta (or t ta when the alternative hypothesis is Ha : md 0) t ta/2 or t ta/2 or when p-value a The critical values of t, ta, and ta/2 are based on (n 1) df. These tabulated values can be found using Table 4 or the Student’s t Probabilities applet. 10.5 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO MEANS: A PAIRED-DIFFERENCE TEST ❍ 413 (1 a)100% SMALL-SAMPLE CONFIDENCE INTERVAL FOR (m1 m2) md, BASED ON A PAIRED-DIFFERENCE EXPERIMENT sd d ta/2 n Assumptions: The experiment is designed as a paired-difference test so that the n differences represent a random sample from a normal population. EXAMPLE 10.8 Do the data in Table 10.3 provide sufficient evidence to indicate a difference in the mean wear for tire types A and B? Test using a .05. Solution You can verify using your calculator that the average and standard deviation of the ﬁve difference measurements are sd .0837 d .48 and Then H0 : md 0 and Ha : md 0 and d 0 8 .4 t 12.8 /5 n s 7 .083 d/ The critical value of t for a two-tailed statistical test, a .05 and 4 df, is 2.776. Certainly, the observed value of t 12.8 is extremely large and highly signiﬁcant. Hence, you can conclude that there is a difference in the mean wear for tire types A and B. EXAMPLE 10.9 Find a 95% conﬁdence interval for (m1 m2) md using the data in Table 10.3. Solution A 95% conﬁdence interval for the difference between the mean levels of wear is sd d ta/2 n .48 2.776 7 83 .0 5 .48 .10 Conﬁdence intervals are always interpreted in the same way! In repeated sampling, intervals constructed in this way enclose the true value of the parameter 100(1 a)% of the time. or .38 (m1 m2) .58. How does the width of this interval compare with the width of an interval you might have constructed if you had designed the experiment in an unpaired manner? It probably would have been of the same magnitude as the interval calculated in Figure 10.14, where the observed data were incorrectly analyzed using the unpaired analysis. This interval, 1.45 (m1 m2) 2.41, is much wider than the paired interval, which indicates that the paired difference design increased the accuracy of our estimate, and we have gained valuable information by using this design. The paired-difference test or matched pairs design used in the tire wear experiment is a simple example of an experimental design called a randomized block design. 414 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Paired difference test: df n 1 When there is a great deal of variability among the experimental units, even before any experimental procedures are implemented, the effect of this variability can be minimized by blocking—that is, comparing the different procedures within groups of relatively similar experimental units called blocks. In this way, the “noise” caused by the large variability does not mask the true differences between the procedures. We will discuss randomized block designs in more detail in Chapter 11. It is important for you to remember that the pairing or blocking occurs when the experiment is planned, and not after the data are collected. An experimenter may choose to use pairs of identical twins to compare two learning methods. A physician may record a patient’s blood pressure before and after a particular medication is given. Once you have used a paired design for an experiment, you no longer have the option of using the unpaired analysis of Section 10.4. The independence assumption has been purposely violated, and your only choice is to use the paired analysis described here! Although pairing was very beneﬁcial in the tire wear experiment, this may not always be the case. In the paired analysis, the degrees of freedom for the t-test are cut in half—from (n n 2) 2(n 1) to (n 1). This reduction increases the critical value of t for rejecting H0 and also increases the width of the conﬁdence interval for the difference in the two means. If pairing is not effective, this increase is not offset by a decrease in the variability, and you may in fact lose rather than gain information by pairing. This, of course, did not happen in the tire experiment—the large reduction in the standard error more than compensated for the loss in degrees of freedom. Except for notation, the paired-difference analysis is the same as the singlesample analysis presented in Section 10.3. However, MINITAB provides a single procedure called Paired t to analyze the differences, as shown in Figure 10.15. The p-value for the paired analysis, .000, indicates a highly signiﬁcant difference in the means. You will ﬁnd instructions for generating this MINITAB output in the “My MINITAB ” section at the end of this chapter. ● F IG URE 10 .1 5 MINITAB output for paired-difference analysis of tire wear data Paired T-Test and CI: Tire A, Tire B Paired T for Tire A - Tire B Tire A Tire B Difference N 5 5 5 Mean 10.240 9.760 0.4800 StDev 1.316 1.328 0.0837 SE Mean 0.589 0.594 0.0374 95% CI for mean difference: (0.3761, 0.5839) T-Test of mean difference = 0 (vs not = 0): T-Value = 12.83 P-Value = 0.000 10.5 EXERCISES BASIC TECHNIQUES 10.35 A paired-difference experiment was conducted using n 10 pairs of observations. a. Test the null hypothesis H0 : (m1 m2) 0 against Ha : (m1 m2) 0 for a .05, d .3, and d .16. Give the approximate p-value for the s2 test. b. Find a 95% conﬁdence interval for (m1 m2). c. How many pairs of observations do you need if you want to estimate (m1 m2) correct to within .1 with probability equal to .95? 10.36 A paired-difference experiment consists of n 18 pairs, d 5.7, and s2 d 256. Suppose you wish to detect md 0. a. Give the null and alternative hypotheses for the test. b. Conduct the test and state your conclusions. 10.5 SMALL-SAMPLE INFERENCES FOR THE DIFFERENCE BETWEEN TWO MEANS: A PAIRED-DIFFERENCE TEST ❍ 415 10.37 A paired-difference experiment was conducted to compare the means of two populations: Pairs Population 1 2 1 1.3 1.2 2 1.6 1.5 3 1.1 1.1 4 1.4 1.2 5 1.7 1.8 a. Do the data provide sufficient evidence to indicate that m1 differs from m2? Test using a .05. b. Find the approximate p-value for the test and inter- pret its value. c. Find a 95% conﬁdence interval for (m1 m2). Compare your interpretation of the conﬁdence interval with your test results in part a. d. What assumptions must you make for your infer- ences to be valid? APPLICATIONS 10.38 Auto Insurance The cost of automobile insurance has become a sore subject in EX1038 California because the rates are dependent on so many variables, such as the city in which you live, the number of cars you insure, and the company with which you are insured. Here are the annual 2006–2007 premiums for a single male, licensed for 6–8 years, who drives a Honda Accord 12,600 to 15,000 miles per year and has no violations or accidents.10 City Allstate 21st Century Long Beach Pomona San Bernardino Moreno Valley $2617 2305 2286 2247 Source: www.insurance.ca.gov $2228 2098 2064 1890 a. Why would you expect these pairs of observations to be dependent? b. Do the data provide sufficient evidence to indicate that there is a difference in the average annual premiums between Allstate and 21st Century insurance? Test using a .01. c. Find the approximate p-value for the test and inter- pret its value. d. Find a 99% conﬁdence interval for the difference in the average annual premiums for Allstate and 21st Century insurance. e. Can we use the information in the table to make valid comparisons between Allstate and 21st Century insurance throughout the United States? Why or why not? 10.39 Runners and Cyclists II Refer to Exercise 10.26. In addition to the compartment pressures, the level of creatine phosphokinase (CPK) in blood samples, a measure of muscle damage, was determined for each of 10 runners and 10 cyclists before and after exercise.7 The data summary—CPK values in units/liter—is as follows: Condition Before exercise After exercise Difference Mean 255.63 284.75 29.13 Runners Cyclists Standard Deviation Mean Standard Deviation 115.48 132.64 21.01 173.8 177.1 3.3 60.69 64.53 6.85 a. Test for a signiﬁcant difference in mean CPK values for runners and cyclists before exercise under the assumption that s 2 2; use a .05. Find a 1 s 2 95% conﬁdence interval estimate for the corresponding difference in means. b. Test for a signiﬁcant difference in mean CPK val- ues for runners and cyclists after exercise under the assumption that s 2 2; use a .05. Find a 95% conﬁdence interval estimate for the corresponding difference in means. 1 s 2 c. Test for a signiﬁcant difference in mean CPK values for runners before and after exercise. d. Find a 95% conﬁdence interval estimate for the difference in mean CPK values for cyclists before and after exercise. Does your estimate indicate that there is no signiﬁcant difference in mean CPK levels for cyclists before and after exercise? EX1040 10.40 America’s Market Basket An advertisement for Albertsons, a supermarket chain in the western United States, claims that Albertsons has had consistently lower prices than four other fullservice supermarkets. As part of a survey conducted by an “independent market basket price-checking company,” the average weekly total, based on the prices of approximately 95 items, is given for two different supermarket chains recorded during 4 consecutive weeks in a particular month. Week Albertsons Ralphs 1 2 3 4 254.26 240.62 231.90 234.13 256.03 255.65 255.12 261.18 a. Is there a signiﬁcant difference in the average prices for these two different supermarket chains? b. What is the approximate p-value for the test con- ducted in part a? 416 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES c. Construct a 99% conﬁdence interval for the difference in the average prices for the two supermarket chains. Interpret this interval. EX1041 1
0.41 No Left Turn An experiment was conducted to compare the mean reaction times to two types of traffic signs: prohibitive (No Left Turn) and permissive (Left Turn Only). Ten drivers were included in the experiment. Each driver was presented with 40 traffic signs, 20 prohibitive and 20 permissive, in random order. The mean time to reaction and the number of correct actions were recorded for each driver. The mean reaction times (in milliseconds) to the 20 prohibitive and 20 permissive traffic signs are shown here for each of the 10 drivers: Driver Prohibitive Permissive 1 2 3 4 5 6 7 8 9 10 824 866 841 770 829 764 857 831 846 759 702 725 744 663 792 708 747 685 742 610 a. Explain why this is a paired-difference experiment and give reasons why the pairing should be useful in increasing information on the difference between the mean reaction times to prohibitive and permissive traffic signs. b. Do the data present sufficient evidence to indicate a difference in mean reaction times to prohibitive and permissive traffic signs? Use the p-value approach. c. Find a 95% conﬁdence interval for the difference in mean reaction times to prohibitive and permissive traffic signs. 10.42 Healthy Teeth II Exercise 10.24 describes a dental experiment conducted to investigate the effectiveness of an oral rinse used to inhibit the growth of plaque on teeth. Subjects were divided into two groups: One group used a rinse with an antiplaque ingredient, and the control group used a rinse containing inactive ingredients. Suppose that the plaque growth on each person’s teeth was measured after using the rinse after 4 hours and then again after 8 hours. If you wish to estimate the difference in plaque growth from 4 to 8 hours, should you use a conﬁdence interval based on a paired or an unpaired analysis? Explain. 10.43 Ground or Air? The earth’s temperature (which affects seed germination, crop survival in bad weather, and many other aspects of agricultural production) can be measured using either ground-based sensors or infrared-sensing devices mounted in aircraft or space satellites. Ground-based sensoring is tedious, requiring many replications to obtain an accurate estimate of ground temperature. On the other hand, airplane or satellite sensoring of infrared waves appears to introduce a bias in the temperature readings. To determine the bias, readings were obtained at ﬁve different locations using both ground- and air-based temperature sensors. The readings (in degrees Celsius) are listed here: Location Ground Air 1 2 3 4 5 46.9 45.4 36.3 31.0 24.7 47.3 48.1 37.9 32.7 26.2 a. Do the data present sufficient evidence to indicate a bias in the air-based temperature readings? Explain. b. Estimate the difference in mean temperatures between ground- and air-based sensors using a 95% conﬁdence interval. c. How many paired observations are required to estimate the difference between mean temperatures for ground- versus air-based sensors correct to within .2°C, with probability approximately equal to .95? 10.44 Red Dye To test the comparative brightness of two red dyes, nine samples of EX1044 cloth were taken from a production line and each sample was divided into two pieces. One of the two pieces in each sample was randomly chosen and red dye 1 applied; red dye 2 was applied to the remaining piece. The following data represent a “brightness score” for each piece. Is there sufficient evidence to indicate a difference in mean brightness scores for the two dyes? Use a .05. Sample Dye 1 Dye 2 1 10 8 2 12 11 3 9 10 4 8 6 5 15 12 6 12 13 7 9 9 8 10 8 9 15 13 10.45 Tax Assessors In response to a complaint that a particular tax assessor (A) was EX1045 biased, an experiment was conducted to compare the assessor named in the complaint with another tax assessor (B) from the same office. Eight properties were selected, and each was assessed by both assessors. The assessments (in thousands of dollars) are shown in the table. Property Assessor A Assessor B 1 2 3 4 5 6 7 8 76.3 88.4 80.2 94.7 68.7 82.8 76.1 79.0 75.1 86.8 77.3 90.6 69.1 81.0 75.3 79.1 Use the MINITAB printout to answer the questions. MINITAB output for Exercise 10.45 Paired T-Test and CI: Assessor A, Assessor B Paired T for Assessor A - Assessor B Assessor A Assessor B Difference N 8 8 8 Mean 80.77 79.29 1.488 StDev 7.99 6.85 1.491 SE Mean 2.83 2.42 0.527 95% lower bound for mean difference: 0.489 T-Test of mean difference = 0 (vs > 0): T-Value = 2.82 P-value = 0.013 a. Do the data provide sufficient evidence to indicate that assessor A tends to give higher assessments than assessor B? b. Estimate the difference in mean assessments for the two assessors. c. What assumptions must you make in order for the inferences in parts a and b to be valid? d. Suppose that assessor A had been compared with a more stable standard—say, the average x of the assessments given by four assessors selected from the tax office. Thus, each property would be assessed by A and also by each of the four other assessors and (xA x) would be calculated. If the test in part a is valid, can you use the paired-difference t-test to test the hypothesis that the bias, the mean difference between A’s assessments and the mean of the assessments of the four assessors, is equal to 0? Explain. 10.6 INFERENCES CONCERNING A POPULATION VARIANCE ❍ 417 10.46 Memory Experiments A psychology class performed an experiment to compare EX1046 whether a recall score in which instructions to form images of 25 words were given is better than an initial recall score for which no imagery instructions were given. Twenty students participated in the experiment with the following results: Student With Imagery Without Imagery Student With Imagery Without Imagery 1 2 3 4 5 6 7 8 9 10 20 24 20 18 22 19 20 19 17 21 5 9 5 9 6 11 8 11 7 9 11 12 13 14 15 16 17 18 19 20 17 20 20 16 24 22 25 21 19 23 8 16 10 12 7 9 21 14 12 13 Does it appear that the average recall score is higher when imagery is used? 10.47 Music in the Workplace Before contracting to have stereo music piped into each of EX1047 his suites of offices, an executive had his office manager randomly select seven offices in which to have the system installed. The average time (in minutes) spent outside these offices per excursion among the employees involved was recorded before and after the music system was installed with the following results. Office Number No Music Music 10 7 7 7 8 Would you suggest that the executive proceed with the installation? Conduct an appropriate test of hypothesis. Find the approximate p-value and interpret your results. INFERENCES CONCERNING A POPULATION VARIANCE 10.6 You have seen in the preceding sections that an estimate of the population variance s 2 is usually needed before you can make inferences about population means. Sometimes, however, the population variance s 2 is the primary objective in an experimental investigation. It may be more important to the experimenter than the population mean! Consider these examples: • Scientiﬁc measuring instruments must provide unbiased readings with a very small error of measurement. An aircraft altimeter that measures the correct 418 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES altitude on the average is fairly useless if the measurements are in error by as much as 1000 feet above or below the correct altitude. • Machined parts in a manufacturing process must be produced with minimum variability in order to reduce out-of-size and hence defective parts. • Aptitude tests must be designed so that scores will exhibit a reasonable amount of variability. For example, an 800-point test is not very discriminatory if all students score between 601 and 605. In previous chapters, you have used S( x)2 xi s2 n 1 as an unbiased estimator of the population variance s 2. This means that, in repeated sampling, the average of all your sample estimates will equal the target parameter, s 2. But how close or far from the target is your estimator s2 likely to be? To answer this question, we use the sampling distribution of s2, which describes its behavior in repeated sampling. Consider the distribution of s2 based on repeated random sampling from a normal distribution with a speciﬁed mean and variance. We can show theoretically that the distribution begins at s2 0 (since the variance cannot be negative) with a mean equal to s 2. Its shape is nonsymmetric and changes with each different sample size and each different value of s 2. Finding critical values for the sampling distribution of s2 would be quite difficult and would require separate tables for each population variance. Fortunately, we can simplify the problem by standardizing, as we did with the z distribution. Definition The standardized statistic x 2 (n s 2 1)s2 is called a chi-square variable and has a sampling distribution called the chi-square probability distribution, with n 1 degrees of freedom. The equation of the density function for this statistic is quite complicated to look at, but it traces the curve shown in Figure 10.16. F IG URE 10 .1 6 A chi-square distribution ● f(χ2) 0 a χ2 a χ2 Certain critical values of the chi-square statistic, which are used for making inferences about the population variance, have been tabulated by statisticians and appear in Table 5 of Appendix I. Since the shape of the distribution varies with the sample 10.6 INFERENCES CONCERNING A POPULATION VARIANCE ❍ 419 size n or, more precisely, the degrees of freedom, n 1, associated with s2, Table 5, partially reproduced in Table 10.5, is constructed in exactly the same way as the t table, with the degrees of freedom in the ﬁrst and last columns. The symbol x 2 a indicates that the tabulated x 2-value has an area a to its right (see Figure 10.16). TABLE 10.5 Testing one variance: df n 1 ● Format of the Chi-Square Table from Table 5 in Appendix 950 .900 .100 .050 df .995 .005 1 2 3 4 5 6 . . . 15 16 17 18 19 . . . .0000393 .0100251 .0717212 .206990 .411740 .0675727 . . . 4.60094 5.14224 5.69724 6.26481 6.
84398 . . . .0039321 .102587 .351846 .710721 1.145476 1.63539 . . . 7.26094 7.96164 8.67176 9.39046 10.1170 . . . .0157908 .210720 .584375 1.063623 1.610310 2.204130 . . . 8.54675 9.31223 10.0852 10.8649 11.6509 . . . 2.70554 4.60517 6.25139 7.77944 9.23635 10.6446 . . . 22.3072 23.5418 24.7690 25.9894 27.2036 . . . 3.84146 5.99147 7.81473 9.48773 11.0705 12.5916 . . . 24.9958 26.2962 27.5871 28.8693 30.1435 . . . 7.87944 10.5966 12.8381 14.8602 16.7496 18.5476 . . . 32.8013 34.2672 35.7185 37.1564 38.5822 . . . df 1 2 3 4 5 6 . . . 15 16 17 18 19 . . . You can see in Table 10.5 that, because the distribution is nonsymmetric and starts at 0, both upper and lower tail areas must be tabulated for the chi-square statistic. For example, the value x 2 .95 is the value that has 95% of the area under the curve to its right and 5% of the area to its left. This value cuts off an area equal to .05 in the lower tail of the chi-square distribution. EXAMPLE 10.10 Check your ability to use Table 5 in Appendix I by verifying the following statements: 1. The probability that x 2, based on n 16 measurements (df 15), exceeds 24.9958 is .05. 2. For a sample of n 6 measurements, 95% of the area under the x 2 distribu- tion lies to the right of 1.145476. These values are shaded in Table 10.5. You can use the Chi-Square Probabilities applet to ﬁnd the x 2-value described in Example 10.10. Since the applet provides x 2-values and their one-tailed probabilities for the degrees of freedom that you select using the slider on the right side of the applet, you should choose df 5 and type .95 in the box marked “prob:” at the bottom of the applet. The applet will provide the value of x 2 that puts .95 in the right tail of the x 2 distribution and hence .05 in the left tail. The applet in Figure 10.17 shows x 2 1.14, which differs only slightly from the value in Example 10.10. We will use this applet for the MyApplet Exercises at the end of the chapter. 420 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES F IGU RE 1 0. 17 Chi-Square Probabilities applet ● The statistical test of a null hypothesis concerning a population variance 0 H0 : s 2 s 2 uses the test statistic x 2 (n 1)s2 s 2 0 Notice that when H0 is true, s2/s 2 0 should be near 1, so x 2 should be close to (n 1), the degrees of freedom. If s 2 is really greater than the hypothesized value 0, the test statistic will tend to be larger than (n 1) and will probably fall toward s 2 the upper tail of the distribution. If s 2 s 2 0, the test statistic will tend to be smaller than (n 1) and will probably fall toward the lower tail of the chi-square distribution. As in other testing situations, you may use either a one- or a two-tailed statistical test, depending on the alternative hypothesis. This test of hypothesis and the (1 a)100% conﬁdence interval for s 2 are both based on the chi-square distribution and are described next. TEST OF HYPOTHESIS CONCERNING A POPULATION VARIANCE 1. Null hypothesis: H0 : s 2 s 2 2. Alternative hypothesis: 0 One-Tailed Test Two-Tailed Test 0 Ha : s 2 s 2 (or Ha : s 2 s 2 0) 3. Test statistic: x 2 (n 1)s2 s 2 0 Ha : s 2 s 2 0 10.6 INFERENCES CONCERNING A POPULATION VARIANCE ❍ 421 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test (1a) when the alternative x 2 x 2 a (or x 2 x 2 hypothesis is Ha : s 2 s 2 x 2 upper- and lower-tail values of x 2 that place a in the tail areas 0), where (1a) are, respectively, the a and x 2 a/2 or x 2 x 2 x 2 x 2 (1a/2), where x 2 a/2 and x 2 (1a/2) are, respectively, the upper- and lower-tail values of x 2 that place a/2 in the tail areas or when p-value a The critical values of x 2 are based on (n 1) df. These tabulated values can be found using Table 5 of Appendix I or the Chi-Square Probabilities applet. 0 α χ2 α α/2 0 χ2 (1 – α/2) α/2 χ2 α/2 (1 a)100% CONFIDENCE INTERVAL FOR s 2 n x )s2 2 s 1 ) a/ 2 ) (/ 2 ( where x 2 a/2 and x 2 of a in each tail of the chi-square distribution. Assumption: The sample is randomly selected from a normal population. (1a/2) are the upper and lower x 2-values, which locate one-half EXAMPLE 10.11 A cement manufacturer claims that concrete prepared from his product has a relatively stable compressive strength and that the strength measured in kilograms per square centimeter (kg/cm2) lies within a range of 40 kg/cm2. A sample of n 10 measurements produced a mean and variance equal to, respectively, x 312 and s2 195 Do these data present sufficient evidence to reject the manufacturer’s claim? Solution In Section 2.5, you learned that the range of a set of measurements should be approximately four standard deviations. The manufacturer’s claim that the range of the strength measurements is within 40 kg/cm2 must mean that the standard deviation of the measurements is roughly 10 kg/cm2 or less. To test his claim, the appropriate hypotheses are H0 : s 2 102 100 versus Ha : s 2 100 If the sample variance is much larger than the hypothesized value of 100, then the test statistic x 2 (n 1)s2 5 17.55 5 1 7 s 2 0 0 1 0 will be unusually large, favoring rejection of H0 and acceptance of Ha. There are two ways to use the test statistic to make a decision for this test. 422 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES • The critical value approach: The appropriate test requires a one-tailed rejection region in the right tail of the x 2 distribution. The critical value for .05 16.9190 from Table 5 in Appendix I. a .05 and (n 1) 9 df is x 2 Figure 10.18 shows the rejection region; you can reject H0 if the test statistic exceeds 16.9190. Since the observed value of the test statistic is x 2 17.55, you can conclude that the null hypothesis is false and that the range of concrete strength measurements exceeds the manufacturer’s claim. F IG URE 10 .1 8 Rejection region and p-value (shaded) for Example 10.11 ● f(χ2) EXAMPLE 10.12 .050 .025 0 16.9190 χ2 19.0228 Reject H0 • The p-value approach: The p-value for a statistical test is the smallest value of a for which H0 can be rejected. It is calculated, as in other one-tailed tests, as the area in the tail of the x 2 distribution to the right of the observed value, x 2 17.55. Although computer packages allow you to calculate this area exactly, Table 5 in Appendix I allows you only to bound the p-value. Since .025 19.0228, the the value 17.55 lies between x 2 p-value lies between .025 and .05. Most researchers would reject H0 and report these results as signiﬁcant at the 5% level, or P .05. Again, you can reject H0 and conclude that the range of measurements exceeds the manufacturer’s claim. .050 16.9190 and x 2 An experimenter is convinced that her measuring instrument had a variability measured by standard deviation s 2. During an experiment, she recorded the measurements 4.1, 5.2, and 10.2. Do these data conﬁrm or disprove her assertion? Test the appropriate hypothesis, and construct a 90% conﬁdence interval to estimate the true value of the population variance. Solution Since there is no preset level of signiﬁcance, you should choose to use the p-value approach in testing these hypotheses: H0 : s 2 4 versus Ha : s 2 4 Use your scientiﬁc calculator to verify that the sample variance is s2 10.57 and the test statistic is x 2 (n 1)s2 .57) 5.285 2(10 s 2 4 0 Since this is a two-tailed test, the rejection region is divided into two parts, half in each tail of the x 2 distribution. If you approximate the area to the right of the observed test statistic, x 2 5.285, you will have only half of the p-value for the test. Since an equally unlikely value of x 2 might occur in the lower tail of the distribution, 10.6 INFERENCES CONCERNING A POPULATION VARIANCE ❍ 423 with equal probability, you must double the upper area to obtain the p-value. With 2 df, the observed value, 5.29, falls between x 2 .10 and x 2 .05 so that .05 1 ( p-value) .10 or 2 .10 p-value .20 Since the p-value is greater than .10, the results are not statistically signiﬁcant. There is insufficient evidence to reject the null hypothesis H0 : s 2 4. )s2 The corresponding 90% conﬁdence interval is (/ 2 ( The values of x 2 (1 a/2) x 2 2 s 1 ) a/ 2 ) (1a/2) and x 2 .95 .102587 a/2 are n x x 2 x 2 a/2 x 2 .05 5.99147 Substituting these values into the formula for the interval estimate, you get ) or 3.53 s 2 206.07 0 ) s 2 2 ( Thus, you can estimate the population variance to fall into the interval 3.53 to 206.07. This very wide conﬁdence interval indicates how little information on the population variance is obtained from a sample of only three measurements. Consequently, it is not surprising that there is insufficient evidence to reject the null hypothesis s 2 4. To obtain more information on s 2, the experimenter needs to increase the sample size. The MINITAB command Stat Basic Statistics 1 Variance allows you to enter raw data or a summary statistic to perform the F-test for a single variance, and calculate a conﬁdence interval. The MINITAB printout corresponding to Example 10.12 is shown in Figure 10.19. F IG URE 10 .1 9 MINITAB output for Example 10.12 ● Chi-Square Method (Normal Distribution) Variable N Variance Measurements 3 10.6 90% CI Chi-Square P 0.142 (3.5, 206.1) 5.28 10.6 EXERCISES BASIC TECHNIQUES 10.48 A random sample of n 25 observations from a normal population produced a sample variance equal to 21.4. Do these data provide sufficient evidence to indicate that s 2 15? Test using a .05. 10.49 A random sample of n 15 observations was selected from a normal population. The sample mean and variance were x 3.91 and s2 .3214. Find a 90% conﬁdence interval for the population variance s 2. 10.50 A random sample of size n 7 from a normal population produced these measurements: 1.4, 3.6, 1.7, 2.0, 3.3, 2.8, 2.9. a. Calculate the sample variance, s2. b. Construct a 95% conﬁdence interval for the popula- tion variance, s 2. c. Test H0 : s 2 .8 versus Ha : s 2 .8 using a .05. State your conclusions. d. What is the approximate p-value for the test in part c? APPLICATIONS 10.51 Instrument Precision A precis
ion instrument is guaranteed to read accurately to within 2 units. A sample of four instrument readings on the same object yielded the measurements 353, 351, 351, and 424 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES 355. Test the null hypothesis that s .7 against the alternative s .7. Use a .05. estimate the variance of the manufacturer’s potency measurements. 10.52 Instrument Precision, continued Find a 90% conﬁdence interval for the population variance in Exercise 10.51. 10.53 Drug Potency To properly treat patients, drugs prescribed by physicians must have a potency that is accurately deﬁned. Consequently, not only must the distribution of potency values for shipments of a drug have a mean value as speciﬁed on the drug’s container, but also the variation in potency must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that his drug is marketed with a potency of 5 .1 milligram per cubic centimeter (mg/cc). A random sample of four containers gave potency readings equal to 4.94, 5.09, 5.03, and 4.90 mg/cc. a. Do the data present sufficient evidence to indicate that the mean potency differs from 5 mg/cc? b. Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits speciﬁed by the manufacturer? [HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as speciﬁed by a manufacturer. Since he implies that the potency values will fall into the interval 5 .1 mg/cc with very high probability—the implication is always—let us assume that the range .2; or (4.9 to 5.1), represents 6s, as suggested by the Empirical Rule. Note that letting the range equal 6s rather than 4s places a stringent interpretation on the manufacturer’s claim. We want the potency to fall into the interval 5 .1 with very high probability.] 10.54 Drug Potency, continued Refer to Exercise 10.53. Testing of 60 additional randomly selected containers of the drug gave a sample mean and variance equal to 5.04 and .0063 (for the total of n 64 containers). Using a 95% conﬁdence interval, 10.55 Hard Hats A manufacturer of hard safety hats for construction workers is concerned about the mean and the variation of the forces helmets transmit to wearers when subjected to a standard external force. The manufacturer desires the mean force transmitted by helmets to be 800 pounds (or less), well under the legal 1000-pound limit, and s to be less than 40. A random sample of n 40 helmets was tested, and the sample mean and variance were found to be equal to 825 pounds and 2350 pounds2, respectively. a. If m 800 and s 40, is it likely that any helmet, subjected to the standard external force, will transmit a force to a wearer in excess of 1000 pounds? Explain. b. Do the data provide sufficient evidence to indicate that when the helmets are subjected to the standard external force, the mean force transmitted by the helmets exceeds 800 pounds? 10.56 Hard Hats, continued Refer to Exercise 10.55. Do the data provide sufficient evidence to indicate that s exceeds 40? 10.57 Light Bulbs A manufacturer of industrial light bulbs likes its bulbs to have a EX1057 mean life that is acceptable to its customers and a variation in life that is relatively small. If some bulbs fail too early in their life, customers become annoyed and shift to competitive products. Large variations above the mean reduce replacement sales, and variation in general disrupts customers’ replacement schedules. A sample of 20 bulbs tested produced the following lengths of life (in hours): 2100 1924 2302 2183 1951 2077 2067 2392 2415 2286 1883 2501 2101 1946 2146 2161 2278 2253 2019 1827 The manufacturer wishes to control the variability in length of life so that s is less than 150 hours. Do the data provide sufficient evidence to indicate that the manufacturer is achieving this goal? Test using a .01. COMPARING TWO POPULATION VARIANCES 10.7 Just as a single population variance is sometimes important to an experimenter, you might also need to compare two population variances. You might need to compare the precision of one measuring device with that of another, the stability of one manufacturing process with that of another, or even the variability in the grading procedure of one college professor with that of another. 10.7 COMPARING TWO POPULATION VARIANCES ❍ 425 of the sample variances, s2 dence to indicate that s 2 small value for s2 One way to compare two population variances, s 2 1/s2 1 and s 2 2, is to use the ratio 2 is nearly equal to 1, you will ﬁnd little evi2 are unequal. On the other hand, a very large or very 2 provides evidence of a difference in the population variances. 2 be for sufficient evidence to exist to reject the fol- How large or small must s2 1 and s 2 2. If s2 1/s2 1/s2 1/s2 lowing null hypothesis? H0 : s 2 1 s 2 2 The answer to this question may be found by studying the distribution of s2 peated sampling. 1/s2 2 in re- When independent random samples are drawn from two normal populations with equal 2 has a probability distribution in repeated variances—that is, s 2 1/s2 sampling that is known to statisticians as an F distribution, shown in Figure 10.20. 2—then s2 1 s 2 FI GUR E 10. 20 An F distribution with df1 10 and df2 10 ● f(F) a 0 1 2 3 Fa 4 5 6 7 8 9 10 F ASSUMPTIONS FOR s2 AN F DISTRIBUTION 1/s 2 2 TO HAVE • Random and independent samples are drawn from each of two normal populations. • The variability of the measurements in the two populations is the same and can be measured by a common variance, s 2; that is, s 2 1 s 2 2 s 2. Testing two variances: df1 n1 1 and df2 n2 1 It is not important for you to know the complex equation of the density function for F. For your purposes, you need only to use the well-tabulated critical values of F given in Table 6 in Appendix I. Critical values of F and p-values for signiﬁcance tests can also be found using the F Probabilities applet shown in Figure 10.21. Like the x 2 distribution, the shape of the F distribution is nonsymmetric and depends on the number of degrees of freedom associated with s2 2, represented as df1 (n1 1) and df2 (n2 1), respectively. This complicates the tabulation of critical values of the F distribution because a table is needed for each different combination of df1, df2, and a. 1 and s2 In Table 6 in Appendix I, critical values of F for right-tailed areas corresponding to a .100, .050, .025, .010, and .005 are tabulated for various combinations of df1 numerator degrees of freedom and df2 denominator degrees of freedom. A portion of Table 6 is reproduced in Table 10.6. The numerator degrees of freedom df1 are listed across the top margin, and the denominator degrees of freedom df2 are listed along the side margin. The values of a are listed in the second column. For a ﬁxed combination of df1 and df2, the appropriate critical values of F are found in the line indexed by the value of a required. 426 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES F IG URE 10 .2 1 F Probabilities applet ● EXAMPLE 10.13 Check your ability to use Table 6 in Appendix I by verifying the following statements: 1. The value of F with area .05 to its right for df1 6 and df2 9 is 3.37. 2. The value of F with area .05 to its right for df1 5 and df2 10 is 3.33. 3. The value of F with area .01 to its right for df1 6 and df2 9 is 5.80. These values are shaded in Table 10.6. TABLE 10.6 ● Format of the F Table from Table 6 in Appendix I df2 1 2 3 . . . 9 10 a .100 .050 .025 .010 .005 .100 .050 .025 .010 .005 .100 .050 .025 .010 .005 . . . .100 .050 .025 .010 .005 .100 .050 .025 .010 .005 1 39.86 161.4 647.8 4052 16211 8.53 18.51 38.51 98.50 198.5 5.54 10.13 17.44 34.12 55.55 3.36 5.12 7.21 10.56 13.61 3.29 4.96 6.94 10.04 12.83 2 49.50 199.5 799.5 4999.5 20000 9.00 19.00 39.00 99.00 199.0 5.46 9.55 16.04 30.82 49.80 3.01 4.26 5.71 8.02 10.11 2.92 4.10 5.46 7.56 9.43 df1 3 53.59 215.7 864.2 5403 21615 9.16 19.16 39.17 99.17 199.2 5.39 9.28 15.44 29.46 47.47 . . . 2.81 3.86 5.08 6.99 8.72 2.73 3.71 4.83 6.55 8.08 4 5 6 55.83 224.6 899.6 5625 22500 9.24 19.25 39.25 99.25 199.2 5.34 9.12 15.10 28.71 46.19 2.69 3.63 4.72 6.42 7.96 2.61 3.48 4.47 5.99 7.34 57.24 230.2 921.8 5764 23056 9.29 19.30 39.30 99.30 199.3 5.31 9.01 14.88 28.24 45.39 2.61 3.48 4.48 6.06 7.47 2.52 3.33 4.24 5.64 6.87 58.20 234.0 937.1 5859 23437 9.33 19.33 39.33 99.33 199.3 5.28 8.94 14.73 27.91 44.84 . . . 2.55 3.37 4.32 5.80 7.13 2.46 3.22 4.07 5.39 6.54 10.7 COMPARING TWO POPULATION VARIANCES ❍ 427 The statistical test of the null hypothesis H0 : s 2 1 s 2 2 uses the test statistic 2 F s 1 2 s 2 When the alternative hypothesis implies a one-tailed test—that is, Ha : s 2 1 s 2 2 you can ﬁnd the right-tailed critical value for rejecting H0 directly from Table 6 in Appendix I. However, when the alternative hypothesis requires a two-tailed test—that is, H0 : s 2 1 s 2 2 the rejection region is divided between the upper and lower tails of the F distribution. These left-tailed critical values are not given in Table 6 for the following reason: You are free to decide which of the two populations you want to call “Population 1.” If you always choose to call the population with the larger sample variance “Population 1,” then the observed value of your test statistic will always be in the right tail of the F distribution. Even though half of the rejection region, the area a/2 to its left, will be in the lower tail of the distribution, you will never need to use it! Remember these points, though, for a two-tailed test: • The area in the right tail of the rejection region is only a/2. • The area to the right of the observed test statistic is only ( p-value)/2. The formal procedures for a test of hypothesis and a (1 a)100% conﬁdence in- terval for two population variances are shown next. TEST OF HYPOTHESIS CONCERNING THE EQUALITY OF TWO POPULATION VARIANCES 1. Null hypothesis: H0 : s 2 2. Altern
ative hypothesis: 1 s 2 2 One-Tailed Test Two-Tailed Test 1 s 2 2 Ha : s 2 (or Ha : s 2 3. Test statistic: 1 s 2 2) Ha : s 2 1 s 2 2 One-Tailed Test Two-Tailed Test s F s where s2 is the larger sample variance 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test F Fa or when p-value a F Fa/2 (continued) 428 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES TEST OF HYPOTHESIS CONCERNING THE EQUALITY OF TWO POPULATION VARIANCES (continued) The critical values of Fa and Fa/2 are based on df1 (n1 1) and df2 (n2 1). These tabulated values, for a .100, .050, .025, .010, and .005, can be found using Table 6 in Appendix I, or the F Probabilities applet. α Fα 0 α/2 0 Fα/2 Assumptions: The samples are randomly and independently selected from normally distributed populations. CONFIDENCE INTERVAL FOR s 2 1/ Fdf 1,df2 2 1 2 2 s s Fdf2,df1 2 1 2 2 EXAMPLE 10.14 where df1 (n1 1) and df2 (n2 1). Fdf1,df2 is the tabulated critical value of F corresponding to df1 and df2 degrees of freedom in the numerator and denominator of F, respectively, with area a/2 to its right. Assumptions: The samples are randomly and independently selected from normally distributed populations. An experimenter is concerned that the variability of responses using two different experimental procedures may not be the same. Before conducting his research, he con1 7.14 and ducts a prestudy with random samples of 10 and 8 responses and gets s2 2 3.21, respectively. Do the sample variances present sufficient evidence to indis2 cate that the population variances are unequal? Solution Assume that the populations have probability distributions that are reasonably mound-shaped and hence satisfy, for all practical purposes, the assumption that the populations are normal. You wish to test these hypotheses: H0 : s 2 1 s 2 2 versus Ha : s 2 1 s 2 2 Using Table 6 in Appendix I for a/2 .025, you can reject H0 when F 4.82 with a .05. The calculated value of the test statistic is .22 2 1 . 3 Because the test statistic does not fall into the rejection region, you cannot reject H0 : s 2 2. Thus, there is insufficient evidence to indicate a difference in the population variances. 1 s 2 EXAMPLE 10.15 FI GUR E 10. 22 MINITAB output for Example 10.14 ● EXAMPLE 10.16 10.7 COMPARING TWO POPULATION VARIANCES ❍ 429 Refer to Example 10.14 and ﬁnd a 90% conﬁdence interval for s 2 Solution The 90% conﬁdence interval for s 2 s s s 1 s Fdf 1,df2 Fdf2,df1 1/s 2 2 is /s 2 2. where 1 7.14 s2 df1 (n1 1) 9 F9,7 3.68 2 3.21 s2 df2 (n2 1) 7 F7,9 3.29 Substituting these values into the formula for the conﬁdence interval, you get 7 . . 3 s 1 s 68 3. 4 1 1 2 7 3.29 or . . .32 .60 1 s 2 2 The calculated interval estimate .60 to 7.32 includes 1.0, the value hypothesized in H0. This indicates that it is quite possible that s 2 2 and therefore agrees with the test conclusions. Do not reject H0 : s 2 1 s 2 2. The MINITAB command Stat Basic Statistics 2 Variances allows you to enter either raw data or summary statistics to perform the F-test for the equality of variances and calculates conﬁdence intervals for the two individual standard deviations (which we have not discussed). The relevant printout, containing the F statistic and its p-value, is shaded in Figure 10.22. 1 s 2 Test for Equal Variances 95% Bonferroni conﬁdence intervals for standard deviations Sample 1 2 Upper 5.38064 4.10374 StDev 2.67208 1.79165 Lower 1.74787 1.12088 N 10 8 F-Test (Normal Distribution) Test statistic = 2.22, p-value = 0.304 The variability in the amount of impurities present in a batch of chemical used for a particular process depends on the length of time the process is in operation. A manufacturer using two production lines, 1 and 2, has made a slight adjustment to line 2, hoping to reduce the variability as well as the average amount of impurities in the chemical. Samples of n1 25 and n2 25 measurements from the two batches yield these means and variances: 1 1.04 s2 2 .51 s2 x1 3.2 x2 3.0 Do the data present sufficient evidence to indicate that the process variability is less for line 2? Solution The experimenter believes that the average levels of impurities are the same for the two production lines but that her adjustment may have decreased the variability of the levels for line 2, as illustrated in Figure 10.23. This adjustment would be good for the company because it would decrease the probability of producing shipments of the chemical with unacceptably high levels of impurities. 430 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES F IG URE 10 .2 3 Distributions of impurity measurements for two production lines ● f(x) Distribution for production line 2 Distribution for production line 1 Level of impurities x To test for a decrease in variability, the test of hypothesis is H0 : s 2 1 s 2 2 versus Ha : s 2 1 s 2 2 and the observed value of the test statistic is 4 2.04 Using the p-value approach, you can bound the one-tailed p-value using Table 6 in Appendix I with df1 df2 (25 1) 24. The observed value of F falls between F.050 1.98 and F.025 2.27, so that .025 p-value .05. The results are judged signiﬁcant at the 5% level, and H0 is rejected. You can conclude that the variability of line 2 is less than that of line 1. The F-test for the difference in two population variances completes the battery of tests you have learned in this chapter for making inferences about population parameters under these conditions: • The sample sizes are small. • The sample or samples are drawn from normal populations. You will ﬁnd that the F and x 2 distributions, as well as the Student’s t distribution, are very important in other applications in the chapters that follow. They will be used for different estimators designed to answer different types of inferential questions, but the basic techniques for making inferences remain the same. In the next section, we review the assumptions required for all of these inference tools, and discuss options that are available when the assumptions do not seem to be reasonably correct. 10.7 EXERCISES BASIC TECHNIQUES 10.58 Independent random samples from two normal populations produced the variances listed here: Sample Size Sample Variance 16 20 55.7 31.4 a. Do the data provide sufficient evidence to indicate that s 2 1 differs from s 2 2? Test using a .05. b. Find the approximate p-value for the test and inter- pret its value. 10.59 Refer to Exercise 10.58 and ﬁnd a 95% conﬁdence interval for s 2 1/s 2 2. 10.60 Independent random samples from two normal populations produced the given variances: b. Find the approximate p-value for the test and inter- pret its value. 10.7 COMPARING TWO POPULATION VARIANCES ❍ 431 Sample Size Sample Variance 13 13 18.3 7.9 that s 2 a. Do the data provide sufficient evidence to indicate 2? Test using a .05. b. Find the approximate p-value for the test and inter- 1 s 2 pret its value. APPLICATIONS 10.61 SAT Scores The SAT subject tests in chemistry and physics11 for two groups of 15 students each electing to take these tests are given below. Chemistry x 629 s 110 n 15 Physics x 643 s 107 n 15 To use the two-sample t-test with a pooled estimate of 2, you must assume that the two population variances are equal. Test this assumption using the F-test for equality of variances. What is the approximate p-value for the test? 10.62 Product Quality The stability of measurements on a manufactured product is important in maintaining product quality. In fact, it is sometimes better to have small variation in the measured value of some important characteristic of a product and have the process mean be slightly off target than to suffer wide variation with a mean value that perfectly ﬁts requirements. The latter situation may produce a higher percentage of defective products than the former. A manufacturer of light bulbs suspected that one of her production lines was producing bulbs with a wide variation in length of life. To test this theory, she compared the lengths of life for n 50 bulbs randomly sampled from the suspect line and n 50 from a line that seemed to be “in control.” The sample means and variances for the two samples were as follows: “Suspect Line” x1 1520 1 92,000 s 2 Line “in Control” x2 1476 2 37,000 s 2 a. Do the data provide sufficient evidence to indicate that bulbs produced by the “suspect line” have a larger variance in length of life than those produced by the line that is assumed to be in control? Test using a .05. 10.63 Construct a 90% conﬁdence interval for the variance ratio in Exercise 10.62. 10.64 Tuna III In Exercise 10.25 and dataset EX1025, you conducted a test to detect a difference in the average prices of light tuna in water versus light tuna in oil. a. What assumption had to be made concerning the population variances so that the test would be valid? b. Do the data present sufficient evidence to indicate that the variances violate the assumption in part a? Test using a .05. 10.65 Runners and Cyclists III Refer to Exercise 10.26. Susan Beckham and colleagues conducted an experiment involving 10 healthy runners and 10 healthy cyclists to determine if there are signiﬁcant differences in pressure measurements within the anterior muscle compartment for runners and cyclists.7 The data—compartment pressure, in millimeters of mercury (Hg)—are reproduced here: Condition Rest 80% maximal O2 consumption Maximal O2 consumption Runners Cyclists Standard Standard Mean Deviation Mean Deviation 14.5 12.2 19.1 3.92 3.49 16.9 11.1 11.5 12.2 3.98 4.95 4.47 For each of the three variables measured in this experiment, test to see whether there is a signiﬁcant difference in the variances for runners versus cyclists. Find the approximate p-values for each of these tests. Will a two-sample t-test with a pooled estimate of s 2 be appropriate for all three of these variables? Explain. 10.66 Impurities A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for bo
th suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, it could affect the quality of the pharmaceutical product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects 10 shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the table. 432 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Supplier A x1 1.89 1 .273 s2 n1 10 Supplier B x2 1.85 2 .094 s 2 n2 10 a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using a .01. Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a 99% conﬁdence interval for s 2 2 and interpret your results. How Do I Decide Which Test to Use? Are you interested in testing means? If the design involves: a. One random sample, use the one-sample t statistic. b. Two independent random samples, are the population variances equal? i. ii. If equal, use the two-sample t statistic with pooled s2. If unequal, use the unpooled t with estimated df. c. Two paired samples with random pairs, use a one-sample t for analyzing differences. Are you interested in testing variances? If the design involves: a. One random sample, use the x 2 test for a single variance. b. Two independent random samples, use the F-test to compare two variances. REVISITING THE SMALL-SAMPLE ASSUMPTIONS 10.8 All of the tests and estimation procedures discussed in this chapter require that the data satisfy certain conditions in order that the error probabilities (for the tests) and the conﬁdence coefficients (for the conﬁdence intervals) be equal to the values you have speciﬁed. For example, if you construct what you believe to be a 95% conﬁdence interval, you want to be certain that, in repeated sampling, 95% (and not 85% or 75% or less) of all such intervals will contain the parameter of interest. These conditions are summarized in these assumptions: ASSUMPTIONS 1. For all tests and conﬁdence intervals described in this chapter, it is assumed that samples are randomly selected from normally distributed populations. 2. When two samples are selected, it is assumed that they are selected in an independent manner except in the case of the paired-difference experiment. 3. For tests or conﬁdence intervals concerning the difference between two population means m1 and m2 based on independent random samples, it is assumed that s 2 1 s 2 2. CHAPTER REVIEW ❍ 433 In reality, you will never know everything about the sampled population. If you did, there would be no need for sampling or statistics. It is also highly unlikely that a population will exactly satisfy the assumptions given in the box. Fortunately, the procedures presented in this chapter give good inferences even when the data exhibit moderate departures from the necessary conditions. A statistical procedure that is not sensitive to departures from the conditions on which it is based is said to be robust. The Student’s t-tests are quite robust for moderate departures from normality. Also, as long as the sample sizes are nearly equal, there is not much difference between the pooled and unpooled t statistics for the difference in two population means. However, if the sample sizes are not clearly equal, and if the population variances are unequal, the pooled t statistic provides inaccurate conclusions. If you are concerned that your data do not satisfy the assumptions, other options are available: • If you can select relatively large samples, you can use one of the largesample procedures of Chapters 8 and 9, which do not rely on the normality or equal variance assumptions. • You may be able to use a nonparametric test to answer your inferential questions. These tests have been developed speciﬁcally so that few or no distributional assumptions are required for their use. Tests that can be used to compare the locations or variability of two populations are presented in Chapter 15. CHAPTER REVIEW Key Concepts and Formulas I. Experimental Designs for Small III. Small-Sample Test Statistics Samples 1. Single random sample: The sampled popula- tion must be normal. 2. Two independent random samples: Both sampled populations must be normal. a. Populations have a common variance s 2. b. Populations have different variances: s 2 1 and s 2 2. 3. Paired-difference or matched pairs design: The samples are not independent. II. Statistical Tests of Significance 1. Based on the t, F, and x 2 distributions 2. Use the same procedure as in Chapter 9 3. Rejection region—critical values and signiﬁcance levels: based on the t, F, or x 2 distributions with the appropriate degrees of freedom 4. Tests of population parameters: a single mean, the difference between two means, a single variance, and the ratio of two variances To test one of the population parameters when the sample sizes are small, use the following test statistics: Parameter m Test Statistic m x 0 t n / s m1 m2 (equal variances) m1 m2 (unequal variances) m1 m2 (paired samples) s 2 s 2 1/s 2 2 t t (x1 x2) (m1 m2x1 x2) (m1 m2n 1)s2 s 2 0 F s 2 1/s 2 2 Degrees of Freedom n 1 n1 n2 2 Satterthwaite’s approximation n 1 n 1 n1 1 and n2 1 434 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Small-Sample Testing and Estimation The tests and conﬁdence intervals for population means based on the Student’s t distribution are found in a MINITAB submenu by choosing Stat Basic Statistics. You will see choices for 1-Sample t, 2-Sample t, and Paired t, which will generate Dialog boxes for the procedures in Sections 10.3, 10.4, and 10.5, respectively. You must choose the columns in which the data are stored and the null and alternative hypotheses to be tested (or the conﬁdence coefficient for a conﬁdence interval). In the case of the two-sample t-test, you must indicate whether the population variances are assumed equal or unequal, so that MINITAB can perform the correct test. We will display some of the Dialog boxes and Session window outputs for the examples in this chapter, beginning with the one-sample t-test of Example 10.3. First, enter the six recorded weights—.46, .61, .52, .48, .57, .54—in column C1 and name them “Weights.” Use Stat Basic Statistics 1-Sample t to generate the Dialog box in Figure 10.24. To test H0 : m .5 versus Ha : m .5, use the list on the left to select “Weights” for the box marked “Samples in Columns.” Check the box marked “Perform hypothesis test.” Then, place your cursor in the box marked “Hypothesized mean:” and enter .5 as the test value. Finally, use Options and the drop-down menu marked “Alternative” to select “greater than.” Click OK twice to obtain the output in Figure 10.25. Notice that MINITAB produces a one- or a two-sided conﬁdence interval for the single population mean, consistent with the alternative hypothesis you have chosen. You can change the conﬁdence coefficient from the default of .95 in the Options box. Also, the Graphs option will produce a histogram, a box plot, or an individual value plot of the data in column C1. Data for a two-sample t-test with independent samples can be entered into the work- sheet in one of two ways: F IG URE 10 .2 4 ● FI GUR E 10. 25 ● MY MINITAB ❍ 435 • Enter measurements from both samples into a single column and enter numbers (1 or 2) in a second column to identify the sample from which the measurement comes. • Enter the samples in two separate columns. If you do not have the raw data, but rather have summary statistics—the sample mean, standard deviation, and sample size—MINITAB 15 will allow you to use these values by selecting the radio button marked “Summarized data” and entering the appropriate values in the boxes. Use the second method and enter the data from Example 10.5 into columns C2 and C3. Then use Stat Basic Statistics 2-Sample t to generate the Dialog box in Figure 10.26. Check “Samples in different columns,” selecting C2 and C3 from the box on the left. Check the “Assume equal variances” box and select the proper alternative hypothesis in the Options box. (Otherwise, MINITAB will perform Satterthwaite’s approximation for unequal variances.) The two-sample output when you click OK twice automatically contains a 95% one- or two-sided conﬁdence interval as well as the test statistic and p-value (you can change the conﬁdence coefficient if you like). The output for Example 10.5 is shown in Figure 10.13. For a paired-difference test, the two samples are entered into separate columns, which we did with the tire wear data in Table 10.3. Use Stat Basic Statistics Paired t FI GUR E 10. 26 ● 436 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES to generate the Dialog box in Figure 10.27. If you have only summary statistics—the sample mean and standard deviation of the differences and sample size—MINITAB will allow you to use these values by selecting the radio button marked “Summarized data” and entering the appropriate values in the boxes. Select C4 and C5 from the box on the left, and use Options to pick the proper alternative hypothesis. You may change the conﬁdence coefficient or the test value (the default value is zero). When you click OK twice, you will obtain the output shown in Figure 10.15. The MINITAB command Stat Basic Statistics 2 Variances allows you to enter either raw data or summary statistics to perform the F-test for the equality of variances, as shown in Figure 10.28. The MINITAB command Stat Basic Statistics 1 Variance will allow you to perform the x 2 test and construct a conﬁdence interval for a single population variance, s 2. F IG URE 10 .2 7 ● F IG URE 10 .2 8 ● Supplementary Exercises 10.67 What assumptions are made when Student’s t-test is used to test a hypothesis concerning a population mean? 10.68 What assumptions are m
ade about the populations from which random samples are obtained when the t distribution is used in making small-sample inferences concerning the difference in population means? 10.69 Why use paired observations to estimate the difference between two population means rather than estimation based on independent random samples selected from the two populations? Is a paired experiment always preferable? Explain. 10.70 Impurities II A manufacturer can tolerate a small amount (.05 milligrams per liter (mg/l)) of impurities in a raw material needed for manufacturing its product. Because the laboratory test for the impurities is subject to experimental error, the manufacturer tests each batch 10 times. Assume that the mean value of the experimental error is 0 and hence that the mean value of the ten test readings is an unbiased estimate of the true amount of the impurities in the batch. For a particular batch of the raw material, the mean of the ten test readings is .058 mg/l, with a standard deviation of .012 mg/l. Do the data provide sufficient evidence to indicate that the amount of impurities in the batch exceeds .05 mg/l? Find the p-value for the test and interpret its value. 10.71 Red Pine The main stem growth measured for a sample of seventeen 4-year-old red pine trees produced a mean and standard deviation equal to 11.3 and 3.4 inches, respectively. Find a 90% conﬁdence interval for the mean growth of a population of 4-year-old red pine trees subjected to similar environmental conditions. 10.72 Sodium Hydroxide The object of a general chemistry experiment is to determine the amount (in milliliters) of sodium hydroxide (NaOH) solution needed to neutralize 1 gram of a speciﬁed acid. This will be an exact amount, but when the experiment is run in the laboratory, variation will occur as the result of experimental error. Three titrations are made using phenolphthalein as an indicator of the neutrality of the solution (pH equals 7 for a neutral solution). The three volumes of NaOH required to attain a pH of 7 in each SUPPLEMENTARY EXERCISES ❍ 437 of the three titrations are as follows: 82.10, 75.75, and 75.44 milliliters. Use a 99% conﬁdence interval to estimate the mean number of milliliters required to neutralize 1 gram of the acid. 10.73 Sodium Chloride Measurements of water intake, obtained from a sample of 17 rats that had been injected with a sodium chloride solution, produced a mean and standard deviation of 31.0 and 6.2 cubic centimeters (cm3), respectively. Given that the average water intake for noninjected rats observed over a comparable period of time is 22.0 cm3, do the data indicate that injected rats drink more water than noninjected rats? Test at the 5% level of signiﬁcance. Find a 90% conﬁdence interval for the mean water intake for injected rats. 10.74 Sea Urchins An experimenter was interested in determining the mean thickness of the cortex of the sea urchin egg. The thickness was measured for n 10 sea urchin eggs. These measurements were obtained: 4.5 5.2 6.1 2.6 3.2 3.7 3.9 4.6 4.7 4.1 Estimate the mean thickness of the cortex using a 95% conﬁdence interval. 10.75 Fabricating Systems A production plant has two extremely complex fabricating systems; one system is twice as old as the other. Both systems are checked, lubricated, and maintained once every 2 weeks. The number of ﬁnished products fabricated daily by each of the systems is recorded for 30 working days. The results are given in the table. Do these data present sufficient evidence to conclude that the variability in daily production warrants increased maintenance of the older fabricating system? Use the p-value approach. New System x1 246 s1 15.6 Old System x2 240 s2 28.2 10.76 Fossils The data in the table are the diameters and heights of ten fossil specimens EX1076 of a species of small shellﬁsh, Rotularia (Annelida) fallax, that were unearthed in a mapping expedition near the Antarctic Peninsula.12 The table gives an identiﬁcation symbol for the fossil specimen, the fossil’s diameter and height in millimeters, and the ratio of diameter to height. 438 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Specimen Diameter Height OSU 36651 OSU 36652 OSU 36653 OSU 36654 OSU 36655 OSU 36656 OSU 36657 OSU 36658 OSU 36659 OSU 36660 x: s: 185 194 173 200 179 213 134 191 177 199 184.5 21.5 78 65 77 76 72 76 75 77 69 65 73 5 D/H 2.37 2.98 2.25 2.63 2.49 2.80 1.79 2.48 2.57 3.06 2.54 .37 a. Find a 95% conﬁdence interval for the mean diam- eter of the species. 10.80 Drug Absorption An experiment was conducted to compare the mean lengths of time required for the bodily absorption of two drugs A and B. Ten people were randomly selected and assigned to receive one of the drugs. The length of time (in minutes) for the drug to reach a speciﬁed level in the blood was recorded, and the data summary is given in the table: Drug A x1 27.2 1 16.36 s2 Drug B x2 33.5 2 18.92 s2 a. Do the data provide sufficient evidence to indicate a difference in mean times to absorption for the two drugs? Test using a .05. b. Find a 95% conﬁdence interval for the mean height b. Find the approximate p-value for the test. Does this of the species. value conﬁrm your conclusions? c. Find a 95% conﬁdence interval for the mean ratio c. Find a 95% conﬁdence interval for the difference of diameter to height. d. Compare the three intervals constructed in parts a, b, and c. Is the average of the ratios the same as the ratio of the average diameter to average height? 10.77 Fossils, continued Refer to Exercise 10.76 and data set EX1076. Suppose you want to estimate the mean diameter of the fossil specimens correct to within 5 millimeters with probability equal to .95. How many fossils do you have to include in your sample? 10.78 Alcohol and Reaction Times To test the effect of alcohol in increasing the reac- EX1078 tion time to respond to a given stimulus, the reaction times of seven people were measured. After consuming 3 ounces of 40% alcohol, the reaction time for each of the seven people was measured again. Do the following data indicate that the mean reaction time after consuming alcohol was greater than the mean reaction time before consuming alcohol? Use a .05. Person Before After 10.79 Cheese, Please Here are the prices per ounce of n 13 different brands of indi- EX1079 vidually wrapped cheese slices: 29.0 28.7 21.6 24.1 28.0 25.9 23.7 23.8 27.4 19.6 18.9 27.5 23.9 Construct a 95% conﬁdence interval estimate of the underlying average price per ounce of individually wrapped cheese slices. in mean times to absorption. Does the interval conﬁrm your conclusions in part a? 10.81 Drug Absorption, continued Refer to Exercise 10.80. Suppose you wish to estimate the difference in mean times to absorption correct to within 1 minute with probability approximately equal to .95. a. Approximately how large a sample is required for each drug (assume that the sample sizes are equal)? b. If conducting the experiment using the sample sizes of part a will require a large amount of time and money, can anything be done to reduce the sample sizes and still achieve the 1-minute margin of error for estimation? 10.82 Ring-Necked Pheasants The weights in grams of 10 males and 10 female EX1082 juvenile ring-necked pheasants are given below. Males Females 1384 1286 1503 1627 1450 1672 1370 1659 1725 1394 1073 1053 1038 1018 1146 1058 1123 1089 1034 1281 a. Use a statistical test to determine if the population variance of the weights of the male birds differs from that of the females. b. Test whether the average weight of juvenile male ring-necked pheasants exceeds that of the females by more than 300 grams. (HINT: The procedure that you use should take into account the results of the analysis in part a.) EX1083 10.83 Bees Insects hovering in ﬂight expend enormous amounts of energy for their size and weight. The data shown here were taken from a much larger body of data collected by T.M. Casey and colleagues.13 They show the wing stroke frequencies (in hertz) for two different species of bees, n1 4 Euglossa mandibularis Friese and n2 6 Euglossa imperialis Cockerell. E. mandibularis Friese E. imperialis Cockerell 235 225 190 188 180 169 180 185 178 182 a. Based on the observed ranges, do you think that a difference exists between the two population variances? b. Use an appropriate test to determine whether a dif- ference exists. c. Explain why a Student’s t-test with a pooled estimator s2 is unsuitable for comparing the mean wing stroke frequencies for the two species of bees. 10.84 Calcium The calcium (Ca) content of a powdered mineral substance was analyzed EX1084 10 times with the following percent compositions recorded: .0271 .0271 .0282 .0281 .0279 .0269 .0281 .0275 .0268 .0276 a. Find a 99% conﬁdence interval for the true cal- cium content of this substance. b. What does the phrase “99% conﬁdent” mean? c. What assumptions must you make about the sam- pling procedure so that this conﬁdence interval will be valid? What does this mean to the chemist who is performing the analysis? 10.85 Sun or Shade? Karl Niklas and T.G. Owens examined the differences in a particular plant, Plantago Major L., when grown in full sunlight versus shade conditions.14 In this study, shaded plants received direct sunlight for less than 2 hours each day, whereas full-sun plants were never shaded. A partial summary of the data based on n1 16 full-sun plants and n2 15 shade plants is shown here: SUPPLEMENTARY EXERCISES ❍ 439 Full Sun Shade x 128.00 46.80 9.75 .90 8.70 5.24 s 43.00 2.21 2.27 .03 1.64 .98 x 78.70 8.10 6.93 .50 8.91 3.41 s 41.70 1.26 1.49 .02 1.23 .61 Leaf Area (cm2) Overlap Area (cm2) Leaf Number Thickness (mm) Length (cm) Width (cm) a. What assumptions are required in order to use the small-sample procedures given in this chapter to compare full-sun versus shade plants? From the summary presented, do you think that any of these assumptions have been violated? b. Do the data present sufficient evidence to indicate a differe
nce in mean leaf area for full-sun versus shade plants? c. Do the data present sufficient evidence to indicate a difference in mean overlap area for full-sun versus shade plants? 10.86 Orange Juice A comparison of the precisions of two machines developed for extracting juice from oranges is to be made using the following data: Machine A s 2 3.1 ounces2 n 25 Machine B s 2 1.4 ounces2 n 25 a. Is there sufficient evidence to indicate that there is a difference in the precision of the two machines at the 5% level of signiﬁcance? b. Find a 95% conﬁdence interval for the ratio of the two population variances. Does this interval conﬁrm your conclusion from part a? Explain. 10.87 At Home or at School? Four sets of identical twins (pairs A, B, C, and D) were selected at random from a computer database of identical twins. One child was selected at random from each pair to form an “experimental group.” These four children were sent to school. The other four children were kept at home as a control group. At the end of the school year, the following IQ scores were obtained: Pair Experimental Group Control Group A B C D 110 125 139 142 111 120 128 135 Does this evidence justify the conclusion that lack of school experience has a depressing effect on IQ scores? Use the p-value approach. 440 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES EX1088 10.88 Dieting Eight obese persons were placed on a diet for 1 month, and their weights, at the beginning and at the end of the month, were recorded: Weights Subjects Initial Final 1 2 3 4 5 6 7 8 310 295 287 305 270 323 277 299 263 251 249 259 233 267 242 265 Estimate the mean weight loss for obese persons when placed on the diet for a 1-month period. Use a 95% conﬁdence interval and interpret your results. What assumptions must you make so that your inference is valid? 10.89 Repair Costs Car manufacturers try to design the bumpers of their automobiles to EX1089 prevent costly damage in parking-lot type accidents. To compare repair costs of front versus back bumpers for several brands of cars, the cars were subject to a front and rear impacts at 5 mph, and the repair costs recorded.15 Vehicle VW Jetta Daewoo Nubira Acura 3.4 RL Dodge Neon Nissan Sentra Front $396 451 1123 687 583 Rear $602 404 968 748 571 Do the data provide sufficient evidence to indicate that there is a signiﬁcant difference in average repair costs for front versus rear bumper repairs costs? Test using a .05. 10.90 Breathing Patterns Research psychologists measured the baseline breathing EX1090 patterns—the total ventilation (in liters of air per minute) adjusted for body size—for each of n 30 patients, so that they could estimate the average total ventilation for patients before any experimentation was done. The data, along with some MINITAB output, are presented here: 5.23 5.54 5.92 4.72 4.67 5.72 4.79 6.04 5.38 5.17 5.77 5.16 5.83 5.48 6.34 4.99 5.84 5.32 5.37 6.58 5.12 4.51 6.19 4.96 4.35 4.82 5.14 5.70 5.58 5.63 MINITAB output for Exercise 10.90 Stem-and-Leaf Display: Ltrs/min Stem-and-leaf of Ltrs/min N = 30 Leaf Unit = 0.10 1 2 5 8 12 (4) 14 11 7 4 2 1 4 3 4 5 4 677 4 899 5 1111 5 2333 5 455 5 6777 5 889 6 01 6 3 6 5 Descriptive Statistics: Ltrs/min Variable Mean 5.3953 Ltrs/min N* 0 N 30 SE Mean 0.0997 StDev 0.5462 Minimum 4.3500 Q1 4.9825 Median 5.3750 Q3 5.7850 Maximum 6.5800 a. What information does the stem and leaf plot give you about the data? Why is this important? b. Use the MINITAB output to construct a 99% conﬁdence interval for the average total ventilation for patients. 10.91 Reaction Times A comparison of reaction times (in seconds) for two different stimuli in a psychological word-association experiment produced the following results when applied to a random sample of 16 people: Stimulus 1 2 1 2 3 1 2 1 3 Stimulus 2 4 2 3 3 1 2 3 3 Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using a .05. 10.92 Reaction Times II Refer to Exercise 10.91. Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: Person Stimulus 1 Stimulus Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using a .05. 10.93 Refer to Exercises 10.91 and 10.92. Calculate a 95% conﬁdence interval for the difference in the two population means for each of these experimental designs. Does it appear that blocking increased the amount of information available in the experiment? 10.94 Impact Strength The following data are readings (in foot-pounds) of the impact EX1094 strengths of two kinds of packaging material: A 1.25 1.16 1.33 1.15 1.23 1.20 1.32 1.28 1.21 B .89 1.01 .97 .95 .94 1.02 .98 1.06 .98 MINITAB output for Exercise 10.94 Two-Sample T-Test and CI: A, B Two-sample T for A vs B N 9 9 Mean 1.2367 0.9778 A B StDev 0.0644 0.0494 SE Mean 0.021 0.016 Difference = mu (A) - mu (B) Estimate for difference: 0.2589 95% CI for difference: (0.2015, 0.3163) T-Test of difference = 0 (vs not =): T-Value = 9.56 P-Value = 0.000 DF = 16 Both use Pooled StDev = 0.0574 a. Use the MINITAB printout to determine whether there is evidence of a difference in the mean strengths for the two kinds of material. b. Are there practical implications to your results? 10.95 Cake Mixes An experiment was conducted to compare the densities (in ounces per cubic inch) of cakes prepared from two different cake mixes. Six cake pans were ﬁlled with batter A, and six were ﬁlled with batter B. Expecting a variation in oven temperature, the experimenter placed a pan ﬁlled with batter A and another with batter B side by side at six different locations in the oven. The six paired observations of densities are as follows: Batter A Batter B .135 .129 .102 .120 .098 .112 .141 .152 .131 .135 .144 .163 SUPPLEMENTARY EXERCISES ❍ 441 a. Do the data present sufficient evidence to indicate a difference between the average densities of cakes prepared using the two types of batter? b. Construct a 95% conﬁdence interval for the difference between the average densities for the two mixes. 10.96 Under what assumptions can the F distribution be used in making inferences about the ratio of population variances? 10.97 Got Milk? A dairy is in the market for a new container-ﬁlling machine and is considering two models, manufactured by company A and company B. Ruggedness, cost, and convenience are comparable in the two models, so the deciding factor is the variability of ﬁlls. The model that produces ﬁlls with the smaller variance is preferred. If you obtain samples of ﬁlls for each of the two models, an F-test can be used to test for the equality of population variances. Which type of rejection region would be most favored by each of these individuals? a. The manager of the dairy—Why? b. A sales representative for company A—Why? c. A sales representative for company B—Why? 10.98 Got Milk II Refer to Exercise 10.97. Wishing to demonstrate that the variability of ﬁlls is less for her model than for her competitor’s, a sales representative for company A acquired a sample of 30 ﬁlls from her company’s model and a sample of 10 ﬁlls from her competitor’s model. The sample variances were A .027 and s2 s2 provide statistical support at the .05 level of signiﬁcance for the sales representative’s claim? B .065, respectively. Does this result 10.99 Chemical Purity A chemical manufacturer claims that the purity of his product never varies by more than 2%. Five batches were tested and given purity readings of 98.2, 97.1, 98.9, 97.7, and 97.9%. a. Do the data provide sufficient evidence to contradict the manufacturer’s claim? (HINT: To be generous, let a range of 2% equal 4s.) b. Find a 90% conﬁdence interval for s 2. 10.100 16-Ounce Cans? A cannery prints “weight 16 ounces” on its label. The quality control supervisor selects nine cans at random and weighs them. She ﬁnds x 15.7 and s .5. Do the data present sufficient evidence to indicate that the mean weight is less than that claimed on the label? 442 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES 10.101 Reaction Time III A psychologist wishes to verify that a certain drug increases the reaction time to a given stimulus. The following reaction times (in tenths of a second) were recorded before and after injection of the drug for each of four subjects: Reaction Time Subject Before After 1 2 3 4 7 2 12 12 13 3 18 13 Test at the 5% level of signiﬁcance to determine whether the drug signiﬁcantly increases reaction time. EX10102 10.102 Food Production At a time when energy conservation is so important, some scientists think closer scrutiny should be given to the cost (in energy) of producing various forms of food. Suppose you wish to compare the mean amount of oil required to produce 1 acre of corn versus 1 acre of cauliﬂower. The readings (in barrels of oil per acre), based on 20-acre plots, seven for each crop, are shown in the table. Use these data to ﬁnd a 90% conﬁdence interval for the difference between the mean amounts of oil required to produce these two crops. Corn Cauliﬂower 5.6 7.1 4.5 6.0 7.9 4.8 5.7 15.9 13.4 17.6 16.8 15.8 16.3 17.1 10.103 Alcohol and Altitude The effect of alcohol consumption on the body appears to be much greater at high altitudes than at sea level. To test this theory, a scientist randomly selects 12 subjects and randomly divides them into two groups of six each. One group is put into a chamber that simulates conditions at an altitude of 12,000 feet, and each subject ingests a drink containing 100 cubic centimeters (cc) of alcohol. The second group receives the same drink in a chamber that simulates conditions at sea level. After 2 hours, the amount of alcohol in the blood (grams per 100 cc) for each subject is measured. The data are shown in the table. Do the data provide sufficient
evidence to support the theory that retention of alcohol in the blood is greater at high altitudes? Sea Level 12,000 Feet .07 .10 .09 .12 .09 .13 .13 .17 .15 .14 .10 .14 10.104 Stock Risks The closing prices of two common stocks were recorded for a period of 15 days. The means and variances are x1 40.33 1 1.54 s2 x2 42.54 2 2.96 s2 a. Do these data present sufficient evidence to indicate a difference between the variabilities of the closing prices of the two stocks for the populations associated with the two samples? Give the p-value for the test and interpret its value. b. Construct a 99% conﬁdence interval for the ratio of the two population variances. 10.105 Auto Design An experiment is conducted to compare two new automobile designs. Twenty people are randomly selected, and each person is asked to rate each design on a scale of 1 (poor) to 10 (excellent). The resulting ratings will be used to test the null hypothesis that the mean level of approval is the same for both designs against the alternative hypothesis that one of the automobile designs is preferred. Do these data satisfy the assumptions required for the Student’s t-test of Section 10.4? Explain. 10.106 Safety Programs The data shown here were collected on lost-time accidents (the ﬁgures given are mean work-hours lost per month over a period of 1 year) before and after an industrial safety program was put into effect. Data were recorded for six industrial plants. Do the data provide sufficient evidence to indicate whether the safety program was effective in reducing lost-time accidents? Test using a .01. 1 Before Program 38 After Program 31 Plant Number 2 64 58 3 42 43 4 70 65 5 58 52 6 30 29 10.107 Two Different Entrees To compare the demand for two different entrees, the EX10107 manager of a cafeteria recorded the number of purchases of each entree on seven consecutive days. The data are shown in the table. Do the data provide sufficient evidence to indicate a greater mean demand for one of the entrees? Use the MINITAB printout. a. Is there sufficient evidence to reject his claim at the a .05 level of signiﬁcance? SUPPLEMENTARY EXERCISES ❍ 443 Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday A 420 374 434 395 637 594 679 B 391 343 469 412 538 521 625 MINITAB output for Exercise 10.107 Paired T-Test and CI: A, B Paired T for A - B A B Difference N 7 7 7 Mean 504.7 471.3 33.4 StDev 127.2 97.4 47.5 SE Mean 48.1 36.8 18.0 b. Find a 95% conﬁdence interval for the variance of the rod diameters. 10.111 Sleep and the College Student How much sleep do you get on a typical school night? A group of 10 college students were asked to report the number of hours that they slept on the previous night with the following results: 7, 6, 7.25, 7, 8.5, 5, 8, 7, 6.75, 6 a. Find a 99% conﬁdence interval for the average number of hours that college students sleep. b. What assumptions are required in order for this conﬁdence interval to be valid? 95% CI for mean difference: (-10.5, 77.4) T-Test of mean difference = 0 (vs not = 0): T-Value = 1.86 P-Value = 0.112 EX10112 10.112 Arranging Objects The following data are the response times in seconds for n 25 ﬁrst graders to arrange three objects by size. 10.108 Pollution Control The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter (mg/l) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is 48 mg/l, but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for n 20 days and found s2 39 (mg/l)2. Find a 90% conﬁdence interval for s 2. Interpret your results. 10.109 Enzymes Two methods were used to measure the speciﬁc activity (in units of enzyme activity per milligram of protein) of an enzyme. One unit of enzyme activity is the amount that catalyzes the formation of 1 micromole of product per minute under speciﬁed conditions. Use an appropriate test or estimation procedure to compare the two methods of measurement. Comment on the validity of any assumptions you need to make. Method 1 Method 2 125 137 137 143 130 151 151 156 142 149 10.110 Connector Rods A producer of machine parts claimed that the diameters of the connector rods produced by his plant had a variance of at most .03 inch2. A random sample of 15 connector rods from his plant produced a sample mean and variance of .55 inch and .053 inch2, respectively. 5.2 4.2 3.1 3.6 4.7 3.8 4.1 2.5 3.9 3.3 5.7 4.3 3.0 4.8 4.2 3.9 4.7 4.4 5.3 3.8 3.7 4.3 4.8 4.2 5.4 Find a 95% conﬁdence interval for the average response time for ﬁrst graders to arrange three objects by size. Interpret this interval. 10.113 Finger-Lickin’ Good! Maybe too good, according to tests performed by the consumer testing division of Good Housekeeping. Nutritional information provided by Kentucky Fried Chicken claims that each small bag of Potato Wedges contains 4.8 ounces of food, for a total of 280 calories. A sample of 10 orders from KFC restaurants in New York and New Jersey averaged 358 calories.16 If the standard deviation of this sample was s 54, is there sufficient evidence to indicate that the average number of calories in small bags of KFC Potato Wedges is greater than advertised? Test at the 1% level of signiﬁcance. 10.114 Mall Rats An article in American Demographics investigated consumer habits at the mall. We tend to spend the most money shopping on the weekends, and, in particular, on Sundays from 4 to 6 P.M. Wednesday morning shoppers spend the least!17 Suppose that a random sample of 20 weekend shoppers and a random sample of 20 weekday shoppers were selected, and the amount spent per trip to the mall was recorded. 444 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Weekends Weekdays Sample Size Sample Mean Sample Standard Deviation 20 $78 $22 20 $67 $20 a. Is it reasonable to assume that the two population variances are equal? Use the F-test to test this hypothesis with a .05. b. Based on the results of part a, use the appropriate test to determine whether there is a difference in the average amount spent per trip on weekends versus weekdays. Use a .05. 10.115 Border Wars As the costs of prescription drugs escalate, more and more senior EX10115 citizens are ordering prescriptions from Canada, or actually crossing the border to buy prescription drugs. The price of a typical prescription for nine best-selling drugs was recorded at randomly selected stores in both the United States and in Canada.18 Drug U.S. Canada Lipitor® Zocor® Prilosec® Norvasc® Zyprexa® Paxil® Prevacid® Celebrex® Zoloft® $290 412 117 139 571 276 484 161 235 $179 211 72 125 396 171 196 67 156 a. Is there sufficient evidence to indicate that the average cost of prescription drugs in the United States is different from the average cost in Canada? Use a .01. b. What is the approximate the p-value for this test? Does this conﬁrm your conclusions in part a? Exercises 10.116 Use the Student’s t Probabilities applet to ﬁnd the following probabilities: a. P(t 1.2) with 5 df b. P(t 2) P(t 2) with 10 df c. P(t 3.3) with 8 df d. P(t .6) with 12 df 10.117 Use the Student’s t Probabilities applet to ﬁnd the following critical values: a. an upper one-tailed rejection region with a .05 and 11 df b. a two-tailed rejection region with a .05 and 7 df c. a lower one-tailed rejection region with a .01 and 15 df 10.118 Refer to the Interpreting Conﬁdence Intervals applet. a. Suppose that you have a random sample of size n 10 from a population with unknown mean m. What formula would you use to construct a 95% conﬁdence interval for the unknown population mean? b. Use the button in the ﬁrst applet to create a single 95% conﬁdence interval for m. Use the formula in part a and the information given in the applet to verify the conﬁdence limits provided. (The applet rounds to the nearest integer.) Did this conﬁdence interval enclose the true value, m 100? 10.119 Refer to the Interpreting Conﬁdence Intervals applet. a. Use the button in the ﬁrst applet to create ten 95% conﬁdence intervals for m. b. Are the widths of these intervals all the same? Explain why or why not. c. How many of the intervals work properly and enclose the true value of m? d. Try this simulation again by clicking the button a few more times and counting the number of intervals that work correctly. Is it close to our 95% conﬁdence level? e. Use the button in the second applet to create ten 99% conﬁdence intervals for m. How many of these intervals work properly? 10.120 Refer to the Interpreting Conﬁdence Intervals applet. a. Use the button to create one hundred 95% conﬁdence intervals for m. How many of the intervals work properly and enclose the true value of m? b. Repeat the instructions of part a to construct 99% conﬁdence intervals. How many of the intervals work properly and enclose the true value of m? c. Try this simulation again by clicking the button a few more times and counting the number of intervals that work correctly. Use both the 95% and 99% conﬁdence intervals. Do the percentage of intervals that work come close to our 95% and 99% conﬁdence levels? 10.121 A random sample of n 12 observations from a normal population produced x 47.1 and s2 4.7. Test the hypothesis H0: m 48 against Ha: m 48. Use the Small-Sample Test of a Population Mean applet and a 5% signiﬁcance level. 10.122 SAT Scores In Exercise 9.73, we reported that the national average SAT scores for the class of 2005 were 508 on the verbal portion and 520 on the math portion. Suppose that we have a small random sample of 15 California students in the class of 2005; their SAT scores are recorded in the following table. Verbal Math Sample Average Sample Standard Deviation 499 98 516 96 a. Use the Small-Sample Test of a Population Mean applet. Do the data provide sufficient evidence to indicate that the average verbal score for all California students in the class of 2005 is differe
nt from the national average? Test using a .05. CASE STUDY ❍ 445 b. Use the Small-Sample Test of a Population Mean applet. Do the data provide sufficient evidence to indicate that the average math score for all California students in the class of 2005 is different from the national average? Test using a .05. 10.123 Surgery Recovery Times The length of time to recovery was recorded for patients randomly assigned and subjected to two different surgical procedures. The data (recorded in days) are as follows: Procedure I Procedure II Sample Average Sample Variance Sample Size 7.3 1.23 11 8.9 1.49 13 Do the data present sufficient evidence to indicate a difference between the mean recovery times for the two surgical procedures? Perform the test of hypothesis, calculating the test statistic and the approximate p-value by hand. Then check your results using the Two-Sample T-Test: Independent Samples applet. 10.124 Stock Prices Refer to Exercise 10.104 in which we reported the closing prices of two common stocks, recorded over a period of 15 days. x1 40.33 1 1.54 s2 x2 42.54 2 2.96 s2 Use the Two-Sample T-Test: Independent Samples applet. Do the data provide sufficient evidence to indicate that the average prices of the two common stocks are different? Use the p-value to access the signiﬁcance of the test. CASE STUDY Flextime How Would You Like a Four-Day Workweek? Will a ﬂexible workweek schedule result in positive beneﬁts for both employer and employee? Is a more rested employee, who spends less time commuting to and from work, likely to be more efficient and take less time off for sick leave and personal leave? A report on the beneﬁts of ﬂexible work schedules that appeared in Environmental Health looked at the records of n 11 employees who worked in a satellite office in a county health department in Illinois under a 4-day workweek schedule.19 Employees worked a conventional workweek in year 1 and a 4-day workweek in year 2. Some statistics for these employees are shown in the following table: 446 ❍ CHAPTER 10 INFERENCE FROM SMALL SAMPLES Personal Leave Sick Leave Employee Year 2 Year 1 Year 2 Year 10 11 26 18 24 19 17 34 19 18 9 36 26 33 37 20 26 1 2 13 22 22 13 18 30 61 59 2 79 63 71 83 35 81 79 37 45 56 9 92 65 21 62 26 73 21 1. A 4-day workweek ensures that employees will have one more day that need not be spent at work. One possible result is a reduction in the average number of personal-leave days taken by employees on a 4-day work schedule. Do the data indicate that this is the case? Use the p-value approach to testing to reach your conclusion. 2. A 4-day workweek schedule might also have an effect on the average number of sick-leave days an employee takes. Should a directional alternative be used in this case? Why or why not? 3. Construct a 95% conﬁdence interval to estimate the average difference in days taken for sick leave between these 2 years. What do you conclude about the difference between the average number of sick-leave days for these two work schedules? 4. Based on the analysis of these two variables, what can you conclude about the advantages of a 4-day workweek schedule? Case Study from “Four-Day Work Week Improves Environment,” by C.S. Catlin, Environmental Health, Vol. 59, No. 7, March 1997. Copyright 1997 National Environmental Health Association. Reprinted by permission. 11 The Analysis of Variance GENERAL OBJECTIVE The quantity of information contained in a sample is affected by various factors that the experimenter may or may not be able to control. This chapter introduces three different experimental designs, two of which are direct extensions of the unpaired and paired designs of Chapter 10. A new technique called the analysis of variance is used to determine how the different experimental factors affect the average response. CHAPTER INDEX ● The analysis of variance (11.2) ● The completely randomized design (11.4, 11.5) ● Factorial experiments (11.9, 11.10) ● The randomized block design (11.7, 11.8) ● Tukey’s method of paired comparisons (11.6) How Do I Know Whether My Calculations Are Accurate? © James Leynse/CORBIS “A Fine Mess” Do you risk a ﬁne by parking your car in red zones or next to ﬁre hydrants? Do you fail to put enough money in a parking meter? If so, you are among the thousands of drivers who receive parking tickets every day in almost every city in the United States. Depending on the city in which you receive a ticket, your ﬁne can be as little as $8 for overtime parking in San Luis Obispo, California, or as high as $340 for illegal parking in a handicapped space in San Diego, California. The case study at the end of this chapter statistically analyzes the variation in parking ﬁnes in southern California cities. 447 448 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE THE DESIGN OF AN EXPERIMENT 11.1 The way that a sample is selected is called the sampling plan or experimental design and determines the amount of information in the sample. Some research involves an observational study, in which the researcher does not actually produce the data but only observes the characteristics of data that already exist. Most sample surveys, in which information is gathered with a questionnaire, fall into this category. The researcher forms a plan for collecting the data—called the sampling plan—and then uses the appropriate statistical procedures to draw conclusions about the population or populations from which the sample comes. Other research involves experimentation. The researcher may deliberately impose one or more experimental conditions on the experimental units in order to determine their effect on the response. Here are some new terms we will use to discuss the design of a statistical experiment. Definition An experimental unit is the object on which a measurement (or measurements) is taken. A factor is an independent variable whose values are controlled and varied by the experimenter. A level is the intensity setting of a factor. A treatment is a speciﬁc combination of factor levels. The response is the variable being measured by the experimenter. A group of people is randomly divided into an experimental and a control group. The control group is given an aptitude test after having eaten a full breakfast. The experimental group is given the same test without having eaten any breakfast. What are the factors, levels, and treatments in this experiment? Solution The experimental units are the people on which the response (test score) is measured. The factor of interest could be described as “meal” and has two levels: “breakfast” and “no breakfast.” Since this is the only factor controlled by the experimenter, the two levels—“breakfast” and “no breakfast”—also represent the treatments of interest in the experiment. Suppose that the experimenter in Example 11.1 began by randomly selecting 20 men and 20 women for the experiment. These two groups were then randomly divided into 10 each for the experimental and control groups. What are the factors, levels, and treatments in this experiment? Solution Now there are two factors of interest to the experimenter, and each factor has two levels: • • “Gender” at two levels: men and women “Meal” at two levels: breakfast and no breakfast In this more complex experiment, there are four treatments, one for each speciﬁc combination of factor levels: men without breakfast, men with breakfast, women without breakfast, and women with breakfast. EXAMPLE 11.1 EXAMPLE 11.2 11.3 THE ASSUMPTIONS FOR AN ANALYSIS OF VARIANCE ❍ 449 In this chapter, we will concentrate on experiments that have been designed in three different ways, and we will use a technique called the analysis of variance to judge the effects of various factors on the experimental response. Two of these experimental designs are extensions of the unpaired and paired designs from Chapter 10. WHAT IS AN ANALYSIS OF VARIANCE? 11.2 The responses that are generated in an experimental situation always exhibit a certain amount of variability. In an analysis of variance, you divide the total variation in the response measurements into portions that may be attributed to various factors of interest to the experimenter. If the experiment has been properly designed, these portions can then be used to answer questions about the effects of the various factors on the response of interest. You can better understand the logic underlying an analysis of variance by looking at a simple experiment. Consider two sets of samples randomly selected from populations 1 () and 2 (), each with identical pairs of means, x1 and x2. The two sets are shown in Figure 11.1. Is it easier to detect the difference in the two means when you look at set A or set B? You will probably agree that set A shows the difference much more clearly. In set A, the variability of the measurements within the groups (s and s) is much smaller than the variability between the two groups. In set B, there is more variability within the groups (s and s), causing the two groups to “mix” together and making it more difficult to see the identical difference in the means. FI GUR E 1 1. 1 Two sets of samples with the same means ● Set A Set B x1 x2 x x1 x2 x The comparison you have just done intuitively is formalized by the analysis of variance. Moreover, the analysis of variance can be used not only to compare two means but also to make comparisons of more than two population means and to determine the effects of various factors in more complex experimental designs. The analysis of variance relies on statistics with sampling distributions that are modeled by the F distribution of Section 10.7. THE ASSUMPTIONS FOR AN ANALYSIS OF VARIANCE 11.3 The assumptions required for an analysis of variance are similar to those required for the Student’s t and F statistics of Chapter 10. Regardless of the experimental design used to generate the data, you must assume that the observations within each treatment group are normally distributed with a common variance s 2. As in Cha
pter 10, the analysis of variance procedures are fairly robust when the sample sizes are equal and when the data are fairly mound-shaped. Violating the assumption of a common variance is more serious, especially when the sample sizes are not nearly equal. 450 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE ASSUMPTIONS FOR ANALYSIS OF VARIANCE TEST AND ESTIMATION PROCEDURES • The observations within each population are normally distributed with a common variance s 2. • Assumptions regarding the sampling procedure are speciﬁed for each design in the sections that follow. This chapter describes the analysis of variance for three different experimental designs. The ﬁrst design is based on independent random sampling from several populations and is an extension of the unpaired t-test of Chapter 10. The second is an extension of the paired-difference or matched pairs design and involves a random assignment of treatments within matched sets of observations. The third is a design that allows you to judge the effect of two experimental factors on the response. The sampling procedures necessary for each design are restated in their respective sections. THE COMPLETELY RANDOMIZED DESIGN: A ONE-WAY CLASSIFICATION 11.4 One of the simplest experimental designs is the completely randomized design, in which random samples are selected independently from each of k populations. This design involves only one factor, the population from which the measurement comes— hence the designation as a one-way classiﬁcation. There are k different levels corresponding to the k populations, which are also the treatments for this one-way classiﬁcation. Are the k population means all the same, or is at least one mean different from the others? Why do you need a new procedure, the analysis of variance, to compare the population means when you already have the Student’s t-test available? In comparing k 3 means, you could test each of three pairs of hypotheses: H0 : m1 m2 H0 : m1 m3 H0 : m2 m3 to ﬁnd out where the differences lie. However, you must remember that each test you perform is subject to the possibility of error. To compare k 4 means, you would need six tests, and you would need 10 tests to compare k 5 means. The more tests you perform on a set of measurements, the more likely it is that at least one of your conclusions will be incorrect. The analysis of variance procedure provides one overall test to judge the equality of the k population means. Once you have determined whether there is actually a difference in the means, you can use another procedure to ﬁnd out where the differences lie. How can you select these k random samples? Sometimes the populations actually exist in fact, and you can use a computerized random number generator or a random number table to randomly select the samples. For example, in a study to compare the average sizes of health insurance claims in four different states, you could use a computer database provided by the health insurance companies to select random samples from the four states. In other situations, the populations may be hypothetical, and responses can be generated only after the experimental treatments have been applied. EXAMPLE 11.3 A researcher is interested in the effects of ﬁve types of insecticides for use in controlling the boll weevil in cotton ﬁelds. Explain how to implement a completely randomized design to investigate the effects of the ﬁve insecticides on crop yield. 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN ❍ 451 Solution The only way to generate the equivalent of ﬁve random samples from the hypothetical populations corresponding to the ﬁve insecticides is to use a method called a randomized assignment. A ﬁxed number of cotton plants are chosen for treatment, and each is assigned a random number. Suppose that each sample is to have an equal number of measurements. Using a randomization device, you can assign the ﬁrst n plants chosen to receive insecticide 1, the second n plants to receive insecticide 2, and so on, until all ﬁve treatments have been assigned. Whether by random selection or random assignment, both of these examples result in a completely randomized design, or one-way classiﬁcation, for which the analysis of variance is used. 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN Suppose you want to compare k population means, m1, m2, . . . , mk, based on independent random samples of size n1, n2, . . . , nk from normal populations with a common variance s 2. That is, each of the normal populations has the same shape, but their locations might be different, as shown in Figure 11.2. ● FI GUR E 1 1. 2 Normal populations with a common variance but different means ... µ 1 µ 2 µ k Partitioning the Total Variation in an Experiment Let xij be the jth measurement ( j 1, 2, . . . , ni) in the ith sample. The analysis of variance procedure begins by considering the total variation in the experiment, which is measured by a quantity called the total sum of squares (TSS): Total SS S(xij x)2 Sx 2 ij (Sx ij )2 n This is the familiar numerator in the formula for the sample variance for the entire set of n n1 n2 nk measurements. The second part of the calculational formula is sometimes called the correction for the mean (CM). If we let G represent the grand total of all n observations, then CM (Sx ij )2 2 G n n This Total SS is partitioned into two components. The ﬁrst component, called the sum of squares for treatments (SST), measures the variation among the k sample means: 2 SST Sni (xi x)2 ST i CM ni 452 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE where Ti is the total of the observations for treatment i. The second component, called the sum of squares for error (SSE), is used to measure the pooled variation within the k samples: SSE (n1 1)s2 1 (n2 1)s 2 2 (nk 1)s2 k This formula is a direct extension of the numerator in the formula for the pooled estimate of s 2 from Chapter 10. We can show algebraically that, in the analysis of variance, Total SS SST SSE Therefore, you need to calculate only two of the three sums of squares—Total SS, SST, and SSE—and the third can be found by subtraction. Each of the sources of variation, when divided by its appropriate degrees of freedom, provides an estimate of the variation in the experiment. Since Total SS involves n squared observations, its degrees of freedom are df (n 1). Similarly, the sum of squares for treatments involves k squared observations, and its degrees of freedom are df (k 1). Finally, the sum of squares for error, a direct extension of the pooled estimate in Chapter 10, has df (n1 1) (n2 1) (nk 1) n k Notice that the degrees of freedom for treatments and error are additive—that is, df(total) df(treatments) df(error) These two sources of variation and their respective degrees of freedom are combined to form the mean squares as MS SS/df. The total variation in the experiment is then displayed in an analysis of variance (or ANOVA) table. ANOVA TABLE FOR k INDEPENDENT RANDOM SAMPLES: COMPLETELY RANDOMIZED DESIGN The column labeled “SS” satisﬁes: Total SS SST SSE. df k 1 n k n 1 SS SST SSE Total SS Source Treatments Error Total where MS MST SST/(k 1) MST/MSE MSE SSE/(n k) F Total SS Sx 2 ij CM (Sum of squares of all x-values) CM The column labeled “df” always adds up to n 1. with CM (Sx ij)2 2 G n n 2 SST ST i CM ni SSE Total SS SST and ST S MST k 1 SE S MSE k n G Grand total of all n observations Ti Total of all observations in sample i ni Number of observations in sample i n n1 n2 nk 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN ❍ 453 EXAMPLE 11.4 In an experiment to determine the effect of nutrition on the attention spans of elementary school students, a group of 15 students were randomly assigned to each of three meal plans: no breakfast, light breakfast, and full breakfast. Their attention spans (in minutes) were recorded during a morning reading period and are shown in Table 11.1. Construct the analysis of variance table for this experiment. TABLE 11.1 ● Attention Spans of Students After Three Meal Plans No Breakfast Light Breakfast Full Breakfast 8 7 9 13 10 T1 47 14 16 12 17 11 T2 70 10 12 16 15 12 T3 65 Solution To use the calculational formulas, you need the k 3 treatment totals together with n1 n2 n3 5, n 15, and Sxij 182. Then 2)2 8 CM (1 2208.2667 5 1 Total SS (82 72 122 ) CM 2338 2208.2667 129.7333 with (n 1) (15 1) 14 degrees of freedom, 02 652 SST 472 7 CM 2266.8 2208.2667 58.5333 5 with (k 1) (3 1) 2 degrees of freedom, and by subtraction, SSE Total SS SST 129.7333 58.5333 71.2 with (n k) (15 3) 12 degrees of freedom. These three sources of variation, their degrees of freedom, sums of squares, and mean squares are shown in the shaded area of the ANOVA table generated by MINITAB and given in Figure 11.3. You will ﬁnd instructions for generating this output in the “My MINITAB ” section at the end of this chapter. ● One-way ANOVA: Span versus Meal Source Meal Error Total DF 2 12 14 SS 58.53 71.20 129.73 MS 29.27 5.93 S = 2.436 R-Sq = 45.12% Level 1 2 3 N 5 5 5 Mean 9.400 14.000 13.000 StDev 2.302 2.550 2.449 Pooled StDev = 2.436 F 4.93 P 0.027 R-Sq(adj) = 35.97% Individual 95% CIs For Mean Based on Pooled StDev --+---------+---------+---------+------(-----------------) (----------------) (--------*--------) --+---------+---------+---------+------7.5 15.0 12.5 10.0 F IGU RE 1 1 .3 MINITAB output for Example 11.4 454 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE The MINITAB output gives some additional information about the variation in the experiment. The second section shows the means and standard deviations for the three meal plans. More important, you can see in the ﬁrst section of the printout two columns marked “F” and “P.” We can use these values to test a hypothesis concerning the equality of the three treatment means. Testing the Equality of the Treatment Means The mean squares in the analysis of variance table can be used to test the null hypothesis MS SS/df H0 : m1 m2 mk versus the al
ternative hypothesis Ha : At least one of the means is different from the others using the following theoretical argument: • Remember that s 2 is the common variance for all k populations. The quantity SE S MSE k n • is a pooled estimate of s 2, a weighted average of all k sample variances, whether or not H0 is true. If H0 is true, then the variation in the sample means, measured by MST [SST/(k 1)], also provides an unbiased estimate of s 2. However, if H0 is false and the population means are different, then MST—which measures the variation in the sample means—will be unusually large, as shown in Figure 11.4. ● FI GUR E 1 1. 4 Sample means drawn from identical versus different populations x1 H0 true H0 false x2 µ 1 = µ x3 2= µ 3 x1 µ 1 µ 2 x2 µ 3 x3 • The test statistic S T F M E S M F-tests for ANOVA tables are always upper (right) tailed. tends to be larger than usual if H0 is false. Hence, you can reject H0 for large values of F, using a right-tailed statistical test. When H0 is true, this test statistic has an F distribution with df1 (k 1) and df2 (n k) degrees of freedom, and right-tailed critical values of the F distribution (from Table 6 in Appendix I) or computer-generated p-values can be used to draw statistical conclusions about the equality of the population means. 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN ❍ 455 F TEST FOR COMPARING k POPULATION MEANS 1. Null hypothesis: H0 : m1 m2 mk 2. Alternative hypothesis: Ha : One or more pairs of population means differ 3. Test statistic: F MST/MSE, where F is based on df1 (k 1) and df2 (n k) 4. Rejection region: Reject H0 if F Fa, where Fa lies in the upper tail of the F distribution (with df1 k 1 and df2 n k) or if the p-value a. f(F) 0 α Fα F Assumptions • The samples are randomly and independently selected from their respective populations. • The populations are normally distributed with means m1, m2, . . . , mk and equal variances. EXAMPLE 11.5 Do the data in Example 11.4 provide sufficient evidence to indicate a difference in the average attention spans depending on the type of breakfast eaten by the student? Solution To test H0 : m1 m2 m3 versus the alternative hypothesis that the average attention span is different for at least one of the three treatments, you use the analysis of variance F statistic, calculated as 7 4.93 and shown in the column marked “F” in Figure 11.3. It will not surprise you to know that the value in the column marked “P” in Figure 11.3 is the exact p-value for this statistical test. The test statistic MST/MSE calculated above has an F distribution with df1 2 and df2 12 degrees of freedom. Using the critical value approach with a .05, you can reject H0 if F F.05 3.89 from Table 6 in Appendix I (see Figure 11.5). Since the observed value, F 4.93, exceeds the critical value, you reject H0. There is sufﬁcient evidence to indicate that at least one of the three average attention spans is different from at least one of the others. 456 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE F IG URE 11. 5 Rejection region for Example 11.5 ● f(F ) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 = .05 F 10 Rejection region 0 5 3.89 Computer printouts give the exact p-value—use the p-value to make your decision. You could have reached this same conclusion using the exact p-value, P .027, given in Figure 11.3. Since the p-value is less than a .05, the results are statistically signiﬁcant at the 5% level. You still conclude that at least one of the three average attention spans is different from at least one of the others. You can use the F Probabilities applet to ﬁnd critical values of F or p-values for the analysis of variance F-test. Look at the two applets in Figure 11.6. Use the sliders on the left and right of the applets to select the appropriate degrees of freedom (df1 and df2). To ﬁnd the critical value for rejection of H0, enter the signiﬁcance level a in the box marked “Prob” and press Enter. To ﬁnd the p-value, enter the observed value of the test statistic in the box marked “F” and press Enter. Can you identify the critical value for rejection and the p-value for Example 11.5? F IG URE 11. 6 F Probabilities applet ● F Distribution F Distribution Estimating Differences in the Treatment Means The next obvious question you might ask involves the nature of the differences in the population means. Which means are different from the others? How can you estimate the difference, or possibly the individual means for each of the three treatments? 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN ❍ 457 In Section 11.6, we will present a procedure that you can use to compare all possible pairs of treatment means simultaneously. However, if you have a special interest in a particular mean or pair of means, you can construct conﬁdence intervals using the small-sample procedures of Chapter 10, based on the Student’s t distribution. For a single population mean, mi, the conﬁdence interval is xi ta/2 s ni where xi is the sample mean for the ith treatment. Similarly, for a comparison of two population means—say, mi and mj—the conﬁdence interval is (xi xj ) ta/2s2 1 n j 1 n i Before you can use these conﬁdence intervals, however, two questions remain: • How do you calculate s or s2, the best estimate of the common variance s 2? • How many degrees of freedom are used for the critical value of t? To answer these questions, remember that in an analysis of variance, the mean square for error, MSE, always provides an unbiased estimator of s 2 and uses information from the entire set of measurements. Hence, it is the best available estimator of s 2, regardless of what test or estimation procedure you are using. You should always use s2 MSE with df (n k) to estimate s 2! You can ﬁnd the positive square root of this estimator, s MSE, on the last line of Figure 11.3 labeled “Pooled StDev.” Degrees of freedom for conﬁdence intervals are the df for error. COMPLETELY RANDOMIZED DESIGN: (1 a)100% CONFIDENCE INTERVALS FOR A SINGLE TREATMENT MEAN AND THE DIFFERENCE BETWEEN TWO TREATMENT MEANS Single treatment mean: xi ta/2 s ni Difference between two treatment means: (xi xj ) ta/2s2 1 n j 1 n i with s s2 MSE SE S k n where n n1 n2 nk and ta/2 is based on (n k) df. EXAMPLE 11.6 The researcher in Example 11.4 believes that students who have no breakfast will have signiﬁcantly shorter attention spans but that there may be no difference between those who eat a light or a full breakfast. Find a 95% conﬁdence interval for the average attention span for students who eat no breakfast, as well as a 95% conﬁdence interval for the difference in the average attention spans for light versus full breakfast eaters. 458 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE Solution For s2 MSE 5.9333 so that s 5.9333 2.436 with df (n k) 12, you can calculate the two conﬁdence intervals: • For no breakfast: s x1 ta/2 n1 9.4 2.179 6 .43 2 5 9.4 2.37 or between 7.03 and 11.77 minutes. • For light versus full breakfast: (x2 x3) ta/2s2 1 (14 13) 2.1795.93331 .36 a difference of between 2.36 and 4.36 minutes. You can see that the second conﬁdence interval does not indicate a difference in the average attention spans for students who ate light versus full breakfasts, as the researcher suspected. If the researcher, because of prior beliefs, wishes to test the other two possible pairs of means—none versus light breakfast, and none versus full breakfast—the methods given in Section 11.6 should be used for testing all three pairs. Some computer programs have graphics options that provide a powerful visual description of data and the k treatment means. One such option in the MINITAB program is shown in Figure 11.7. The treatment means are indicated by the symbol ⊕ and are connected with straight lines. Notice that the “no breakfast” mean appears to be somewhat different from the other two means, as the researcher suspected, although there is a bit of overlap in the box plots. In the next section, we present a formal procedure for testing the signiﬁcance of the differences between all pairs of treatment means. F IG URE 11. 7 Box plots for Example 11.6 ● Boxplot of Span by Meal n a p S 18 16 14 12 10 8 6 1 2 Meal 3 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN ❍ 459 How Do I Know Whether My Calculations Are Accurate? The following suggestions apply to all the analyses of variance in this chapter: 1. When calculating sums of squares, be certain to carry at least six signiﬁcant ﬁgures before performing subtractions. 2. Remember, sums of squares can never be negative. If you obtain a negative sum of squares, you have made a mistake in arithmetic. 3. Always check your analysis of variance table to make certain that the degrees of freedom sum to the total degrees of freedom (n 1) and that the sums of squares sum to Total SS. 11.5 EXERCISES BASIC TECHNIQUES 11.1 Suppose you wish to compare the means of six populations based on independent random samples, each of which contains 10 observations. Insert, in an ANOVA table, the sources of variation and their respective degrees of freedom. 11.2 The values of Total SS and SSE for the experiment in Exercise 11.1 are Total SS 21.4 and SSE 16.2. a. Complete the ANOVA table for Exercise 11.1. b. How many degrees of freedom are associated with the F statistic for testing H0 : m1 m2 m6? c. Give the rejection region for the test in part b for a .05. d. Do the data provide sufficient evidence to indicate differences among the population means? e. Estimate the p-value for the test. Does this value conﬁrm your conclusions in part d? 11.3 The sample means corresponding to populations 1 and 2 in Exercise 11.1 are x1 3.07 and x2 2.52. a. Find a 95% conﬁdence interval for m1. b. Find a 95% conﬁdence interval for the difference (m1 m2). 11.4 Suppose you wish to compare the means of four populations based on independent random samples, each of which contains six observations. Insert, in an ANOVA table, the sources of variation and their respective degr
ees of freedom. 11.5 The values of Total SS and SST for the experiment in Exercise 11.4 are Total SS 473.2 and SST 339.8. a. Complete the ANOVA table for Exercise 11.4. b. How many degrees of freedom are associated with the F statistic for testing H0 : m1 m2 m3 m4? c. Give the rejection region for the test in part b for a .05. d. Do the data provide sufficient evidence to indicate differences among the population means? e. Approximate the p-value for the test. Does this con- ﬁrm your conclusions in part d? 11.6 The sample means corresponding to populations 1 and 2 in Exercise 11.4 are x1 88.0 and x2 83.9. a. Find a 90% conﬁdence interval for m1. b. Find a 90% conﬁdence interval for the difference (m1 m2). 11.7 These data are observations collected using a completely randomized design: EX1107 Sample 1 Sample 2 Sample . Calculate CM and Total SS. b. Calculate SST and MST. c. Calculate SSE and MSE. d. Construct an ANOVA table for the data. 460 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE e. State the null and alternative hypotheses for an analysis of variance F-test. f. Use the p-value approach to determine whether there is a difference in the three population means. 11.8 Refer to Exercise 11.7 and data set EX1107. Do the data provide sufficient evidence to indicate a difference between m2 and m3? Test using the t-test of Section 10.4 with a .05. 11.9 Refer to Exercise 11.7 and data set EX1107. a. Find a 90% conﬁdence interval for m1. b. Find a 90% conﬁdence interval for the difference (m1 m3). APPLICATIONS 11.10 Reducing Hostility A clinical psychologist wished to compare three methods for reducing hostility levels in university students using a certain psychological test (HLT). High scores on this test were taken to indicate great hostility. Eleven students who got high and nearly equal scores were used in the experiment. Five were selected at random from among the 11 problem cases and treated by method A, three were taken at random from the remaining six students and treated by method B, and the other three students were treated by method C. All treatments continued throughout a semester, when the HLT test was given again. The results are shown in the table. Method Scores on the HLT Test A B C 68 80 73 54 79 83 74 95 76 71 87 a. Perform an analysis of variance for this experiment. b. Do the data provide sufficient evidence to indicate a difference in mean student response to the three methods after treatment? 11.11 Hostility, continued Refer to Exercise 11.10. Let mA and mB, respectively, denote the mean scores at the end of the semester for the populations of extremely hostile students who were treated throughout that semester by method A and method B. a. Find a 95% conﬁdence interval for mA. b. Find a 95% conﬁdence interval for mB. c. Find a 95% conﬁdence interval for (mA mB). d. Is it correct to claim that the conﬁdence intervals found in parts a, b, and c are jointly valid? 11.12 Assembling Electronic Equipment An experiment was conducted to compare the EX1112 effectiveness of three training programs, A, B, and C, in training assemblers of a piece of electronic equipment. Fifteen employees were randomly assigned, ﬁve each, to the three programs. After completion of the courses, each person was required to assemble four pieces of the equipment, and the average length of time required to complete the assembly was recorded. Several of the employees resigned during the course of the program; the remainder were evaluated, producing the data shown in the accompanying table. Use the MINITAB printout to answer the questions. Training Program Average Assembly Time (min) A B C 59 52 58 64 58 65 57 54 71 62 63 64 a. Do the data provide sufficient evidence to indicate a difference in mean assembly times for people trained by the three programs? Give the p-value for the test and interpret its value. b. Find a 99% conﬁdence interval for the difference in mean assembly times between persons trained by programs A and B. c. Find a 99% conﬁdence interval for the mean assem- bly times for persons trained in program A. d. Do you think the data will satisfy (approximately) the assumption that they have been selected from normal populations? Why? MINITAB output for Exercise 11.12 One-way ANOVA: Time versus Program Source DF SS MS F P Program 2 170.5 85.2 5.70 0.025 Error 9 134.5 14.9 Total 11 304.9 S = 3.865 R-Sq = 55.90% R-Sq(adj) = 46.10% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+---------+----1 4 60.500 3.109 (----------------) 2 3 54.667 3.055 (------------------) 3 5 64.200 4.658 (-------------) -+---------+---------+---------+----- Pooled StDev = 3.865 50.0 55.0 60.0 65.0 EX1113 11.13 Swampy Sites An ecological study was conducted to compare the rates of growth of vegetation at four swampy undeveloped sites and to determine the cause of any differences that might be observed. Part of the study involved measuring the leaf lengths of a particular plant species on a preselected date in May. Six plants were randomly selected at each of the four sites to be used in the comparison. The data in the table are the mean leaf length per plant (in centimeters) for a random sample of ten leaves per plant. The MINITAB analysis of variance computer printout for these data is also provided. Location Mean Leaf Length (cm) 1 2 3 4 5.7 6.2 5.4 3.7 6.3 5.3 5.0 3.2 6.1 5.7 6.0 3.9 6.0 6.0 5.6 4.0 5.8 5.2 4.9 3.5 6.2 5.5 5.2 3.6 MINITAB output for Exercise 11.13 One-way ANOVA: Length versus Location Source DF SS MS F P Location 3 19.740 6.580 57.38 0.000 Error 20 2.293 0.115 Total 23 22.033 S = 0.3386 R-Sq = 89.59% R-Sq(adj) = 88.03% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev --------+---------+---------+---------+1 6 6.0167 0.2317 (-----) 2 6 5.6500 0.3937 (-----) 3 6 5.3500 0.4087 (-----) 4 6 3.6500 0.2881 (-----) --------+---------+---------+---------+ 6.40 Pooled StDev = 0.3386 4.00 4.80 5.60 a. You will recall that the test and estimation procedures for an analysis of variance require that the observations be selected from normally distributed (at least, roughly so) populations. Why might you feel reasonably conﬁdent that your data satisfy this assumption? b. Do the data provide sufficient evidence to indicate a difference in mean leaf length among the four locations? What is the p-value for the test? c. Suppose, prior to seeing the data, you decided to compare the mean leaf lengths of locations 1 and 4. Test the null hypothesis m1 m4 against the alternative m1 m4. d. Refer to part c. Construct a 99% conﬁdence interval for (m1 m4). e. Rather than use an analysis of variance F-test, it would seem simpler to examine one’s data, select the two locations that have the smallest and largest sample mean lengths, and then compare these two means using a Student’s t-test. If there is evidence to indicate a difference in these means, there is clearly evidence of a difference among the four. (If you were to use this logic, there would be no need for the analysis of variance F-test.) Explain why this procedure is invalid. EX1114 11.14 Dissolved O2 Content Water samples were taken at four different locations in a river to determine whether the quantity of dissolved oxygen, a measure of water pollution, varied from one location to another. Locations 1 and 2 were selected above an industrial plant, one near the shore and the other in midstream; location 3 was adjacent to the industrial water discharge for the plant; and location 4 was slightly downriver in midstream. Five water specimens 11.5 THE ANALYSIS OF VARIANCE FOR A COMPLETELY RANDOMIZED DESIGN ❍ 461 were randomly selected at each location, but one specimen, corresponding to location 4, was lost in the laboratory. The data and a MINITAB analysis of variance computer printout are provided here (the greater the pollution, the lower the dissolved oxygen readings). Location Mean Dissolved Oxygen Content 1 2 3 4 5.9 6.3 4.8 6.0 6.1 6.6 4.3 6.2 6.3 6.4 5.0 6.1 6.1 6.4 4.7 5.8 6.0 6.5 5.1 MINITAB output for Exercise 11.14 One-way ANOVA: Oxygen versus Location Source DF SS MS F P Location 3 7.8361 2.6120 63.66 0.000 Error 15 0.6155 0.0410 Total 18 8.4516 S = 0.2026 R-Sq = 92.72% R-Sq(adj) = 91.26% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ----+---------+---------+---------+-1 5 6.0800 0.1483 (-----) 2 5 6.4400 0.1140 (-----) 3 5 4.7800 0.3114 (-----) 4 4 6.0250 0.1708 (--*---) ----+---------+---------+---------+-- Pooled StDev = 0.2026 4.80 5.40 6.00 6.60 a. Do the data provide sufficient evidence to indicate a difference in the mean dissolved oxygen contents for the four locations? b. Compare the mean dissolved oxygen content in midstream above the plant with the mean content adjacent to the plant (location 2 versus location 3). Use a 95% conﬁdence interval. 11.15 Calcium The calcium content of a powdered mineral substance was analyzed ﬁve EX1115 times by each of three methods, with similar standard deviations: Method Percent Calcium 1 2 3 .0279 .0268 .0280 .0276 .0274 .0279 .0270 .0267 .0282 .0275 .0263 .0278 .0281 .0267 .0283 Use an appropriate test to compare the three methods of measurement. Comment on the validity of any assumptions you need to make. 11.16 Tuna Fish In Exercise 10.6, we reported the estimated average prices for a EX1116 6-ounce can or a 7.06-ounce pouch of tuna ﬁsh, based on prices paid nationally for a variety of different brands of tuna.1 462 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE Light Tuna White Tuna White Tuna in Water in Water in Oil 1.27 1.22 1.19 1.22 .99 1.92 1.23 .85 .65 .69 .60 .53 1.41 1.12 .63 .67 .60 .66 1.49 1.29 1.27 1.35 1.29 1.00 1.27 1.28 Light Tuna in Oil 2.56 1.92 1.30 1.79 1.23 .62 .66 .62 .65 .60 .67 State 1 2 3 4 $241 235 238 247 250 $216 220 205 213 220 $230 225 235 228 240 $245 250 238 255 255 Source: From “Pricing of Tuna” Copyright 2001 by Consumers Union of U.S., Inc., Yonkers, NY 10703-1057, a nonproﬁt organization. Reprinted with permission from th
e June 2001 issue of Consumer Reports® for educational purposes only. No commercial use or reproduction permitted. www.ConsumerReports.org®. a. Use an analysis of variance for a completely ran- domized design to determine if there are signiﬁcant differences in the prices of tuna packaged in these four different ways. Can you reject the hypothesis of no difference in average price for these packages at the a .05 level of signiﬁcance? At the a .01 level of signiﬁcance? b. Find a 95% conﬁdence interval estimate of the difference in price between light tuna in water and light tuna in oil. Does there appear to be a signiﬁcant difference in the price of these two kinds of packaged tuna? c. Find a 95% conﬁdence interval estimate of the difference in price between white tuna in water and white tuna in oil. Does there appear to be a signiﬁcant difference in the price of these two kinds of packaged tuna? d. What other conﬁdence intervals might be of interest to the researcher who conducted this experiment? 11.17 The Cost of Lumber A national home builder wants to compare the prices per EX1117 1,000 board feet of standard or better grade Douglas ﬁr framing lumber. He randomly selects ﬁve suppliers in each of the four states where the builder is planning to begin construction. The prices are given in the table. a. What type of experimental design has been used? b. Construct the analysis of variance table for this data. c. Do the data provide sufficient evidence to indicate that the average price per 1000 board feet of Douglas ﬁr differs among the four states? Test using a .05. EX1118 11.18 Good at Math? Twenty third graders were randomly separated into four equal groups, and each group was taught a mathematical concept using a different teaching method. At the end of the teaching period, progress was measured by a unit test. The scores are shown below (one child in group 3 was absent on the day that the test was administered). Group 1 112 92 124 89 97 2 111 129 102 136 99 3 140 121 130 106 4 101 116 105 126 119 a. What type of design has been used in this experiment? b. Construct an ANOVA table for the experiment. c. Do the data present sufficient evidence to indicate a difference in the average scores for the four teaching methods? Test using a .05. RANKING POPULATION MEANS 11.6 Many experiments are exploratory in nature. You have no preconceived notions about the results and have not decided (before conducting the experiment) to make speciﬁc treatment comparisons. Rather, you want to rank the treatment means, determine which means differ, and identify sets of means for which no evidence of difference exists. 11.6 RANKING POPULATION MEANS ❍ 463 One option might be to order the sample means from the smallest to the largest and then to conduct t-tests for adjacent means in the ordering. If two means differ by more than ta/2s2 1 n 2 1 n 1 you conclude that the pair of population means differ. The problem with this procedure is that the probability of making a Type I error—that is, concluding that two means differ when, in fact, they are equal—is a for each test. If you compare a large number of pairs of means, the probability of detecting at least one difference in means, when in fact none exists, is quite large. A simple way to avoid the high risk of declaring differences when they do not exist is to use the studentized range, the difference between the smallest and the largest in a set of k sample means, as the yardstick for determining whether there is a difference in a pair of population means. This method, often called Tukey’s method for paired comparisons, makes the probability of declaring that a difference exists between at least one pair in a set of k treatment means, when no difference exists, equal to a. Tukey’s method for making paired comparisons is based on the usual analysis of variance assumptions. In addition, it assumes that the sample means are independent and based on samples of equal size. The yardstick that determines whether a difference exists between a pair of treatment means is the quantity v (Greek lowercase omega), which is presented next. YARDSTICK FOR MAKING PAIRED COMPARISONS v qa(k, df ) s nt where k Number of treatments s2 MSE Estimator of the common variance s 2 and s s2 df Number of degrees of freedom for s2 nt Common sample size—that is, the number of observations in each of the k treatment means qa(k, df ) Tabulated value from Tables 11(a) and 11(b) in Appendix I, for a .05 and .01, respectively, and for various combinations of k and df Rule: Two population means are judged to differ if the corresponding sample means differ by v or more. Tables 11(a) and 11(b) in Appendix I list the values of qa(k, df ) for a .05 and .01, respectively. To illustrate the use of the tables, refer to the portion of Table 11(a) reproduced in Table 11.2. Suppose you want to make pairwise comparisons of k 5 means with a .05 for an analysis of variance, where s2 possesses 9 df. The tabulated value for k 5, df 9, and a .05, shaded in Table 11.2, is q.05(5, 9) 4.76. 464 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE A Partial Reproduction of Table 11(a) in Appendix I; TABLE 11.2 EXAMPLE 11.7 ● Upper 5% Points 2 df 10 11 12 17.97 6.08 4.50 3.93 3.64 3.46 3.34 3.26 3.20 3.15 3.11 3.08 26.98 8.33 5.91 5.04 4.60 4.34 4.16 4.04 3.95 3.88 3.82 3.77 32.82 9.80 6.82 5.76 5.22 4.90 4.68 4.53 4.41 4.33 4.26 4.20 5 37.08 10.88 7.50 6.29 5.67 5.30 5.06 4.89 4.76 4.65 4.57 4.51 6 40.41 11.74 8.04 6.71 6.03 5.63 5.36 5.17 5.02 4.91 4.82 4.75 7 43.12 12.44 8.48 7.05 6.33 5.90 5.61 5.40 5.24 5.12 5.03 4.95 8 45.40 13.03 8.85 7.35 6.58 6.12 5.82 5.60 5.43 5.30 5.20 5.12 9 47.36 13.54 9.18 7.60 6.80 6.32 6.00 5.77 5.59 5.46 5.35 5.27 10 49.07 13.99 9.46 7.83 6.99 6.49 6.16 5.92 5.74 5.60 5.49 5.39 11 12 50.59 14.39 9.72 8.03 7.17 6.65 6.30 6.05 5.87 5.72 5.61 5.51 51.96 14.75 9.95 8.21 7.32 6.79 6.43 6.18 5.98 5.83 5.71 5.61 Refer to Example 11.4, in which you compared the average attention spans for students given three different “meal” treatments in the morning: no breakfast, a light breakfast, or a full breakfast. The ANOVA F-test in Example 11.5 indicated a signiﬁcant difference in the population means. Use Tukey’s method for paired comparisons to determine which of the three population means differ from the others. Solution For this example, there are k 3 treatment means, with s MSE 2.436. Tukey’s method can be used, with each of the three samples containing nt 5 measurements and (n k) 12 degrees of freedom. Consult Table 11 in Appendix I to ﬁnd q.05(k, df ) q.05(3, 12) 3.77 and calculate the “yardstick” as 6 .43 2 4.11 5 s 3.77 nt v q.05(3, 12) The three treatment means are arranged in order from the smallest, 9.4, to the largest, 14.0, in Figure 11.8. The next step is to check the difference between every pair of means. The only difference that exceeds v 4.11 is the difference between no breakfast and a light breakfast. These two treatments are thus declared signiﬁcantly different. You cannot declare a difference between the other two pairs of treatments. To indicate this fact visually, Figure 11.8 shows a line under those pairs of means that are not signiﬁcantly different. F IGU RE 1 1 .8 Ranked means for Example 11.7 ● None 9.4 Full 13.0 Light 14.0 The results here may seem confusing. However, it usually helps to think of ranking the means and interpreting nonsigniﬁcant differences as our inability to distinctly rank those means underlined by the same line. For this example, the light breakfast deﬁnitely ranked higher than no breakfast, but the full breakfast could not be ranked higher than no breakfast, or lower than the light breakfast. The probability that we make at least one error among the three comparisons is at most a .05. 11.6 RANKING POPULATION MEANS ❍ 465 If zero is not in the interval, there is evidence of a difference between the two methods. Most computer programs provide an option to perform paired comparisons, including Tukey’s method. The MINITAB output in Figure 11.9 shows its form of Tukey’s test, which differs slightly from the method we have presented. The three intervals that you see in the printout marked “Lower” and “Upper” represent the difference in the two sample means plus or minus the yardstick v. If the interval contains the value 0, the two means are judged to be not signiﬁcantly different. You can see that only means 1 and 2 (none versus light) show a signiﬁcant difference. FI GUR E 1 1. 9 MINITAB output for Example 11.7 ● Tukey's 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Meal Individual confidence level = 97.94% Meal = 1 subtracted from: Meal Lower Center Upper -----+---------+---------+---------+---2 0.493 4.600 8.707 (----------------------) 3 -0.507 3.600 7.707 (---------------------) -----+---------+---------+---------+---- -3.5 Meal = 2 subtracted from: 0.0 3.5 7.0 Meal Lower Center Upper -----+---------+---------+---------+---3 -5.107 -1.000 3.107 (-----------*-----------) -----+---------+---------+---------+---- -3.5 0.0 3.5 7.0 As you study two more experimental designs in the next sections of this chapter, remember that, once you have found a factor to be signiﬁcant, you should use Tukey’s method or another method of paired comparisons to ﬁnd out exactly where the differences lie! 11.6 EXERCISES BASIC TECHNIQUES 11.19 Suppose you wish to use Tukey’s method of paired comparisons to rank a set of population means. In addition to the analysis of variance assumptions, what other property must the treatment means satisfy? 11.20 Consult Tables 11(a) and 11(b) in Appendix I and ﬁnd the values of qa(k, df ) for these cases: a. a .05, k 5, df 7 b. a .05, k 3, df 10 c. a .01, k 4, df 8 d. a .01, k 7, df 5 11.21 If the sample size for each treatment is nt and if s 2 is based on 12 df, ﬁnd v in these cases: a. a .05, k 4, nt 5 b. a .01, k 6, nt 8 11.22 An independent random sampling design was used to compare the means of six treatments based on samples of four
observations per treatment. The pooled estimator of s 2 is 9.12, and the sample means follow: x1 101.6 x4 92.9 a. Give the value of v that you would use to make pairwise comparisons of the treatment means for a .05. x2 98.4 x5 104.2 x3 112.3 x6 113.8 b. Rank the treatment means using pairwise comparisons. APPLICATIONS 11.23 Swamp Sites, again Refer to Exercise 11.13 and data set EX1113. Rank the mean leaf growth for the four locations. Use a .01. 11.24 Calcium Refer to Exercise 11.15 and data set EX1115. The paired comparisons option in MINITAB generated the output provided here. What do these results tell you about the differences in the population 466 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE means? Does this conﬁrm your conclusions in Exercise 11.15? MINITAB output for Exercise 11.24 Tukey's 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Method Individual confidence level = 97.94% Method = 1 subtracted from: Method Lower Center Upper 2 -0.0014377 -0.0008400 -0.0002423 3 -0.0001777 0.0004200 0.0010177 Method --------+---------+---------+---------+ 2 (----------) 3 (----------) --------+---------+---------+---------+ -0.0010 0.0000 0.0010 0.0020 Method = 2 subtracted from: Method Lower Center Upper 3 0.0006623 0.0012600 0.0018577 Method --------+---------+---------+---------+ 3 (-----*-----) --------+---------+---------+---------+ -0.0010 0.0000 0.0010 0.0020 EX1125 11.25 Glucose Tolerance Physicians depend on laboratory test results when managing medical problems such as diabetes or epilepsy. In a uniformity test for glucose tolerance, three different laboratories were each sent nt 5 identical blood samples from a person who had drunk 50 milligrams (mg) of glucose dissolved in water. The laboratory results (in mg/dl) are listed here: Lab 1 Lab 2 Lab 3 120.1 110.7 108.9 104.2 100.4 98.3 112.1 107.7 107.9 99.2 103.0 108.5 101.1 110.0 105.4 a. Do the data indicate a difference in the average readings for the three laboratories? b. Use Tukey’s method for paired comparisons to rank the three treatment means. Use a .05. 11.26 The Cost of Lumber, continued The analysis of variance F-test in Exercise 11.17 (and data set EX1117) determined that there was indeed a difference in the average cost of lumber for the four states. The following information from Exercise 11.17 is given in the table: Sample Means x1 242.2 MSE x2 214.8 x3 231.6 x4 248.6 Error df : ni : k: 41.25 16 5 4 Use Tukey’s method for paired comparisons to determine which means differ signiﬁcantly from the others at the a .01 level. EX1127 11.27 GRE Scores The Graduate Record Examination (GRE) scores were recorded for students admitted to three different graduate programs at a local university. Graduate Program 1 532 548 619 509 627 2 670 590 640 710 690 3 502 607 549 524 542 a. Do these data provide sufficient evidence to indi- cate a difference in the mean GRE scores for applicants admitted to the three programs? b. Find a 95% conﬁdence interval for the difference in mean GRE scores for programs 1 and 2. c. If you ﬁnd a signiﬁcant difference in the average GRE scores for the three programs, use Tukey’s method for paired comparisons to determine which means differ signiﬁcantly from the others. Use a .05. THE RANDOMIZED BLOCK DESIGN: A TWO-WAY CLASSIFICATION 11.7 The completely randomized design introduced in Section 11.4 is a generalization of the two independent samples design presented in Section 10.4. It is meant to be used when the experimental units are quite similar or homogeneous in their makeup and when there is only one factor—the treatment—that might inﬂuence the response. Any other variation in the response is due to random variation or experimental error. b blocks k treatments n bk 11.8 THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN ❍ 467 Sometimes it is clear to the researcher that the experimental units are not homogeneous. Experimental subjects or animals, agricultural ﬁelds, days of the week, and other experimental units often add their own variability to the response. Although the researcher is not really interested in this source of variation, but rather in some treatment he chooses to apply, he may be able to increase the information by isolating this source of variation using the randomized block design—a direct extension of the matched pairs or paired-difference design in Section 10.5. In a randomized block design, the experimenter is interested in comparing k treatment means. The design uses blocks of k experimental units that are relatively similar, or homogeneous, with one unit within each block randomly assigned to each treatment. If the randomized block design involves k treatments within each of b blocks, then the total number of observations in the experiment is n bk. A production supervisor wants to compare the mean times for assembly-line operators to assemble an item using one of three methods: A, B, or C. Expecting variation in assembly times from operator to operator, the supervisor uses a randomized block design to compare the three methods. Five assembly-line operators are selected to serve as blocks, and each is assigned to assemble the item three times, once for each of the three methods. Since the sequence in which the operator uses the three methods may be important (fatigue or increasing dexterity may be factors affecting the response), each operator should be assigned a random sequencing of the three methods. For example, operator 1 might be assigned to perform method C ﬁrst, followed by A and B. Operator 2 might perform method A ﬁrst, then C and B. To compare four different teaching methods, a group of students might be divided into blocks of size 4, so that the groups are most nearly matched according to academic achievement. To compare the average costs for three different cellular phone companies, costs might be compared at each of three usage levels: low, medium, and high. To compare the average yields for three species of fruit trees when a variation in yield is expected because of the ﬁeld in which the trees are planted, a researcher uses ﬁve ﬁelds. She divides each ﬁeld into three plots on which the three species of fruit trees are planted. Matching or blocking can take place in many different ways. Comparisons of treatments are often made within blocks of time, within blocks of people, or within similar external environments. The purpose of blocking is to remove or isolate the block-to-block variability that might otherwise hide the effect of the treatments. You will ﬁnd more examples of the use of the randomized block design in the exercises at the end of the next section. THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN 11.8 The randomized block design identiﬁes two factors: treatments and blocks—both of which affect the response. Partitioning the Total Variation in the Experiment Let xij be the response when the ith treatment (i 1, 2, . . . , k) is applied in the jth block ( j 1, 2, . . . , b). The total variation in the n bk observations is ij (Sx ij)2 n Total SS S(xij x)2 Sx 2 468 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE This is partitioned into three (rather than two) parts in such a way that Total SS SSB SST SSE where • SSB (sum of squares for blocks) measures the variation among the block means. • SST (sum of squares for treatments) measures the variation among the treatment means. • SSE (sum of squares for error) measures the variation of the differences among the treatment observations within blocks, which measures the experimental error. The calculational formulas for the four sums of squares are similar in form to those you used for the completely randomized design in Section 11.5. Although you can simplify your work by using a computer program to calculate these sums of squares, the formulas are given next. CALCULATING THE SUMS OF SQUARES FOR A RANDOMIZED BLOCK DESIGN, k TREATMENTS IN b BLOCKS 2 CM G n where G Sxij Total of all n bk observations Total SS Sx 2 ij CM (Sum of squares of all x-values) CM 2 SST ST i CM b 2 SSB SB j CM k Total SS SST SSB SSE SSE Total SS SST SSB with Ti Total of all observations receiving treatment i, i 1, 2, . . . , k Bj Total of all observations in block j, j 1, 2, . . . , b Each of the three sources of variation, when divided by the appropriate degrees of freedom, provides an estimate of the variation in the experiment. Since Total SS involves n bk squared observations, its degrees of freedom are df (n 1). Similarly, SST involves k squared totals, and its degrees of freedom are df (k 1), while SSB involves b squared totals and has (b 1) degrees of freedom. Finally, since the degrees of freedom are additive, the remaining degrees of freedom associated with SSE can be shown algebraically to be df (b 1)(k 1). These three sources of variation and their respective degrees of freedom are combined to form the mean squares as MS SS/df, and the total variation in the experiment is then displayed in an analysis of variance (or ANOVA) table as shown here: Degrees of freedom are additive. EXAMPLE 11.8 11.8 THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN ❍ 469 ANOVA TABLE FOR A RANDOMIZED BLOCK DESIGN, k TREATMENTS AND b BLOCKS Source Treatments Blocks Error Total df k 1 b 1 (b 1)(k 1) n 1 bk 1 MS SS SST MST SST/(k 1) SSB MSB SSB/(b 1) SSE MSE SSE/(b 1)(k 1) F MST/MSE MSB/MSE The cellular phone industry is involved in a ﬁerce battle for customers, with each company devising its own complex pricing plan to lure customers. Since the cost of a cell phone minute varies drastically depending on the number of minutes per month used by the customer, a consumer watchdog group decided to compare the average costs for four cellular phone companies using three different usage levels as blocks. The monthly costs (in dollars) computed by the cell phone companies for peak-time callers at low (20 minutes per month), middle (150 minutes per month), and high (1000 minutes per month) usage levels are given in Table 11.3. Construct the analysis
of variance table for this experiment. TABLE 11.3 ● Monthly Phone Costs of Four Companies at Three Usage Levels Company Usage Level A B C D Low Middle High Totals 27 68 308 T1 403 24 76 326 T2 426 31 65 312 T3 408 23 67 300 T4 390 Totals B1 105 B2 276 B3 1246 G 1627 Blocks contain experimental units that are relatively the same. Solution The experiment is designed as a randomized block design with b 3 usage levels (blocks) and k 4 companies (treatments), so there are n bk 12 observations and G 1627. Then 2 72 CM G 2 16 220,594.0833 n 1 2 Total SS (272 242 3002) CM 189,798.9167 SST SSB 4032 3902 3 1052 2762 12462 4 CM 222.25 CM 189,335.1667 and by subtraction, SSE Total SS SST SSB 241.5 These four sources of variation, their degrees of freedom, sums of squares, and mean squares are shown in the shaded area of the analysis of variance table, generated by MINITAB and given in Figure 11.10. You will ﬁnd instructions for generating this output in the section “My MINITAB ” at the end of this chapter. 470 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE F IG URE 11. 10 MINITAB output for Example 11.8 ● Two-way ANOVA: Dollars versus Usage, Company Source Usage Company Error Total DF 2 3 6 11 SS 189335 222 242 189799 MS 94667.6 74.1 40.3 F 2351.99 1.84 P 0.000 0.240 S = 6.344 R-Sq = 99.87% R-Sq(adj) = 99.77% Notice that the MINITAB ANOVA table shows two different F statistics and p-values. It will not surprise you to know that these statistics are used to test hypotheses concerning the equality of both the treatment and block means. Testing the Equality of the Treatment and Block Means The mean squares in the analysis of variance table can be used to test the null hypotheses H0 : No difference among the k treatment means or H0 : No difference among the b block means versus the alternative hypothesis Ha : At least one of the means is different from at least one other using a theoretical argument similar to the one we used for the completely randomized design. • Remember that s 2 is the common variance for the observations in all bk block-treatment combinations. The quantity S E S MSE k 1) )( 1 (b is an unbiased estimate of s 2, whether or not H0 is true. • The two mean squares, MST and MSB, estimate s 2 only if H0 is true and tend to be unusually large if H0 is false and either the treatment or block means are different. • The test statistics B S and are used to test the equality of treatment and block means, respectively. Both statistics tend to be larger than usual if H0 is false. Hence, you can reject H0 for large values of F, using right-tailed critical values of the F distribution with the appropriate degrees of freedom (see Table 6 in Appendix I) or computer-generated p-values to draw statistical conclusions about the equality of the population means. As an alternative, you can use the F Probabilities applet to ﬁnd either critical values of F or p-values. 11.8 THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN ❍ 471 TESTS FOR A RANDOMIZED BLOCK DESIGN For comparing treatment means: 1. Null hypothesis: H0 : The treatment means are equal 2. Alternative hypothesis: Ha : At least two of the treatment means differ 3. Test statistic: F MST/MSE, where F is based on df1 (k 1) and df2 (b 1)(k 1) 4. Rejection region: Reject if F Fa, where Fa lies in the upper tail of the F distribution (see the ﬁgure), or when the p-value a For comparing block means: 1. Null hypothesis: H0 : The block means are equal 2. Alternative hypothesis: Ha : At least two of the block means differ 3. Test statistic: F MSB/MSE, where F is based on df1 (b 1) and df2 (b 1)(k 1) 4. Rejection region: Reject if F Fa, where Fa lies in the upper tail of the F distribution (see the ﬁgure), or when the p-value a f(F) 0 α Fα F EXAMPLE 11.9 Do the data in Example 11.8 provide sufficient evidence to indicate a difference in the average monthly cell phone cost depending on the company the customer uses? Solution The cell phone companies represent the treatments in this randomized block design, and the differences in their average monthly costs are of primary interest to the researcher. To test H0 : No difference in the average cost among companies versus the alternative that the average cost is different for at least one of the four companies, you use the analysis of variance F statistic, calculated as 4 1 . 7 S T F M 1.84 3 . 0 4 E S M and shown in the column marked “F” and the row marked “Company” in Figure 11.10. The exact p-value for this statistical test is also given in Figure 11.10 as .240, which is too large to allow rejection of H0. The results do not show a signiﬁcant difference in the treatment means. That is, there is insufficient evidence to indicate a difference in the average monthly costs for the four companies. 472 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE The researcher in Example 11.9 was fairly certain in using a randomized block design that there would be a signiﬁcant difference in the block means—that is, a signiﬁcant difference in the average monthly costs depending on the usage level. This suspicion is justiﬁed by looking at the test of equality of block means. Notice that the observed test statistic is F 2351.99 with P .000, showing a highly signiﬁcant difference, as expected, in the block means. Identifying Differences in the Treatment and Block Means Once the overall F-test for equality of the treatment or block means has been performed, what more can you do to identify the nature of any differences you have found? As in Section 11.5, you can use Tukey’s method of paired comparisons to determine which pairs of treatment or block means are signiﬁcantly different from one another. However, if the F-test does not indicate a signiﬁcant difference in the means, there is no reason to use Tukey’s procedure. If you have a special interest in a particular pair of treatment or block means, you can estimate the difference using a (1 a)100% conﬁdence interval.† The formulas for these procedures, shown next, follow a pattern similar to the formulas for the completely randomized design. Remember that MSE always provides an unbiased estimator of s 2 and uses information from the entire set of measurements. Hence, it is the best available estimator of s 2, regardless of what test or estimation procedure you are using. You will again use s2 MSE with df (b 1)(k 1) to estimate s 2 in comparing the treatment and block means. COMPARING TREATMENT AND BLOCK MEANS Tukey’s yardstick for comparing block means: v qa(b, df ) s k Tukey’s yardstick for comparing treatment means: v qa(k, df ) s b Degrees of freedom for Tukey’s test and for conﬁdence intervals are error df. k (1 a)100% conﬁdence interval for the difference in two block means: 1 (Bi Bj) ta/2s21 k where Bi is the average of all observations in block i (1 a)100% conﬁdence interval for the difference in two treatment means: 1 (Ti Tj) ta/2s21 b where Ti is the average of all observations in treatment i. b †You cannot construct a conﬁdence interval for a single mean unless the blocks have been randomly selected from among the population of all blocks. The procedure for constructing intervals for single means is beyond the scope of this book. 11.8 THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN ❍ 473 Note: The values qa(*, df ) from Table 11 in Appendix I, ta/2 from Table 4 in Appendix I, and s 2 MSE all depend on df (b 1)(k 1) degrees of freedom. EXAMPLE 11.10 Identify the nature of any differences you found in the average monthly cell phone costs from Example 11.8. Solution Since the F-test did not show any signiﬁcant differences in the average costs for the four companies, there is no reason to use Tukey’s method of paired comparisons. Suppose, however, that you are an executive for company B and your major competitor is company C. Can you claim a signiﬁcant difference in the two average costs? Using a 95% conﬁdence interval, you can calculate (T2 T3) t.025MSE2 b You cannot form a conﬁdence interval or test an hypothesis about a single treatment mean in a randomized block design! 42 6 40 3 3 3 8 2.44740.32 6 12.68 so the difference between the two average costs is estimated as between $6.68 and $18.68. Since 0 is contained in the interval, you do not have evidence to indicate a signiﬁcant difference in your average costs. Sorry! Some Cautionary Comments on Blocking Here are some important points to remember: • A randomized block design should not be used when treatments and blocks both correspond to experimental factors of interest to the researcher. In designating one factor as a block, you may assume that the effect of the treatment will be the same, regardless of which block you are using. If this is not the case, the two factors—blocks and treatments—are said to interact, and your analysis could lead to incorrect conclusions regarding the relationship between the treatments and the response. When an interaction is suspected between two factors, you should analyze the data as a factorial experiment, which is introduced in the next section. • Remember that blocking may not always be beneﬁcial. When SSB is removed from SSE, the number of degrees of freedom associated with SSE gets smaller. For blocking to be beneﬁcial, the information gained by isolating the block variation must outweigh the loss of degrees of freedom for error. Usually, though, if you suspect that the experimental units are not homogeneous and you can group the units into blocks, it pays to use the randomized block design! • Finally, remember that you cannot construct conﬁdence intervals for individual treatment means unless it is reasonable to assume that the b blocks have been randomly selected from a population of blocks. If you construct such an interval, the sample treatment mean will be biased by the positive and negative effects that the blocks have on the response. 474 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE 11.8 EXERCISES BASIC TECHNIQUES 11.28 A randomized block design was used to compare the means of three treatments within six blo
cks. Construct an ANOVA table showing the sources of variation and their respective degrees of freedom. 11.34 The data shown here are observations collected from an experiment that compared EX1134 three treatments, A, B, and C, within each of ﬁve blocks, using a randomized block design: Block 11.29 Suppose that the analysis of variance calculations for Exercise 11.28 are SST 11.4, SSB 17.1, and Total SS 42.7. Complete the ANOVA table, showing all sums of squares, mean squares, and pertinent F-values. Treatment A B C Total 1 2.1 3.4 3.0 8.5 2 2.6 3.8 3.6 10.0 3 1.9 3.6 3.2 8.7 4 3.2 4.1 3.9 5 2.7 3.9 3.9 11.2 10.5 Total 12.5 18.8 17.6 48.9 11.30 Do the data of Exercise 11.28 provide sufficient evidence to indicate differences among the treatment means? Test using a .05. 11.31 Refer to Exercise 11.28. Find a 95% conﬁdence interval for the difference between a pair of treatment means A and B if xA 21.9 and xB 24.2. 11.32 Do the data of Exercise 11.28 provide sufficient evidence to indicate that blocking increased the amount of information in the experiment about the treatment means? Justify your answer. EX1133 11.33 The data that follow are observations collected from an experiment that compared four treatments, A, B, C, and D, within each of three blocks, using a randomized block design. Treatment Block 1 2 3 Total A 6 4 12 22 B 10 9 15 34 C 8 5 14 27 D 9 7 14 30 Total 33 25 55 113 a. Do the data present sufficient evidence to indicate differences among the treatment means? Test using a .05. b. Do the data present sufficient evidence to indicate differences among the block means? Test using a .05. c. Rank the four treatment means using Tukey’s method of paired comparisons with a .01. d. Find a 95% conﬁdence interval for the difference in means for treatments A and B. e. Does it appear that the use of a randomized block design for this experiment was justiﬁed? Explain. MINITAB output for Exercise 11.34 Two-way ANOVA: Response versus Trts, Blocks Source DF SS MS F P Trts 2 4.476 2.238 79.93 0.000 Blocks 4 1.796 0.449 16.04 0.001 Error 8 0.224 0.028 Total 14 6.496 S = 0.1673 R-Sq = 96.55% R-Sq(adj) = 93.97% Use the MINITAB ouput to analyze the experiment. Investigate possible differences in the block and/or treatment means and, if any differences exist, use an appropriate method to speciﬁcally identify where the differences lie. Has blocking been effective in this experiment? Present your results in the form of a report. 11.35 The partially completed ANOVA table for a randomized block design is presented here: Source Treatments Blocks Error Total df 4 24 34 SS MS F 14.2 18.9 41.9 a. How many blocks are involved in the design? b. How many observations are in each treatment total? c. How many observations are in each block total? d. Fill in the blanks in the ANOVA table. e. Do the data present sufficient evidence to indicate differences among the treatment means? Test using a .05. f. Do the data present sufficient evidence to indicate differences among the block means? Test using a .05. 11.8 THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN ❍ 475 APPLICATIONS EX1136 11.36 Gas Mileage A study was conducted to compare automobile gasoline mileage for three formulations of gasoline. A was a non-leaded 87 octane formulation, B was a non-leaded 91 octane formulation, and C was a non-leaded 87 octane formulation with 15% ethanol. Four automobiles, all of the same make and model, were used in the experiment, and each formulation was tested in each automobile. Using each formulation in the same automobile has the effect of eliminating (blocking out) automobile-toautomobile variability. The data (in miles per gallon) follow. Formulation A B C Automobile 1 25.7 27.2 26.1 2 27.0 28.1 27.5 3 27.3 27.9 26.8 4 26.1 27.7 27.8 a. Do the data provide sufficient evidence to indicate a difference in mean mileage per gallon for the three gasoline formulations? b. Is there evidence of a difference in mean mileage for the four automobiles? c. Suppose that prior to looking at the data, you had decided to compare the mean mileage per gallon for formulations A and B. Find a 90% conﬁdence interval for this difference. d. Use an appropriate method to identify the pairwise differences, if any, in the average mileages for the three formulations. 11.37 Water Resistance in Textiles An experiment was conducted to compare the EX1137 effects of four different chemicals, A, B, C, and D, in producing water resistance in textiles. A strip of material, randomly selected from a bolt, was cut into four pieces, and the four pieces were randomly assigned to receive one of the four chemicals, A, B, C, or D. This process was replicated three times, thus producing a randomized block design. The design, with moistureresistance measurements, is as shown in the ﬁgure (low readings indicate low moisture penetration). Analyze the experiment using a method appropriate for this randomized block design. Identify the blocks and treatments, and investigate any possible differences in treatment means. If any differences exist, use an appropriate method to speciﬁcally identify where the differences lie. What are the practical implications for the chemical producers? Has blocking been effective in this experiment? Present your results in the form of a report. Illustration for Exercise 11.37 Blocks (bolt samples) 1 C 9.9 A 10.1 B 11.4 D 12.1 2 D 13.4 B 12.9 A 12.2 C 12.3 3 B 12.7 D 12.9 C 11.4 A 11.9 11.38 Glare in Rearview Mirrors An experiment was conducted to compare the glare characteristics of four types of automobile rearview mirrors. Forty drivers were randomly selected to participate in the experiment. Each driver was exposed to the glare produced by a headlight located 30 feet behind the rear window of the experimental automobile. The driver then rated the glare produced by the rearview mirror on a scale of 1 (low) to 10 (high). Each of the four mirrors was tested by each driver; the mirrors were assigned to a driver in random order. An analysis of variance of the data produced this ANOVA table: Source df SS MS F Mirrors Drivers Error Total 46.98 8.42 638.61 a. Fill in the blanks in the ANOVA table. b. Do the data present sufficient evidence to indicate differences in the mean glare ratings of the four rearview mirrors? Calculate the approximate p-value and use it to make your decision. c. Do the data present sufficient evidence to indicate that the level of glare perceived by the drivers varied from driver to driver? Use the p-value approach. d. Based on the results of part b, what are the practical implications of this experiment for the manufacturers of the rearview mirrors? 11.39 Slash Pine Seedings An experiment was conducted to determine the effects of three EX1139 methods of soil preparation on the ﬁrst-year growth of slash pine seedlings. Four locations (state forest lands) were selected, and each location was divided into three plots. Since it was felt that soil fertility within a location was more homogeneous than between locations, a randomized block design was employed using locations 476 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE as blocks. The methods of soil preparation were A (no preparation), B (light fertilization), and C (burning). Each soil preparation was randomly applied to a plot within each location. On each plot, the same number of seedlings were planted and the average ﬁrst-year growth of the seedlings was recorded on each plot. Use the MINITAB printout to answer the questions. Soil Preparation A B C Location 1 11 15 10 2 13 17 15 3 16 20 13 4 10 12 10 a. Conduct an analysis of variance. Do the data provide evidence to indicate a difference in the mean growths for the three soil preparations? b. Is there evidence to indicate a difference in mean rates of growth for the four locations? c. Use Tukey’s method of paired comparisons to rank the mean growths for the three soil preparations. Use a .01. d. Use a 95% conﬁdence interval to estimate the difference in mean growths for methods A and B. MINITAB output for Exercise 11.39 Two-way ANOVA: Growth versus Soil Prep, Location Source DF SS MS F P Soil Prep 2 38.000 19.0000 10.06 0.012 Location 3 61.667 20.5556 10.88 0.008 Error 6 11.333 1.8889 Total 11 111.000 S = 1.374 R-Sq = 89.79% R-Sq(adj) = 81.28% Individual 95% CIs For Mean Based on Pooled StDev Soil Prep Mean ---------+---------+---------+---------+-1 12.5 (--------------) 2 16.0 (--------------) 3 12.0 (--------------) ---------+---------+---------+---------+ 12.0 14.0 16.0 18.0 Individual 95% CIs For Mean Based on Pooled StDev Location Mean ------+---------+---------+---------+----1 12.0000 (--------------) 2 15.0000 (--------------) 3 16.3333 (-------------) 4 10.6667 (-------*------) -----+---------+---------+---------+---- 10.0 12.5 15.0 17.5 EX1140 11.40 Digitalis and Calcium Uptake A study was conducted to compare the effects of three levels of digitalis on the levels of calcium in the heart muscles of dogs. Because general level of calcium uptake varies from one animal to another, the tissue for a heart muscle was regarded as a block, and comparisons of the three digitalis levels (treatments) were made within a given animal. The calcium uptakes for the three levels of digitalis, A, B, and C, were compared based on the heart muscles of four dogs and the results are given in the table. Use the MINITAB printout to answer the questions. Dogs 2 C 1698 B 1387 A 1140 3 B 1296 A 1029 C 1549 4 A 1150 C 1579 B 1319 1 A 1342 B 1608 C 1881 a. How many degrees of freedom are associated with SSE? b. Do the data present sufficient evidence to indicate a difference in the mean uptakes of calcium for the three levels of digitalis? c. Use Tukey’s method of paired comparisons with a .01 to rank the mean calcium uptakes for the three levels of digitalis. d. Do the data indicate a difference in the mean uptakes of calcium for the four heart muscles? e. Use Tukey’s method of paired comparisons with a .01 to rank the mean calcium uptakes for the he
art muscles of the four dogs used in the experiment. Are these results of any practical value to the researcher? f. Give the standard error of the difference between the mean calcium uptakes for two levels of digitalis. g. Find a 95% conﬁdence interval for the difference in mean responses between treatments A and B. MINITAB output for Exercise 11.40 Two-way ANOVA: Uptake versus Digitalis, Dog Source DF SS MS F P Digitalis 2 542177 262089 258.24 0.000 Dog 3 173415 57805 56.96 0.000 Error 6 6090 1015 Total 11 703682 S = 31.86 R-Sq = 99.13% R-Sq(adj) = 98.41% Individual 95% CIs For Mean Based on Pooled StDev Digitalis Mean -----+---------+---------+---------+---1 1165.25 (---) 2 1402.50 (---) 3 1676.75 (---) -----+---------+---------+---------+--- 1200 1350 1500 1650 Individual 95% CIs For Mean Based on Pooled StDev Dog Mean ------+---------+---------+---------+--1 1610.33 (------) 2 1408.33 (-----) 3 1291.33 (-----) 4 1349.33 (--*---) ------+---------+---------+---------+-- 1320 1440 1560 1680 11.41 Bidding on Construction Jobs A building contractor employs three construction EX1141 engineers, A, B, and C, to estimate and bid on jobs. 11.8 THE ANALYSIS OF VARIANCE FOR A RANDOMIZED BLOCK DESIGN ❍ 477 To determine whether one tends to be a more conservative (or liberal) estimator than the others, the contractor selects four projected construction jobs and has each estimator independently estimate the cost (in dollars per square foot) of each job. The data are shown in the table: Construction Job Estimator 1 A B C 35.10 37.45 36.30 2 34.50 34.60 35.10 Total 108.85 104.20 3 29.25 33.10 32.45 94.80 4 31.60 34.40 32.90 98.90 Total 130.45 139.55 136.75 406.75 Analyze the experiment using the appropriate methods. Identify the blocks and treatments, and investigate any possible differences in treatment means. If any differences exist, use an appropriate method to speciﬁcally identify where the differences lie. Has blocking been effective in this experiment? What are the practical implications of the experiment? Present your results in the form of a report. EX1142 11.42 “In Good Hands” The cost of automobile insurance varies by location, ages of the drivers, and type of coverage. The following are estimates for the annual 2006–2007 premium for a single male, licensed for 6–8 years, who drives a Honda Accord 12,600 to 15,000 miles per year and has no violations or accidents. These estimates are provided by the California Department of Insurance for the year 2006–2007 on its website (http://www.insurance .ca.gov).2 Insurance Company Location 21st Century Allstate AAA Fireman’s State Farm Fund Riverside San Bernardino Hollywood Long Beach $1870 2064 3542 2228 Source: www.insurance.ca.gov $2250 2286 3773 2617 $2154 2316 3235 2681 $2324 2005 3360 3279 $3053 3151 3883 3396 a. What type of design was used in collecting these data? b. Is there sufficient evidence to indicate that insurance premiums for the same type of coverage differs from company to company? c. Is there sufficient evidence to indicate that insurance premiums vary from location to location? d. Use Tukey’s procedure to determine which insurance companies listed here differ from others in the premiums they charge for this typical client. Use a .05. e. Summarize your ﬁndings. 11.43 Warehouse Shopping Warehouse stores such as Costco and Sam’s Club are the EX1143 shopping choice of many Americans because of the low cost associated with bulk shopping. When a new warehouse grocery store called WinCo Foods was opened in Moreno Valley, California, an advertising mailer claimed that they were the area’s “low price leader.”3 They compared their prices with those of four other grocery stores for a number of items purchased on the same day. A partial list of the items and their prices is given in the following table. Items WinCo Albertsons Ralphs Bros Less Stores Stater Food-4- Salad mix, 1 lb. bag Hillshire Farm® Smoked Sausage, 16 oz. Kellogg’s Raisin Bran®, 25.5 oz. 0.88 2.48 2.48 Kraft® Philadelphia® 1.18 Cream Cheese, 8 oz. Kraft® Ranch Dressing, 16 oz. Langers® Apple Juice, 128 oz. Dial® Bar Soap, Gold, 8–4.5 oz. Jif® Peanut Butter, Creamy, 28 oz. 1.58 1.98 3.48 2.58 1.99 4.29 4.99 1.50 3.89 4.99 5.79 4.89 1.79 1.89 0.98 2.50 3.00 3.68 4.69 3.49 3.38 1.99 1.89 1.97 2.69 2.50 1.68 4.59 3.79 2.98 4.19 3.99 4.58 3.99 3.89 2.68 a. What are the blocks and treatments in this experiment? b. Do the data provide evidence to indicate that there are signiﬁcant differences in prices from store to store? Support your answer statistically using the ANOVA printout that follows. c. Are there signiﬁcant differences from block to block? Was blocking effective? d. The advertisement includes the following statement: “Though this list is not intended to represent a typical weekly grocery order or a random list of grocery items, WinCo continues to be the area’s low price leader.” How might this statement affect the reliability of your conclusions in part b? Two-way ANOVA: Price versus Item, Store SS Source 38.2360 Item 16.6644 Store 7.8862 Error 62.7866 Total MS 5.46228 4.16610 0.28165 DF 7 4 28 39 F P 19.39 0.000 14.79 0.000 S = 0.5307 R-Sq = 87.44% R-Sq(adj) = 82.51% 478 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE 11.44 Warehouse Shopping, continued Refer to Exercise 11.43. The printout that follows provides the average costs of the selected items for the k 5 stores. Store Albertsons Food-4-Less Ralphs Stater Bros WinCo Mean 4.04125 2.74125 3.30375 3.05500 2.08000 a. What is the appropriate value of q.05(k, df ) for test- ing for differences among stores? SE? b. What is the value of v q.05(k, df )M b c. Use Tukey’s pairwise comparison test among stores used to determine which stores differ signiﬁcantly in average prices of the selected items. THE a b FACTORIAL EXPERIMENT: A TWO-WAY CLASSIFICATION 11.9 Suppose the manager of a manufacturing plant suspects that the output (in number of units produced per shift) of a production line depends on two factors: • Which of two supervisors is in charge of the line • Which of three shifts—day, swing, or night—is being measured That is, the manager is interested in two factors: “supervisor” at two levels and “shift” at three levels. Can you use a randomized block design, designating one of the two factors as a block factor? In order to do this, you would need to assume that the effect of the two supervisors is the same, regardless of which shift you are considering. This may not be the case; maybe the ﬁrst supervisor is most effective in the morning, and the second is more effective at night. You cannot generalize and say that one supervisor is better than the other or that the output of one particular shift is best. You need to investigate not only the average output for the two supervisors and the average output for the three shifts, but also the interaction or relationship between the two factors. Consider two different examples that show the effect of interaction on the responses in this situation. Suppose that the two supervisors are each observed on three randomly selected days for each of the three different shifts. The average outputs for the three shifts are shown in Table 11.4 for each of the supervisors. Look at the relationship between the two factors in the line chart for these means, shown in Figure 11.11. Notice that supervisor 2 always produces a higher output, regardless of the shift. The two factors behave independently; that is, the output is always about 100 units higher for supervisor 2, no matter which shift you look at. EXAMPLE 11.11 TABLE 11.4 ● Average Outputs for Two Supervisors on Three Shifts Shift Supervisor Day Swing Night 1 2 487 602 498 602 550 637 11.9 THE a b FACTORIAL EXPERIMENT: A TWO-WAY CLASSIFICATION ❍ 479 FI GUR E 1 1. 11 Interaction plot for means in Table 11.4 ● Interaction Plot (data means) for Response Supervisor 1 2 650 625 600 575 550 525 500 n a e M Day Swing Shift Night Now consider another set of data for the same situation, shown in Table 11.5. There is a deﬁnite difference in the results, depending on which shift you look at, and the interaction can be seen in the crossed lines of the chart in Figure 11.12. TABLE 11.5 ● Average Outputs for Two Supervisors on Three Shifts Shift Supervisor Day Swing Night 1 2 602 487 498 602 450 657 FI GUR E 1 1. 12 Interaction plot for means in Table 11.5 ● Interaction Plot (data means) for Response Supervisor 1 2 650 600 n a e M 550 500 450 Day Swing Shift Night 480 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE When the effect of one factor on the response changes, depending on the level at which the other factor is measured, the two factors are said to interact. This situation is an example of a factorial experiment in which there are a total of 2 3 possible combinations of the levels for the two factors. These 2 3 6 combinations form the treatments, and the experiment is called a 2 3 factorial experiment. This type of experiment can actually be used to investigate the effects of three or more factors on a response and to explore the interactions between the factors. However, we conﬁne our discussion to two factors and their interaction. When you compare treatment means for a factorial experiment (or for any other experiment), you will need more than one observation per treatment. For example, if you obtain two observations for each of the factor combinations of a complete factorial experiment, you have two replications of the experiment. In the next section on the analysis of variance for a factorial experiment, you can assume that each treatment or combination of factor levels is replicated the same number of times r. THE ANALYSIS OF VARIANCE FOR AN a b FACTORIAL EXPERIMENT 11.10 An analysis of variance for a two-factor factorial experiment replicated r times follows the same pattern as the previous designs. If the letters A and B are used to identify the two factors, the total variation in the experiment Total SS S(x x)2 Sx2 CM is partitioned into four parts in such a w
ay that Total SS SSA SSB SS(AB) SSE where • SSA (sum of squares for factor A) measures the variation among the factor A means. • SSB (sum of squares for factor B) measures the variation among the factor B means. • SS(AB) (sum of squares for interaction) measures the variation among the different combinations of factor levels. • SSE (sum of squares for error) measures the variation of the differences among the observations within each combination of factor levels—the experimental error. Sums of squares SSA and SSB are often called the main effect sums of squares, to distinguish them from the interaction sum of squares. Although you can simplify your work by using a computer program to calculate these sums of squares, the calculational formulas are given next. You can assume that there are: a levels of factor A b levels of factor B r replications of each of the ab factor combinations • • • • A total of n abr observations 11.10 THE ANALYSIS OF VARIANCE FOR AN a b FACTORIAL EXPERIMENT ❍ 481 CALCULATING THE SUMS OF SQUARES FOR A TWO-FACTOR FACTORIAL EXPERIMENT 2 CM G n Total SS Sx2 CM 2 2 A B j CM SSA S i CM SSB S a r b r )2 SS(AB) S(AB ij CM SSA SSB r where G Sum of all n abr observations Ai Total of all observations at the ith level of factor A, Bj Total of all observations at the jth level of factor B, i 1, 2, . . . , a j 1, 2, . . . , b (AB)ij Total of the r observations at the ith level of factor A and the jth level of factor B Each of the ﬁve sources of variation, when divided by the appropriate degrees of freedom, provides an estimate of the variation in the experiment. These estimates are called mean squares—MS SS/df—and are displayed along with their respective sums of squares and df in the analysis of variance (or ANOVA) table. ANOVA TABLE FOR r REPLICATIONS OF A TWO-FACTOR FACTORIAL EXPERIMENT: FACTOR A AT a LEVELS AND FACTOR B AT b LEVELS Source df SS MS A B a 1 b 1 SSA SSB SA S MSA 1 a SB S MSB AB (a 1)(b 1) SS(AB) B) S( A M (A ) B SS MS(AB) E S M 1) (b 1) (a Error ab (r 1) SSE S E S MSE 1) r ab( Total abr 1 Total SS Finally, the equality of means for various levels of the factor combinations (the interaction effect) and for the levels of both main effects, A and B, can be tested using the ANOVA F-tests, as shown next. 482 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE TESTS FOR A FACTORIAL EXPERIMENT • For interaction: 1. Null hypothesis: H0 : Factors A and B do not interact 2. Alternative hypothesis: Ha : Factors A and B interact 3. Test statistic: F MS(AB)/MSE, where F is based on df1 (a 1)(b 1) and df2 ab(r 1) 4. Rejection region: Reject H0 when F Fa, where Fa lies in the upper tail of the F distribution (see the ﬁgure), or when the p-value a • For main effects, factor A: 1. Null hypothesis: H0 : There are no differences among the factor A means 2. Alternative hypothesis: Ha : At least two of the factor a means differ 3. Test statistic: F MSA/MSE, where F is based on df1 (a 1) and df2 ab(r 1) 4. Rejection region: Reject H0 when F Fa (see the ﬁgure) or when the p-value a • For main effects, factor B: 1. Null hypothesis: H0 : There are no differences among the factor B means 2. Alternative hypothesis: Ha : At least two of the factor B means differ 3. Test statistic: F MSB/MSE, where F is based on df1 (b 1) and df2 ab(r 1) 4. Rejection region: Reject H0 when F Fa (see the ﬁgure) or when the p-value a f(F) 0 α Fα F EXAMPLE 11.12 Table 11.6 shows the original data used to generate Table 11.5 in Example 11.11. That is, the two supervisors were each observed on three randomly selected days for each of the three different shifts, and the production outputs were recorded. Analyze these data using the appropriate analysis of variance procedure. TABLE 11.6 ● Outputs for Two Supervisors on Three Shifts 11.10 THE ANALYSIS OF VARIANCE FOR AN a b FACTORIAL EXPERIMENT ❍ 483 Shift Supervisor Day Swing Night 1 2 571 610 625 480 516 465 480 474 540 625 600 581 470 430 450 630 680 661 Solution The computer output in Figure 11.13 was generated using the two-way analysis of variance procedure in the MINITAB software package. You can verify the quantities in the ANOVA table using the calculational formulas presented earlier, or you may choose just to use the results and interpret their meaning. FI GUR E 1 1. 13 MINITAB output for Example 11.12 ● Two-way ANOVA: Output versus Supervisor, Shift Source DF SS MS F P Supervisor 1 19208 19208.0 26.68 0.000 Shift 2 247 123.5 0.17 0.844 Interaction 2 81127 40563.5 56.34 0.000 Error 12 8640 720.0 Total 17 109222 S = 26.83 R-Sq = 92.09% R-Sq(adj) = 88.79% Individual 95% CIs For Mean Based on Pooled StDev Supervisor Mean ----+---------+---------+---------+----1 516.667 (-------------) 2 582.000 (--------------) ----+---------+---------+---------+---- 510 540 570 600 Individual 95% CIs For Mean Based on Pooled StDev Shift Mean ---+---------+---------+---------+-------Day 544.5 (------------------------------) Swing 550.0 (------------------------------) Night 553.5 (---------------*---------------) ---+---------+---------+---------+------- 525 540 555 570 If the interaction is not signiﬁcant, test each of the factors individually. At this point, you have undoubtedly discovered the familiar pattern in testing the signiﬁcance of the various experimental factors with the F statistic and its p-value. The small p-value (P .000) in the row marked “Supervisor” means that there is sufficient evidence to declare a difference in the mean levels for factor A—that is, a difference in mean outputs per supervisor. This fact is visually apparent in the nonoverlapping conﬁdence intervals for the supervisor means shown in the printout. But this is overshadowed by the fact that there is strong evidence (P .000) of an interaction between factors A and B. This means that the average output for a given shift depends on the supervisor on duty. You saw this effect clearly in Figure 11.11. The three largest mean outputs occur when supervisor 1 is on the day shift and when supervisor 2 is on either the swing or night shift. As a practical result, the manager should schedule supervisor 1 for the day shift and supervisor 2 for the night shift. 484 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE If the interaction effect is signiﬁcant, the differences in the treatment means can be further studied, not by comparing the means for factor A or B individually but rather by looking at comparisons for the 2 3 (AB) factor-level combinations. If the interaction effect is not signiﬁcant, then the signiﬁcance of the main effect means should be investigated, ﬁrst with the overall F-test and next with Tukey’s method for paired comparisons and/or speciﬁc conﬁdence intervals. Remember that these analysis of variance procedures always use s2 MSE as the best estimator of s 2 with degrees of freedom equal to df ab(r 1). For example, using Tukey’s yardstick to compare the average outputs for the two supervisors on each of the three shifts, you could calculate v q.05(6, 12) s 4.75 r 0 72 73.59 3 Since all three pairs of means—602 and 487 on the day shift, 498 and 602 on the swing shift, and 450 and 657 on the night shift—differ by more than v, our practical conclusions have been conﬁrmed statistically. 11.10 EXERCISES BASIC TECHNIQUES 11.45 Suppose you were to conduct a two-factor factorial experiment, factor A at four levels and factor B at ﬁve levels, with three replications per treatment. a. How many treatments are involved in the experiment? b. How many observations are involved? c. List the sources of variation and their respective degrees of freedom. 11.46 The analysis of variance table for a 3 4 factorial experiment, with factor A at three levels and factor B at four levels, and with two observations per treatment, is shown here: Source Total df 2 3 6 12 23 SS MS F 5.3 9.1 24.5 43.7 a. Fill in the missing items in the table. b. Do the data provide sufficient evidence to indicate that factors A and B interact? Test using a .05. What are the practical implications of your answer? c. Do the data provide sufficient evidence to indicate that factors A and B affect the response variable x? Explain. 11.47 Refer to Exercise 11.46. The means of two of the factor-level combinations—say, A1B1 and A2B1— are x1 8.3 and x2 6.3, respectively. Find a 95% conﬁdence interval for the difference between the two corresponding population means. 11.48 The table gives data for a 3 3 factorial experiment, with two replications per EX1148 treatment: Levels of Factor A Levels of Factor B 1 2 3 1 2 3 5, 7 8, 7 14, 11 9, 7 12, 13 8, 9 4, 6 7, 10 12, 15 a. Perform an analysis of variance for the data, and present the results in an analysis of variance table. b. What do we mean when we say that factors A and B interact? c. Do the data provide sufficient evidence to indicate interaction between factors A and B? Test using a .05. d. Find the approximate p-value for the test in part c. e. What are the practical implications of your results in part c? Explain your results using a line graph similar to the one in Figure 11.11. 11.49 2 2 Factorial The table gives data for a 2 2 factorial experiment, with four EX1149 replications per treatment. 11.10 THE ANALYSIS OF VARIANCE FOR AN a b FACTORIAL EXPERIMENT ❍ 485 Levels of Factor A APPLICATIONS Levels of Factor B 1 2 1 2 2.1, 2.7, 2.4, 2.5 3.1, 3.6, 3.4, 3.9 3.7, 3.2, 3.0, 3.5 2.9, 2.7, 2.2, 2.5 a. The accompanying graph was generated by MINITAB. Verify that the four points that connect the two lines are the means of the four observations within each factor-level combination. What does the graph tell you about the interaction between factors A and B? MINITAB interaction plot for Exercise 11.49 Interaction Plot (data means) for Response Factor A 1 2 3.50 3.25 3.00 n a e M 2.75 2.50 1 Factor B 2 b. Use the MINITAB output to test for a signiﬁcant interaction between A and B. Does this conﬁrm your conclusions in part a? MINITAB output for Exercise 11.49 Two-way ANOVA: Response versus Factor A, Factor B Source DF SS MS F P Factor
A 1 0.0000 0.00000 0.00 1.000 Factor B 1 0.0900 0.09000 1.00 0.338 Interaction 1 3.4225 3.42250 37.85 0.000 Error 12 1.0850 0.09042 Total 15 4.5975 S = 0.3007 R-Sq = 76.40% R-Sq(adj) = 70.50% c. Considering your results in part b, how can you explain the fact that neither of the main effects is signiﬁcant? d. If a signiﬁcant interaction is found, is it necessary to test for signiﬁcant main effect differences? Explain. e. Write a short paragraph summarizing the results of this experiment. 11.50 Demand for Diamonds A chain of jewelry stores conducted an experiment to EX1150 investigate the effect of price and location on the demand for its diamonds. Six small-town stores were selected for the study, as well as six stores located in large suburban malls. Two stores in each of these locations were assigned to each of three item percentage markups. The percentage gain (or loss) in sales for each store was recorded at the end of 1 month. The data are shown in the accompanying table. Markup Location Small towns Suburban malls 1 10 4 14 18 2 3 7 8 3 3 10 24 4 3 a. Do the data provide sufficient evidence to indicate an interaction between markup and location? Test using a .05. b. What are the practical implications of your test in part a? c. Draw a line graph similar to Figure 11.11 to help visualize the results of this experiment. Summarize the results. d. Find a 95% conﬁdence interval for the difference in mean change in sales for stores in small towns versus those in suburban malls if the stores are using price markup 3. 11.51 Terrain Visualization A study was conducted to determine the effect of two factors on terrain visualization training for soldiers.4 During the training programs, participants viewed contour maps of various terrains and then were permitted to view a computer reconstruction of the terrain as it would appear from a speciﬁed angle. The two factors investigated in the experiment were the participants’ spatial abilities (abilities to visualize in three dimensions) and the viewing procedures (active or passive). Active participation permitted participants to view the computer-generated reconstructions of the terrain from any and all angles. Passive participation gave the participants a set of preselected reconstructions of the terrain. Participants were tested according to spatial ability, and from the test scores 20 were categorized as possessing high spatial ability, 20 medium, and 20 low. Then 10 participants within each of these groups were assigned to each of the two training modes, active or passive. The 486 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE accompanying tables are the ANOVA table computed the researchers and the table of the treatment means. MINITAB output for Exercise 11.52 Two-way ANOVA: Cost versus City, Distance Source df MS Error df F p Main effects: Training condition Ability Interaction: Training condition Ability Within cells 1 2 2 54 103.7009 760.5889 124.9905 28.3015 54 54 54 3.66 26.87 .0610 .0005 4.42 .0167 Source DF SS MS F P City 2 201.33 100.667 3.06 0.084 Distance 3 1873.33 624.444 18.97 0.000 Interaction 6 303.67 50.611 1.54 0.247 Error 12 395.00 32.917 Total 23 2773.33 S = 5.737 R-Sq = 85.76% R-Sq(adj) = 72.70% Individual 95% CIs For Mean Based on Pooled StDev Distance Mean ------+---------+---------+---------+--1 32.1667 (-----+------) 2 19.1667 (-----+-----) 3 11.8333 (------+-----) 4 9.5000 (-----+-----) ------+---------+---------+---------+-- 10 20 30 40 Training Condition Spatial Ability Active Passive High Medium Low Note: Maximum score 36. 17.895 5.031 1.728 9.508 5.648 1.610 a. Explain how the authors arrived at the degrees of freedom shown in the ANOVA table. b. Are the F-values correct? c. Interpret the test results. What are their practical implications? d. Use Table 6 in Appendix I to approximate the pvalues for the F statistics shown in the ANOVA table. Source: H.F. Barsam and Z.M. Simutis, “Computer-Based Graphics for Terrain Visualization Training,” Human Factors, no. 26, 1984. Copyright 1984 by the Human Factors Society, Inc. Reproduced by permission. 11.52 The Cost of Flying In an attempt to determine what factors affect airfares, a EX1152 researcher recorded a weighted average of the costs per mile for two airports in each of three major U.S. cities for each of four different travel distances.5 The results are shown in the table. City Distance 300 miles 301–750 miles 751–1500 miles 1500 miles New York Houston Chicago 40, 48 19, 26 10, 14 9, 10 20, 26 15, 17 10, 13 8, 11 19, 40 14, 24 9, 15 7, 12 Use the MINITAB output to analyze the experiment with the appropriate method. Identify the two factors, and investigate any possible effect due to their interaction or the main effects. What are the practical implications of this experiment? Explain your conclusions in the form of a report. MINITAB plots for Exercise 11.52 Interaction Plot (data means) for Cost City Chicago Houston NY 1 2 3 Distance 4 Main Effects Plot (data means) for Cost City Distance n a e M t s o C 45 40 35 30 25 20 15 10 35 30 25 20 15 10 Chicago Houston NY 1 2 3 4 11.53 Fourth-Grade Test Scores A local school board was interested in comparing test EX1153 scores on a standarized reading test for fourth-grade students in their district. They selected a random sample of ﬁve male and ﬁve female fourth grade students at each of four different elementary schools in the district and recorded the test scores. The results are shown in the table below. 11.11 REVISITING THE ANALYSIS OF VARIANCE ASSUMPTIONS ❍ 487 Gender School 1 School 2 School 3 School 4 a. What are the experimental units in this exper- Male Female 631 566 620 542 560 669 644 600 610 559 642 710 649 596 660 722 769 723 649 766 651 611 755 693 620 709 545 657 722 711 350 565 543 509 494 505 498 474 470 463 a. What type of experimental design is this? What are the experimental units? What are the factors and levels of interest to the school board? b. Perform the appropriate analysis of variance for this experiment. iment? b. What are the two factors considered in the experi- ment? c. What are the levels of each factor? d. How many treatments are there in the experiment? e. What type of experimental design has been used? 11.55 Management Training, continued Refer to Exercise 11.54. The data for this EX1155 experiment are shown in the table. Situation (B) Trained Not Trained Totals Training (A) c. Do the data indicate that effect of gender on the Standard average test score is different depending on the student’s school? Test the appropriate hypothesis using a .05. d. Plot the average scores using an interaction plot. How would you describe the effect of gender and school on the average test scores? e. Do the data indicate that either of the main effects is signiﬁcant? If the main effect is signiﬁcant, use Tukey’s method of paired comparisons to examine the differences in detail. Use a .01. 11.54 Management Training An experiment was conducted to investigate the effect of management training on the decision-making abilities of supervisors in a large corporation. Sixteen supervisors were selected, and eight were randomly chosen to receive managerial training. Four trained and four untrained supervisors were then randomly selected to function in a situation in which a standard problem arose. The other eight supervisors were presented with an emergency situation in which standard procedures could not be used. The response was a management behavior rating for each supervisor as assessed by a rating scheme devised by the experimenter. Emergency 519 473 85 91 80 78 76 67 82 71 53 49 38 45 40 52 46 39 Totals 630 362 992 a. Construct the ANOVA table for this experiment. b. Is there a signiﬁcant interaction between the presence or absence of training and the type of decision-making situation? Test at the 5% level of signiﬁcance. c. Do the data indicate a signiﬁcant difference in behavior ratings for the two types of situations at the 5% level of signiﬁcance? d. Do behavior ratings differ signiﬁcantly for the two types of training categories at the 5% level of signiﬁcance. e. Plot the average scores using an interaction plot. How would you describe the effect of training and emergency situation on the decision-making abilities of the supervisors? REVISITING THE ANALYSIS OF VARIANCE ASSUMPTIONS 11.11 In Section 11.3, you learned that the assumptions and test procedures for the analysis of variance are similar to those required for the t and F-tests in Chapter 10—namely, that observations within a treatment group must be normally distributed with common variance s 2. You also learned that the analysis of variance procedures are fairly 488 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE robust when the sample sizes are equal and the data are fairly mound-shaped. If this is the case, one way to protect yourself from inaccurate conclusions is to try when possible to select samples of equal sizes! There are some quick and simple ways to check the data for violation of assumptions. Look ﬁrst at the type of response variable you are measuring. You might immediately see a problem with either the normality or common variance assumption. It may be that the data you have collected cannot be measured quantitatively. For example, many responses, such as product preferences, can be ranked only as “A is better than B” or “C is the least preferable.” Data that are qualitative cannot have a normal distribution. If the response variable is discrete and can assume only three values— say, 0, 1, or 2—then it is again unreasonable to assume that the response variable is normally distributed. Suppose that the response variable is binomial—say, the proportion p of people who favor a particular type of investment. Although binomial data can be approximately mound-shaped under certain conditions, they violate the equal variance assumption. The variance of a sample proportion is p) q p(1 s 2 p n n so that the variance changes depending on the value of p. As the treatment means change
, the value of p changes and so does the variance s 2. A similar situation occurs when the response variable is a Poisson random variable—say, the number of industrial accidents per month in a manufacturing plant. Since the variance of a Poisson random variable is s 2 m, the variance changes exactly as the treatment mean changes. If you cannot see any ﬂagrant violations in the type of data being measured, look at the range of the data within each treatment group. If these ranges are nearly the same, then the common variance assumption is probably reasonable. To check for normality, you might make a quick dotplot or stem and leaf plot for a particular treatment group. However, quite often you do not have enough measurements to obtain a reasonable plot. If you are using a computer program to analyze your experiment, there are some valuable diagnostic tools you can use. These procedures are too complicated to be performed using hand calculations, but they are easy to use when the computer does all the work! Residual Plots In the analysis of variance, the total variation in the data is partitioned into several parts, depending on the factors identiﬁed as important to the researcher. Once the effects of these sources of variation have been removed, the “leftover” variability in each observation is called the residual for that data point. These residuals represent experimental error, the basic variability in the experiment, and should have an approximately normal distribution with a mean of 0 and the same variation for each treatment group. Most computer packages will provide options for plotting these residuals: • The normal probability plot of residuals is a graph that plots the residuals for each observation against the expected value of that residual had it come from a normal distribution. If the residuals are approximately normal, the plot will closely resemble a straight line, sloping upward to the right. EXAMPLE 11.13 11.11 REVISITING THE ANALYSIS OF VARIANCE ASSUMPTIONS ❍ 489 • The plot of residuals versus ﬁt or residuals versus variables is a graph that plots the residuals against the expected value of that observation using the experimental design we have used. If no assumptions have been violated and there are no “leftover” sources of variation other than experimental error, this plot should show a random scatter of points around the horizontal “zero error line” for each treatment group, with approximately the same vertical spread. The data from Example 11.4 involving the attention spans of three groups of elementary students were analyzed using MINITAB. The graphs in Figure 11.14, generated by MINITAB, are the normal probability plot and the residuals versus ﬁt plot for this experiment. Look at the straight-line pattern in the normal probability plot, which indicates a normal distribution in the residuals. In the other plot, the residuals are plotted against the estimated expected values, which are the sample averages for each of the three treatments in the completely randomized design. The random scatter around the horizontal “zero error line” and the constant spread indicate no violations in the constant variance assumption. FI GUR E 1 1. 14 MINITAB diagnostic plots for Example 11.13 ● 99 95 90 80 70 60 50 40 30 20 10 5 1 t n e c r e P Normal Probability Plot of the Residuals (response is Span) Residuals versus the Fitted Values (response is Span.0 2.5 0.0 Residual 2.5 5.0 9 10 11 12 Fitted Value 13 14 2 3 EXAMPLE 11.14 A company plans to promote a new product by using one of three advertising campaigns. To investigate the extent of product recognition from these three campaigns, 15 market areas were selected and ﬁve were randomly assigned to each advertising plan. At the end of the ad campaigns, random samples of 400 adults were selected in each area and the proportions who were familiar with the new product were recorded, as in Table 11.7. Have any of the analysis of variance assumptions been violated in this experiment? TABLE 11.7 ● Proportions of Product Recognition for Three Advertising Campaigns Campaign 1 Campaign 2 Campaign 3 .33 .29 .21 .32 .25 .28 .41 .34 .39 .27 .21 .30 .26 .33 .31 Solution The experiment is designed as a completely randomized design, but the response variable is a binomial sample proportion. This indicates that both the normality and the common variance assumptions might be invalid. Look at the normal probability plot of the residuals and the plot of residuals versus ﬁt generated as an option in the MINITAB analysis of variance procedure and shown in Figure 11.15. The 490 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE curved pattern in the normal probability plot indicates that the residuals do not have a normal distribution. In the residual versus ﬁt plot, you can see three vertical lines of residuals, one for each of the three ad campaigns. Notice that two of the lines (campaigns 1 and 3) are close together and have similar spread. However, the third line (campaign 2) is farther to the right, which indicates a larger sample proportion and consequently a larger variance in this group. Both analysis of variance assumptions are suspect in this experiment. F IG URE 11. 15 MINITAB diagnostic plots for Example 11.14 ● t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 Normal Probability Plot of the Residuals (response is Proportion) Residuals versus the Fitted Values (response is Proportion.08 0.06 0.04 0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.05 0.00 Residual 0.05 0.10 0.15 0.28 0.29 0.30 0.31 Fitted Value 0.32 0.33 0.34 What can you do when the ANOVA assumptions are not satisﬁed? The constant variance assumption can often be remedied by transforming the response measurements. That is, instead of using the original measurements, you might use their square roots, logarithms, or some other function of the response. Transformations that tend to stabilize the variance of the response also tend to make their distributions more nearly normal. When nothing can be done to even approximately satisfy the ANOVA assumptions or if the data are rankings, you should use nonparametric testing and estimation procedures, presented in Chapter 15. We have mentioned these procedures before; they are almost as powerful in detecting treatment differences as the tests presented in this chapter when the data are normally distributed. When the parametric ANOVA assumptions are violated, the nonparametric tests are generally more powerful. A BRIEF SUMMARY 11.12 We presented three different experimental designs in this chapter, each of which can be analyzed using the analysis of variance procedure. The objective of the analysis of variance is to detect differences in the mean responses for experimental units that have received different treatments—that is, different combinations of the experimental factor levels. Once an overall test of the differences is performed, the nature of these differences (if any exist) can be explored using methods of paired comparisons and/or interval estimation procedures. The three designs presented in this chapter represent only a brief introduction to the subject of analyzing designed experiments. Designs are available for experiments that involve several design variables, as well as more than two treatment factors and other more complex designs. Remember that design variables are factors whose effect you want to control and hence remove from experimental error, whereas treatment CHAPTER REVIEW ❍ 491 variables are factors whose effect you want to investigate. If your experiment is properly designed, you will be able to analyze it using the analysis of variance. Experiments in which the levels of a variable are measured experimentally rather than controlled or preselected ahead of time may be analyzed using linear or multiple regression analysis—the subject of Chapters 12 and 13. CHAPTER REVIEW Key Concepts and Formulas I. Experimental Designs III. Interpreting an Analysis of Variance 1. Experimental units, factors, levels, treatments, response variables. 2. Assumptions: Observations within each treat- ment group must be normally distributed with a common variance s 2. 3. One-way classiﬁcation—completely randomized design: Independent random samples are selected from each of k populations. 4. Two-way classiﬁcation—randomized block design: k treatments are compared within b relatively homogeneous groups of experimental units called blocks. 5. Two-way classiﬁcation—a b factorial ex- periment: Two factors, A and B, are compared at several levels. Each factor–level combination is replicated r times to allow for the investigation of an interaction between the two factors. II. Analysis of Variance 1. The total variation in the experiment is divided into variation (sums of squares) explained by the various experimental factors and variation due to experimental error (unexplained). 2. If there is an effect due to a particular factor, its mean square (MS SS/df ) is usually large and F MS(factor)/MSE is large. 3. Test statistics for the various experimental factors are based on F statistics, with appropriate degrees of freedom (df2 Error degrees of freedom). 1. For the completely randomized and randomized block design, each factor is tested for signiﬁcance. 2. For the factorial experiment, ﬁrst test for a signiﬁcant interaction. If the interaction is signiﬁcant, main effects need not be tested. The nature of the differences in the factor–level combinations should be further examined. 3. If a signiﬁcant difference in the population means is found, Tukey’s method of pairwise comparisons or a similar method can be used to further identify the nature of the differences. 4. If you have a special interest in one population mean or the difference between two population means, you can use a conﬁdence interval estimate. (For a randomized block design, conﬁdence intervals do not provide unbiased estimates for single population means.) IV. Checking the Analysis of Variance Assumptions 1. To check for normal
ity, use the normal probability plot for the residuals. The residuals should exhibit a straight-line pattern, increasing upwards toward the right. 2. To check for equality of variance, use the residuals versus ﬁt plot. The plot should exhibit a random scatter, with the same vertical spread around the horizontal “zero error line.” 492 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE Analysis of Variance Procedures The statistical procedures used to perform the analysis of variance for the three different experimental designs in this chapter are found in a MINITAB submenu by choosing Stat ANOVA. You will see choices for One-way, One-way (Unstacked), and Two-way that will generate Dialog boxes used for the completely randomized, randomized block, and factorial designs, respectively. You must properly store the data and then choose the columns corresponding to the necessary factors in the experiment. We will display some of the Dialog boxes and Session window outputs for the examples in this chapter, beginning with a one-way classiﬁcation—the completely randomized breakfast study in Example 11.4. First, enter the 15 recorded attention spans in column C1 of a MINITAB worksheet and name them “Span.” Next, enter the integers 1, 2, and 3 into a second column C2 to identify the meal assignment (treatment) for each observation. You can let MINITAB set this pattern for you using Calc Make Patterned Data Simple Set of Numbers and entering the appropriate numbers, as shown in Figure 11.16. Then use Stat ANOVA One-way to generate the Dialog box in Figure 11.17.† You must select the column of observations for the “Response” box and the column of treatment indicators for the “Factor” box. Then you have several options. Under Comparisons, you can select “Tukey’s family error rate” (which has a default level of 5%) to obtain paired comparisons output. Under Graphs, you can select individual value plots and/or box plots to compare the three meal assignments, and you can generate residual plots (use “Normal plot of residuals” and/or “Residuals versus ﬁts”) to verify the validity of the ANOVA assumptions. Click OK from the main dialog box to obtain the output in Figure 11.3 in the text. The Stat ANOVA Two-way command can be used for both the randomized block and the factorial designs. You must ﬁrst enter all of the observations into a single column and then integers or descriptive names to indicate either of these cases: • The block and treatment for each of the measurements in a randomized block design • The levels of factors A and B for the factorial experiment. MINITAB will recognize a number of replications within each factor-level combination in the factorial experiment and will break out the sum of squares for interaction (as long as you do not check the box “Fit additive model”). Since these two designs involve the same sequence of commands, we will use the data from Example 11.12 to generate the analysis of variance for the factorial experiment. The data are entered into the worksheet in Figure 11.18. See if you can use the Calc Make Patterned Data Simple Set of Numbers to enter the data into columns C2–C3. Once the data have been entered, use Stat ANOVA Two-way to generate the Dialog box in Figure 11.19. Choose “Output” for the “Response” box, and “Supervisor” and “Shift” for the “Row factor” and “Column factor,” respectively. You may choose to display the main effect means along with 95% conﬁdence intervals by checking “Display means,” and you may select residual plots if you wish. Click OK to obtain the ANOVA printout in Figure 11.13. †If you had entered each of the three samples into separate columns, the proper command would have been Stat ANOVA One-way (Unstacked). FI GUR E 1 1. 16 ● MY MINITAB ❍ 493 FI GUR E 1 1. 17 ● 494 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE F IG URE 11. 18 ● Data Display Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Output Supervisor 571 610 625 480 516 465 480 474 540 625 600 581 470 430 450 630 680 661 Shift IG URE 11. 19 ● Since the interaction between supervisors and shifts is highly signiﬁcant, you may want to explore the nature of this interaction by plotting the average output for each supervisor at each of the three shifts. Use Stat ANOVA Interactions Plot and choose the appropriate response and factor variables. The plot is generated by MINITAB and shown in Figure 11.20. You can see the strong difference in the behaviors of the mean outputs for the two supervisors, indicating a strong interaction between the two factors. FI GUR E 1 1. 20 ● SUPPLEMENTARY EXERCISES ❍ 495 Supplementary Exercises 11.56 Reaction Times vs. Stimuli Twentyseven people participated in an experiment to EX1156 MINITAB output for Exercise 11.56 One-way ANOVA: Time versus Stimulus compare the effects of ﬁve different stimuli on reaction time. The experiment was run using a completely randomized design, and, regardless of the results of the analysis of variance, the experimenters wanted to compare stimuli A and D. The results of the experiment are given here. Use the MINITAB printout to complete the exercise. Source DF SS MS F P Stimulus 4 1.2118 0.3030 11.67 0.000 Error 22 0.5711 0.0260 Total 26 1.7830 S = 0.1611 R-Sq = 67.97% R-Sq(adj) = 62.14% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -------+---------+---------+---------+-A 4 0.6250 0.1258 (------------) B 7 0.6714 0.1496 (--------) C 6 1.0667 0.1966 (---------) D 5 0.9200 0.1483 (----------) E 5 0.4800 0.1643 (----------) -------+---------+---------+---------+-- Pooled StDev = 0.1611 0.50 0.75 1.00 1.25 Stimulus Reaction Time (sec) Total Mean A B C D E .8 .7 1.2 1.0 .6 .6 .8 1.0 .9 .4 .6 .5 .9 .9 .4 .5 .5 1.2 1.1 .7 .6 1.3 .7 .3 .7 .9 .8 2.5 4.7 6.4 4.6 2.4 .625 .671 1.067 .920 .480 a. Conduct an analysis of variance and test for a difference in the mean reaction times due to the ﬁve stimuli. b. Compare stimuli A and D to see if there is a differ- ence in mean reaction times. 496 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE 11.57 Refer to Exercise 11.56. Use this MINITAB output to identify the differences in the treatment means. MINITAB output for Exercise 11.57 Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Stimulus Individual confidence level = 99.29% Stimulus = A subtracted from: Stimulus Lower Center Upper --------+---------+---------+---------+B -0.2535 0.0464 0.3463 (----------) C 0.1328 0.4417 0.7505 (----------) D -0.0260 0.2950 0.6160 (----------) E -0.4660 -0.1450 0.1760 (----------) --------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Stimulus = B subtracted from: Stimulus Lower Center Upper --------+---------+---------+---------+C 0.1290 0.3952 0.6615 (----------) D -0.0316 0.2486 0.5288 (----------) E -0.4716 -0.1914 0.0888 (----------) --------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Stimulus = C subtracted from: Stimulus Lower Center Upper --------+---------+---------+---------+D -0.4364 -0.1467 0.1431 (----------) E -0.8764 -0.5867 -0.2969 (----------) --------+---------+---------+---------+ -0.50 0.00 0.50 1.00 Stimulus = D subtracted from: Stimulus Lower Center Upper --------+---------+---------+---------+E -0.7426 -0.4400 -0.1374 (----------) --------+---------+---------+---------+ -0.50 0.00 0.50 1.00 11.58 Refer to Exercise 11.56. What do the normal probability plot and the residuals versus ﬁt plot tell you about the validity of your analysis of variance results? MINITAB diagnostic plots for Exercise 11.58 Normal Probability Plot of the Residuals (response is Time) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 0.4 0.3 0.2 0.1 0.0 Residual 0.1 0.2 0.3 0.4 Residuals versus the Fitted Values (response is Time.3 0.2 0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Fitted Value 0.9 1.0 1.1 11.59 Reaction Times II The experiment in Exercise 11.56 might have been conducted EX1159 more effectively using a randomized block design with people as blocks, since you would expect mean reaction time to vary from one person to another. Hence, four people were used in a new experiment, and each person was subjected to each of the ﬁve stimuli in a random order. The reaction times (in seconds) are listed here: Stimulus Subject 1 2 3 4 A .7 .6 .9 .6 B .8 .6 1.0 .8 C 1.0 1.1 1.2 .9 D 1.0 1.0 1.1 1.0 E .5 .6 .6 .4 MINITAB output for Exercise 11.59 Two-way ANOVA: Time versus Subject, Stimulus Source DF SS MS F P Subject 3 0.140 0.046667 6.59 0.007 Stimulus 4 0.787 0.196750 27.78 0.000 Error 12 0.085 0.007083 Total 19 1.012 S = 0.08416 R-Sq = 91.60% R-Sq(adj) = 86.70% Individual 95% CIs For Mean Based on Pooled StDev Stimulus Mean ---------+---------+---------+---------+A 0.700 (--------) B 0.800 (--------) C 1.050 (-------) D 1.025 (-------) E 0.525 (-------) ---------+---------+---------+---------+ 0.60 0.80 1.00 1.20 a. Use the MINITAB printout to analyze the data and test for differences in treatment means. b. Use Tukey’s method of paired comparisons to identify the signiﬁcant pairwise differences in the stimuli. c. Does it appear that blocking was effective in this experiment? EX1160 11.60 Heart Rate and Exercise An experiment was conducted to examine the effect of age on heart rate when a person is subjected to a speciﬁc amount of exercise. Ten male subjects were randomly selected from four age groups: 10–19, 20–39, 40–59, and 60–69. Each subject walked on a treadmill at a ﬁxed grade for a period of 12 minutes, and the increase in heart rate, the difference before and after exercise, was recorded (in beats per minute): 10–19 20–39 40–59 60–69 29 33 26 27 39 35 33 29 36 22 24 27 33 31 21 28 24 34 21 32 37 25 22 33 28 26 30 34 27 33 28 29 34 36 21 20 25 24 33 32 Total 309 275 295 282 Use an appropriate computer program to answer these questions: a. Do the data provide sufficient evidence to indicate a difference in mean increase in heart rate among the four age groups? Test by using a .05. b. Find a 90% conﬁdence interval for the difference in mean increase in heart rate betwee
n age groups 10–19 and 60–69. c. Find a 90% conﬁdence interval for the mean increase in heart rate for the age group 20–39. d. Approximately how many people would you need in each group if you wanted to be able to estimate a group mean correct to within two beats per minute with probability equal to .95? EX1161 11.61 Learning to Sell A company wished to study the effects of four training programs on the sales abilities of their sales personnel. Thirtytwo people were randomly divided into four groups of equal size, and each group was then subjected to one of the different sales training programs. Because there were some dropouts during the training programs due to illness, vacations, and so on, the number of trainees completing the programs varied from group to group. At the end of the training programs, each salesperson was randomly assigned a sales area from a group of sales areas that were judged to have equivalent sales potentials. The sales made by each of the four groups of salespeople during the ﬁrst week after completing the training program are listed in the table: 1 78 84 86 92 69 73 Training Program 2 99 86 90 93 94 85 97 91 3 74 87 80 83 78 4 81 63 71 65 86 79 73 70 Total 482 735 402 588 SUPPLEMENTARY EXERCISES ❍ 497 Analyze the experiment using the appropriate method. Identify the treatments or factors of interest to the researcher and investigate any signiﬁcant effects. What are the practical implications of this experiment? Write a paragraph explaining the results of your analysis. 11.62 4 2 Factorial Suppose you were to conduct a two-factor factorial experiment, factor A at four levels and factor B at two levels, with r replications per treatment. a. How many treatments are involved in the experiment? b. How many observations are involved? c. List the sources of variation and their respective degrees of freedom. 11.63 2 3 Factorial The analysis of variance table for a 2 3 factorial experiment, factor A at two levels and factor B at three levels, with ﬁve observations per treatment, is shown in the table. MS F Source df A B AB Error Total SS 1.14 2.58 .49 8.41 a. Do the data provide sufficient evidence to indicate an interaction between factors A and B? Test using a .05. What are the practical implications of your answer? b. Give the approximate p-value for the test in part a. c. Do the data provide sufficient evidence to indicate that factor A affects the response? Test using a .05. d. Do the data provide sufficient evidence to indicate that factor B affects the response? Test using a .05. 11.64 Refer to Exercise 11.63. The means of all observations, at the factor A levels A1 and A2 are x1 3.7 and x2 1.4, respectively. Find a 95% conﬁdence interval for the difference in mean response for factor levels A1 and A2. EX1165 11.65 The Whiteﬂy in California The whiteﬂy, which causes defoliation of shrubs and trees and a reduction in salable crop yields, has emerged as a pest in Southern California. In a study to determine factors that affect the life cycle of the whiteﬂy, an experiment was conducted in which whiteﬂies were placed on two different types of plants at three 498 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE different temperatures. The observation of interest was the total number of eggs laid by caged females under one of the six possible treatment combinations. Each treatment combination was run using ﬁve cages. Temperature Plant 70°F 77°F 82°F Cotton Cucumber 37 21 36 43 31 50 53 25 37 48 34 54 40 42 16 59 53 31 69 51 46 32 41 36 38 43 62 71 49 59 MINITAB output for Exercise 11.65 Two-way ANOVA: Eggs versus Plant, Temperature Source DF SS MS F P Plant 1 1512.30 1512.30 12.29 0.002 Temperature 2 487.47 243.73 1.98 0.160 Interaction 2 111.20 55.60 0.45 0.642 Error 24 2952.40 123.02 Total 29 5063.37 S = 11.09 R-Sq = 41.69% R-Sq(adj) = 29.54% a. What type of experimental design has been used? b. Do the data provide sufficient evidence to indicate that the effect of temperature on the number of eggs laid is different depending on the type of plant? Use the MINITAB printout to test the appropriate hypothesis. c. Plot the treatment means for cotton as a function of temperature. Plot the treatment means for cucumber as a function of temperature. Comment on the similarity or difference in these two plots. d. Find the mean number of eggs laid on cotton and cucumber based on 15 observations each. Calculate a 95% conﬁdence interval for the difference in the underlying population means. 11.66 Pollution from Chemical Plants Four chemical plants, producing the same EX1166 product and owned by the same company, discharge effluents into streams in the vicinity of their locations. To check on the extent of the pollution created by the effluents and to determine whether this varies from plant to plant, the company collected random samples of liquid waste, ﬁve specimens for each of the four plants. The data are shown in the table: Plant Polluting Effluents (lb/gal of waste) A B C D 1.65 1.70 1.40 2.10 1.72 1.85 1.75 1.95 1.50 1.46 1.38 1.65 1.37 2.05 1.65 1.88 1.60 1.80 1.55 2.00 a. Do the data provide sufficient evidence to indicate a difference in the mean amounts of effluents discharged by the four plants? b. If the maximum mean discharge of effluents is 1.5 lb/gal, do the data provide sufficient evidence to indicate that the limit is exceeded at plant A? c. Estimate the difference in the mean discharge of effluents between plants A and D, using a 95% conﬁdence interval. 11.67 America’s Market Basket Exer- EX1167 cise 10.40 examined an advertisement for Albertsons, a supermarket chain in the western United States. The advertiser claims that Albertsons has consistently had lower prices than four other full-service supermarkets. As part of a survey conducted by an “independent market basket price-checking company,” the average weekly total based on the prices of approximately 95 items is given for ﬁve different supermarket chains recorded during 4 consecutive weeks.6 Albertsons Ralphs Vons Alpha Beta Lucky Week 1 $254.26 240.62 Week 2 231.90 Week 3 234.13 Week 4 $256.03 255.65 255.12 261.18 $267.92 251.55 245.89 254.12 $260.71 251.80 246.77 249.45 $258.84 242.14 246.80 248.99 a. What type of design has been used in this experiment? b. Conduct an analysis of variance for the data. c. Is there sufficient evidence to indicate that there is a difference in the average weekly totals for the ﬁve supermarkets? Use a .05. d. Use Tukey’s method for paired comparisons to determine which of the means are signiﬁcantly different from each other. Use a .05. EX1168 11.68 Yield of Wheat The yields of wheat (in bushels per acre) were compared for ﬁve different varieties, A, B, C, D, and E, at six different locations. Each variety was randomly assigned to a plot at each location. The results of the experiment are shown in the accompanying table, along with a MINITAB printout of the analysis of variance. Analyze the experiment using the appropriate method. Identify the treatments or factors of interest to the researcher and investigate any effects that exist. Use the diagnostic plots to comment on the validity of the analysis of variance assumptions. What are the practical implications of this experiment? Write a paragraph explaining the results of your analysis. Location Variety A B C D E 1 35.3 30.7 38.2 34.9 32.4 2 31.0 32.2 33.4 36.1 28.9 3 32.7 31.4 33.6 35.2 29.2 MINITAB output for Exercise 11.68 4 36.8 31.7 37.1 38.3 30.7 5 37.2 35.0 37.3 40.2 33.9 6 33.1 32.7 38.2 36.0 32.1 Two-way ANOVA: Yield versus Varieties, Location Source DF SS MS F P Varieties 4 142.670 35.6675 18.61 0.000 Locations 5 68.142 13.6283 7.11 0.001 Error 20 38.303 1.9165 Total 29 249.142 S = 1.384 R-Sq = 84.62% R-Sq(adj) = 77.69% Individual 95% CIs For Mean Based on Pooled StDev Varieties Mean +---------+---------+---------+--------A 34.3500 (----------) B 32.2833 (---------) C 36.3000 (---------) D 36.7833 (----------) E 31.2000 (-----*-----) +---------+---------+---------+-------- 30.0 32.0 34.0 36.0 SUPPLEMENTARY EXERCISES ❍ 499 to assess cardiorespiratory ﬁtness levels in youth aged 12 to 19 years.7 Attaining ﬁtness standards is a common prerequisite for entry into occupations such as law enforcement, ﬁreﬁghting, and the military, as well as other jobs that involve physically demanding labor. Estimated maximum oxygen uptake (VO2max) was used to measure a person’s cardiorespiratory level. The focus of our study investigates the relationship between levels of physical activity (more than others, same as others, or less than others) and gender on VO2max. The data that follows are based on this study. Males Females Physical Activity More Same Less 50.1 47.2 49.7 50.4 41.2 39.8 41.5 38.2 45.7 44.2 46.8 44.9 37.2 39.4 38.6 37.8 40.9 41.3 39.2 40.9 36.5 35.0 37.2 35.4 MINITAB diagnostic plots for Exercise 11.68 block design? Explain. a. Is this a factorial experiment or a randomized Normal Probability Plot of the Residuals (response is Yield) 3 2 1 0 Residual 1 2 3 Residuals versus the Fitted Values (response is Yield) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 30 32 34 Fitted Value 36 38 40 EX1169 11.69 Physical Fitness Researchers Russell R. Pate and colleagues analyzed the results of the National Health and Nutrition Examination Survey b. Is there a signiﬁcant interaction between levels of physical activity and gender? Are there signiﬁcant differences between males and females? Levels of physical activity? c. If the interaction is signiﬁcant, use Tukey’s pairwise procedure to investigate differences among the six cell means. Comment on the results found using this procedure. Use a .05. 11.70 In a study of starting salaries of assistant professors,8 ﬁve male assistant professors EX1170 and ﬁve female assistant professors at each of three types of institutions granting doctoral degrees were polled and their initial starting salaries were recorded under the condition of anonymity. The results of the survey in $1000 are given in the following table. Ge
nder Public Universities Private/Independent Church-Related Males Females $57.3 57.9 56.5 76.5 62.0 47.4 56.7 69.0 63.2 65.3 $85.8 75.2 66.9 73.0 73.0 62.1 69.1 66.5 61.8 76.7 $78.9 69.3 69.7 58.2 61.2 60.4 62.1 59.8 71.9 61.6 Source: Based on “Average Salary for Men and Women Faculty by Category, Affiliation, and Academic Rank, 2005–2006.” 500 ❍ CHAPTER 11 THE ANALYSIS OF VARIANCE a. What type of design was used in collecting these data? b. Use an analysis of variance to test if there are signiﬁcant differences in gender, in type of institution, and to test for a signiﬁcant interaction of gender type of institution. c. Find a 95% conﬁdence interval estimate for the difference in starting salaries for male assistant professors and female assistant professors. Interpret this interval in terms of a gender difference in starting salaries. d. Use Tukey’s procedure to investigate differences in assistant professor salaries for the three types of institutions. Use a .01. e. Summarize the results of your analysis. 11.71 Pottery in the United Kingdom An article in Archaeometry involved an analysis EX1171 of 26 samples of Romano-British pottery, found at four different kiln sites in the United Kingdom.9 Since one site only yielded two samples, consider the samples found at the other three sites. The samples were analyzed to determine their chemical composition and the percentage of iron oxide is shown below. Llanederyn Island Thorns Ashley Rails 7.00 7.08 7.09 6.37 7.06 6.26 4.26 5.78 5.49 6.92 6.13 6.64 6.69 6.44 1.28 2.39 1.50 1.88 1.51 1.12 1.14 .92 2.74 1.64 a. What type of experimental design is this? b. Use an analysis of variance to determine if there is a difference in the average percentage of iron oxide at the three sites. Use a .01. c. If you have access to a computer program, gener- ate the diagnostic plots for this experiment. Does it appear that any of the analysis of variance assumptions have been violated? Explain. EX1172 11.72 Cell Phones How satisﬁed are you with your current mobile-phone service provider? Surveys done by Consumer Reports indicate that there is a high level of dissatisfaction among consumers, resulting in high customer turnover rates.10 The following table shows the overall satisfaction scores, based on a maximum score of 100, for four wireless providers in four different cities. Chicago Dallas Philadelphia Francisco San AT&T Wireless Cingular Wireless Sprint Verizon Wireless 63 67 60 71 66 67 68 75 61 64 60 73 64 60 61 73 a. What type of experimental design was used in this article? If the design used is a randomized block design, what are the blocks and what are the treatments? b. Conduct an analysis of variance for the data. c. Are there signiﬁcant differences in the average satisfaction scores for the four wireless providers considered here? d. Are there signiﬁcant differences in the average satisfaction scores for the four cities? 11.73 Cell Phones, continued Refer to Exercise 11.72. The diagnostic plots for this experiment are shown below. Does it appear that any of the analysis of variance assumptions have been violated? Explain. Normal Probability Plot of the Residuals (response is Score) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 Residual 1 2 3 4 Residuals versus the Fitted Values (response is Score) 60 62 64 66 68 70 Fitted Value 72 74 76 78 EX1174 11.74 Professor’s Salaries II Each year, the American Association of University Professors reports on salaries of academic professors at universities and colleges in the United States.8 The following data (in thousands of dollars), adapted from this report, are based on samples of n 10 in each of three professorial ranks, for both male and female professors. Gender Male Female Assistant Professor $63.9 63.9 64.8 68.3 67.5 56.6 57.6 53.5 64.4 62.6 $64.4 62.2 64.2 64.9 67.5 59.0 58.6 54.9 62.9 59.8 Rank Associate Professor Full Professor $70.0 77.7 77.1 76.0 70.1 65.4 71.9 65.9 67.9 73.6 $74.4 77.2 76.3 78.8 73.1 66.3 74.6 73.0 69.4 71.0 $109.4 111.3 112.5 111.6 118.3 110.3 97.0 91.5 103.5 95.6 $110.5 104.4 106.3 106.9 109.9 100.9 102.8 102.0 96.7 97.8 Source: Based on “Average Salary for Men and Women Faculty by Category, Affiliation, and Academic Rank, 2005–2006.” CASE STUDY ❍ 501 a. Identify the design used in this survey. b. Use the appropriate analysis of variance for these data. c. Do the data indicate that the salary at the different ranks vary by gender? d. If there is no interaction, determine whether there are differences in salaries by rank, and whether there are differences by gender. Discuss your results. e. Plot the average salaries using an interaction plot. If the main effect of ranks is signiﬁcant, use Tukey’s method of pairwise comparisons to determine if there are signiﬁcant differences among the ranks. Use a .01. CASE STUDY Tickets “A Fine Mess” Do you risk a parking ticket by parking where you shouldn’t or forgetting how much time you have left on the parking meter? Do the ﬁnes associated with various parking infractions vary depending on the city in which you receive a parking ticket? To look at this issue, the ﬁnes imposed for overtime parking, parking in a red zone, and parking next to a ﬁre hydrant were recorded for 13 cities in southern California.11 City Overtime Parking Red Zone Fire Hydrant Long Beach Bakersﬁeld Orange San Bernardino Riverside San Luis Obispo Beverly Hills Palm Springs Laguna Beach Del Mar Los Angeles San Diego Newport Beach $17 17 22 20 21 8 23 22 22 25 20 35 32 $30 33 30 30 30 20 38 28 22 40 55 60 42 $30 33 32 78 30 75 30 46 32 55 30 60 30 Source: From “A Fine Mess,” by R. McGarvey, Avenues, July/August 1994. Reprinted by permission of the author. 1. Identify the design used for the data collection in this case study. 2. Analyze the data using the appropriate analysis. What can you say about the variation among the cities in this study? Among ﬁnes for the three types of violations? Can Tukey’s procedure be of use in further delineating any signiﬁcant differences you may ﬁnd? Would conﬁdence interval estimates be useful in your analysis? 3. Summarize the results of your analysis of these data. 12 Linear Regression and Correlation GENERAL OBJECTIVES In this chapter, we consider the situation in which the mean value of a random variable y is related to another variable x. By measuring both y and x for each experimental unit, thereby generating bivariate data, you can use the information provided by x to estimate the average value of y and to predict values of y for preassigned values of x. CHAPTER INDEX ● Analysis of variance for linear regression (12.4) ● Correlation analysis (12.8) ● Diagnostic tools for checking the regression assumptions (12.6) ● Estimation and prediction using the ﬁtted line (12.7) ● The method of least squares (12.3) ● A simple linear probabilistic model (12.2) ● Testing the usefulness of the linear regression model: inferences about b, the ANOVA F-test, and r 2 (12.5) How Do I Make Sure That My Calculations Are Correct? © Justin Sullivan/Getty Images Is Your Car “Made in the U.S.A.”? The phrase “made in the U.S.A.” has become a battle cry in the past few years as American workers try to protect their jobs from overseas competition. In the case study at the end of this chapter, we explore the changing attitudes of American consumers toward automobiles made outside the United States, using a simple linear regression analysis. 502 12.2 A SIMPLE LINEAR PROBABILISTIC MODEL ❍ 503 INTRODUCTION 12.1 High school seniors, freshmen entering college, their parents, and a university administration are concerned about the academic achievement of a student after he or she has enrolled in a university. Can you estimate or predict a student’s grade point average (GPA) at the end of the freshman year before the student enrolls in the university? At ﬁrst glance this might seem like a difficult problem. However, you would expect highly motivated students who have graduated with a high class rank from a high school with superior academic standards to achieve a high GPA at the end of the college freshman year. On the other hand, students who lack motivation or who have achieved only moderate success in high school are not expected to do so well. You would expect the college achievement of a student to be a function of several variables: • Rank in high school class • High school’s overall rating • High school GPA • SAT scores This problem is of a fairly general nature. You are interested in a random variable y (college GPA) that is related to a number of independent variables. The objective is to create a prediction equation that expresses y as a function of these independent variables. Then, if you can measure the independent variables, you can substitute these values into the prediction equation and obtain the prediction for y—the student’s college GPA in our example. But which variables should you use as predictors? How strong is their relationship to y? How do you construct a good prediction equation for y as a function of the selected predictor variables? We will answer these questions in the next two chapters. In this chapter, we restrict our attention to the simple problem of predicting y as a linear function of a single predictor variable x. This problem was originally addressed in Chapter 3 in the discussion of bivariate data. Remember that we used the equation of a straight line to describe the relationship between x and y and we described the strength of the relationship using the correlation coefficient r. We rely on some of these results as we revisit the subject of linear regression and correlation. 12.2 A SIMPLE LINEAR PROBABILISTIC MODEL Consider the problem of trying to predict the value of a response y based on the value of an independent variable x. The best-ﬁtting line of Chapter 3, y a bx was based on a sample of n bivariate observations drawn from a larger population of measurements. The line that describes the relationship between y and x in the population is
similar to, but not the same as, the best-ﬁtting line from the sample. How can you construct a population model to describe the relationship between a random variable y and a related independent variable x? You begin by assuming that the variable of interest, y, is linearly related to an independent variable x. To describe the linear relationship, you can use the deterministic model y a bx 504 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION where a is the y-intercept—the value of y when x 0—and b is the slope of the line, deﬁned as the change in y for a one-unit change in x, as shown in Figure 12.1. This model describes a deterministic relationship between the variable of interest y, sometimes called the response variable, and the independent variable x, often called the predictor variable. That is, the linear equation determines an exact value of y when the value of x is given. Is this a realistic model for an experimental situation? Consider the following example. F IG URE 12. 1 The y-intercept and slope for a line ● y slope change in y for a 1-unit change in x y-intercept value of y when x 0 y-intercept = α Slope = β 0 1 2 x Table 12.1 displays the mathematics achievement test scores for a random sample of n 10 college freshmen, along with their ﬁnal calculus grades. A bivariate plot of these scores and grades is given in Figure 12.2. You can use the Building a Scatterplot applet to refresh your memory as to how this plot is drawn. Notice that the points do not lie exactly on a line but rather seem to be deviations about an underlying line. A simple way to modify the deterministic model is to add a random error component to explain the deviations of the points about the line. A particular response y is described using the probabilistic model y a bx e Mathematics Achievement Test Scores and Final Calculus Grades TABLE 12.1 ● for College Freshmen ● Mathematics Achievement Test Score Final Calculus Grade Student 1 2 3 4 5 6 7 8 9 10 39 43 21 64 57 47 28 75 34 52 65 78 52 82 92 89 73 98 56 75 12.2 A SIMPLE LINEAR PROBABILISTIC MODEL ❍ 505 FI GUR E 1 2. 2 Scatterplot of the data in Table 12.1 ● e d a r G 100 90 80 70 60 50 20 30 40 50 Score 60 70 80 The ﬁrst part of the equation, a bx—called the line of means—describes the average value of y for a given value of x. The error component e allows each individual response y to deviate from the line of means by a small amount. In order to use this probabilistic model for making inferences, you need to be more speciﬁc about this “small amount,” e. ASSUMPTIONS ABOUT THE RANDOM ERROR e Assume that the values of e satisfy these conditions: • Are independent in the probabilistic sense • Have a mean of 0 and a common variance equal to s 2 • Have a normal probability distribution These assumptions about the random error e are shown in Figure 12.3 for three ﬁxed values of x—say, x1, x2, and x3. Notice the similarity between these assumptions and the assumptions necessary for the tests in Chapters 10 and 11. We will revisit these assumptions later in this chapter and provide some diagnostic tools for you to use in checking their validity. FI GUR E 1 2. 3 Linear probabilistic model ● y x1 x2 x3 x 506 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION Remember that this model is created for a population of measurements that is generally unknown to you. However, you can use sample information to estimate the values of a and b, which are the coefficients of the line of means, E(y) a bx. These estimates are used to form the best-ﬁtting line for a given set of data, called the least squares line or regression line. We review how to calculate the intercept and the slope of this line in the next section. 12.3 THE METHOD OF LEAST SQUARES The statistical procedure for ﬁnding the best-ﬁtting line for a set of bivariate data does mathematically what you do visually when you move a ruler until you think you have minimized the vertical distances, or deviations, from the ruler to a set of points. The formula for the best-ﬁtting line is slope coefficient of x y-intercept constant term yˆ a bx where a and b are the estimates of the intercept and slope parameters a and b, respectively. The ﬁtted line for the data in Table 12.1 is shown in the Method of Least Squares applet, Figure 12.4. The red vertical lines (light blue in Figure 12.4) drawn from the prediction line to each point (xi, yi ) represent the deviations of the points from the line. F IG URE 12. 4 Method of Least Squares applet ● To minimize the distances from the points to the ﬁtted line, you can use the prin- ciple of least squares. PRINCIPLE OF LEAST SQUARES The line that minimizes the sum of squares of the deviations of the observed values of y from those predicted is the best-ﬁtting line. The sum of squared deviations is commonly called the sum of squares for error (SSE) and deﬁned as SSE S( yi yˆi)2 S( yi a bxi)2 12.3 THE METHOD OF LEAST SQUARES ❍ 507 Look at the regression line and the data points in Figure 12.4. SSE is the sum of the squared distances represented by the area of the yellow squares (light blue in Figure 12.4). Finding the values of a and b, the estimates of a and b, uses differential calculus, which is beyond the scope of this text. Rather than derive their values, we will simply present formulas for calculating the values of a and b—called the least-squares estimators of a and b. We will use notation that is based on the sums of squares for the variables in the regression problem, which are similar in form to the sums of squares used in Chapter 11. These formulas look different from the formulas presented in Chapter 3, but they are in fact algebraically identical! You should use the data entry method for your scientiﬁc calculator to enter the sample data. • • If your calculator has only a one-variable statistics function, you can still save some time in ﬁnding the necessary sums and sums of squares. If your calculator has a two-variable statistics function, or if you have a graphing calculator, the calculator will automatically store all of the sums and sums of squares as well as the values of a, b, and the correlation coefficient r. • Make sure you consult your calculator manual to ﬁnd the easiest way to obtain the least squares estimators. LEAST-SQUARES ESTIMATORS OF a AND b b S S and a y bx x y x x where the quantities Sxy and Sxx are deﬁned as (Syi) Sxy S(xi x)( yi y) Sxi yi (Sxi) n and Sxx S(xi x)2 Sx 2 i (S xi)2 n Notice that the sum of squares of the x-values is found using the computing formula given in Section 2.3 and the sum of the cross-products is the numerator of the covariance deﬁned in Section 3.4. EXAMPLE 12.1 Find the least-squares prediction line for the calculus grade data in Table 12.1. Solution Use the data in Table 12.2 and the data entry method in your scientiﬁc calculator to ﬁnd the following sums of squares: Sxx Sx 2 i (S 0)2 xi)2 6 23,634 (4 2474 n 0 1 (Syi) 36,854 (460 Sxy Sxiyi (Sxi) 760) 1894 )( n 0 1 S yi 7 0 76 6 y n 1 0 S xi 4 0 46 6 x n 0 1 508 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION TABLE 12.2 ● Calculations for the Data in Table 12.1 yi xi 65 78 52 82 92 89 73 98 56 75 39 43 21 64 57 47 28 75 34 52 x 2 i 1521 1849 441 4096 3249 2209 784 5625 1156 2704 xi yi 2535 3354 1092 5248 5244 4183 2044 7350 1904 3900 y 2 i 4225 6084 2704 6724 8464 7921 5329 9604 3136 5625 Sum 760 460 23,634 36,854 59,816 Then 76556 and a y bx 76 (.76556)(46) 40.78424 4 7 4 2 x x You can predict y for a given value of x by substituting x into the equation to ﬁnd yˆ. The least-squares regression line is then yˆ a bx 40.78424 .76556x The graph of this line is shown in Figure 12.4. It can now be used to predict y for a given value of x—either by referring to Figure 12.4 or by substituting the proper value of x into the equation. For example, if a freshman scored x 50 on the achievement test, the student’s predicted calculus grade is (using full decimal accuracy) yˆ a b(50) 40.78424 (.76556)(50) 79.06 How Do I Make Sure That My Calculations Are Correct? • Be careful of rounding errors. Carry at least six signiﬁcant ﬁgures, and round off only in reporting the end result. • Use a scientiﬁc or graphing calculator to do all the work for you. Most of these calculators will calculate the values for a and b if you enter the data properly. • Use a computer software program if you have access to one. • Always plot the data and graph the line. If the line does not ﬁt through the points, you have probably made a mistake! You can use the Method of Least Squares applet to ﬁnd the values of a and b that determine the best ﬁtting line, yˆ a bx. The horizontal line that you see is the line y y. Use your mouse to drag the line and watch the yellow squares change size. The object is to make SSE—the total area of the yellow squares (light blue in Figure 12.4)—as small as possible. The value of SSE is the red portion of the bar on the left of the applet (dark blue in Figure 12.4) marked SSE . When you think button and see how well you did! that you have minimized SSE, click the 12.4 AN ANALYSIS OF VARIANCE FOR LINEAR REGRESSION ❍ 509 12.4 AN ANALYSIS OF VARIANCE FOR LINEAR REGRESSION In Chapter 11, you used the analysis of variance procedures to divide the total variation in the experiment into portions attributed to various factors of interest to the experimenter. In a regression analysis, the response y is related to the independent variable x. Hence, the total variation in the response variable y, given by Total SS Syy S(yi y)2 Sy 2 i (S yi)2 n is divided into two portions: • SSR (sum of squares for regression) measures the amount of variation explained by using the regression line with one independent variable x • SSE (sum of squares for error) measures the “residual” variation in the data that is not explained by the independent variable x so that Total SS SSR SSE For a particular value of the response yi, you can visualize this breakdown in the variation using the vertical distances illustrated in Figure 12.5. You can
see that SSR is the sum of the squared deviations of the differences between the estimated response without using x ( y) and the estimated response using x (the regression line, yˆ); SSE is the sum of the squared differences between the regression line ( yˆ) and the point y. FI GUR E 1 2. 5 Deviations from the ﬁtted line ● y 100 e d a r G 90 80 70 60 50 SSE { } SSR ^ y = 40.7842 + 0.76556x 20 30 40 50 Score x 60 70 80 It is not too hard to show algebraically that SSR S( yˆi yi)2 S(a bxi y)2 S( y bx bxi y)2 b2S(xi x)2 S x y 2 S x x )2 Sxx (S x y S x x Since Total SS SSR SSE, you can complete the partition by calculating )2 SSE Total SS SSR Syy (S x y S x x 510 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION Remember from Chapter 11 that each of the various sources of variation, when divided by the appropriate degrees of freedom, provides an estimate of the variation in the experiment. These estimates are called mean squares—MS SS/df—and are displayed in an ANOVA table. In examining the degrees of freedom associated with each of these sums of squares, notice that the total degrees of freedom for n measurements is (n 1). Since estimating the regression line, yˆ a bxi y bx bxi , involves estimating one additional parameter b, there is one degree of freedom associated with SSR, leaving (n 2) degrees of freedom with SSE. As with all ANOVA tables we have discussed, the mean square for error, SE S MSE s 2 n 2 is an unbiased estimator of the underlying variance s 2. The analysis of variance table is shown in Table 12.3. TABLE 12.3 ● Analysis of Variance for Linear Regression Source df SS MS Regression 1 Error Total n 2 n 1 Syy MSR )2 (S x y S x x )2 Syy (S MSE x y S x x For the data in Table 12.1, you can calculate i (S 0)2 yi)2 6 59,816 (7 2056 n 0 1 Total SS Syy Sy 2 SSR (S x S y )2 )2 9 4 (1 8 1449.9741 4 7 4 2 x x so that SSE Total SS SSR 2056 1449.9741 606.0259 and 0259 75.7532 606. SE S MSE n 8 2 The analysis of variance table, part of the linear regression output generated by MINITAB, is the lower shaded section in the printout in Figure 12.6. The ﬁrst two lines give the equation of the least-squares line, yˆ 40.8 .766x. The least-squares estimates a and b are given with greater accuracy in the column labeled “Coef.” You can ﬁnd instructions for generating this output in the section “My MINITAB ” at the end of this chapter. 12.4 AN ANALYSIS OF VARIANCE FOR LINEAR REGRESSION ❍ 511 FI GUR E 1 2. 6 MINITAB output for the data of Table 12.1 ● Regression Analysis: y versus x The regression equation is y = 40.8 + 0.766 x Predictor Coef SE Coef T P Constant 40.784 8.507 4.79 0.001 x 0.7656 0.1750 4.38 0.002 S = 8.70363 R-Sq = 70.5% R-Sq(adj) = 66.8% Analysis of Variance Source DF SS MS F P Regression 1 1450.0 1450.0 19.14 0.002 Residual Error 8 606.0 75.8 Total 9 2056.0 Look for a and b in the column called “Coef.” The MINITAB output also gives some information about the variation in the experiment. Each of the least-squares estimates, a and b, has an associated standard error, labeled “SE Coef” in Figure 12.6. In the middle of the printout, you will ﬁnd the best unbiased estimate of s—S MSE 75.7532 8.70363—which measures the residual error, the unexplained or “leftover” variation in the experiment. It will not surprise you to know that the t and F statistics and their p-values found in the printout are used to test statistical hypotheses. We explain these entries in the next section. 12.4 EXERCISES BASIC TECHNIQUES 12.1 Graph the line corresponding to the equation y 2x 1 by graphing the points corresponding to x 0, 1, and 2. Give the y-intercept and slope for the line. 12.2 Graph the line corresponding to the equation y 2x 1 by graphing the points corresponding to x 0, 1, and 2. Give the y-intercept and slope for the line. How is this line related to the line y 2x 1 of Exercise 12.1? 12.3 Give the equation and graph for a line with y-intercept equal to 3 and slope equal to 1. 12.4 Give the equation and graph for a line with y-intercept equal to 3 and slope equal to 1. 12.5 What is the difference between deterministic and probabilistic mathematical models? 12.6 You are given ﬁve points with these coordinates. Use the data entry method on your scientiﬁc or graphing calculator to enter the n 5 observations. Find the sums of squares and cross-products, Sxx, Sxy, and Syy. b. Find the least-squares line for the data. c. Plot the ﬁve points and graph the line in part b. Does the line appear to provide a good ﬁt to the data points? d. Construct the ANOVA table for the linear regres- sion. 12.7 Six points have these coordinates: x y 1 5.6 2 4.6 3 4.5 4 3.7 5 3.2 6 2.7 a. Find the least-squares line for the data. b. Plot the six points and graph the line. Does the line appear to provide a good ﬁt to the data points? c. Use the least-squares line to predict the value of y when x 3.5. d. Fill in the missing entries in the MINITAB analysis of variance table. 512 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION MINITAB ANOVA table for Exercise 12.7 Analysis of Variance Source DF SS MS Regression * *** 5.4321 Residual Error * 0.1429 *** Total * 5.5750 c. Plot the points and the ﬁtted line. Does the assumption of a linear relationship appear to be reasonable? d. Use the regression line to predict the peak current generated when a solution containing 100 ppb of nickel is added to the buffer. e. Construct the ANOVA table for the linear regres- APPLICATIONS sion. 12.8 Professor Asimov Professor Isaac Asimov was one of the most proliﬁc writers of all time. Prior to his death, he wrote nearly 500 books during a 40-year career. In fact, as his career progressed, he became even more productive in terms of the number of books written within a given period of time.1 The data give the time in months required to write his books in increments of 100: Number of Books, x Time in Months, y 100 237 200 350 300 419 400 465 490 507 a. Assume that the number of books x and the time in months y are linearly related. Find the least-squares line relating y to x. b. Plot the time as a function of the number of books written using a scatterplot, and graph the leastsquares line on the same paper. Does it seem to provide a good ﬁt to the data points? c. Construct the ANOVA table for the linear regres- sion. EX1209 12.9 A Chemical Experiment Using a chemical procedure called differential pulse polarography, a chemist measured the peak current generated (in microamperes) when a solution containing a given amount of nickel (in parts per billion) is added to a buffer:2 x Ni (ppb) y Peak Current (mA) 19.1 38.2 57.3 76.2 95 114 131 150 170 .095 .174 .256 .348 .429 .500 .580 .651 .722 a. Use the data entry method for your calculator to calculate the preliminary sums of squares and crossproducts, Sxx, Syy, and Sxy. 12.10 Sleep Deprivation A study was conducted to determine the effects of sleep EX1210 deprivation on people’s ability to solve problems without sleep. A total of 10 subjects participated in the study, two at each of ﬁve sleep deprivation levels—8, 12, 16, 20, and 24 hours. After his or her speciﬁed sleep deprivation period, each subject was administered a set of simple addition problems, and the number of errors was recorded. These results were obtained: Number of Errors, y 8, 6 6, 10 8, 14 Number of Hours without Sleep, x 8 12 16 Number of Errors, y 14, 12 16, 12 Number of Hours without Sleep, x 20 24 a. How many pairs of observations are in the experi- ment? b. What are the total number of degrees of freedom? c. Complete the MINITAB printout. MINITAB output for Exercise 12.10 Regression Analysis: y versus x The regression equation is y = 3.00 + 0.475 x Predictor Coef SE Coef T P Constant 3.000 2.127 1.41 0.196 x * 0.1253 3.79 0.005 S = 2.24165 R-Sq = 64.2% R-Sq(adj) = 59.8% Analysis of Variance Source DF SS MS F P Regression 72.200 72.200 14.37 0.005 Residual Error * 5.025 Total * d. What is the least-squares prediction equation? e. Use the prediction equation to predict the number of errors for a person who has not slept for 10 hours. b. Calculate the least-squares regression line. EX1211 12.11 Achievement Tests The Academic Performance Index (API) is a measure of school achievement based on the results of the Stanford 9 Achievement test. Scores range from 200 to 1000, with 800 considered a long-range goal for schools. The following table shows the API for eight elementary schools in Riverside County, California, along with the percent of students at that school who are considered English Language Learners (ELL).3 School 1 API ELL 588 58 2 659 22 3 4 710 14 657 30 5 669 11 6 641 26 7 557 39 8 743 6 a. Which of the two variables is the independent variable and which is the dependent variable? Explain your choice. b. Use a scatterplot to plot the data. Is the assumption of a linear relationship between x and y reasonable? c. Assuming that x and y are linearly related, calculate the least-squares regression line. d. Plot the line on the scatterplot in part b. Does the line ﬁt through the data points? 12.12 How Long Is It? How good are you at estimating? To test a subject’s ability to EX1212 estimate sizes, he was shown 10 different objects and asked to estimate their length or diameter. The object was then measured, and the results were recorded in the table below. Object Estimated (inches) Actual (inches) Pencil Dinner plate Book 1 Cell phone Photograph Toy Belt Clothespin Book 2 Calculator 7.00 9.50 7.50 4.00 14.50 3.75 42.00 2.75 10.00 3.50 6.00 10.25 6.75 4.25 15.75 5.00 41.50 3.75 9.25 4.75 a. Find the least-squares regression line for predicting the actual measurement as a function of the estimated measurement. b. Plot the points and the ﬁtted line. Does the assumption of a linear relationship appear to be reasonable? 12.13 Test Interviews Of two personnel evaluation techniques available, the ﬁrst EX1213 requires a two-hour test interview while the second 12.4 AN ANALYSIS OF VARIANCE FOR LINEAR REGRESSION ❍ 513 can be completed in less
than an hour. The scores for each of the 15 individuals who took both tests are given in the next table. Applicant Test 1 (x) Test 2(y 10 11 12 13 14 15 75 89 60 71 92 105 55 87 73 77 84 91 75 82 76 38 56 35 45 59 70 31 52 48 41 51 58 45 49 47 a. Construct a scatterplot for the data. Does the assumption of linearity appear to be reasonable? b. Find the least-squares line for the data. c. Use the regression line to predict the score on the second test for an applicant who scored 85 on Test 1. 12.14 Test Interviews, continued Refer to Exercise 12.13. Construct the ANOVA table for the linear regression relating y, the score on Test 2, to x, the score on Test 1. 12.15 Armspan and Height Leonardo da Vinci (1452–1519) drew a sketch of a man, EX1215 indicating that a person’s armspan (measuring across the back with your arms outstretched to make a “T”) is roughly equal to the person’s height. To test this claim, we measured eight people with the following results: Person Armspan (inches) Height (inches) Person Armspan (inches) Height (inches) 1 68 69 5 68 67 2 62.25 62 6 69 67 3 65 65 7 62 63 4 69.5 70 8 60.25 62 514 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION b. If da Vinci is correct, and a person’s armspan is roughly the same as the person’s height, what should the slope of the regression line be? c. Calculate the regression line for predicting height based on a person’s armspan. Does the value of the slope b conﬁrm your conclusions in part b? d. If a person has an armspan of 62 inches, what would you predict the person’s height to be? 12.16 Strawberries The following data were obtained in an experiment relating the EX1216 dependent variable, y (texture of strawberries), with x (coded storage temperature). x 2 2 y 3.5 4.0 0 2.0 2 0.5 2 0.0 a. Find the least-squares line for the data. b. Plot the data points and graph the least-squares line as a check on your calculations. c. Construct the ANOVA table. a. Draw a scatterplot for armspan and height. Use the same scale on both the horizontal and vertical axes. Describe the relationship between the two variables. TESTING THE USEFULNESS OF THE LINEAR REGRESSION MODEL In considering linear regression, you may ask two questions: 12.5 • • Is the independent variable x useful in predicting the response variable y? If so, how well does it work? This section examines several statistical tests and measures that will help you reach some answers. Once you have determined that the model is working, you can then use the model for predicting the response y for a given value of x. Inferences Concerning b, the Slope of the Line of Means Is the least-squares regression line useful? That is, is the regression equation that uses information provided by x substantially better than the simple predictor y that does not rely on x? If the independent variable x is not useful in the population model y a bx e, then the value of y does not change for different values of x. The only way that this happens for all values of x is when the slope b of the line of means equals 0. This would indicate that the relationship between y and x is not linear, so that the initial question about the usefulness of the independent variable x can be restated as: Is there a linear relationship between x and y? You can answer this question by using either a test of hypothesis or a conﬁdence interval for b. Both of these procedures are based on the sampling distribution of b, 12.5 TESTING THE USEFULNESS OF THE LINEAR REGRESSION MODEL ❍ 515 the sample estimator of the slope b. It can be shown that, if the assumptions about the random error e are valid, then the estimator b has a normal distribution in repeated sampling with mean E(b) b and standard error given by 2 s SE where s 2 is the variance of the random error e. Since the value of s 2 is estimated with s2 MSE, you can base inferences on the statistic given by S x x b b t /Sxx M SE which has a t distribution with df (n 2), the degrees of freedom associated with MSE. TEST OF HYPOTHESIS CONCERNING THE SLOPE OF A LINE 1. Null hypothesis: H0 : b b0 2. Alternative hypothesis: One-Tailed Test Ha : b b0 (or b b0) b b 0 3. Test statistic: t xx /S SE M Two-Tailed Test Ha : b b0 When the assumptions given in Section 12.2 are satisﬁed, the test statistic will have a Student’s t distribution with (n 2) degrees of freedom. 4. Rejection region: Reject H0 when One-Tailed Test t ta (or t ta when the alternative hypothesis is Ha : b b0) or when p-value a Two-Tailed Test t ta/2 or t ta/2 α α/2 0 αt –t α/2 0 α/2 α/2t The values of ta and ta/2 can be found using Table 4 in Appendix I or the t Probabilities applet. Use the values of t corresponding to (n 2) degrees of freedom. 516 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION EXAMPLE 12.2 Determine whether there is a signiﬁcant linear relationship between the calculus grades and test scores listed in Table 12.1. Test at the 5% level of signiﬁcance. Solution The hypotheses to be tested are H0 : b 0 versus Ha : b 0 and the observed value of the test statistic is calculated as .38 474 32 /Sxx M 7 /2 5 5 S E 7 . with (n 2) 8 degrees of freedom. With a .05, you can reject H0 when t 2.306 or t 2.306. Since the observed value of the test statistic falls into the rejection region, H0 is rejected and you can conclude that there is a signiﬁcant linear relationship between the calculus grades and the test scores for the population of college freshmen. You can use the t-Test for the Slope applet shown in Figure 12.7 to ﬁnd p-values or rejection regions for this test. You must ﬁrst calculate the standard error SE MSE/Sxx, type its value into the box marked “Std Error,” and press “Enter.” F IG URE 12. 7 t-Test for the Slope applet ● • • If you enter the value of b into the formula at the top of the applet and press “Enter,” the applet will calculate the test statistic and its one- or two-tailed p-value. If you enter the signiﬁcance level a in the box marked “prob:” and select the “Area to Right” or “Two Tails” option in the drop-down list, the applet will calculate the positive value of t necessary for rejecting H0. (You could also use the Student’s t Probabilities applet to ﬁnd the critical values.) What is the p-value for the test performed in Example 12.2? Does this p-value conﬁrm our conclusions? Another way to make inferences about the value of b is to construct a conﬁdence interval for b and examine the range of possible values for b. 12.5 TESTING THE USEFULNESS OF THE LINEAR REGRESSION MODEL ❍ 517 A (1 a)100% CONFIDENCE INTERVAL FOR b b ta/2(SE) where ta/2 is based on (n 2) degrees of freedom and SE M E 2 s S x x S S xx EXAMPLE 12.3 Find a 95% conﬁdence interval estimate of the slope b for the calculus grade data in Table 12.1. Solution Substituting previously calculated values into b t.025M E S S xx you have .766 2.30675 .766 .404 32 5 7 . 4 7 4 2 The resulting 95% conﬁdence interval is .362 to 1.170. Since the interval does not contain 0, you can conclude that the true value of b is not 0, and you can reject the null hypothesis H0 : b 0 in favor of Ha : b 0, a conclusion that agrees with the ﬁndings in Example 12.2. Furthermore, the conﬁdence interval estimate indicates that there is an increase from as little as .4 to as much as 1.2 points in a calculus test score for each 1-point increase in the achievement test score. If you are using computer software to perform the regression analysis, you will ﬁnd the t statistic and its p-value on the printout. Look at the MINITAB regression analysis printout reproduced in Figure 12.8. In the second portion of the printout, you will ﬁnd the least-squares estimates a (“Constant”) and b (“x”) in the column marked “Coef,” their standard errors (“SE Coef”), the calculated value of the t statistic (“T”) used for testing the hypothesis that the parameter equals 0, and its p-value (“P”). The t-test for signiﬁcant regression, H0 : b 0, has a p-value of P .002, and the null hypothesis is rejected, as in Example 12.2. Does this agree with the p-value found using the t-Test for the Slope applet in Figure 12.7? In any event, there is a signiﬁcant linear relationship between x and y. FI GUR E 1 2. 8 MINITAB output for the calculus grade data ● Regression Analysis: y versus x The regression equation is y = 40.8 + 0.766 x Look for the standard error of b in the column marked “SE Coef.” Predictor Coef SE Coef T P Constant 40.784 8.507 4.79 0.001 x 0.7656 0.1750 4.38 0.002 S = 8.70363 R-Sq = 70.5% R-Sq(adj) = 66.8% Analysis of Variance Source DF SS MS F P Regression 1 1450.0 1450.0 19.14 0.002 Residual Error 8 606.0 75.8 Total 9 2056.0 518 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION ANOVA F-tests are always one-tailed (upper-tail). The Analysis of Variance F-Test The analysis of variance portion of the printout in Figure 12.8 shows an F statistic given by R S F M 19.14 S E M with 1 numerator degree of freedom and (n 2) 8 denominator degrees of freedom. This is an equivalent test statistic that can also be used for testing the hypothesis H0 : b 0. Notice that, within rounding error, the value of F is equal to t 2 with the identical p-value. In this case, if you use ﬁve-decimal-place accuracy prior to rounding, you ﬁnd that t 2 (.76556/.17498)2 (4.37513)2 19.14175 19.14 F as given in the printout. This is no accident and results from the fact that the square of a t statistic with df degrees of freedom has the same distribution as an F-statistic with 1 numerator and df denominator degrees of freedom. The F-test is a more general test of the usefulness of the model and can be used when the model has more than one independent variable. Measuring the Strength of the Relationship: The Coefficient of Determination How well does the regression model ﬁt? To answer this question, you can use a measure related to the correlation coefficient r, introduced in Chapter 3. Remember that S s x y for 1 r 1 r y x Syy S s s y x x x where sxy, sx, and sy were deﬁned in Chapter 3 and the various sums of squares were deﬁned in Se
ction 12.4. The sum of squares for regression, SSR, in the analysis of variance measures the portion of the total variation, Total SS Syy, that can be explained by the regression of y on x. The remaining portion, SSE, is the “unexplained” variation attributed to random error. One way to measure the strength of the relationship between the response variable y and the predictor variable x is to calculate the coefficient of determination—the proportion of the total variation that is explained by the linear regression of y on x. For the calculus grade data, this proportion is equal to On computer printouts, r 2 is often given as a percentage rather than a proportion. R S S 0 5 1 4 .705 or 70.5% ta SS l To 0 5 6 2 )2 Since Total SS Syy and SSR (S , you can write x y S x x S 2 (S R S S y) x x Syy S S S l SS ta To xx x y x y y 2 r2 Therefore, the coefficient of determination, which was calculated as SSR/Total SS, is simply the square of the correlation coefficient r. It is the entry labeled “R-Sq” in Figure 12.8. Remember that the analysis of variance table isolates the variation due to regression (SSR) from the total variation in the experiment. Doing so reduces the amount of random variation in the experiment, now measured by SSE rather than Total SS. In this context, the coefficient of determination, r2, can be deﬁned as follows: 12.5 TESTING THE USEFULNESS OF THE LINEAR REGRESSION MODEL ❍ 519 r 2 is called “R-Sq” on the MINITAB printout. Definition The coefficient of determination r 2 can be interpreted as the percent reduction in the total variation in the experiment obtained by using the regression line yˆ a bx, instead of ignoring x and using the sample mean y to predict the response variable y. For the calculus grade data, a reduction of r 2 .705 or 70.5% is substantial. The regression model is working very well! Interpreting the Results of a Significant Regression Once you have performed the t-test or F-test to determine the signiﬁcance of the linear regression, you must interpret your results carefully. The slope b of the line of means is estimated based on data from only a particular region of observation. Even if you do not reject the null hypothesis that the slope of the line equals 0, it does not necessarily mean that y and x are unrelated. It may be that you have committed a Type II error—falsely declaring that the slope is 0 and that x and y are unrelated. Fitting the Wrong Model It may happen that y and x are perfectly related in a nonlinear way, as shown in Figure 12.9. Here are three possibilities: FI GUR E 1 2. 9 Curvilinear relationship ● y L in e 1 Line 2 a b c d f x • • • If observations were taken only within the interval b x c, the relationship would appear to be linear with a positive slope. If observations were taken only within the interval d x f, the relationship would appear to be linear with a negative slope. If the observations were taken over the interval c x d, the line would be ﬁtted with a slope close to 0, indicating no linear relationship between y and x. For the example shown in Figure 12.9, no straight line accurately describes the true relationship between x and y, which is really a curvilinear relationship. In this case, we have chosen the wrong model to describe the relationship. Sometimes this type of mistake can be detected using residual plots, the subject of Section 12.7. It is dangerous to try to predict values of y outside of the range of the ﬁtted data. Extrapolation One serious problem is to apply the results of a linear regression analysis to values of x that are not included within the range of the ﬁtted data. This is called extrapolation and can lead to serious errors in prediction, as shown for line 1 in Figure 12.9. 520 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION Prediction results would be good over the interval b x c but would seriously overestimate the values of y for x c. Causality When there is a signiﬁcant regression of y and x, it is tempting to conclude that x causes y. However, it is possible that one or more unknown variables that you have not even measured and that are not included in the analysis may be causing the observed relationship. In general, the statistician reports the results of an analysis but leaves conclusions concerning causality to scientists and investigators who are experts in these areas. These experts are better prepared to make such decisions! 12.5 EXERCISES BASIC TECHNIQUES 12.17 Refer to Exercise 12.6. The data are reproduced below. Do the data present sufficient evidence to indicate that y and x are linearly related? Test the hypothesis that b 0 at the 5% level of signiﬁcance. b. Use the ANOVA table from Exercise 12.6 to calculate F MSR/MSE. Verify that the square of the t statistic used in part a is equal to F. c. Compare the two-tailed critical value for the t-test in part a with the critical value for F with a .05. What is the relationship between the critical values? 12.18 Refer to Exercise 12.17. Find a 95% conﬁdence interval for the slope of the line. What does the phrase “95% conﬁdent” mean? 12.19 Refer to Exercise 12.7. The data, along with the MINITAB analysis of variance table are reproduced below. x y 1 5.6 2 4.6 3 4.5 4 3.7 5 3.2 6 2.7 MINITAB ANOVA table for Exercise 12.19 Regression Analysis: y versus x Analysis of Variance Source DF SS MS F P Regression 1 5.4321 5.4321 152.10 0.000 Residual Error 4 0.1429 0.0357 Total 5 5.5750 a. Do the data provide sufficient evidence to indicate that y and x are linearly related? Use the informa- tion in the MINITAB printout to answer this question at the 1% level of signiﬁcance. b. Calculate the coefficient of determination r 2. What information does this value give about the usefulness of the linear model? APPLICATIONS EX1220 12.20 Air Pollution An experiment was designed to compare several different types of air pollution monitors.4 The monitor was set up, and then exposed to different concentrations of ozone, ranging between 15 and 230 parts per million (ppm) for periods of 8–72 hours. Filters on the monitor were then analyzed, and the amount (in micrograms) of sodium nitrate (NO3) recorded by the monitor was measured. The results for one type of monitor are given in the table. Ozone, x (ppm/hr) NO3, y (mg) .8 2.44 1.3 5.21 1.7 6.07 2.2 8.98 2.7 2.9 10.82 12.16 a. Find the least-squares regression line relating the monitor’s response to the ozone concentration. b. Do the data provide sufficient evidence to indicate that there is a linear relationship between the ozone concentration and the amount of sodium nitrate detected? c. Calculate r 2. What does this value tell you about the effectiveness of the linear regression analysis? 12.21 The Cost of Flying How is the cost of a plane ﬂight related to the length of the EX1221 trip? The table shows the average round-trip coach airfare paid by customers of American Airlines on each of 18 heavily traveled U.S. air routes.5 12.5 TESTING THE USEFULNESS OF THE LINEAR REGRESSION MODEL ❍ 521 Route Dallas–Austin Houston–Dallas Chicago–Detroit Chicago–St. Louis Chicago–Cleveland Chicago–Atlanta New York–Miami New York–San Juan New York–Chicago Chicago–Denver Dallas–Salt Lake New York–Dallas Chicago–Seattle Los Angeles–Chicago Los Angeles–Atlanta New York–Los Angeles Los Angeles–Honolulu New York–San Francisco Distance (miles) 178 232 238 262 301 593 1092 1608 714 901 1005 1374 1736 1757 1946 2463 2556 2574 Cost $125 123 148 136 129 162 224 264 287 256 365 459 424 361 309 444 323 513 a. If you want to estimate the cost of a ﬂight based on the distance traveled, which variable is the response variable and which is the independent predictor variable? b. Assume that there is a linear relationship between cost and distance. Calculate the least-squares regression line describing cost as a linear function of distance. c. Plot the data points and the regression line. Does it appear that the line ﬁts the data? d. Use the appropriate statistical tests and measures to explain the usefulness of the regression model for predicting cost. 12.22 Professor Asimov, continued Refer to the data in Exercise 12.8, relating x, the number of books written by Professor Isaac Asimov, to y, the number of months he took to write his books (in increments of 100). The data are reproduced below. Number of Books, x Time in Months, y 100 237 200 350 300 419 400 465 490 507 a. Do the data support the hypothesis that b 0? Use the p-value approach, bounding the p-value using Table 4 of Appendix I or ﬁnding the exact p-value using the t-Test for the Slope applet. Explain your conclusions in practical terms. b. Use the ANOVA table in Exercise 12.8, part c, to calculate the coefficient of determination r 2. What percentage reduction in the total variation is achieved by using the linear regression model? c. Plot the data or refer to the plot in Exercise 12.8, part b. Do the results of parts a and b indicate that the model provides a good ﬁt for the data? Are there any assumptions that may have been violated in ﬁtting the linear model? 12.23 Refer to the sleep deprivation experiment described in Exercise 12.10 and data set EX1210. The data and the MINITAB printout are reproduced here. Number of Errors, y 8, 6 6, 10 8, 14 Number of Hours without Sleep, x 8 12 16 Number of Errors, y 14, 12 16, 12 Number of Hours without Sleep, x 20 24 MINITAB output for Exercise 12.23 Regression Analysis: y versus x The regression equation is y = 3.00 + 0.475 x Predictor Coef SE Coef T P Constant 3.000 2.127 1.41 0.196 x 0.4750 0.1253 3.79 0.005 S = 2.24165 R-Sq = 64.2% R-Sq(adj) = 59.8% Analysis of Variance Source DF SS MS F P Regression 1 72.200 72.200 14.37 0.005 Residual Error 8 40.200 5.025 Total 9 112.400 a. Do the data present sufficient evidence to indicate that the number of errors is linearly related to the number of hours without sleep? Identify the two test statistics in the printout that can be used to answer this question. b. Would you expect the relationship between y and x to
be linear if x varied over a wider range (say, x 4 to x 48)? c. How do you describe the strength of the relation- ship between y and x? d. What is the best estimate of the common population variance s 2? e. Find a 95% conﬁdence interval for the slope of the line. 12.24 Strawberries II The following data (Exercise 12.16 and data set EX1216) were obtained in an experiment relating the dependent variable, y (texture of strawberries), with x (coded storage temperature). Use the information from Exercise 12.16 to answer the following questions: x 2 2 y 3.5 4.0 2.0 0.0 0.5 2 2 0 a. What is the best estimate of s 2, the variance of the random error ? 522 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION b. Do the data indicate that texture and storage tem- perature are linearly related? Use a .05. c. Calculate the coefficient of determination, r2. d. Of what value is the linear model in increasing the accuracy of prediction as compared to the predictor, y? 12.25 Laptops and Learning In Exercise 1.61 we described an informal experiment EX1225 conducted at McNair Academic High School in Jersey City, New Jersey. Two freshman algebra classes were studied, one of which used laptop computers at school and at home, while the other class did not. In each class, students were given a survey at the beginning and end of the semester, measuring his or her technological level. The scores were recorded for the end of semester survey (x) and the ﬁnal examination (y) for the laptop group.6 The data and the MINITAB printout are shown here. Student Posttest Final Exam Student Posttest Final Exam 1 2 3 4 5 6 7 8 9 10 100 96 88 100 100 96 80 68 92 96 98 97 88 100 100 78 68 47 90 94 11 12 13 14 15 16 17 18 19 20 88 92 68 84 84 88 72 88 72 88 84 93 57 84 81 83 84 93 57 83 Regression Analysis: y versus x The regression equation is y = -26.8 + 1.26 x Predictor Coef SE Coef T P Constant -26.82 14.76 -1.82 0.086 x 1.2617 0.1685 7.49 0.000 S = 7.61912 R-Sq = 75.7% R-Sq(adj) = 74.3% Analysis of Variance Source DF SS MS F P Regression 1 3254.0 3254.0 56.05 0.000 Residual Error 18 1044.9 58.1 Total 19 4299.0 a. Construct a scatterplot for the data. Does the assumption of linearity appear to be reasonable? b. What is the equation of the regression line used for predicting ﬁnal exam score as a function of the posttest score? c. Do the data present sufficient evidence to indicate that ﬁnal exam score is linearly related to the posttest score? Use a .01. d. Find a 99% conﬁdence interval for the slope of the regression line. 12.26 Laptops and Learning, continued Refer to Exercise 12.25. a. Use the MINITAB printout to ﬁnd the value of the coefficient of determination, r 2. Show that r 2 SSR/Total SS. b. What percentage reduction in the total variation is achieved by using the linear regression model? 12.27 Armspan and Height II In Exercise 12.15 (data set EX1215), we measured the armspan and height of eight people with the following results: Person Armspan (inches) Height (inches) Person Armspan (inches) Height (inches) 1 68 69 5 68 67 2 62.25 62 6 69 67 3 65 65 7 62 63 4 69.5 70 8 60.25 62 a. Does the data provide sufficient evidence to indicate that there is a linear relationship between armspan and height? Test at the 5% level of signiﬁcance. b. Construct a 95% conﬁdence interval for the slope of the line of means, b. c. If Leonardo da Vinci is correct, and a person’s armspan is roughly the same as the person’s height, the slope of the regression line is approximately equal to 1. Is this supposition conﬁrmed by the conﬁdence interval constructed in part b? Explain. 12.6 DIAGNOSTIC TOOLS FOR CHECKING THE REGRESSION ASSUMPTIONS Even though you have determined—using the t-test for the slope (or the ANOVA F-test) and the value of r 2—that x is useful in predicting the value of y, the results of a regression analysis are valid only when the data satisfy the necessary regression assumptions. 12.6 DIAGNOSTIC TOOLS FOR CHECKING THE REGRESSION ASSUMPTIONS ❍ 523 REGRESSION ASSUMPTIONS • The relationship between y and x must be linear, given by the model y a bx e • The values of the random error term e (1) are independent, (2) have a mean of 0 and a common variance s 2, independent of x, and (3) are normally distributed. Since these assumptions are quite similar to those presented in Chapter 11 for an analysis of variance, it should not surprise you to ﬁnd that the diagnostic tools for checking these assumptions are the same as those we used in that chapter. These tools involve the analysis of the residual error, the unexplained variation in each observation once the variation explained by the regression model has been removed. Dependent Error Terms The error terms are often dependent when the observations are collected at regular time intervals. When this is the case, the observations make up a time series whose error terms are correlated. This in turn causes bias in the estimates of model parameters. Time series data should be analyzed using time series methods. You will ﬁnd a discussion of time-series analysis in the text Statistics for Management and Economics, 7th edition, by Mendenhall, Beaver, and Beaver. Residual Plots The other regression assumptions can be checked using residual plots, which are fairly complicated to construct by hand but easy to use once a computer has graphed them for you! In simple linear regression, you can use the plot of residuals versus ﬁt to check for a constant variance as well as to make sure that the linear model is in fact adequate. This plot should be free of any patterns. It should appear as a random scatter of points about 0 on the vertical axis with approximately the same vertical spread for all values of yˆ. One property of the residuals is that they sum to 0 and therefore have a sample mean of 0. The plot of the residuals versus ﬁt for the calculus grade example is shown in Figure 12.10. There are no apparent patterns in this residual plot, which indicates that the model assumptions appear to be satisﬁed for these data. FI GUR E 1 2. 10 Plot of the residuals versus yˆ for Example 12.1 ● Residuals versus Fitted Value (response is y) l a u d i s e R 15 10 5 0 5 10 60 70 80 Fitted Value 90 100 524 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION Residuals vs. ﬁts ⇔ random scatter Normal plot ⇔ straight line, sloping up Recall from Chapter 11 that the normal probability plot is a graph that plots the residuals against the expected value of that residual if it had come from a normal distribution. When the residuals are normally distributed or approximately so, the plot should appear as a straight line, sloping upward. The normal probability plot for the residuals in Example 12.1 is given in Figure 12.11. With the exception of the fourth and ﬁfth plotted points, the remaining points appear to lie approximately on a straight line. This plot is not unusual and does not indicate underlying nonnormality. The most serious violations of the normality assumption usually appear in the tails of the distribution because this is where the normal distribution differs most from other types of distributions with a similar mean and measure of spread. Hence, curvature in either or both of the two ends of the normal probability plot is indicative of nonnormality. F IG URE 12. 11 Normal probability plot of residuals for Example 12.1 ● Normal Probability Plot of the Residuals (response is y) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 20 10 0 Residual 10 20 12.6 EXERCISES BASIC TECHNIQUES 12.28 What diagnostic plot can you use to determine whether the data satisfy the normality assumption? What should the plot look like for normal residuals? 12.29 What diagnostic plot can you use to determine whether the incorrect model has been used? What should the plot look like if the correct model has been used? 12.30 What diagnostic plot can you use to determine whether the assumption of equal variance has been violated? What should the plot look like when the variances are equal for all values of x? 12.31 Refer to the data in Exercise 12.7. The normal probability plot and the residuals versus ﬁtted values plots generated by MINITAB are shown here. Does it appear that any regression assumptions have been violated? Explain. MINITAB output for Exercise 12.31 Normal Probability Plot of the Residuals (response is y) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 0.4 0.3 0.2 0.1 0.0 Residual 01. 02. 0.3 0.4 12.6 DIAGNOSTIC TOOLS FOR CHECKING THE REGRESSION ASSUMPTIONS ❍ 525 Residuals versus the Fitted Value (response is y) Isaac Asimov are related to the number of months y he took to write them. A plot of the data is shown.2 0.1 0.0 0.1 0.2 0.3 2.5 3.0 3.5 4.0 Fitted Value 4.5 5.0 5.5 APPLICATIONS 12.32 Air Pollution Refer to Exercise 12.20, in which an air pollution monitor’s response to ozone was recorded for several different concentrations of ozone. Use the MINITAB residual plots to comment on the validity of the regression assumptions. MINITAB output for Exercise 12.32 Residuals versus the Fitted Value (response is NO3) 0.50 0.25 0.00 l a u d i s e R 0.25 0.50 2 4 6 8 10 12 Fitted Value Normal Probability Plot of the Residuals (response is NO3) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 1.0 0.5 0.0 Residual 0.5 1.0 12.33 Professor Asimov, again Refer to Exercise 12.8, in which the number of books x written by ) 500 450 400 350 300 250 200 100 200 300 400 500 x(Number of Books) a. Can you see any pattern other than a linear relation- ship in the original plot? b. The value of r 2 for these data is .959. What does this tell you about the ﬁt of the regression line? c. Look at the accompanying diagnostic plots for these data. Do you see any pattern in the residuals? Does this suggest that the relationship between number of months and number of books written is something other than linear? Residuals versus the Fitted Values (response is y) l a u d i s e R 20 10 0 10 20 30 99 95 90 80 70 60 50 40 30 20 10 5 1 t n e c r e P 250 300 350 400 Fit
ted Value 450 500 550 Normal Probability Plot of the Residuals (response is y) 50 25 0 Residual 25 50 526 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION 12.34 Laptops and Learning, again Refer to the data given in Exercise 12.25. The MINITAB printout is reproduced here. Regression Analysis: y versus x The regression equation is y = -26.8 + 1.26 x Predictor Coef SE Coef T P Constant -26.82 14.76 -1.82 0.086 x 1.2617 0.1685 7.49 0.000 S = 7.61912 R-Sq = 75.7% R-Sq(adj) = 74.3% Analysis of Variance Source DF SS MS F P Regression 1 3254.0 3254.0 56.05 0.000 Residual Error 18 1044.9 58.1 Total 19 4299.0 a. What assumptions must be made about the distribu- tion of the random error, ? b. What is the best estimate of s 2, the variance of the random error, ? c. Use the diagnostic plots for these data to comment on the validity of the regression assumptions. Normal Probability Plot of the Residuals (response is y) 99 95 90 80 70 60 50 40 30 20 10 5 1 t n e c r e P 20 10 0 Residual 10 20 Residuals versus the Fitted Values (response is y) EX1235 12.35 HDTVs In Exercise 3.19, Consumer Reports gave the prices for the top 10 LCD high deﬁnition TVs (HDTVs) in the 30- to 40-inch category: Does the price of an LCD TV depend on the size of the screen? The table below shows the ten costs again, along with the screen size in inches.7 Brand JVC LT-40FH96 Sony Bravia KDL-V32XBR1 Sony Bravia KDL-V40XBR1 Toshiba 37HLX95 Sharp Aquos LC-32DA5U Sony Bravia KLV-S32A10 Panasonic Viera TC-32LX50 JVC LT-37X776 LG 37LP1D Samsung LN-R328W Price $2900 1800 2600 3000 1300 1500 1350 2000 2200 1200 Size 40 32 40 37 32 32 32 37 37 32 Does the price of an HDTV depend on the size of the screen? Suppose we assume that the relationship between x and y is linear, and perform a linear regression, resulting in a value of r 2 .787. a. What does the value of r 2 tell you about the strength of the relationship between price and screen size? b. The residual plot for this data, generated by MINITAB, is shown below. Does this plot reveal any outliers in the data set? If so, which point is the outlier? Residuals versus the Fitted Values (response is Price) l a u d i s e R 800 600 400 200 0 200 400 l a u d i s e R 20 10 0 10 20 60 70 80 Fitted Value 90 100 1500 1750 2000 2250 2500 2750 3000 Fitted Value c. Plot the values of x and y using a scatterplot. Does this plot conﬁrm your suspicions in part b? Which HDTV does the outlier represent? Is this a faulty measurement that should be removed from the data set? Explain. 12.7 ESTIMATION AND PREDICTION USING THE FITTED LINE ❍ 527 ESTIMATION AND PREDICTION USING THE FITTED LINE Now that you have 12.7 • • tested the ﬁtted regression line, yˆ a bx, to make sure that it is useful for prediction and used the diagnostic tools to make sure that none of the regression assumptions have been violated you are ready to use the line for one of its two purposes: • Estimating the average value of y for a given value of x • Predicting a particular value of y for a given value of x The sample of n pairs of observations have been chosen from a population in which the average value of y is related to the value of the predictor variable x by the line of means, E( y) a bx an unknown line, shown as a broken line in Figure 12.12. Remember that for a ﬁxed value of x—say, x0—the particular values of y deviate from the line of means. These values of y are assumed to have a normal distribution with mean equal to a bx0 and variance s 2, as shown in Figure 12.12. FI GUR E 1 2. 12 Distribution of y for x x0 ● y Line of means E(y) = α + βx x = x0 x Since the computed values of a and b vary from sample to sample, each new sample produces a different regression line yˆ a bx, which can be used either to estimate the line of means or to predict a particular value of y. Figure 12.13 shows one of the possible conﬁgurations of the ﬁtted line (blue), the unknown line of means (gray), and a particular value of y (the blue dot). 528 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION F IG URE 12. 13 Error in estimating E (y) and in predicting y ● y Error of estimating E(y) x0 Actual value of y you are attempting to predict ) Predicted value of y x How far will our estimator yˆ a bx0 be from the quantity to be estimated or predicted? This depends, as always, on the variability in our estimator, measured by its standard error. It can be shown that yˆ a bx0 the estimated value of y when x x0, is an unbiased estimator of the line of means, a bx0, and that yˆ is normally distributed with the standard error of yˆ estimated by SE(yˆ) MSE1 (x0 x)2 n S xx For a given value of x, the prediction interval is always wider than the conﬁdence interval. Estimation and testing are based on the statistic y) E t yˆ ( yˆ E( ) S which has a t distribution with (n 2) degrees of freedom. To form a (1 a)100% conﬁdence interval for the average value of y when x x0, measured by the line of means, a bx0, you can use the usual form for a conﬁdence interval based on the t distribution: yˆ ta/2SE( yˆ) If you choose to predict a particular value of y when x x0, however, there is some additional error in the prediction because of the deviation of y from the line of means. If you examine Figure 12.13, you can see that the error in prediction has two components: • The error in using the ﬁtted line to estimate the line of means • The error caused by the deviation of y from the line of means, measured by s 2 The variance of the difference between y and yˆ is the sum of these two variances and forms the basis for the standard error of ( y yˆ) used for prediction: SE( y yˆ) MSE1 1 x)2 (x0 S n xx and the (1 a)100% prediction interval is formed as yˆ ta/2SE( y yˆ) 12.7 ESTIMATION AND PREDICTION USING THE FITTED LINE ❍ 529 (1 a)100% CONFIDENCE AND PREDICTION INTERVALS • For estimating the average value of y when x x0: x)2 yˆ ta/2MSE1 (x0 n S xx • For predicting a particular value of y when x x0: (x0 yˆ ta/2MSE1 1 x)2 n S x x EXAMPLE 12.4 where ta/2 is the value of t with (n 2) degrees of freedom and area a/2 to its right. Use the information in Example 12.1 to estimate the average calculus grade for students whose achievement score is 50, with a 95% conﬁdence interval. Solution The point estimate of E(yx0 50), the average calculus grade for students whose achievement score is 50, is yˆ 40.78424 .76556(50) 79.06 The standard error of yˆ is MSE1 (x0 75.7532 x)2 474 (50 1 0 1 2 2.840 46)2 n S xx EXAMPLE 12.5 and the 95% conﬁdence interval is 79.06 2.306(2.840) 79.06 6.55 Our results indicate that the average calculus grade for students who score 50 on the achievement test will lie between 72.51 and 85.61. A student took the achievement test and scored 50 but has not yet taken the calculus test. Using the information in Example 12.1, predict the calculus grade for this student with a 95% prediction interval. Solution The predicted value of y is yˆ 79.06, as in Example 12.4. However, the error in prediction is measured by SE( y yˆ), and the 95% prediction interval is 474 (50 79.06 2.30675.75321 79.06 2.306(9.155) 79.06 21.11 46)2 1 0 1 2 or from 57.95 to 100.17. The prediction interval is wider than the conﬁdence interval in Example 12.4 because of the extra variability in predicting the actual value of the response y. 530 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION EXAMPLE 12.6 One particular point on the line of means is often of interest to experimenters, the y-intercept a—the average value of y when x0 0. Prior to ﬁtting a line to the calculus grade-achievement score data, you may have thought that a score of 0 on the achievement test would predict a grade of 0 on the calculus test. This implies that we should ﬁt a model with a equal to 0. Do the data support the hypothesis of a 0 intercept? Solution You can answer this question by constructing a 95% conﬁdence interval for the y-intercept a, which is the average value of y when x 0. The estimate of a is yˆ 40.784 .76556(0) 40.784 a and the 95% conﬁdence interval is n yˆ ta/2MSE1 40.784 2.30675.7532 40.784 19.617 x)2 (x0 474 (0 1 0 1 S 2 xx 6)2 4 or from 21.167 to 60.401, an interval that does not contain the value a 0. Hence, it is unlikely that the y-intercept is 0. You should include a nonzero intercept in the model y a bx e. For this special situation in which you are interested in testing or estimating the y-intercept a for the line of means, the inferences involve the sample estimate a. The test for a 0 intercept is given in Figure 12.14 in the shaded line labeled “Constant.” The coefficient given as 40.784 is a, with standard error given in the column labeled “SE Coef ” as 8.507, which agrees with the value calculated in Example 12.6. The value of t 4.79 is found by dividing a by its standard error with p-value .001. F IG URE 12. 14 Portion of the MINITAB output for Example 12.6 ● Predictor Coef SE Coef T P Constant 40.784 8.507 4.79 0.001 x 0.7656 0.1750 4.38 0.002 You can see that it is quite time-consuming to calculate these estimation and prediction intervals by hand. Moreover, it is difficult to maintain accuracy in your calculations. Fortunately, computer programs can perform these calculations for you. The MINITAB regression command provides an option for either estimation or prediction when you specify the necessary value(s) of x. The printout in Figure 12.15 gives the values of yˆ 79.06 labeled “Fit,” the standard error of yˆ, SE( yˆ), labeled “SE Fit,” the conﬁdence interval for the average value of y when x 50, labeled “95.0% CI,” and the prediction interval for y when x 50, labeled “95.0% PI.” F IG URE 12. 15 MINITAB option for estimation and prediction ● Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 79.06 2.84 (72.51, 85.61) (57.95, 100.17) Values of Predictors for New Observations New Obs x 1 50.0 12.7 ESTIMATION AND PREDICTION USING THE FITTED LINE ❍ 531 The conﬁdence bands and prediction bands generated by MINITAB for the calculus grades data are shown in Figure 12.16. Notice that in general the conﬁd
ence bands are narrower than the prediction bands for every value of the achievement test score x. Certainly you would expect predictions for an individual value to be much more variable than estimates of the average value. Also notice that the bands seem to get wider as the value of x0 gets farther from the mean x. This is because the standard errors used in the conﬁdence and prediction intervals contain the term (x0 x)2, which gets larger as the two values diverge. In practice, this means that estimation and prediction are more accurate when x0 is near the center of the range of the x-values. You can locate the calculated conﬁdence and prediction intervals when x 50 in Figure 12.16. ● FI GUR E 1 2. 16 Conﬁdence and prediction intervals for the data in Table 12.1 e d a r G 120 110 100 90 80 70 60 50 40 30 Fitted Line Plot y 40.78 0.7656 x Regression 95% CI 95% PI S R-Sq R-Sq(adj) 8.70363 70.5% 66.8% 20 30 40 50 Score 60 70 80 12.7 EXERCISES BASIC TECHNIQUES 12.36 Refer to Exercise 12.6. a. Estimate the average value of y when x 1, using a 90% conﬁdence interval. b. Find a 90% prediction interval for some value of y to be observed in the future when x 1. 12.37 Refer to Exercise 12.7. Portions of the MINITAB printout are shown here. MINITAB output for Exercise 12.37 Regression Analysis: y versus x The regression equation is y = 6.00 - 0.557 x Predictor Coef SE Coef T P Constant 6.0000 0.1759 34.10 0.000 x -0.55714 0.04518 -12.33 0.000 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 4.8857 0.1027 (4.6006, 5.1708) (4.2886, 5.4829) 2 1.5429 0.2174 (0.9392, 2.1466) (0.7430, 2.3427) X X denotes a point that is an outlier in the predictors. Values of Predictors for New Observations New Obs x 1 2.00 2 8.00 a. Find a 95% conﬁdence interval for the average value of y when x 2. b. Find a 95% prediction interval for some value of y to be observed in the future when x 2. c. The last line in the third section of the printout indicates a problem with one of the ﬁtted values. What value of x corresponds to the ﬁtted value yˆ 1.5429? What problem has the MINITAB program detected? APPLICATIONS 12.38 What to Buy? A marketing research experiment was conducted to study the rela- EX1238 tionship between the length of time necessary for a buyer to reach a decision and the number of alternative package designs of a product presented. Brand names were eliminated from the packages to reduce the effects of brand preferences. The buyers made their selections using the manufacturer’s product descriptions on the packages as the only buying guide. The length of time necessary to reach a decision was 532 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION recorded for 15 participants in the marketing research study. MINITAB output for Exercise 12.39 5, 8, 8, 7, 9 7, 9, 8, 9, 10 10, 11, 10, 12, 9 Regression Analysis: y versus x The regression equation is y = 251206 + 27.4 x Length of Decision Time, y (sec) Number of Alternatives, x 2 3 4 a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Calculate s2. d. Do the data present sufficient evidence to indicate that the length of decision time is linearly related to the number of alternative package designs? (Test at the a .05 level of signiﬁcance.) e. Find the approximate p-value for the test and inter- pret its value. f. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. g. Estimate the average length of time necessary to reach a decision when three alternatives are presented, using a 95% conﬁdence interval. EX1239 12.39 Housing Prices If you try to rent an apartment or buy a house, you ﬁnd that real estate representatives establish apartment rents and house prices on the basis of square footage of heated ﬂoor space. The data in the table give the square footages and sales prices of n 12 houses randomly selected from those sold in a small city. Use the MINITAB printout to answer the questions. Square Feet, x Price, y Square Feet, x Price, y 1460 2108 1743 1499 1864 2391 $288,700 309,300 301,400 291,100 302,400 314,900 1977 1610 1530 1759 1821 2216 $305,400 297,000 292,400 298,200 304,300 311,700 Plot of data for Exercise 12.39 315,000 310,000 305,000 300,000 ) e c i r P ( y 295,000 290,000 Predictor Coef SE Coef T P Constant 251206 3389 74.13 0.000 x 27.406 1.828 14.99 0.000 S = 1792.72 R-Sq = 95.7% R-Sq(adj) = 95.3% Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 299989 526 (298817, 301161) (295826, 304151) 2 306018 602 (304676, 307360) (301804, 310232) Values of Predictors for New Observations New Obs x 1 1780 2 2000 a. Can you see any pattern other than a linear relation- ship in the original plot? b. The value of r 2 for these data is .957. What does this tell you about the ﬁt of the regression line? c. Look at the accompanying diagnostic plots for these data. Do you see any pattern in the residuals? Does this suggest that the relationship between price and square feet is something other than linear? MINITAB output for Exercise 12.39 Normal Probability Plot of the Residuals (response is y) 99 95 90 80 70 60 50 40 30 20 10 5 t n e c r e P 1 5000 4000 3000 2000 1000 0 Residual 1000 2000 3000 4000 Residuals versus the Fitted Values (response is y) 3000 2000 1000 0 1000 2000 3000 l a u d i s e R 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 290,000 295,000 300,000 x (Square Feet) 305,000 Fitted Value 310,000 315,000 320,000 12.40 Housing Prices II Refer to Exercise 12.39 and data set EX1239. a. Estimate the average increase in the price for an increase of 1 square foot for houses sold in the city. Use a 99% conﬁdence interval. Interpret your estimate. b. A real estate salesperson needs to estimate the average sales price of houses with a total of 2000 square feet of heated space. Use a 95% conﬁdence interval and interpret your estimate. c. Calculate the price per square foot for each house and then calculate the sample mean. Why is this estimate of the average cost per square foot not equal to the answer in part a? Should it be? Explain. d. Suppose that a house with 1780 square feet of heated ﬂoor space is offered for sale. Construct a 95% prediction interval for the price at which the house will sell. 12.41 Strawberries III The following data (Exercises 12.16 and 12.24) were obtained in an experiment relating the dependent variable, y (texture of strawberries), with x (coded storage temperature). x 2 2 y 3.5 4.0 0 2.0 2 0.5 2 0.0 a. Estimate the expected strawberry texture for a coded storage temperature of x 1. Use a 99% conﬁdence interval. b. Predict the particular value of y when x 1 with a 99% prediction interval. c. At what value of x will the width of the prediction interval for a particular value of y be a minimum, assuming n remains ﬁxed? 12.42 Tom Brady The number of passes completed and the total number of passing EX1242 yards for Tom Brady, quarterback for the New England Patriots, were recorded for the 16 regular games in the 2006 football season.8 Week 6 was a bye and no data was reported. 12.8 CORRELATION ANALYSIS ❍ 533 Week Completions Total Yards 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 11 15 31 15 16 * 18 29 20 24 20 22 27 12 16 28 15 163 220 320 188 140 * 195 372 201 253 244 267 305 78 109 249 225 a. What is the least-squares line relating the total passing yards to the number of pass completions for Tom Brady? b. What proportion of the total variation is explained by the regression of total passing yards ( y) on the number of pass completions (x)? c. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. 12.43 Tom Brady, continued Refer to Exercise 12.42. a. Estimate the average number of passing yards for games in which Brady throws 20 completed passes using a 95% conﬁdence interval. b. Predict the actual number of passing yards for games in which Brady throws 20 completed passes using a 95% conﬁdence interval. c. Would it be advisable to use the least-squares line from Exercise 12.42 to predict Brady’s total number of passing yards for a game in which he threw only 5 completed passes? Explain. 12.8 CORRELATION ANALYSIS In Chapter 3, we introduced the correlation coefficient as a measure of the strength of the linear relationship between two variables. The correlation coefficient, r— formally called the Pearson product moment sample coefficient of correlation—is deﬁned next. 534 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION PEARSON PRODUCT MOMENT COEFFICIENT OF CORRELATION S s x y for 1 r 1 r x y Syy S s s x y x x r is always between 1 and 1. S s 2 sxy xy The variances and covariance can be found by direct calculation, by using a calculator with a two-variable statistics capacity, or by using a statistical package such as MINITAB. The variances and covariance are calculated as S s 2 x xx n S y yy n 1 and use Sxy, Sxx, and Syy, the same quantities used in regression analysis earlier in this chapter. In general, when a sample of n individuals or experimental units is selected and two variables are measured on each individual or unit so that both variables are random, the correlation coefficient r is the appropriate measure of linearity for use in this situation. 1 1 n EXAMPLE 12.7 The heights and weights of n 10 offensive backﬁeld football players are randomly selected from a county’s football all-stars. Calculate the correlation coefficient for the heights (in inches) and weights (in pounds) given in Table 12.4. TABLE 12.4 ● Heights and Weights of n 10 Backfield All-Stars Player Height, x Weight 10 73 71 75 72 72 75 67 69 71 69 185 175 200 210 190 195 150 170 180 175 Solution You should use the appropriate data entry method of your scientiﬁc calculator to verify the calculations for the sums of squares and cross-products, Syy 2610 Sxx 60.4 Sxy 328 using the calculational formulas given earlier in this chapter. Then 2 8 3 r .8261 (2610)
(60. ) 4 or r .83. This value of r is fairly close to 1, the largest possible value of r, which indicates a fairly strong positive linear relationship between height and weight. There is a direct relationship between the calculational formulas for the correlation coefficient r and the slope of the regression line b. Since the numerator of both quantities is Sxy, both r and b have the same sign. Therefore, the correlation coefficient has these general properties: • When r 0, the slope is b 0, and there is no linear relationship between x and y. • When r is positive, so is b, and there is a positive linear relationship between x and y. The sign of r is always the same as the sign of the slope b. • When r is negative, so is b, and there is a negative linear relationship between 12.8 CORRELATION ANALYSIS ❍ 535 x and y. In Section 12.5, we showed that SSE S Tota S R S S r 2 SS al t o ta SS l To l T In this form, you can see that r 2 can never be greater than 1, so that 1 r 1. Moreover, you can see the relationship between the random variation (measured by SSE) and r 2. • • If there is no random variation and all the points fall on the regression line, then SSE 0 and r 2 1. If the points are randomly scattered and there is no variation explained by regression, then SSR 0 and r 2 0. You can use the Exploring Correlation shown in Figure 12.17 applet to visualize the connection between the value of r and the pattern of points shown in the scatterplot. Use your mouse to move the slider at the bottom of the scatterplot. You will see the value of r change as the pattern of the points changes. Try to reproduce the patterns described above for r 2 1 and r 2 0. FI GUR E 1 2. 17 Exploring Correlation applet ● Figure 12.18 shows four typical scatterplots and their associated correlation coefﬁcients. Notice that in scatterplot (d) there appears to be a curvilinear relationship between x and y, but r is approximately 0, which reinforces the fact that r is a measure of a linear (not curvilinear) relationship between two variables. 536 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION F IG URE 12. 18 Some typical scatterplots with approximate values of r ● y (a) Strong positive linear correlation; r is near 1 y y y x x (b) Strong negative linear correlation; r is near –1 x x (c) No apparent linear correlation; r is near 0 (d) Curvilinear, but not linear, correlation; r is near 0 Consider a population generated by measuring two random variables on each experimental unit. In this bivariate population, the population correlation coefficient r (Greek lowercase rho) is calculated and interpreted as it is in the sample. In this situation, the experimenter can test the hypothesis that there is no correlation between the variables x and y using a test statistic that is exactly equivalent to the test of the slope b in Section 12.5. The test procedure is shown next. TEST OF HYPOTHESIS CONCERNING THE CORRELATION COEFFICIENT r 1. Null hypothesis: H0 : r 0 2. Alternative hypothesis: One-Tailed Test Ha : r 0 (or r 0) Two-Tailed Test Ha : r 0 3. Test statistic: t r 22 n 1 r When the assumptions given in Section 12.2 are satisﬁed, the test statistic will have a Student’s t distribution with (n 2) degrees of freedom. 4. Rejection region: Reject H0 when Two-Tailed Test t ta/2 or t ta/2 One-Tailed Test t ta (or t ta when the alternative hypothesis is Ha : r 0) or when p-value a You can prove that t r 2 2 r n 1 0 b . Sxx M S E/ 12.8 CORRELATION ANALYSIS ❍ 537 The values of ta and ta/2 can be found using Table 4 in Appendix I or the t-Probabilities applet. Use the values of t corresponding to (n 2) degrees of freedom. EXAMPLE 12.8 Refer to the height and weight data in Example 12.7. The correlation of height and weight was calculated to be r .8261. Is this correlation signiﬁcantly different from 0? The t-value and p-value for testing H0 : r 0 will be identical to the t- and p-value for testing H0 : b 0. Solution To test the hypotheses versus Ha : r 0 H0 : r 0 the value of the test statistic is t r .8261 10 1 (. 826 4.15 2 1)2 2 2 r n 1 which for n 10 has a t distribution with 8 degrees of freedom. Since this value is greater than t.005 3.355, the two-tailed p-value is less than 2(.005) .01, and the correlation is declared signiﬁcant at the 1% level (P .01). The value r 2 .82612 .6824 means that about 68% of the variation in one of the variables is explained by the other. The MINITAB printout in Figure 12.19 displays the correlation r and the exact p-value for testing its signiﬁcance. FI GUR E 1 2. 19 MINITAB output for Example 12.8 ● Correlations: x, y Pearson correlation of x and y = 0.826 P-Value = 0.003 If the linear coefficients of correlation between y and each of two variables x1 and x2 are calculated to be .4 and .5, respectively, it does not follow that a predictor using both variables will account for [(.4)2 (.5)2] .41, or a 41% reduction in the sum of squares of deviations. Actually, x1 and x2 might be highly correlated and therefore contribute virtually the same information for the prediction of y. Finally, remember that r is a measure of linear correlation and that x and y could be perfectly related by some curvilinear function when the observed value of r is equal to 0. The problem of estimating or predicting y using information given by several independent variables, x1, x2, . . . , xk, is the subject of Chapter 13. 12.8 EXERCISES BASIC TECHNIQUES 12.44 How does the coefficient of correlation measure the strength of the linear relationship between two variables y and x? 12.45 Describe the signiﬁcance of the algebraic sign and the magnitude of r. 12.46 What value does r assume if all the data points fall on the same straight line in these cases? a. The line has positive slope. b. The line has negative slope. 12.47 You are given these data. Plot the data points. Based on your graph, what will be the sign of the sample correlation coefficient? b. Calculate r and r 2 and interpret their values. 538 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION 12.48 You are given these data. Plot the six points on graph paper. b. Calculate the sample coefficient of correlation r and 0 interpret. c. By what percentage was the sum of squares of deviations reduced by using the least-squares predictor yˆ a bx rather than y as a predictor of y? 12.49 Reverse the slope of the line in Exercise 12.48 by reordering the y observations, as follows Repeat the steps of Exercise 12.48. Notice the change in the sign of r and the relationship between the values of r 2 of Exercise 12.48 and this exercise. APPLICATIONS EX1250 12.50 Lobster The table gives the numbers of Octolasmis tridens and O. lowei barnacles on each of 10 lobsters.9 Does it appear that the barnacles compete for space on the surface of a lobster? Lobster Field Number O. tridens O. lowei AO61 AO62 AO66 AO70 AO67 AO69 AO64 AO68 AO65 AO63 645 320 401 364 327 73 20 221 3 5 6 23 40 9 24 5 86 0 109 350 a. If they do compete, do you expect the number x of O. tridens and the number y of O. lowei barnacles to be positively or negatively correlated? Explain. b. If you want to test the theory that the two types of barnacles compete for space by conducting a test of the null hypothesis “the population correlation coefﬁcient r equals 0,” what is your alternative hypothesis? c. Conduct the test in part b and state your conclusions. 12.51 Social Skills Training A social skills training program was implemented with seven EX1251 mildly challenged students in a study to determine whether the program caused improvement in pre/post measures and behavior ratings. For one such test, the pre- and posttest scores for the seven students are given in the table.10 Subject Pretest Posttest Earl Ned Jasper Charlie Tom Susie Lori 101 89 112 105 90 91 89 113 89 121 99 104 94 99 a. What type of correlation, if any, do you expect to see between the pre- and posttest scores? Plot the data. Does the correlation appear to be positive or negative? b. Calculate the correlation coefficient, r. Is there a signiﬁcant positive correlation? 12.52 Hockey G. W. Marino investigated the variables related to a hockey player’s ability to make a fast start from a stopped position.11 In the experiment, each skater started from a stopped position and attempted to move as rapidly as possible over a 6-meter distance. The correlation coefficient r between a skater’s stride rate (number of strides per second) and the length of time to cover the 6-meter distance for the sample of 69 skaters was .37. a. Do the data provide sufficient evidence to indicate a correlation between stride rate and time to cover the distance? Test using a .05. b. Find the approximate p-value for the test. c. What are the practical implications of the test in part a? 12.53 Hockey II Refer to Exercise 12.52. Marino calculated the sample correlation coefficient r for the stride rate and the average acceleration rate for the 69 skaters to be .36. Do the data provide sufficient evidence to indicate a correlation between stride rate and average acceleration for the skaters? Use the p-value approach. EX1254 12.54 Geothermal Power Geothermal power is an important source of energy. Since the amount of energy contained in 1 pound of water is a function of its temperature, you might wonder whether water obtained from deeper wells contains more energy per pound. The data in the table are reproduced from an article on geothermal systems by A.J. Ellis.12 Location of Well El Tateo, Chile Ahuachapan, El Salvador Namafjall, Iceland Larderello (region), Italy Matsukawa, Japan Cerro Prieto, Mexico Wairakei, New Zealand Kizildere, Turkey The Geysers, United States Average (max.) Drill Hole Depth (m) Average (max.) Temperature (°C) 650 1000 1000 600 1000 800 800 700 1500 230 230 250 200 220 300 230 190 250 Is there a signiﬁcant positive correlation between average maximum drill hole depth and average maximum temperature? 12.55 Cheese, Please! The demand for healthy foods that are low in fat and calories has
resulted in a large number of “low-fat” or “fat-free” products. The table shows the number of calories and the amount of sodium (in milligrams) per slice for ﬁve different brands of fat-free American cheese. Brand Sodium (mg) Calories Kraft Fat Free Singles Ralphs Fat Free Singles Borden® Fat Free Healthy Choice® Fat Free Smart Beat® American 300 300 320 290 180 30 30 30 30 25 a. Should you use the methods of linear regression analysis or correlation analysis to analyze the data? Explain. b. Analyze the data to determine the nature of the relationship between sodium and calories in fat-free American cheese. Use any statistical tests that are appropriate. 12.56 Body Temperature and Heart Rate Is there any relationship between these two EX1256 variables? To ﬁnd out, we randomly selected 12 people from a data set constructed by Allen Shoemaker (Journal of Statistics Education) and recorded their body temperature and heart rate.13 12.8 CORRELATION ANALYSIS ❍ 539 Person Temperature (degrees) Heart Rate (beats per minute) Person Temperature (degrees) Heart Rate (beats per minute) 1 2 3 4 5 6 96.3 97.4 98.9 99.0 99.0 96.8 70 68 80 75 79 75 7 8 9 10 11 12 98.4 98.4 98.8 98.8 99.2 99.3 74 84 73 84 66 68 a. Find the correlation coefficient r, relating body temperature to heart rate. b. Is there sufficient evidence to indicate that there is a correlation between these two variables? Test at the 5% level of signiﬁcance. EX1257 12.57 Baseball Stats Does a team’s batting average depend in any way on the number of home runs hit by the team? The data in the table show the number of team home runs and the overall team batting average for eight selected major league teams for the 2006 season.14 Team Total Home Runs Team Batting Average Atlanta Braves Baltimore Orioles Boston Red Sox Chicago White Sox Houston Astros Philadelphia Phillies New York Giants Seattle Mariners Source: ESPN.com 222 164 192 236 174 216 163 172 .270 .227 .269 .280 .255 .267 .259 .272 a. Plot the points using a scatterplot. Does it appear that there is any relationship between total home runs and team batting average? b. Is there a signiﬁcant positive correlation between total home runs and team batting average? Test at the 5% level of signiﬁcance. c. Do you think that the relationship between these two variables would be different if we had looked at the entire set of major league franchises? 540 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION CHAPTER REVIEW Key Concepts and Formulas I. A Linear Probabilistic Model 1. When the data exhibit a linear relationship, the appropriate model is y a bx e. 2. The random error e has a normal distribution with mean 0 and variance s 2. II. Method of Least Squares 1. Estimates a and b, for a and b, are chosen to minimize SSE, the sum of squared deviations about the regression line, yˆ a bx. 2. The least-squares estimates are b Sxy/Sxx and a y bx. 3. Use residual plots to check for nonnormality, inequality of variances, or an incorrectly ﬁt model. 4. Conﬁdence intervals can be constructed to estimate the intercept a and slope b of the regression line and to estimate the average value of y, E( y), for a given value of x. 5. Prediction intervals can be constructed to predict a particular observation, y, for a given value of x. For a given x, prediction intervals are always wider than conﬁdence intervals. III. Analysis of Variance V. Correlation Analysis 1. Total SS SSR SSE, where Total SS Syy and SSR (Sxy)2/Sxx. 2. The best estimate of s 2 is MSE SSE/(n 2). 1. Use the correlation coefficient to measure the relationship between x and y when both variables are random: IV. Testing, Estimation, and Prediction 1. A test for the signiﬁcance of the linear regression—H0 : b 0—can be implemented using one of two test statistics: b S R or F M t E/Sxx MS E S M 2. The strength of the relationship between x and y can be measured using R S S r 2 l SS ta To S x y r Syy S x x 2. The sign of r indicates the direction of the relationship; r near 0 indicates no linear relationship, and r near 1 or 1 indicates a strong linear relationship. 3. A test of the signiﬁcance of the correlation coefficient uses the statistic t r 22 n 1 r which gets closer to 1 as the relationship gets stronger. and is identical to the test of the slope b. Linear Regression Procedures In Chapter 3, we used some of the linear regression procedures available in MINITAB to obtain a graph of the best-ﬁtting least-squares regression line and to calculate the correlation coefficient r for a bivariate data set. Now that you have studied the testing and estimation techniques for a simple linear regression analysis, more MINITAB options are available to you. MY MINITAB ❍ 541 Consider the relationship between x mathematics achievement test score and y ﬁnal calculus grade, which was used as an example throughout this chapter. Enter the data into the ﬁrst two columns of a MINITAB worksheet. If you use Graph Scatterplot Simple, you can generate the scatterplot for the data, as shown in Figure 12.2. However, the main inferential tools for linear regression analysis are generated using Stat Regression Regression. (You will use this same sequence of commands in Chapter 13 when you study multiple regression analysis.) The Dialog box for the Regression command is shown in Figure 12.20. Select y for the “Response” variable and x for the “Predictor” variable. You may now choose to generate some residual plots to check the validity of your regression assumptions before you use the model for estimation or prediction. Choose Graphs to display the Dialog box in Figure 12.21. FI GUR E 1 2. 20 ● FI GUR E 1 2. 21 ● 542 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION We have used Regular residual plots, checking the boxes for “Normal plot of residuals” and “Residuals versus ﬁts.” Click OK to return to the main Dialog box. If you now choose Options, you can obtain conﬁdence and prediction intervals for either of these cases: • A single value of x (typed in the box marked “Prediction intervals for new observations”) • Several values of x stored in a column (say, C3) of the worksheet Enter the value x 50 in Figure 12.22 to match the output given in Figure 12.15. When you click OK twice, the regression output is generated as shown in Figure 12.23. The two diagnostic plots will appear in separate graphics windows. F IG URE 12. 22 ● F IG URE 12. 23 ● SUPPLEMENTARY EXERCISES ❍ 543 If you wish, you can now plot the data points, the regression line, and the upper and lower conﬁdence and prediction limits (see Figure 12.16) using Stat Regression Fitted Line Plot. Select y and x for the response and predictor variables and click “Display conﬁdence interval” and “Display prediction interval” in the Options Dialog box. Make sure that Linear is selected as the “Type of Regression Model,” so that you will obtain a linear ﬁt to the data. Recall that in Chapter 3, we used the command Stat Basic Statistics Correlation to obtain the value of the correlation coefficient r. Make sure that the box marked “Display p-values” is checked. The output for this command (using the test/grade data) is shown in Figure 12.24. Notice that the p-value for the test of H0 : r 0 is identical to the p-value for the test of H0 : b 0 because the tests are exactly equivalent! FI GUR E 1 2. 24 ● Supplementary Exercises 12.58 Potency of an Antibiotic An experiment was conducted to observe the effect of EX1258 an increase in temperature on the potency of an antibiotic. Three 1-ounce portions of the antibiotic were stored for equal lengths of time at each of these temperatures: 30°, 50°, 70°, and 90°. The potency readings observed at each temperature of the experimental period are listed here: Potency Readings, y 38, 43, 29 32, 26, 33 19, 27, 23 14, 19, 21 Temperature, x 30° 50° 70° 90° Use an appropriate computer program to answer these questions: a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Construct the ANOVA table for linear regression. d. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. e. Estimate the change in potency for a 1-unit change in temperature. Use a 95% conﬁdence interval. f. Estimate the average potency corresponding to a temperature of 50°. Use a 95% conﬁdence interval. g. Suppose that a batch of the antibiotic was stored at 50° for the same length of time as the experimental period. Predict the potency of the batch at the end of the storage period. Use a 95% prediction interval. 12.59 Plant Science An experiment was conducted to determine the effect of soil appli- EX1259 cations of various levels of phosphorus on the inorganic phosphorus levels in a particular plant. The data in the table represent the levels of inorganic phosphorus in micromoles (mmol) per gram dry weight of Sudan grass roots grown in the greenhouse for 28 days, in the absence of zinc. Use the MINITAB output to answer the questions. 544 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION Phosphorus Applied, x .5 mmol .25 mmol .10 mmol Phosphorus in Plant, y 204 195 247 245 159 127 95 144 128 192 84 71 a. Plot the data. Do the data appear to exhibit a linear Use an appropriate computer software package to analyze these data. State any conclusions you can draw. 12.61 Nematodes Some varieties of nematodes, roundworms that live in the soil and fre- EX1261 quently are so small as to be invisible to the naked eye, feed on the roots of lawn grasses and other plants. This pest, which is particularly troublesome in warm climates, can be treated by the application of nematicides. Data collected on the percent kill of nematodes for various rates of application (dosages given in pounds per acre of active ingredient) are as follows: Rate of Application, x 2 Percent Kill, y 3 4 5 50, 56, 48 63, 69, 71 86, 82, 76 94, 99, 97 relationship? MINITAB diagnostic plots for Exercise 12.61 b. Find the least-squares line relating the plant p
hosphorus levels y to the amount of phosphorus applied to the soil x. Graph the least-squares line as a check on your answer. c. Do the data provide sufficient evidence to indicate that the amount of phosphorus present in the plant is linearly related to the amount of phosphorus applied to the soil? d. Estimate the mean amount of phosphorus in the plant if .20 mmol of phosphorus is applied to the soil, in the absence of zinc. Use a 90% conﬁdence interval. MINITAB output for Exercise 12.59 Regression Analysis: y versus x The regression equation is y = 80.9 + 271 x Predictor Coef SE Coef T P Constant 80.85 22.40 3.61 0.005 x 270.82 68.31 3.96 0.003 S = 39.0419 R-Sq = 61.1% R-Sq(adj) = 57.2% Predicted Values for New Observations New Obs Fit SE Fit 90.0% CI 90.0% PI 1 135.0 12.6 (112.1, 157.9) (60.6, 209.4) Values of Predictors for New Observations New Obs x 1 0.200 Normal Probability Plot of the Residuals (response is y) 99 95 90 80 70 60 50 40 30 20 10 5 t n e c r e P 1 10 5 0 Residual 5 10 Residuals versus the Fitted Values (response is y) 5.0 2.5 0.0 2.5 5.0 l a u d i s e R EX1260 12.60 Track Stats! An experiment was conducted to investigate the effect of a training program on the length of time for a typical male college student to complete the 100-yard dash. Nine students were placed in the program. The reduction y in time to complete the 100-yard dash was measured for three students at the end of 2 weeks, for three at the end of 4 weeks, and for three at the end of 6 weeks of training. The data are given in the table. Reduction in Time, y (sec) 1.6, .8, 1.0 2.1, 1.6, 2.5 3.8, 2.7, 3.1 Length of Training, x (wk) 2 4 6 50 60 70 80 90 100 Fitted Value Use an appropriate computer printout to answer these questions: a. Calculate the coefficient of correlation r between rates of application x and percent kill y. b. Calculate the coefficient of determination r 2 and interpret. c. Fit a least-squares line to the data. d. Suppose you wish to estimate the mean percent kill for an application of 4 pounds of the nematicide per acre. What do the diagnostic plots generated by MINITAB tell you about the validity of the regression assumptions? Which assumptions may have been violated? Can you explain why? sizes. The table that follows gives the actual and estimated lengths of the speciﬁed objects. Object Estimated (inches) Actual (inches) SUPPLEMENTARY EXERCISES ❍ 545 12.62 Knee Injuries Athletes and others suffering the same type of injury to the knee often require anterior and posterior ligament reconstruction. In order to determine the proper length of bone-patellar tendonbone grafts, experiments were done using three imaging techniques to determine the required length of the grafts, and these results were compared to the actual length required. A summary of the results of a simple linear regression analysis for each of these three methods is given in the following table.15 Imaging Technique Radiographs Standard MRI 3-dimensional MRI Coefficient of Determination, r 2 0.80 0.43 0.65 Intercept 3.75 20.29 1.80 p-value Slope 1.031 0.0001 0.497 0.011 0.977 0.0001 a. What can you say about the signiﬁcance of each of the three regression analyses? b. How would you rank the effectiveness of the three regression analyses? What is the basis of your decision? c. How do the values of r 2 and the p-values compare in determining the best predictor of actual graft lengths of ligament required? EX1263 12.63 Achievement Tests II Refer to Exercise 12.11 and data set EX1211 regarding the relationship between the Academic Performance Index (API), a measure of school achievement based on the results of the Stanford 9 Achievement test, and the percentage of students who are considered English Language Learners (ELL). The following table shows the API for eight elementary schools in Riverside County, California, along with the percentage of students at that school who are considered English Language Learners.3 School 1 API ELL 588 58 2 659 22 3 4 710 14 657 30 5 669 11 6 641 26 7 557 39 8 743 6 a. Use an appropriate program to analyze the relation- ship between API and ELL. b. Explain all pertinent details of your analysis. 12.64 How Long Is It? Refer to Exercise 12.12 and data set EX1212 regarding a subject’s ability to estimate Pencil Dinner plate Book 1 Cell phone Photograph Toy Belt Clothespin Book 2 Calculator 7.00 9.50 7.50 4.00 14.50 3.75 42.00 2.75 10.00 3.50 6.00 10.25 6.75 4.25 15.75 5.00 41.50 3.75 9.25 4.75 a. Use an appropriate program to analyze the relationship between the actual and estimated lengths of the listed objects. b. Explain all pertinent details of your analysis. 12.65 Tennis, Anyone? If you play tennis, you know that tennis racquets vary in their EX1265 physical characteristics. The data in the accompanying table give measures of bending stiffness and twisting stiffness as measured by engineering tests for 12 tennis racquets: Racquet Bending Stiffness, x Twisting Stiffness 10 11 12 419 407 363 360 257 622 424 359 346 556 474 441 227 231 200 211 182 304 384 194 158 225 305 235 a. If a racquet has bending stiffness, is it also likely to have twisting stiffness? Do the data provide evidence that x and y are positively correlated? b. Calculate the coefficient of determination r 2 and interpret its value. 12.66 Avocado Research Movement of avocados into the United States from certain areas is EX1266 prohibited because of the possibility of bringing fruit ﬂies into the country with the avocado shipments. However, certain avocado varieties supposedly are resistant to fruit ﬂy infestation before they soften as a result of ripening. The data in the table resulted from an experiment in 546 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION which avocados ranging from 1 to 9 days after harvest were exposed to Mediterranean fruit ﬂies. Penetrability of the avocados was measured on the day of exposure, and the percentage of the avocado fruit infested was assessed. Days after Harvest Penetrability Percentage Infected 1 2 4 5 6 7 9 .91 .81 .95 1.04 1.22 1.38 1.77 30 40 45 57 60 75 100 Use the MINITAB printout of the regression of percentage infected ( y) on days after harvest (x) to analyze the relationship between these two variables. Explain all pertinent parts of the printout and interpret the results of any tests. MINITAB output for Exercise 12.66 Regression Analysis: Percent versus x The regression equation is Percent = 18.4 + 8.18 x Predictor Coef SE Coef T P Constant 18.427 5.110 3.61 0.015 x 8.1768 0.9285 8.81 0.000 S = 6.35552 R-Sq = 93.9% R-Sq(adj) = 92.7% Analysis of Variance Source DF SS MS F P Regression 1 3132.9 3132.9 77.56 0.000 Residual Error 5 202.0 40.4 Total 6 3334.9 12.67 Avocados II Refer to Exercise 12.66. Suppose the experimenter wants to examine the relationship between the penetrability and the number of days after harvest. Does the method of linear regression discussed in this chapter provide an appropriate method of analysis? If not, what assumptions have been violated? Use the MINITAB diagnostic plots provided. MINITAB diagnostic plots for Exercise 12.67 Normal Probability Plot of the Residuals (response is Penetrability) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 0.3 0.2 0.1 0.0 Residual 0.1 0.2 0.3 Residuals versus the Fitted Values (response is Penetrability) 0.20 0.15 0.10 0.05 0.00 0.05 0.10 l a u d i s e R 0.6 0.8 1.0 1.2 1.4 1.6 Fitted Value 12.68 Metabolism and Weight Gain Why is it that one person may tend to gain weight, EX1268 even if he eats no more and exercises no less than a slim friend? Recent studies suggest that the factors that control metabolism may depend on your genetic makeup. One study involved 11 pairs of identical twins fed about 1000 calories per day more than needed to maintain initial weight. Activities were kept constant, and exercise was minimal. At the end of 100 days, the changes in body weight (in kilograms) were recorded for the 22 twins.16 Is there a signiﬁcant positive correlation between the changes in body weight for the twins? Can you conclude that this similarity is caused by genetic similarities? Explain. Pair Twin A Twin 10 11 4.2 5.5 7.1 7.0 7.8 8.2 8.2 9.1 11.5 11.2 13.0 7.3 6.5 5.7 7.2 7.9 6.4 6.5 8.2 6.0 13.7 11.0 12.69 Movie Reviews How many weeks can a movie run and still make a reasonable EX1269 proﬁt? The data that follow show the number of weeks in release (x) and the gross to date (y) for the top 10 movies during a recent week.17 Movie 1. The Prestige 2. The Departed 3. Flags of Our Fathers 4. Open Season 5. Flicka 6. The Grudge 2 7. Man of the Year 8. Marie Antoinette 9. The Texas Chainsaw Massacre: The Beginning 10. The Marine Source: Entertainment Weekly Gross to Date (in millions) Weeks in Release $14.8 $77.1 $10.2 $69.6 $ 7.7 $31.4 $22.5 $ 5.3 $36.0 $12.. Plot the points in a scatterplot. Does it appear that the relationship between x and y is linear? How would you describe the direction and strength of the relationship? b. Calculate the value of r 2. What percentage of the overall variation is explained by using the linear model rather than y to predict the response variable y? c. What is the regression equation? Do the data provide evidence to indicate that x and y are linearly related? Test using a 5% signiﬁcance level. d. Given the results of parts b and c, is it appropriate to use the regression line for estimation and prediction? Explain your answer. 12.70 In addition to increasingly large bounds on error, why should an experimenter refrain from predicting y for values of x outside the experimental region? 12.71 If the experimenter stays within the experimental region, when will the error in predicting a particular value of y be maximum? EX1272 12.72 Oatmeal, Anyone? An agricultural experimenter, investigating the effect of the amount of nitrogen x applied in 100 pounds per acre on the yield of oats y measured in bushels per acre, collected the following data: x y 1 22 19 2 38 41 3 57 54 4 68 65 a. Find the least-squares line for the
data. b. Construct the ANOVA table. c. Is there sufficient evidence to indicate that the yield of oats is linearly related to the amount of nitrogen applied? Use a .05. d. Predict the expected yield of oats with 95% conﬁ- dence if 250 pounds of nitrogen per acre are applied. SUPPLEMENTARY EXERCISES ❍ 547 e. Estimate the average increase in yield for an increase of 100 pounds of nitrogen per acre with 99% conﬁdence. f. Calculate r 2 and explain its signiﬁcance in terms of predicting y, the yield of oats. 12.73 Fresh Roses A horticulturalist devised a scale to measure the freshness EX1273 of roses that were packaged and stored for varying periods of time before transplanting. The freshness measurement y and the length of time in days that the rose is pack-aged and stored before transplanting x are given below. x y 5 15.3 16.8 10 13.6 13.8 15 9.8 8.7 20 5.5 4.7 25 1.8 1.0 a. Fit a least-squares line to the data. b. Construct the ANOVA table. c. Is there sufficient evidence to indicate that freshness is linearly related to storage time? Use a .05. d. Estimate the mean rate of change in freshness for a 1-day increase in storage time usig a 98% conﬁdence interval. e. Estimate the expected freshness measurement for a storage time of 14 days with a 95% conﬁdence interval. f. Of what value is the linear model in reference to y in predicting freshness? EX1274 12.74 Lexus, Inc. The makers of the Lexus automobile have steadily increased their sales since their U.S. launch in 1989. However, the rate of increase changed in 1996 when Lexus introduced a line of trucks. The sales of Lexus from 1996 to 2005 are shown in the table:18 Year 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 80 100 155 180 210 224 234 260 288 303 Sales (thousands of vehicles) Source: Adapted from: Automotive News, 26 January 2004 and 22 May 2006 a. Plot the data using a scatterplot. How would you describe the relationship between year and sales of Lexus? b. Find the least-squares regression line relating the sales of Lexus to the year being measured? c. Is there sufficient evidence to indicate that sales are linearly related to year? Use a .05. 548 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION d. Predict the sales of Lexus for the year 2006 using Data Display (continued) a 95% prediction interval. e. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. f. If you were to predict the sales of Lexus in the year 2015, what problems might arise with your prediction? EX1275 12.75 Starbucks Here is some nutritional data for a sampling of Starbucks products (16 ﬂuid ounces), taken from the company website, www.starbucks.com.19 The complete data set (starbucks.mtp) can be found with the other data sets on the Student Companion Website. Data Display Row Product 1 CaffèMocha-nowhip 2 CaramelFrappuccino® BlendedCoffee-nowhip 3 ChocolateBrownie Frappuccino® BlendedCoffee-nowhip 4 ChocolateMalt Frappuccino® BlendedCrème-whip 5 EggnogLatte-nowhip 6 HotChocolate-nowhip 7 IcedCaffèMocha-whip 8 IcedWhiteChocolate Mocha-whip 9 MochaFrappuccino® BlendedCoffee-whip 10 PeppermintMocha-nowhip 11 Tazo®ChaiCrème Frappuccino® BlendedTea-nowhip 12 ToffeeNutCrème-whip 13 ToffeNutLatte-nowhip 14 VanillaFrappuccino® BlendedCrème-whip 15 WhiteHotChocolate-whip Calories 300 280 Fat Calories 110 30 370 80 610 200 410 340 350 490 420 370 370 460 330 480 580 180 140 180 210 150 110 40 220 120 150 250 Total Fat (g) 12.0 3.5 9.0 22.0 20.0 15.0 20.0 24.0 16.0 12.0 4.5 24.0 13.0 17.0 28.0 Saturated Fat (g) 7 2 6 11 12 8 12 16 10 6 1 15 8 9 19 Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Cholesterol (mg) 40 15 15 65 115 50 75 75 65 40 5 100 50 55 95 Row Total Carbs (g) 41 57 69 90 41 42 37 58 61 59 69 45 41 66 65 Sugar (g) 31 48 56 72 38 35 27 54 51 49 64 44 38 62 64 Sodium (mg) 150 250 310 430 240 190 105 220 260 150 370 380 340 380 310 Protein (g) 13 5 7 15 17 15 9 11 6 13 15 14 13 15 17 Fiber (g 10 2 11 0 12 0 13 0 14 0 15 0 Use the appropriate statistical methods to analyze the relationships between some of the nutritional variables given in the table. Write a summary report explaining any conclusions that you can draw from your analysis. MYAPPLET EXERCISES ❍ 549 Exercises You can refresh your memory about regression lines and the correlation coefficient by doing the MyApplet Exercises at the end of Chapter 3. 12.76 a. Graph the line corresponding to the equation y 0.5x 3 by graphing the points corresponding to x 0, 1, and 2. Give the y-intercept and slope for the line. b. Check your graph using the How a Line Works applet. 12.77 a. Graph the line corresponding to the equation y 0.5x 3 by graphing the points corresponding to x 0, 1, and 2. Give the y-intercept and slope for the line. b. Check your graph using the How a Line Works applet. c. How is this line related to the line y 0.5x 3 of Exercise 12.76? 12.78 The MINITAB printout for the data in Table 12.1 is shown below. MINITAB output for Exercise 12.78. Regression Analysis: y versus x The regression equation is y = 40.8 + 0.766 x Predictor Coef SE Coef T P Constant 40.784 8.507 4.79 0.001 x 0.7656 0.1750 4.38 0.002 b. Find the values of SSE and r 2 on the Method of Least Squares applet. Find these values on the MINITAB printout and conﬁrm that they are the same. c. Use the values of b and its standard error SE(b) from the MINITAB printout along with the t-Test for the Slope applet to verify the value of the t statistic and its p-value, given in the printout. 12.79 Use the ﬁrst applet in Building a Scatterplot to create a scatterplot for the data in Table 12.1. Verify your plot using Figure 12.2. 12.80 Walking Shoes Is your overall satisfaction with your new pair of walking shoes EX1280 correlated with the cost of the shoes? Satisfaction scores and prices were recorded for nine different styles and brands of men’s walking shoes, with the following results:20 Brand and Style Price Score New Balance MW 791 Saucony Grid Omni Walker Asics Gel-Walk Tech New Balance MW 557 Etonic Lite Walker Nike Air Max Healthwalker V Rockport Astride Rockport WT Classic Reebok Move DMX Max $75 90 60 60 60 70 90 90 60 89 84 83 83 79 78 75 72 67 S = 8.70363 R-Sq = 70.5% R-Sq(adj) = 66.8% Source: “Ratings: Walking Shoes,” Consumer Reports, October 2006, p. 52. Analysis of Variance Source DF SS MS F P Regression 1 1450.0 1450.0 19.14 0.002 Residual Error 8 606.0 75.8 Total 9 2056.0 a. Use the Method of Least Squares applet to ﬁnd the values of a and b that determine the best ﬁtting line, yˆ a bx. When you think that you have minimized SSE, click the button and see how well you did. What is the equation of the line? Does it match the regression equation given in the MINITAB printout? a. Calculate the correlation coefficient r between price and overall score. How would you describe the relationship between price and overall score? b. Use the applet called Correlation and the Scatterplot to plot the nine data points. What is the correlation coefficient shown on the applet? Compare with the value you calculated in part a. c. Describe the pattern that you see in the scatterplot. Are there any outliers? If so, how would you explain them? 550 ❍ CHAPTER 12 LINEAR REGRESSION AND CORRELATION CASE STUDY Foreign Cars Is Your Car “Made in the U.S.A.”? The phrase “made in the U.S.A.” has become a familiar battle cry as U.S. workers try to protect their jobs from overseas competition. For the past few decades, a major trade imbalance in the United States has been caused by a ﬂood of imported goods that enter the country and are sold at lower cost than comparable American-made goods. One prime concern is the automotive industry, in which the number of imported cars steadily increased during the 1970s and 1980s. The U.S. automobile industry has been besieged with complaints about product quality, worker layoffs, and high prices, and has spent billions in advertising and research to produce an American-made car that will satisfy consumer demands. Have they been successful in stopping the ﬂood of imported cars purchased by American consumers? The data in the table represent the numbers of imported cars y sold in the United States (in millions) for the years 1969–2005.21 To simplify the analysis, we have coded the year using the coded variable x Year 1969. Year (Year 1969), x Number of Imported Cars, y Year (Year 1969), x Number of Imported Cars, y 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 10 11 12 13 14 15 16 17 1.1 1.3 1.6 1.6 1.8 1.4 1.6 1.5 2.1 2.0 2.3 2.4 2.3 2.2 2.4 2.4 2.8 3.2 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 3.1 3.1 2.8 2.5 2.1 2.0 1.8 1.8 1.6 1.4 1.4 1.4 1.8 2.1 2.2 2.3 2.2 2.2 2.3 1. Using a scatterplot, plot the data for the years 1969–1988. Does there appear to be a linear relationship between the number of imported cars and the year? 2. Use a computer software package to ﬁnd the least-squares line for predicting the number of imported cars as a function of year for the years 1969–1988. 3. Is there a signiﬁcant linear relationship between the number of imported cars and the year? 4. Use the computer program to predict the number of cars that will be imported us- ing 95% prediction intervals for each of the years 2003, 2004, and 2005. 5. Now look at the actual data points for the years 2003–2005. Do the predictions obtained in step 4 provide accurate estimates of the actual values observed in these years? Explain. 6. Add the data for 1989–2005 to your database, and recalculate the regression line. What effect have the new data points had on the slope? What is the effect on SSE? 7. Given the form of the scatterplot for the years 1969–2005, does it appear that a straight line provides an accurate model for the data? What other type of model might be more appropriate? (Use residual plots to help answer this question.) 13 Multiple Regression Analysis GENERA
L OBJECTIVES In this chapter, we extend the concepts of linear regression and correlation to a situation where the average value of a random variable y is related to several independent variables—x1, x2, . . . , xk—in models that are more ﬂexible than the straight-line model of Chapter 12. With multiple regression analysis, we can use the information provided by the independent variables to ﬁt various types of models to the sample data, to evaluate the usefulness of these models, and ﬁnally to estimate the average value of y or predict the actual value of y for given values of x1, x2, . . . , xk. CHAPTER INDEX ● Adjusted R 2 (13.3) ● The analysis of variance F-test (13.3) ● Analysis of variance for multiple regression (13.3) ● Causality and multicollinearity (13.9) ● The coefficient of determination R 2 (13.3) ● Estimation and prediction using the regression model (13.3) ● The general linear model and assumptions (13.2) ● The method of least squares (13.3) ● Polynomial regression model (13.4) ● Qualitative variables in a regression model (13.5) ● Residual plots (13.3) ● Sequential sums of squares (13.3) ● Stepwise regression analysis (13.8) ● Testing the partial regression coefficients (13.3) ● Testing sets of regression coefficients (13.6) © Will & Deni McIntyre/CORBIS “Made in the U.S.A.”— Another Look In Chapter 12, we used simple linear regression analysis to try to predict the number of cars imported into the United States over a period of years. Unfortunately, the number of imported cars does not really follow a linear trend pattern, and our predictions were far from accurate. We reexamine the same data at the end of this chapter, using the methods of multiple regression analysis. 551 552 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS INTRODUCTION 13.1 Multiple linear regression is an extension of simple linear regression to allow for more than one independent variable. That is, instead of using only a single independent variable x to explain the variation in y, you can simultaneously use several independent (or predictor) variables. By using more than one independent variable, you should do a better job of explaining the variation in y and hence be able to make more accurate predictions. For example, a company’s regional sales y of a product might be related to three factors: • • • x1—the amount spent on television advertising x2—the amount spent on newspaper advertising x3—the number of sales representatives assigned to the region A researcher would collect data measuring the variables y, x1, x2, and x3, and then use these sample data to construct a prediction equation relating y to the three predictor variables. Of course, several questions arise, just as they did with simple linear regression: • How well does the model ﬁt? • How strong is the relationship between y and the predictor variables? • Have any important assumptions been violated? • How good are estimates and predictions? The methods of multiple regression analysis—which are almost always done with a computer software program—can be used to answer these questions. This chapter provides a brief introduction to multiple regression analysis and the difficult task of model building—that is, choosing the correct model for a practical application. THE MULTIPLE REGRESSION MODEL 13.2 The general linear model for a multiple regression analysis describes a particular response y using the model given next. GENERAL LINEAR MODEL AND ASSUMPTIONS y b0 b1x1 b2x2 bkxk e where y is the response variable that you want to predict. • • b0, b1, b2, . . . , bk are unknown constants. • • x1, x2, . . . , xk are independent predictor variables that are measured without error. e is the random error, which allows each response to deviate from the average value of y by the amount e. You must assume that the values of e (1) are independent; (2) have a mean of 0 and a common variance s 2 for any set x1, x2, . . . , xk; and (3) are normally distributed. When these assumptions about e are met, the average value of y for a given set of values x1, x2, . . . , xk is equal to the deterministic part of the model: E(y) b0 b1x1 b2x2 bkxk 13.3 A MULTIPLE REGRESSION ANALYSIS ❍ 553 You will notice that the multiple regression model and assumptions are very similar to the model and assumptions used for linear regression. It will probably not surprise you that the testing and estimation procedures are also extensions of those used in Chapter 12. Multiple regression models are very ﬂexible and can take many forms, depending on the way in which the independent variables x1, x2, . . . , xk are entered into the model. We begin with a simple multiple regression model, explaining the basic concepts and procedures with an example. As you become more familiar with the multiple regression procedures, we increase the complexity of the examples, and you will see that the same procedures can be used for models of different forms, depending on the particular application. EXAMPLE 13.1 Suppose you want to relate a random variable y to two independent variables x1 and x2. The multiple regression model is y b0 b1x1 b2x2 e with the mean value of y given as E(y) b0 b1x1 b2x2 This equation is a three-dimensional extension of the line of means from Chapter 12 and traces a plane in three-dimensional space (see Figure 13.1). The constant b0 is called the intercept—the average value of y when x1 and x2 are both 0. The coefficients b1 and b2 are called the partial slopes or partial regression coefficients. The partial slope bi (for i 1 or 2) measures the change in y for a one-unit change in xi when all other independent variables are held constant. The value of the partial regression coefficient—say, b1—with x1 and x2 in the model is generally not the same as the slope when you ﬁt a line with x1 alone. These coefficients are the unknown constants, which must be estimated using sample data to obtain the prediction equation. Instead of x and y plotted in two-dimensional space, y and x1, x2, . . . , xk have to be plotted in (k 1) dimensions. F IGU RE 1 3. 1 Plane of means for Example 13.1 ● E(y) x1 x2 A MULTIPLE REGRESSION ANALYSIS 13.3 A multiple regression analysis involves estimation, testing, and diagnostic procedures designed to ﬁt the multiple regression model E( y) b0 b1x1 b2x2 bkxk 554 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS to a set of data. Because of the complexity of the calculations involved, these procedures are almost always implemented with a regression program from one of several computer software packages. All give similar output in slightly different forms. We follow the basic patterns set in simple linear regression, beginning with an outline of the general procedures and illustrated with an example. The Method of Least Squares The prediction equation yˆ b0 b1x1 b2x2 bkxk is the line that minimizes SSE, the sum of squares of the deviations of the observed values y from the predicted values yˆ. These values are calculated using a regression program. How do real estate agents decide on the asking price for a newly listed condominium? A computer database in a small community contains the listed selling price y (in thousands of dollars), the amount of living area x1 (in hundreds of square feet), and the numbers of ﬂoors x2, bedrooms x3, and bathrooms x4, for n 15 randomly selected condos currently on the market. The data are shown in Table 13.1. EXAMPLE 13.2 TABLE 13.1 ● Data on 15 Condominiums Observation List Price, y Living Area, x1 Floors, x2 Bedrooms, x3 Baths, x4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 169.0 218.5 216.5 225.0 229.9 235.0 239.9 247.9 260.0 269.9 234.9 255.0 269.9 294.5 309.9 6 10 10 11 13 13 13 17 19 18 13 18 17 20 21 .7 2.5 2 2.5 2 2 2 2 3 3 3 The multiple regression model is E(y) b0 b1x1 b2x2 b3x3 b4x4 which is ﬁt using the MINITAB software package. You can ﬁnd instructions for generating this output in the section “My MINITAB” at the end of this chapter. The ﬁrst portion of the regression output is shown in Figure 13.2. You will ﬁnd the ﬁtted regression equation in the ﬁrst two lines of the printout: yˆ 119 6.27x1 16.2x2 2.67x3 30.3x4 The partial regression coefficients are shown with slightly more accuracy in the second section. The columns list the name given to each independent predictor variable, its estimated regression coefficient, its standard error, and the t- and p-values that are used to test its signiﬁcance in the presence of all the other predictor variables. We explain these tests in more detail in a later section. 13.3 A MULTIPLE REGRESSION ANALYSIS ❍ 555 F IG URE 13 .2 A portion of the MINITAB printout for Example 13.2 ● Regression Analysis: List Price versus Square Feet, Number of Floors, Bedrooms, Baths The regression equation is List Price = 119 + 6.27 Square Feet - 16.2 Number of Floors - 2.67 Bedrooms + 30.3 Baths Predictor Coef SE Coef T P Constant 118.763 9.207 12.90 0.000 Square Feet 6.2698 0.7252 8.65 0.000 Number of Floors -16.203 6.212 -2.61 0.026 Bedrooms -2.673 4.494 -0.59 0.565 Baths 30.271 6.849 4.42 0.001 The Analysis of Variance for Multiple Regression The analysis of variance divides the total variation in the response variable y, Total SS Syi 2 (S yi)2 n into two portions: • SSR (sum of squares for regression) measures the amount of variation explained by using the regression equation. • SSE (sum of squares for error) measures the residual variation in the data that is not explained by the independent variables. so that Total SS SSR SSE The degrees of freedom for these sums of squares are found using the following argument. There are (n 1) total degrees of freedom. Estimating the regression line requires estimating k unknown coefficients; the constant b0 is a function of y and the other estimates. Hence, there are k regression degrees of freedom, leaving (n 1) k degrees of freedom for error. As in previous chapters, the mean squares are calculated as MS SS/df. The ANOVA table for the real estate data in Table 13.1 is shown in the second portion
of the MINITAB printout in Figure 13.3. There are n 15 observations and k 4 independent predictor variables. You can verify that the total degrees of freedom, (n 1) 14, is divided into k 4 for regression and (n k 1) 10 for error. FI GUR E 13. 3 A portion of the MINITAB printout for Example 13.2 ● S = 6.84930 R-Sq = 97.1% R-Sq(adj) = 96.0% Analysis of Variance Source DF SS MS F P Regression 4 15913.0 3978.3 84.80 0.000 Residual Error 10 469.1 46.9 Total 14 16382.2 Source DF Seq SS Square Feet 1 14829.3 Number of Floors 1 0.9 Bedrooms 1 166.4 Baths 1 916.5 The best estimate of the random variation s 2 in the experiment—the variation that is unexplained by the predictor variables—is as usual given by E SS 46.9 s2 MSE 1 n k 556 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS from the ANOVA table. The ﬁrst line of Figure 13.3 also shows s s2 6.84930 using computer accuracy. The computer uses these values internally to produce test statistics, conﬁdence intervals, and prediction intervals, which we discuss in subsequent sections. The last section of Figure 13.3 shows a decomposition of SSR 15,913.0 in which the conditional contribution of each predictor variable given the variables already entered into the model is shown for the order of entry that you specify in your regression program. For the real estate example, the MINITAB program entered the variables in this order: square feet, then numbers of ﬂoors, bedrooms, and baths. These conditional or sequential sums of squares each account for one of the k 4 regression degrees of freedom. It is interesting to notice that the predictor variable x1 alone accounts for 14,829.3/15,913.0 .932 or 93.2% of the total variation explained by the regression model. However, if you change the order of entry, another variable may account for the major part of the regression sum of squares! Testing the Usefulness of the Regression Model Recall in Chapter 12 that you tested to see whether y and x were linearly related by testing H0 : b 0 with either a t-test or an equivalent F-test. In multiple regression, there is more than one partial slope—the partial regression coefficients. The t- and F-tests are no longer equivalent. The Analysis of Variance F-Test Is the regression equation that uses information provided by the predictor variables x1, x2, . . . , xk substantially better than the simple predictor y that does not rely on any of the x-values? This question is answered using an overall F-test with the hypotheses: The overall F-test (for the signiﬁcance of the model) in multiple regression is one-tailed. H0 : b1 b2 bk 0 versus Ha : At least one of b1, b2, . . . , bk is not 0 The test statistic is found in the ANOVA table (Figure 13.3) as .3 84.80 7 39 9 which has an F distribution with df1 k 4 and df2 (n k 1) 10. Since the exact p-value, P .000, is given in the printout, you can declare the regression to be highly signiﬁcant. That is, at least one of the predictor variables is contributing signiﬁcant information for the prediction of the response variable y. The Coefficient of Determination, R 2 How well does the regression model ﬁt? The regression printout provides a statistical measure of the strength of the model in the coefficient of determination, R2—the proportion of the total variation that is explained by the regression of y on x1, x2, . . . , xk—deﬁned as MINITAB printouts report R 2 as a percentage rather than a proportion. 9 1 , R S S R2 3 , ta SS l To 5 6 1 3 1 .971 or 97.1% 8 2 0 2 . . 13.3 A MULTIPLE REGRESSION ANALYSIS ❍ 557 R 2 is the multivariate equivalent of r 2, used in linear regression. The coefficient of determination is sometimes called multiple R2 and is found in the ﬁrst line of Figure 13.3, labeled “R-Sq.” Hence, for the real estate example, 97.1% of the total variation has been explained by the regression model. The model ﬁts very well! It may be helpful to know that the value of the F statistic is related to R2 by the formula F R2/k (1 R2)/(n k 1) so that when R2 is large, F is large, and vice versa. Interpreting the Results of a Significant Regression Testing the Signiﬁcance of the Partial Regression Coefficients Once you have determined that the model is useful for predicting y, you should explore the nature of the “usefulness’’ in more detail. Do all of the predictor variables add important information for prediction in the presence of other predictors already in the model? The individual t-tests in the ﬁrst section of the regression printout are designed to test the hypotheses H0 : bi 0 versus Ha : bi 0 for each of the partial regression coefficients, given that the other predictor variables are already in the model. These tests are based on the Student’s t statistic given by b t b i i (b S i) E You can show that S R F M M E S R 2/k (1 R 2)/(n k 1) which has df (n k 1) degrees of freedom. The procedure is identical to the one used to test a hypothesis about the slope b in the simple linear regression model.† Figure 13.4 shows the t-tests and p-values from the upper portion of the MINITAB printout. By examining the p-values in the last column, you can see that all the variables except x3, the number of bedrooms, add very signiﬁcant information for predicting y, even with all the other independent variables already in the model. Could the model be any better? It may be that x3 is an unnecessary predictor variable. One option is to remove this variable and reﬁt the model with a new set of data! ● Predictor Coef SE Coef T P Constant 118.763 9.207 12.90 0.000 Square Feet 6.2698 0.7252 8.65 0.000 Number of Floors -16.203 6.212 -2.61 0.026 Bedrooms -2.673 4.494 -0.59 0.565 Baths 30.271 6.849 4.42 0.001 Test for the signiﬁcance of the individual coefficient bi, using t-tests. F IG URE 13 .4 A portion of the MINITAB printout for Example 13.2 The Adjusted Value of R 2 Notice from the deﬁnition of R2 SSR/Total SS that its value can never decrease with the addition of more variables into the regression model. Hence, R2 can be artiﬁcially inﬂated by the inclusion of more and more predictor variables. †Some packages use the t statistic just described, whereas others use the equivalent F statistic (F t2), since the square of a t statistic with v degrees of freedom is equal to an F statistic with 1 df in the numerator and v degrees of freedom in the denominator. 558 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS An alternative measure of the strength of the regression model is adjusted for de- grees of freedom by using mean squares rather than sums of squares: R2(adj) 1 100% E S M n 1) /( SS Total Use R 2(adj) for comparing one or more possible models. For the real estate data in Figure 13.3, R2(adj) 1 9 . 6 4 2/14 . 2 8 16,3 100% 96.0% is found in the ﬁrst line of the printout. The value “R-Sq(adj) 96.0%” represents the percentage of variation in the response y explained by the independent variables, corrected for degrees of freedom. The adjusted value of R2 is mainly used to compare two or more regression models that use different numbers of independent predictor variables. Checking the Regression Assumptions Before using the regression model for its main purpose—estimation and prediction of y—you should look at computer-generated residual plots to make sure that all the regression assumptions are valid. The normal probability plot and the plot of residuals versus ﬁt are shown in Figure 13.5 for the real estate data. There appear to be three observations that do not ﬁt the general pattern. You can see them as outliers in both graphs. These three observations should probably be investigated; however, they do not provide strong evidence that the assumptions are violated. FI GUR E 13. 5 MINITAB diagnostic plots ● Residuals versus the Fitted Values (response is List Price) 10 10 15 99 95 90 80 70 60 50 40 30 20 10 5 t n e c r e P 150 175 200 225 250 275 300 Fitted Value Normal Probability Plot of the Residuals (response is List Price) 1 15 10 5 0 Residual 5 10 15 For given values of x1, x2, . . . , xk, the prediction interval will always be wider than the conﬁdence interval. 13.4 A POLYNOMIAL REGRESSION MODEL ❍ 559 Using the Regression Model for Estimation and Prediction Finally, once you have determined that the model is effective in describing the relationship between y and the predictor variables x1, x2, . . . , xk, the model can be used for these purposes: • Estimating the average value of y—E(y)—for given values of x1, x2, . . . , xk • Predicting a particular value of y for given values of x1, x2, . . . , xk The values of x1, x2, . . . , xk are entered into the computer, and the computer generates the ﬁtted value yˆ together with its estimated standard error and the conﬁdence and prediction intervals. Remember that the prediction interval is always wider than the conﬁdence interval. Let’s see how well our prediction works for the real estate data, using another house from the computer database—a house with 1000 square feet of living area, one ﬂoor, three bedrooms, and two baths, which was listed at $221,500. The printout in Figure 13.6 shows the conﬁdence and prediction intervals for these values. The actual value falls within both intervals, which indicates that the model is working very well! F IG URE 13 .6 Conﬁdence and prediction intervals for Example 13.2 ● Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 217.78 3.11 (210.86, 224.70) (201.02, 234.54) Values of Predictors for New Observations New Square Number of Obs Feet Floors Bedrooms Baths 1 10.0 1.00 3.00 2.00 A POLYNOMIAL REGRESSION MODEL 13.4 In Section 13.3, we explained in detail the various portions of the multiple regression printout. When you perform a multiple regression analysis, you should use a step-bystep approach: 1. Obtain the ﬁtted prediction model. 2. Use the analysis of variance F-test and R2 to determine how well the model ﬁts the data. 3. Check the t-tests for the partial regression coefficients to see which ones are 4. contributing signiﬁcant information in the presenc
e of the others. If you choose to compare several different models, use R2(adj) to compare their effectiveness. 5. Use computer-generated residual plots to check for violation of the regression assumptions. A quadratic equation is y a bx cx2. The graph forms a parabola. Once all of these steps have been taken, you are ready to use your model for estimation and prediction. The predictor variables x1, x2, . . . , xk used in the general linear model do not have to represent different predictor variables. For example, if you suspect that one independent variable x affects the response y, but that the relationship is curvilinear rather than linear, then you might choose to ﬁt a quadratic model: y b0 b1x b2 x2 e 560 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS The quadratic model is an example of a second-order model because it involves a term whose exponents sum to 2 (in this case, x2).† It is also an example of a polynomial model—a model that takes the form y a bx cx 2 dx3 To ﬁt this type of model using the multiple regression program, observed values of y, x, and x2 are entered into the computer, and the printout can be generated as in Section 13.3. In a study of variables that affect productivity in the retail grocery trade, W.S. Good uses value added per work-hour to measure the productivity of retail grocery outlets.1 He deﬁnes “value added” as “the surplus [money generated by the business] available to pay for labor, furniture and ﬁxtures, and equipment.” Data consistent with the relationship between value added per work-hour y and the size x of a grocery outlet described in Good’s article are shown in Table 13.2 for 10 ﬁctitious grocery outlets. Choose a model to relate y to x. EXAMPLE 13.3 TABLE 13.2 ● Data on Store Size and Value Added Store Value Added per Work-Hour, y Size of Store (thousand square feet), 10 $4.08 3.40 3.51 3.09 2.92 1.94 4.11 3.16 3.75 3.60 21.0 12.0 25.2 10.4 30.9 6.8 19.6 14.5 25.0 19.1 Solution You can investigate the relationship between y and x by looking at the plot of the data points in Figure 13.7. The graph suggests that productivity, y, increases as the size of the grocery outlet, x, increases until an optimal size is reached. Above that size, productivity tends to decrease. The relationship appears to be curvilinear, and a quadratic model, E(y) b0 b1x b2x2 ● FI GUR E 13. 7 Plot of store size x and value added y for Example 13.3 4.0 3.5 y 3.0 2.5 2.0 10 15 20 x 25 30 †The order of a term is determined by the sum of the exponents of variables making up that term. Terms involving x1 or x2 are ﬁrst-order. Terms involving x2 2, or x1x2 are second-order. 1, x2 EXAMPLE 13.4 13.4 A POLYNOMIAL REGRESSION MODEL ❍ 561 may be appropriate. Remember that, in choosing to use this model, we are not saying that the true relationship is quadratic, but only that it may provide more accurate estimations and predictions than, say, a linear model. Refer to the data on grocery retail outlet productivity and outlet size in Example 13.3. MINITAB was used to ﬁt a quadratic model to the data and to graph the quadratic prediction curve, along with the plotted data points. Discuss the adequacy of the ﬁtted model. Solution From the printout in Figure 13.8, you can see that the regression equation is yˆ .159 .392x .00949x2 The graph of this quadratic equation together with the data points is shown in Figure 13.9. FI GUR E 13. 8 MINITAB printout for Example 13.4 ● Regression Analysis: y versus x, x-sq The regression equation is y = - 0.159 + 0.392 x - 0.00949 x-sq Look at the computer printout and ﬁnd the labels for “Predictor.” This will tell you what variables have been used in the model. Predictor Coef St Coef T P Constant -0.1594 0.5006 -0.32 0.760 x 0.39193 0.05801 6.76 0.000 x-sq -0.009495 0.001535 -6.19 0.000 S = 0.250298 R-Sq = 87.9% R-Sq(adj) = 84.5% Analysis of Variance Source DF SS MS F P Regression 2 3.1989 1.5994 25.53 0.001 Residual Error 7 0.4385 0.0626 Total 9 3.6374 Source DF Seq SS x 1 0.8003 x-sq 1 2.3986 FI GU RE 1 3 .9 Fitted quadratic regression line for Example 13.4 ● Fitted Line Plot y 0.1594 0.3919 x 0.009495 x**2 S R-Sq R-Sq(adj) 0.250298 87.9% 84.5% 4.0 3.5 y 3.0 2.5 2.0 10 15 20 x 25 30 To assess the adequacy of the quadratic model, the test of H0 : b1 b2 0 versus Ha : Either b1 or b2 is not 0 is given in the printout as R S F M 25.53 S E M 562 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS with p-value .001. Hence, the overall ﬁt of the model is highly signiﬁcant. Quadratic regression accounts for R2 87.9% of the variation in y [R2(adj) 84.5%]. From the t-tests for the individual variables in the model, you can see that both b1 and b2 are highly signiﬁcant, with p-values equal to .000. Notice from the sequential sum of squares section that the sum of squares for linear regression is .8003, with an additional sum of squares of 2.3986 when the quadratic term is added. It is apparent that the simple linear regression model is inadequate in describing the data. One last look at the residual plots generated by MINITAB in Figure 13.10 ensures that the regression assumptions are valid. Notice the relatively linear appearance of the normal plot and the relative scatter of the residuals versus ﬁts. The quadratic model provides accurate predictions for values of x that lie within the range of the sampled values of x. F IG URE 13 .1 0 MINITAB diagnostic plots for Example 13.4 ● Normal Probability Plot of the Residuals (response is y) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 .3 0.2 0.1 0.0 0.1 0.2 0.3 0.4 0.50 0.25 0.00 Residual 0.25 0.50 Residuals versus the Fitted Values (response is y) 10 15 20 Fitted Value 25 30 13.4 EXERCISES BASIC TECHNIQUES 13.1 Suppose that E(y) is related to two predictor variables, x1 and x2, by the equation b. What relationship do the lines in part a have to one another? E(y) 3 x1 2x2 a. Graph the relationship between E(y) and x1 when x2 2. Repeat for x2 1 and for x2 0. 13.2 Refer to Exercise 13.1. a. Graph the relationship between E(y) and x2 when x1 0. Repeat for x1 1 and for x1 2. b. What relationship do the lines in part a have to one another? c. Suppose, in a practical situation, you want to model the relationship between E(y) and two predictor variables x1 and x2. What is the implication of using the ﬁrst-order model E(y) b0 b1x1 b2x2? 13.3 Suppose that you ﬁt the model E(y) b0 b1x1 b2x2 b3x3 to 15 data points and found F equal to 57.44. a. Do the data provide sufficient evidence to indicate that the model contributes information for the prediction of y? Test using a 5% level of signiﬁcance. b. Use the value of F to calculate R2. Interpret its value. 13.4 The computer output for the multiple regression analysis for Exercise 13.3 provides this information: b0 1.04 b2 2.72 SE(b2) .65 b1 1.29 SE(b1) .42 b3 .41 SE(b3) .17 a. Which, if any, of the independent variables x1, x2, and x3 contribute information for the prediction of y? b. Give the least-squares prediction equation. c. On the same sheet of graph paper, graph y versus x1 when x2 1 and x3 0 and when x2 1 and x3 .5. What relationship do the two lines have to each other? d. What is the practical interpretation of the parameter b1? 13.5 Suppose that you ﬁt the model E(y) b0 b1x b2x2 13.4 A POLYNOMIAL REGRESSION MODEL ❍ 563 a. What type of model have you chosen to ﬁt the data? b. How well does the model ﬁt the data? Explain. c. Do the data provide sufficient evidence to indicate that the model contributes information for the prediction of y? Use the p-value approach. 13.6 Refer to Exercise 13.5. a. What is the prediction equation? b. Graph the prediction equation over the interval 0 x 6. 13.7 Refer to Exercise 13.5. a. What is your estimate of the average value of y when x 0? b. Do the data provide sufficient evidence to indicate that the average value of y differs from 0 when x 0? 13.8 Refer to Exercise 13.5. a. Suppose that the relationship between E(y) and x is a straight line. What would you know about the value of b 2? b. Do the data provide sufficient evidence to indicate curvature in the relationship between y and x? 13.9 Refer to Exercise 13.5. Suppose that y is the proﬁt for some business and x is the amount of capital invested, and you know that the rate of increase in proﬁt for a unit increase in capital invested can only decrease as x increases. You want to know whether the data provide sufficient evidence to indicate a decreasing rate of increase in proﬁt as the amount of capital invested increases. a. The circumstances described imply a one-tailed sta- tistical test. Why? b. Conduct the test at the 1% level of signiﬁcance. State your conclusions. to 20 data points and obtained the accompanying MINITAB printout. APPLICATIONS MINITAB output for Exercise 13.5 Regression Analysis: y versus x, x-sq The regression equation is y = 10.6 + 4.44 x - 0.648 x-sq Predictor Coef SE Coef T P Constant 10.5638 0.6951 15.20 0.000 x 4.4366 0.5150 8.61 0.000 x-sq -0.64754 0.07988 -8.11 0.000 S = 1.191 R-Sq = 81.5% R-Sq(adj) = 79.3% Analysis of Variance Source DF SS MS F P Regression 2 106.072 53.036 37.37 0.000 Residual Error 17 24.128 1.419 Total 19 130.200 EX1310 13.10 College Textbooks A publisher of college textbooks conducted a study to relate proﬁt per text y to cost of sales x over a 6-year period when its sales force (and sales costs) were growing rapidly. These inﬂation-adjusted data (in thousands of dollars) were collected: Proﬁt per Text, y 16.5 22.4 24.9 28.8 31.5 35.8 Sales Cost per Text, x 5.0 5.6 6.1 6.8 7.4 8.6 Expecting proﬁt per book to rise and then plateau, the publisher ﬁtted the model E(y) b0 b1x b2x 2 to the data. 564 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS MINITAB output for Exercise 13.10 Regression Analysis: y versus x, x-sq The regression equation is y = - 44.2 + 16.3 x - 0.820 x-sq Predictor Coef SE Coef T P Constant -44.192 8.287 -5.33 0.013 x 16.334 2.490 6.56 0.007 x-sq -0.8198 0.1824 -4.49 0.021 S = 0.594379 R-Sq = 99.6% R-Sq(adj) = 99.3% Analysis of Variance Source DF SS MS F P R
egression 2 234.96 117.48 332.53 0.000 Residual Error 3 1.06 0.35 Total 5 236.02 Source DF Seq SS x 1 227.82 x-sq 1 7.14 a. Plot the data points. Does it look as though the quadratic model is necessary? b. Find s on the printout. Conﬁrm that kE s n 1 SS c. Do the data provide sufficient evidence to indicate that the model contributes information for the prediction of y? What is the p-value for this test, and what does it mean? d. How much of the regression sum of squares is accounted for by the quadratic term? The linear term? e. What sign would you expect the actual value of b2 to have? Find the value of b2 in the printout. Does this value conﬁrm your expectation? f. Do the data indicate a signiﬁcant curvature in the relationship between y and x? Test at the 5% level of signiﬁcance. g. What conclusions can you draw from the accompa- nying residual plots? MINITAB plots for Exercise 13.10 Residuals versus the Fitted Values (response is y.00 0.75 0.50 0.25 0.00 0.25 0.50 15 20 25 Fitted Value 30 35 13.11 College Textbooks II Refer to Exercise 13.10. a. Use the values of SSR and Total SS to calculate R2. Compare this value with the value given in the printout. b. Calculate R2(adj). When would it be appropriate to use this value rather than R2 to assess the ﬁt of the model? c. The value of R2(adj) was 95.7% when a simple linear model was ﬁt to the data. Does the linear or the quadratic model ﬁt better? 13.12 Veggie Burgers You have a hot grill and an empty hamburger bun, but you have EX1312 sworn off greasy hamburgers. Would a meatless hamburger do? The data in the table record a ﬂavor and texture score (between 0 and 100) for 12 brands of meatless hamburgers along with the price, number of calories, amount of fat, and amount of sodium per burger.2 Some of these brands try to mimic the taste of meat, while others do not. The MINITAB printout shows the regression of the taste score y on the four predictor variables: price, calories, fat, and sodium. Normal Probability Plot of the Residuals (response is y) Brand Score, y Price, x1 Calories, x2 Fat, x3 Sodium, x4 t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 5 1 1.0 0.5 0.0 Residual 0.5 1. 10 11 12 70 45 43 41 39 30 68 56 40 34 30 26 91 68 92 75 88 67 73 92 71 67 92 95 110 90 80 120 90 140 120 170 130 110 100 130 310 420 280 370 410 440 430 520 180 180 330 340 MINITAB output for Exercise 13.12 Regression Analysis: y versus x1, x2, x3, x4 The regression equation is y = 59.8 + 0.129 x1 - 0.580 x2 + 8.50 x3 + 0.0488 x4 Predictor Coef SE Coef T P Constant 59.85 35.68 1.68 0.137 x1 0.1287 0.3391 0.38 0.716 x2 -0.5805 0.2888 -2.01 0.084 x3 8.498 3.472 2.45 0.044 x4 0.04876 0.04062 1.20 0.269 S = 12.7199 R-Sq = 49.9% R-Sq(adj) = 21.3% Analysis of Variance Source DF SS MS F P Regression 4 1128.4 282.1 1.74 0.244 Residual Error 7 1132.6 161.8 Total 11 2261.0 Source DF Seq SS x1 1 11.2 x2 1 19.6 x3 1 864.5 x4 1 233.2 a. Comment on the ﬁt of the model using the statistical test for the overall ﬁt and the coefficient of determination, R2. b. If you wanted to reﬁt the model by eliminating one of the independent variables, which one would you eliminate? Why? 13.13 Veggie Burgers II Refer to Exercise 13.12. A command in the MINITAB regression menu provides output in which R2 and R2(adj) are calculated for all possible subsets of the four independent variables. The printout is provided here. MINITAB output for Exercise 13.13 Best Subsets Regression: y versus x1, x2, x3, x4 Response is y R-Sq Mallows x x x x Vars R-Sq (adj) C-p s 1 2 3 4 1 17.0 8.7 3.6 13.697 X 1 6.9 0.0 5.0 14.506 X 2 37.2 23.3 2.8 12.556 X X 2 20.3 2.5 5.1 14.153 X X 3 48.9 29.7 3.1 12.020 X X X 3 39.6 16.9 4.4 13.066 X X X 4 49.9 21.3 5.0 12.720 X X X X a. If you had to compare these models and choose the best one, which model would you choose? Explain. b. Comment on the usefulness of the model you chose in part a. Is your model valuable in predicting a taste score based on the chosen predictor variables? 13.14 Air Pollution III An experiment was designed to compare several different types of EX1314 air pollution monitors.3 Each monitor was set up and then exposed to different concentrations of ozone, ranging between 15 and 230 parts per million (ppm), for periods of 8–72 hours. Filters on the monitor were then analyzed, and the response of the monitor was 13.4 A POLYNOMIAL REGRESSION MODEL ❍ 565 measured. The results for one type of monitor showed a linear pattern (see Exercise 12.14). The results for another type of monitor are listed in the table. Ozone (ppm/hr), x Relative Fluorescence Density, y .06 .12 .18 .31 .57 .65 .68 1.29 8 18 27 33 42 47 52 61 a. Plot the data. What model would you expect to provide the best ﬁt to the data? Write the equation of that model. b. Use a computer software package to ﬁt the model from part a. c. Find the least-squares regression line relating the monitor’s response to the ozone concentration. d. Does the model contribute signiﬁcant information for the prediction of the monitor’s response based on ozone exposure? Use the appropriate p-value to make your decision. e. Find R2 on the printout. What does this value tell you about the effectiveness of the multiple regression analysis? 13.15 Corporate Proﬁts In order to study the relationship of advertising and capital EX1315 investment with corporate proﬁts, the following data, recorded in units of $100,000, were collected for 10 medium-sized ﬁrms in the same year. The variable y represents proﬁt for the year, x1 represents capital investment, and x2 represents advertising expenditures. y 15 16 2 3 12 x1 25 1 6 30 29 x2 4 5 3 1 2 y 1 16 18 13 2 x1 x2 20 12 15 6 16 0 4 5 4 2 a. Using the model y b0 b1x b2x2 and an appropriate computer software package, ﬁnd the least-squares prediction equation for these data. b. Use the overall F-test to determine whether the model contributes signiﬁcant information for the prediction of y. Use a .01. c. Does advertising expenditure x2 contribute signiﬁcant information for the prediction of y, given that x1 is already in the model? Use a .01. d. Calculate the coefficient of determination, R2. What percentage of the overall variation is explained by the model? 566 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS 13.16 YouTube The video-sharing site YouTube attracted 19.6 million visitors in EX1316 June 2006, an almost 300% increase from January of that same year. Despite YouTube’s phenomenal growth, some analysts have questioned whether the site can transition from a free service to one that can make money. The growth trend for YouTube from August 2005 to June 2006 is given the following table.4 Time 8/2005 9/2005 10/2005 11/2005 12/2005 1/2006 2/2006 3/2006 4/2006 5/2006 6/2006 Coded Time Total Unique Visitors (000 10 11 72 100 750 990 1600 2800 4100 5700 6600 12,600 12,800 Linear and quadratic ﬁtted plots for these data follow. Fitted Line Plot y 886 691.9 x 166.1 x**2 S R-Sq R-Sq(adj) 944.385 96.7% 95.9% y 14,000 12,000 10,000 8000 6000 4000 2000 0 0 2 4 6 x 8 10 12 a. Based upon the summary statistics in the line plots, which of the two models better ﬁts the data? b. Write the equation for the quadratic model. c. Use the following printout to determine if the quadratic term contributes signiﬁcant information to the prediction of y, in the presence of the linear term. Fitted Line Plot y 3432 1301 x Regression Analysis: Number versus Time, Time-sq The regression equation is Number = 886 - 692 Time + 166 Time-sq S R-Sq R-Sq(adj) 1849.88 85.8% 84.2% Predictor Coef SE Coef T P Constant 886 1037 0.85 0.418 Time -691.9 397.2 -1.74 0.120 Time-sq 166.07 32.24 5.15 0.001 S = 944.381 R-Sq = 96.7% R-Sq(adj) = 95.9% Analysis of Variance Source DF SS MS F P Regression 2 209848944 104924472 117.65 0.000 Residual Error 8 7134840 891855 Total 10 216983783 15,000 10,000 y 5000 0 0 2 4 6 x 8 10 12 USING QUANTITATIVE AND QUALITATIVE PREDICTOR VARIABLES IN A REGRESSION MODEL 13.5 One reason multiple regression models are very ﬂexible is that they allow for the use of both qualitative and quantitative predictor variables. For the multiple regression methods used in this chapter, the response variable y must be quantitative, measuring a numerical random variable that has a normal distribution (according to the assumptions of Section 13.2). However, each independent predictor variable can be either a quantitative variable or a qualitative variable, whose levels represent qualities or characteristics and can only be categorized. 13.5 USING QUANTITATIVE AND QUALITATIVE PREDICTOR VARIABLES IN A REGRESSION MODEL ❍ 567 Quantitative and qualitative variables enter the regression model in different ways. To make things more complicated, we can allow a combination of different types of variables in the model, and we can allow the variables to interact, a concept that may be familiar to you from the factorial experiment of Chapter 11. We consider these options one at a time. A quantitative variable x can be entered as a linear term, x, or to some higher power such as x2 or x3, as in the quadratic model in Example 13.3. When more than one quantitative variable is necessary, the interpretation of the possible models becomes more complicated. For example, with two quantitative variables x1 and x2, you could use a ﬁrst-order model such as E(y) b0 b1x1 b2x2 which traces a plane in three-dimensional space (see Figure 13.1). However, it may be that one of the variables—say, x2—is not related to y in the same way when x1 1 as it is when x1 2. To allow x2 to behave differently depending on the value of x1, we add an interaction term, x1x2, and allow the two-dimensional plane to twist. The model is now a second-order model: E(y) b0 b1x1 b2x2 b3x1x2 The models become complicated quickly when you allow curvilinear relationships and interaction for the two variables. One way to decide on the type of model you need is to plot some of the data—perhaps y versus x1, y versus x2, and y versus x2 for various values of x1. In contrast to quantitative p
redictor variables, qualitative predictor variables are entered into a regression model through dummy or indicator variables. For example, in a model that relates the mean salary of a group of employees to a number of predictor variables, you may want to include the employee’s ethnic background. If each employee included in your study belongs to one of three ethnic groups—say, A, B, or C—you can enter the qualitative variable “ethnicity” into your model using two dummy variables: Enter quantitative variables as • a single x • a higher power, x 2 or x 3 • an interaction with another variable x1 1 0 if group B if not 1 if group C 0 if not Look at the effect these two variables have on the model E(y) b0 b1x1 b2x2: For employees in group A, x2 E(y) b0 b1(0) b2(0) b0 for employees in group B, E(y) b0 b1(1) b2(0) b0 b1 and for those in group C, E(y) b0 b1(0) b2(1) b0 b2 The model allows a different average response for each group. b1 measures the difference in the average responses between groups B and A, while b2 measures the difference between groups C and A. When a qualitative variable involves k categories or levels, (k 1) dummy variables should be added to the regression model. This model may contain other predictor variables—quantitative or qualitative—as well as cross-products (interactions) of the dummy variables with other variables that appear in the model. As you can see, the process of model building—deciding on the appropriate terms to enter into the regression model—can be quite complicated. However, you can become more proﬁcient Qualitative variables are entered as dummy variables—one fewer than the number of categories or levels. 568 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS EXAMPLE 13.5 at model building, gaining experience with the chapter exercises. The next example involves one quantitative and one qualitative variable that interact. A study was conducted to examine the relationship between university salary y, the number of years of experience of the faculty member, and the gender of the faculty member. If you expect a straight-line relationship between mean salary and years of experience for both men and women, write the model that relates mean salary to the two predictor variables: years of experience (quantitative) and gender of the professor (qualitative). Solution Since you may suspect the mean salary lines for women and men to be different, your model for mean salary E(y) may appear as shown in Figure 13.11. A straight-line relationship between E(y) and years of experience x1 implies the model E(y) b0 b1x1 (graphs as a straight line) F IG URE 13 .1 1 Hypothetical relationship for mean salary E(y), years of experience (x1), and gender (x2) for Example 13.5 ● E(y Wom en 0 1 2 3 4 5 x1 Years of Experience The qualitative variable “gender” involves k 2 categories, men and women. Therefore, you need (k 1) 1 dummy variable, x2, deﬁned as x2 1 0 if a man if a woman and the model is expanded to become E(y) b0 b1x1 b2x2 (graphs as two parallel lines) The fact that the slopes of the two lines may differ means that the two predictor variables interact; that is, the change in E(y) corresponding to a change in x1 depends on whether the professor is a man or a woman. To allow for this interaction (difference in slopes), the interaction term x1x2 is introduced into the model. The complete model that characterizes the graph in Figure 13.11 is dummy variable for gender E(y) b0 b1x1 b2x2 b3x1x2 interaction years of experience 13.5 USING QUANTITATIVE AND QUALITATIVE PREDICTOR VARIABLES IN A REGRESSION MODEL ❍ 569 where x1 Years of experience x2 if a man if a woman 1 0 You can see how the model works by assigning values to the dummy variable x2. When the faculty member is a woman, the model is E(y) b0 b1x1 b2(0) b3x1(0) b0 b1x1 which is a straight line with slope b1 and intercept b0. When the faculty member is a man, the model is E(y) b0 b1x1 b2(1) b3x1(1) (b0 b2) (b1 b3)x1 which is a straight line with slope (b1 b3) and intercept (b0 b2). The two lines have different slopes and different intercepts, which allows the relationship between salary y and years of experience x1 to behave differently for men and women. EXAMPLE 13.6 Random samples of six female and six male assistant professors were selected from among the assistant professors in a college of arts and sciences. The data on salary and years of experience are shown in Table 13.3. Note that each of the two samples (male and female) contained two professors with 3 years of experience, but no male professor had 2 years of experience. Interpret the output of the MINITAB regression printout and graph the predicted salary lines. TABLE 13.3 ● Salary versus Gender and Years of Experience Years of Experience, x1 Salary for Men, y Salary for Women60,710 — 63,160 63,210 64,140 65,760 65,590 $59,510 60,440 61,340 61,760 62,750 63,200 — Solution The MINITAB regression printout for the data in Table 13.3 is shown in Figure 13.12. You can use a step-by-step approach to interpret this regression analysis, beginning with the ﬁtted prediction equation, yˆ 58,593 969x1 867x2 260x1x2. By substituting x2 0 or 1 into this equation, you get two straight lines— one for women and one for men—to predict the value of y for a given x1. These lines are Women: Men: yˆ 58,593 969x1 yˆ 59,460 1229x1 and are graphed in Figure 13.13. 570 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS F IG URE 13 .1 2 MINITAB output for Example 13.6 ● Regression Analysis: y versus x1, x2, x1x2 The regression equation is y = 58593 + 969 x1 + 867 x2 + 260 x1x2 Predictor Coef SE Coef T P Constant 58593.0 207.9 281.77 0.000 x1 969.00 63.67 15.22 0.000 x2 866.7 305.3 2.84 0.022 x1x2 260.13 87.06 2.99 0.017 S = 201.344 R-Sq = 99.2% R-Sq(adj) = 98.9% Analysis of Variance Source DF SS MS F P Regression 3 42108777 14036259 346.24 0.000 Residual Error 8 324315 40539 Total 11 42433092 Source DF Seq SS x1 1 33294036 x2 1 8452797 x1x2 1 361944 Next, consider the overall ﬁt of the model using the analysis of variance F-test. Since the observed test statistic in the ANOVA portion of the printout is F 346.24 with P .000, you can conclude that at least one of the predictor variables is contributing information for the prediction of y. The strength of this model is further measured by the coefficient of determination, R2 99.2%. You can see that the model appears to ﬁt very well. ● F IG URE 13 .1 3 A graph of the faculty salary prediction lines for Example 13.6 y 56 55 54 53 52 51 50 49 48 ) en x1 Years of Experience To explore the effect of the predictor variables in more detail, look at the individual t-tests for the three predictor variables. The p-values for these tests—.000, .022, and .017, respectively—are all signiﬁcant, which means that all of the predictor variables add signiﬁcant information to the prediction with the other two variables already in 13.5 USING QUANTITATIVE AND QUALITATIVE PREDICTOR VARIABLES IN A REGRESSION MODEL ❍ 571 the model. Finally, check the residual plots to make sure that there are no strong violations of the regression assumptions. These plots, which behave as expected for a properly ﬁt model, are shown in Figure 13.14. FI GUR E 13. 14 MINITAB residual plots for Example 13.6 ● Normal Probability Plot of the Residuals (response is y) Residuals versus the Fitted Values (response is y) 99 95 90 80 70 60 50 40 30 20 10 5 t n e c r e P 1 500 400 300 200 100 0 Residual l a u d i s e R 300 200 100 0 100 200 300 100 200 300 400 59,000 60,000 61,000 62,000 63,000 64,000 65,000 66,000 Fitted Value EXAMPLE 13.7 Refer to Example 13.6. Do the data provide sufficient evidence to indicate that the annual rate of increase in male junior faculty salaries exceeds the annual rate of increase in female junior faculty salaries? That is, do the data provide sufficient evidence to indicate that the slope of the men’s faculty salary line is greater than the slope of the women’s faculty salary line? Solution Since b3 measures the difference in slopes, the slopes of the two lines will be identical if b3 0. Therefore, you want to test the null hypothesis H0 : b3 0 —that is, the slopes of the two lines are identical—versus the alternative hypothesis Ha : b3 0 —that is, the slope of the men’s faculty salary line is greater than the slope of the women’s faculty salary line. The calculated value of t corresponding to b3, shown in the row labeled “x1x2” in Figure 13.12, is 2.99. Since the MINITAB regression output provides p-values for twotailed signiﬁcance tests, the p-value in the printout, .017, is twice what it would be for a one-tailed test. For this one-tailed test, the p-value is .017/2 .0085, and the null hypothesis is rejected. There is sufficient evidence to indicate that the annual rate of increase in men’s faculty salaries exceeds the rate for women.† †If you want to determine whether the data provide sufficient evidence to indicate that male faculty members start at higher salaries, you would test H0 : b2 0 versus the alternative hypothesis Ha : b2 0. 572 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS 13.5 EXERCISES BASIC TECHNIQUES APPLICATIONS 13.17 Production Yield Suppose you wish to predict production yield y as a function of several independent predictor variables. Indicate whether each of the following independent variables is qualitative or quantitative. If qualitative, deﬁne the appropriate dummy variable(s). a. The prevailing interest rate in the area b. The price per pound of one item used in the produc- tion process c. The plant (A, B, or C) at which the production yield is measured d. The length of time that the production machine has been in operation e. The shift (night or day) in which the yield is mea- sured 13.18 Suppose E(y) is related to two predictor variables x1 and x2 by the equation E(y) 3 x1 2x2 x1x2 a. Graph the relationship between E(y) and x1 when x2 0. Repeat for x2 2 and for x2 2. b. Repeat the instructions of part a for the model E(y) 3 x1 2x2 c. Note that the equation for part a is
exactly the same as the equation in part b except that we have added the term x1x2. How does the addition of the x1x2 term affect the graphs of the three lines? d. What ﬂexibility is added to the ﬁrst-order model E(y) b0 b1x1 b2x2 by the addition of the term b3x1x2, using the model E(y) b0 b1x1 b2x2 b3x1x2? 13.19 A multiple linear regression model involving one qualitative and one quantitative independent variable produced this prediction equation: yˆ 12.6 .54x1 1.2x1x2 3.9x2 2 a. Which of the two variables is the quantitative vari- able? Explain. EX1320 13.20 Less Red Meat! Americans are very vocal about their attempts to improve personal well-being by “eating right and exercising more.” One desirable dietary change is to reduce the intake of red meat and to substitute poultry or ﬁsh. Researchers tracked beef and chicken consumption, y (in annual pounds per person), and found the consumption of beef declining and the consumption of chicken increasing over a period of seven years. A summary of their data is shown in the table. Year Beef Chicken 1 2 3 4 5 6 7 85 89 76 76 68 67 60 37 36 47 47 62 74 79 Consider ﬁtting the following model, which allows for simultaneously ﬁtting two simple linear regression lines: E(y) b0 b1x1 b2x2 b3x1x2 where y is the annual meat (either beef or chicken) consumption per person per year, x1 1 if beef 0 if chicken and x2 Year MINITAB output for Exercise 13.20 Regression Analysis: y versus x1, x2, x1x2 The regression equation is y = 23.6 + 69.0 x1 + 7.75 x2 - 12.3 x1x2 Predictor Coef SE Coef T P Constant 23.571 3.522 6.69 0.000 x1 69.000 4.981 13.85 0.000 x2 7.7500 0.7875 9.84 0.000 x1x2 -12.286 1.114 -11.03 0.000 S = 4.16705 R-Sq = 95.4% R-Sq(adj) = 94.1% Analysis of Variance Source DF SS MS F P Regression 3 3637.9 1212.6 69.83 0.000 Residual Error 10 173.6 17.4 Total 13 3811.5 Source DF Seq SS x1 1 1380.1 x2 1 144.6 x1x2 1 2113.1 b. If x1 can take only the values 0 or 1, ﬁnd the two possible prediction equations for this experiment. c. Graph the two equations found in part b. Compare the shapes of the two curves. Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 56.29 3.52 (48.44, 64.13) (44.13, 68.44) Values of Predictors for New Observations New Obs x1 x2 x1x2 1 1.00 8.00 8.00 13.5 USING QUANTITATIVE AND QUALITATIVE PREDICTOR VARIABLES IN A REGRESSION MODEL ❍ 573 MINITAB diagnostic plots for Exercise 13.20 Normal Probability Plot of the Residuals (response is y) 99 95 90 80 70 60 50 40 30 20 10 5 t n e c r e P 1 10 5 0 Residual 5 10 Residuals versus the Fitted Values (response is y.0 2.5 0.0 2.5 5.0 7.5 30 40 50 60 Fitted Value 70 80 90 a. How well does the model ﬁt? Use any relevant statistics and diagnostic tools from the printout to answer this question. b. Write the equations of the two straight lines that describe the trend in consumption over the period of 7 years for beef and for chicken. c. Use the prediction equation to ﬁnd a point estimate of the average per-person beef consumption in year 8. Compare this value with the value labeled “Fit” in the printout. d. Use the printout to ﬁnd a 95% conﬁdence interval for the average per-person beef consumption in year 8. What is the 95% prediction interval for the perperson beef consumption in year 8? Is there any problem with the validity of the 95% conﬁdence level for these intervals? 13.21 Cotton versus Cucumber In Exercise 11.65, you used the analysis of variance EX1321 procedure to analyze a 2 3 factorial experiment in which each factor–level combination was replicated ﬁve times. The experiment involved the number of eggs laid by caged female whiteﬂies on two different plants at three different temperature levels. Suppose that several of the whiteﬂies died before the experiment was completed, so that the number of replications was no longer the same for each treatment. The analysis of variance formulas of Chapter 11 can no longer be used, but the experiment can be analyzed using a multiple regression analysis. The results of this 2 3 factorial experiment with unequal replications are shown in the table. Cotton Cucumber 70° 77° 82° 70° 77° 82° 37 21 36 43 31 46 32 41 34 54 40 42 50 53 25 37 48 59 53 31 69 51 43 62 71 49 a. Write a model to analyze this experiment. Make sure to include a term for the interaction between plant and temperature. b. Use a computer software package to perform the multiple regression analysis. c. Do the data provide sufficient evidence to indicate that the effect of temperature on the number of eggs laid is different depending on the type of plant? d. Based on the results of part c, do you suggest reﬁtting a different model? If so, rerun the regression analysis using the new model and analyze the printout. e. Write a paragraph summarizing the results of your analyses. 13.22 Achievement Scores III The Academic Performance Index (API), described in EX1322 Exercise 12.11, is a measure of school achievement based on the results of the Stanford 9 Achievement Test. The API scores for eight elementary schools in Riverside County, California, are shown below, along with several other independent variables.5 School API Score y Awards x1 % Meals x2 1 2 3 4 5 6 7 8 588 659 710 657 669 641 557 743 Yes No Yes No No No No Yes 58 62 66 36 40 51 73 22 574 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS School % ELL x3 % Emergency x4 Previous Year’s API x5 1 2 3 4 5 6 7 8 34 22 14 30 11 26 39 6 16 5 19 14 13 2 14 4 533 655 695 680 670 636 532 705 The variables are deﬁned as x1 1 if the school was given a ﬁnancial award for meeting growth goals, 0 if not. x2 % of students who qualify for free or reduced price meals x3 % of students who are English Language Learners x4 % of teachers on emergency credentials x5 API score in 2000 The MINITAB printout for a ﬁrst-order regression model is given below. Regression Analysis: y versus x1, x2, x3, x4, x5 The regression equation is y = 269 + 33.2 x1 - 0.003 x2 - 1.02 x3 - 1.00 x4 + 0.636 x5 Predictor Coef STDev T P Constant 269.03 41.55 6.48 0.023 x1 33.227 4.373 7.60 0.017 x2 -0.0027 0.1396 -0.02 0.987 x3 -1.0159 0.3237 -3.14 0.088 x4 -1.0032 0.3391 -2.96 0.098 x5 0.63560 0.05209 12.20 0.007 S = 4.73394 R-Sq = 99.8% R-Sq(adj) = 99.4% Analysis of Variance Source DF SS MS F P Regression 5 25197.2 5039.4 224.87 0.004 Residual Error 2 44.8 22.4 Total 7 25242.0 a. What is the model that has been ﬁt to this data? What is the least-squares prediction equation? b. How well does the model ﬁt? Use any relevant statistics from the printout to answer this question. c. Which, if any, of the independent variables are useful in predicting the API, given the other independent variables already in the model? Explain. d. Use the values of R2 and R2(adj) in the following printout to choose the best model for prediction. Would you be conﬁdent in using the chosen model for predicting the API score for next year based on a model containing similar variables? Explain. Best Subsets Regression: y versus x1, x2, x3, x4, x5 Response is y R-Sq Mallows x x x x x Vars R-Sq (adj) C-p S 1 2 3 4 5 1 87.9 85.8 132.7 22.596 X 1 84.5 81.9 170.7 25.544 X 2 97.4 96.4 27.1 11.423 X X 2 94.6 92.4 58.8 16.512 X X 3 99.0 98.2 11.8 8.1361 X X X 3 98.9 98.2 11.9 8.1654 X X X 4 99.8 99.6 4.0 3.8656 X X X X 4 99.0 97.8 12.8 8.9626 X X X X 5 99.8 99.4 6.0 4.7339 X X X X X 13.23 Particle Board A quality control engineer is interested in predicting the strength of particle board y as a function of the size of the particles x1 and two types of bonding compounds. If the basic response is expected to be a quadratic function of particle size, write a linear model that incorporates the qualitative variable “bonding compound” into the predictor equation. 13.24 Construction Projects In a study to examine the relationship between the time EX1324 required to complete a construction project and several pertinent independent variables, an analyst compiled a list of four variables that might be useful in predicting the time to completion. These four variables were size of the contract, x1 (in $1000 unit), number of workdays adversely affected by the weather x2, number of subcontractors involved in the project x4, and a variable x3 that measured the presence (x3 1) or absence (x3 0) of a workers’ strike during the construction. Fifteen construction projects were randomly chosen, and each of the four variables as well as the time to completion were measured. y 29 15 60 10 70 15 75 30 45 90 7 21 28 50 30 x1 60 80 100 50 200 50 500 75 750 1200 70 80 300 2600 110 x2 7 10 8 14 12 4 15 5 10 20 5 3 8 14 7 x3 x4 7 8 10 5 11 3 12 6 10 12 3 6 8 13 4 An analysis of these data using a ﬁrst-order model in x1, x2, x3, and x4 produced the following printout. Give a complete analysis of the printout and interpret your results. What can you say about the apparent contribution of x1 and x2 in predicting y? 13.6 TESTING SETS OF REGRESSION COEFFICIENTS ❍ 575 Residuals versus the Fitted Values (response is y) 10 20 30 40 Fitted Value 50 60 70 80 Regression Analysis: y versus x1, x2, x3, x4 The regression equation is y = -1.6 - 0.00784 x1 + 0.68 x2 + 28.0 x3 + 3.49 x4 Predictor Coef SE Coef T P Constant -1.59 11.66 -0.14 0.894 x1 -0.007843 0.006230 -1.26 0.237 x2 -0.6753 0.9998 0.68 0.515 x3 28.01 11.37 2.46 0.033 x4 3.489 1.935 1.80 0.102 S = 11.8450 R-Sq = 84.7% R-Sq(adj) = 78.6% Analysis of Variance Source DF SS MS F P Regression 4 7770.3 1942.6 13.85 0.000 Residual Error 10 1403.0 140.3 Total 14 9173.3 l a u d i s e R 20 10 0 10 20 Source Seq SS DF X1 1 1860.9 x2 1 2615.3 x3 1 2838.0 x4 1 456.0 Normal Probability Plot of the Residuals (response is y) 99 95 90 80 70 60 50 40 30 20 10 5 t n e c r e P 1 30 20 10 0 Residual 10 20 30 TESTING SETS OF REGRESSION COEFFICIENTS 13.6 In the preceding sections, you have tested the complete set of partial regression coefﬁcients using the F-test for the overall ﬁt of the model, and you have tested the partial regression coefficients individually using the Student’s t-test. Besides these two importan
t tests, you might want to test hypotheses about some subsets of these regression coefficients. For example, suppose a company suspects that the demand y for some product could be related to as many as ﬁve independent variables, x1, x2, x3, x4, and x5. The cost of obtaining measurements on the variables x3, x4, and x5 is very high. If, in a small pilot study, the company could show that these three variables contribute little or no information for the prediction of y, they can be eliminated from the study at great savings to the company. If all ﬁve variables, x1, x2, x3, x4, and x5, are used to predict y, the regression model would be written as y b0 b1x1 b2x2 b3x3 b4x4 b5x5 e 576 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS However, if x3, x4, and x5 contribute no information for the prediction of y, then they would not appear in the model—that is, b3 b4 b5 0—and the reduced model would be y b0 b1x1 b2x2 e Hence, you want to test the null hypothesis H0 : b3 b4 b5 0 —that is, the independent variables x3, x4, and x5 contribute no information for the prediction of y—versus the alternative hypothesis Ha : At least one of the parameters b3, b4, or b5 differs from 0 —that is, at least one of the variables x3, x4, or x5 contributes information for the prediction of y. Thus, in deciding whether the complete model is preferable to the reduced model in predicting demand, you are led to a test of hypothesis about a set of three parameters, b3, b4, and b5. A test of hypothesis concerning a set of model parameters involves two models: Model 1 (reduced model) E(y) b0 b1x1 b2x2 brxr Model 2 (complete model) E(y) b0 b1x1 b2x2 brxr br1xr1 br2xr2 bkxk 1444442444443 144444424444443 terms in model 1 additional terms in model 2 Suppose you ﬁt both models to the data set and calculated the sum of squares for error for both regression analyses. If model 2 contributes more information for the prediction of y than model 1, then the errors of prediction for model 2 should be smaller than the corresponding errors for model 1, and SSE2 should be smaller than SSE1. In fact, the greater the difference between SSE1 and SSE2, the greater is the evidence to indicate that model 2 contributes more information for the prediction of y than model 1. The test of the null hypothesis H0 : br1 br2 bk 0 versus the alternative hypothesis Ha : At least one of the parameters br1, br2, . . . , bk differs from 0 uses the test statistic F (SSE1 SSE2)/(k r) MSE2 where F is based on df1 (k r) and df2 n (k 1). Note that the (k r) parameters involved in H0 are those added to model 1 to obtain model 2. The numerator degrees of freedom df1 always equals (k r), the number of parameters involved in H0. The denominator degrees of freedom df2 is the number of degrees of freedom associated with the sum of squares for error, SSE2, for the complete model. The rejection region for the test is identical to the rejection region for all of the analysis of variance F-tests—namely, F Fa 13.6 TESTING SETS OF REGRESSION COEFFICIENTS ❍ 577 EXAMPLE 13.8 Refer to the real estate data of Example 13.2 that relate the listed selling price y to the square feet of living area x1, the number of ﬂoors x2, the number of bedrooms x3, and the number of bathrooms, x4. The realtor suspects that the square footage of living area is the most important predictor variable and that the other variables might be eliminated from the model without loss of much prediction information. Test this claim with a .05. ● FI GUR E 13. 15 Portions of the MINITAB regression printouts for (a) complete and (b) reduced models for Example 13.8 Solution The hypothesis to be tested is H0 : b2 b3 b4 0 versus the alternative hypothesis that at least one of b2, b3, or b4 is different from 0. The complete model 2, given as y b0 b1x1 b2x2 b3x3 b4x4 e was ﬁtted in Example 13.2. A portion of the MINITAB printout from Figure 13.3 is reproduced in Figure 13.15 along with a portion of the MINITAB printout for the simple linear regression analysis of the reduced model 1, given as y b0 b1x1 e Regression Analysis: (a) List Price versus Square Feet, Number of Floors, Bedrooms and Baths S = 6.84930 R-Sq = 97.1% R-Sq(adj) = 96.0% Analysis of Variance Source DF SS MS F P Regression 4 15913.0 3978.3 84.80 0.000 Residual Error 10 469.1 46.9 Total 14 16382.2 Regression Analysis: (b) List Price versus Square Feet S = 10.9294 R-Sq = 90.5% R-Sq(adj) = 89.8% Analysis of Variance Source DF SS MS F P Regression 1 14829 14829 124.14 0.000 Residual Error 13 1553 119 Total 14 16382 Then SSE1 1553 from Figure 13.15(b) and SSE2 469.1 and MSE2 46.9 from Figure 13.15(a). The test statistic is F (SSE1 SSE2)/(k r) MSE2 (1553 469.1)/(4 1) 46.9 7.70 The critical value of F with a .05, df1 3, and df2 n (k 1) 15 (4 1) 10 is F.05 3.71. Hence, H0 is rejected. There is evidence to indicate that at least one of the three variables—number of ﬂoors, bedrooms, or bathrooms—is contributing signiﬁcant information for predicting the listed selling price. 578 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS INTERPRETING RESIDUAL PLOTS 13.7 Once again, you can use residual plots to discover possible violations in the assumptions required for a regression analysis. There are several common patterns you should recognize because they occur frequently in practical applications. The variance of some types of data changes as the mean changes: • Poisson data exhibit variation that increases with the mean. • Binomial data exhibit variation that increases for values of p from .0 to .5, and then decreases for values of p from .5 to 1.0. Residual plots for these types of data have a pattern similar to that shown in Figure 13.16. F IG URE 13 .1 6 Plots of residuals against yˆ ● 1 – 50 100 y (a) Poisson Data (b) Binomial Percentages If the range of the residuals increases as yˆ increases and you know that the data are measurements on Poisson variables, you can stabilize the variance of the response by running the regression analysis on y* y. Or if the percentages are calculated from binomial data, you can use the arcsin transformation, y* sin1y.† Even if you are not sure why the range of the residuals increases as yˆ increases, you can still use a transformation of y that affects larger values of y more than smaller values—say, y* y or y* ln y. These transformations have a tendency both to stabilize the variance of y* and to make the distribution of y* more nearly normal when the distribution of y is highly skewed. Plots of the residuals versus the ﬁts yˆ or versus the individual predictor variables often show a pattern that indicates you have chosen an incorrect model. For example, if E(y) and a single independent variable x are linearly related—that is, E(y) b0 b1x and you ﬁt a straight line to the data, then the observed y-values should vary in a random manner about yˆ, and a plot of the residuals against x will appear as shown in Figure 13.17. ● F IG URE 13 .1 7 Residual plot when the model provides a good approximation to reality In Chapter 11 and earlier chapters, we represented the response variable by the symbol x. In the chapters on regression analysis, Chapters 12 and 13, the response variable is represented by the symbol y. 13.8 STEPWISE REGRESSION ANALYSIS ❍ 579 In Example 13.3, you ﬁt a quadratic model relating productivity y to store size x. If you had incorrectly used a linear model to ﬁt these data, the residual plot in Figure 13.18 would show that the unexplained variation exhibits a curved pattern, which suggests that there is a quadratic effect that has not been included in the model. Residuals versus the Fitted Values (response is y) ● FI GUR E 13. 18 Residual plot for linear ﬁt of store size and productivity data in Example 13..5 0.0 0.5 1.0 3.0 3.2 3.4 3.6 3.8 Fitted Value For the data in Example 13.6, the residuals of a linear regression of salary with years of experience x1 without including gender, x2, would show one distinct set of positive residuals corresponding to the men and a set of negative residuals corresponding to the women (see Figure 13.19). This pattern signals that the “gender” variable was not included in the model. Residuals versus the Fitted Values (response is y) ● FI GUR E 13. 19 Residual plot for linear ﬁt of salary data in Example 13.6 l a u d i s e R 1000 500 0 500 1000 1500 60,000 61,000 62,000 63,000 64,000 65,000 Fitted Value Unfortunately, not all residual plots give such a clear indication of the problem. You should examine the residual plots carefully, looking for nonrandomness in the pattern of residuals. If you can ﬁnd an explanation for the behavior of the residuals, you may be able to modify your model to eliminate the problem. STEPWISE REGRESSION ANALYSIS 13.8 Sometimes there are a large number of independent predictor variables that might have an effect on the response variable y. For example, try to list all the variables that might affect a college freshman’s GPA: • Grades in high school courses, high school GPA, SAT score, ACT score • Major, number of units carried, number of courses taken • Work schedule, marital status, commute or live on campus 580 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS Which of this large number of independent variables should be included in the model? Since the number of terms could quickly get unmanageable, you might choose to use a procedure called a stepwise regression analysis, which is implemented by computer and is available in most statistical packages. Suppose you have data available on y and a number of possible independent variables, x1, x2, . . . , xk. A stepwise regression analysis ﬁts a variety of models to the data, adding and deleting variables as their signiﬁcance in the presence of the other variables is either signiﬁcant or nonsigniﬁcant, respectively. Once the program has performed a sufficient number of iterations and no more variables are signiﬁcant when added to the model, and none of the variables in the model are nonsigniﬁcant when removed, the procedure stops. A stepwise regress
ion analysis is an easy way to locate some variables that contribute information for predicting y, but it is not foolproof. Since these programs always ﬁt ﬁrst-order models of the form E(y) b0 b1x1 b2x2 bkxk they are not helpful in detecting curvature or interaction in the data. The stepwise regression analysis is best used as a preliminary tool for identifying which of a large number of variables should be considered in your model. You must then decide how to enter these variables into the actual model you will use for prediction. MISINTERPRETING A REGRESSION ANALYSIS 13.9 Several misinterpretations of the output of a regression analysis are common. We have already mentioned the importance of model selection. If a model does not ﬁt a set of data, it does not mean that the variables included in the model contribute little or no information for the prediction of y. The variables may be very important contributors of information, but you may have entered the variables into the model in the wrong way. For example, a second-order model in the variables might provide a very good ﬁt to the data when a ﬁrst-order model appears to be completely useless in describing the response variable y. Causality You must be careful not to conclude that changes in x cause changes in y. This type of causal relationship can be detected only with a carefully designed experiment. For example, if you randomly assign experimental units to each of two levels of a variable x—say, x 5 and x 10—and the data show that the mean value of y is larger when x 10, then you can say that the change in the level of x caused a change in the mean value of y. But in most regression analyses, in which the experiments are not designed, there is no guarantee that an important predictor variable—say, x1— caused y to change. It is quite possible that some variable that is not even in the model causes both y and x1 to change. Multicollinearity Neither the size of a regression coefficient nor its t-value indicates the importance of the variable as a contributor of information. For example, suppose you intend to predict y, a college student’s calculus grade, based on x1 high school mathematics average and x2 score on mathematics aptitude test. Since these two variables contain 13.9 MISINTERPRETING A REGRESSION ANALYSIS ❍ 581 much of the same or shared information, it will not surprise you to learn that, once one of the variables is entered into the model, the other contributes very little additional information. The individual t-value is small. If the variables were entered in the reverse order, however, you would see the size of the t-values reversed. The situation described above is called multicollinearity, and it occurs when two or more of the predictor variables are highly correlated with one another. When multicollinearity is present in a regression problem, it can have these effects on the analysis: • The estimated regression coefficients will have large standard errors, causing imprecision in conﬁdence and prediction intervals. • Adding or deleting a predictor variable may cause signiﬁcant changes in the values of the other regression coefficients. How can you tell whether a regression analysis exhibits multicollinearity? Look for these clues: • The value of R2 is large, indicating a good ﬁt, but the individual t-tests are nonsigniﬁcant. • The signs of the regression coefficients are contrary to what you would intuitively expect the contributions of those variables to be. • A matrix of correlations, generated by computer, shows you which predictor variables are highly correlated with each other and with the response y. Figure 13.20 displays the matrix of correlations generated for the real estate data from Example 13.2. The ﬁrst column of the matrix shows the correlations of each predictor variable with the response variable y. They are all signiﬁcantly nonzero, but the ﬁrst variable, x1 living area, is the most highly correlated. The last three columns of the matrix show signiﬁcant correlations between all but one pair of predictor variables. This is a strong indication of multicollinearity. If you try to eliminate one of the variables in the model, it may drastically change the effects of the other three! Another clue can be found by examining the coefficients of the prediction line, ListPrice = 119 + 6.27 Square Feet - 16.2 Number of Floors - 2.67 Bedrooms + 30.3 Baths Correlations: List Price, Square Feet, Number of Floors, Bedrooms, Baths ListPrice SqFeet Numflrs Bdrms Square Feet 0.951 0.000 Number of Fl 0.605 0.630 0.017 0.012 Bedrooms 0.746 0.711 0.375 0.001 0.003 0.168 Baths 0.834 0.720 0.760 0.675 0.000 0.002 0.001 0.006 Cell Contents: Pearson Correlation P-Value You would expect more ﬂoors and bedrooms to increase the list price, but their coefficients are negative. Since multicollinearity exists to some extent in all regression problems, you should think of the individual terms as information contributors, rather than try to measure the practical importance of each term. The primary decision to be made is whether a term contributes sufficient information to justify its inclusion in the model. ● FI GUR E 13. 20 Correlation matrix for the real estate data in Example 13.2 582 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS STEPS TO FOLLOW WHEN BUILDING A MULTIPLE REGRESSION MODEL 13.10 The ultimate objective of a multiple regression analysis is to develop a model that will accurately predict y as a function of a set of predictor variables x1, x2, . . . , xk. The step-by-step procedure for developing this model was presented in Section 13.4 and is restated next with some additional detail. If you use this approach, what may appear to be a complicated problem can be made simpler. As with any statistical procedure, your conﬁdence will grow as you gain experience with multiple regression analysis in a variety of practical situations. 1. Select the predictor variables to be included in the model. Since some of these variables may contain shared information, you can reduce the list by running a stepwise regression analysis (see Section 13.8). Keep the number of predictors small enough to be effective yet manageable. Be aware that the number of observations in your data set must exceed the number of terms in your model; the greater the excess, the better! 2. Write a model using the selected predictor variables. If the variables are qualitative, it is best to begin by including interaction terms. If the variables are quantitative, it is best to start with a second-order model. Unnecessary terms can be deleted later. Obtain the ﬁtted prediction model. 3. Use the analysis of variance F-test and R2 to determine how well the model ﬁts the data. 4. Check the t-tests for the partial regression coefficients to see which ones are contributing signiﬁcant information in the presence of the others. If some terms appear to be nonsigniﬁcant, consider deleting them. If you choose to compare several different models, use R2(adj) to compare their effectiveness. 5. Use computer-generated residual plots to check for violation of the regression assumptions. CHAPTER REVIEW Key Concepts and Formulas I. The General Linear Model III. Analysis of Variance 1. y b0 b1x1 b2x2 bkxk e 2. The random error e has a normal distribution with mean 0 and variance s 2. II. Method of Least Squares 1. Total SS SSR SSE, where Total SS Syy. The ANOVA table is produced by computer. 2. Best estimate of s 2 is E SS MSE 1 n k 1. Estimates b0, b1, . . . , bk, for b0, b1, . . . , bk, are chosen to minimize SSE, the sum of squared deviations about the regression line, yˆ b0 b1x1 b2x2 bkxk. IV. Testing, Estimation, and Prediction 1. A test for the signiﬁcance of the regression, H0 : b1 b2 bk 0, can be implemented using the analysis of variance F-test: 2. Least-squares estimates are produced by com- puter. R S F M S E M 2. The strength of the relationship between x and y can be measured using S R S R 2 SS l ta To which gets closer to 1 as the relationship gets stronger. 3. Use residual plots to check for nonnormality, inequality of variances, and an incorrectly ﬁt model. 4. Signiﬁcance tests for the partial regression coefficients can be performed using the Student’s t-test: b t b i with error df n k 1 i (b S i) E 5. Conﬁdence intervals can be generated by computer to estimate the average value of y, E(y), for given values of x1, x2, . . . , xk. Computergenerated prediction intervals can be used to MY MININTAB ❍ 583 predict a particular observation y for given values of x1, x2, . . . , xk. For given x1, x2, . . . , xk, prediction intervals are always wider than conﬁdence intervals. V. Model Building 1. The number of terms in a regression model cannot exceed the number of observations in the data set and should be considerably less! 2. To account for a curvilinear effect in a quantitative variable, use a second-order polynomial model. For a cubic effect, use a third-order polynomial model. 3. To add a qualitative variable with k categories, use (k 1) dummy or indicator variables. 4. There may be interactions between two quantitative variables or between a quantitative and qualitative variable. Interaction terms are entered as bxixj. 5. Compare models using R 2(adj). Multiple Regression Procedures In Chapter 12, you used the linear regression procedures available in MINITAB to perform estimation and testing for a simple linear regression analysis. You obtained a graph of the best-ﬁtting least-squares regression line and calculated the correlation coefficient r and the coefficient of determination r 2. The testing and estimation techniques for a multiple regression analysis are also available with MINITAB and involve almost the same set of commands. You might want to review the section “My MINITAB ” at the end of Chapter 12 before continuing this section. For a response variable y that is related to several predictor variables, x1, x2, . . . , xk, the observed values of y and each of the k predictor variables
must be entered into the ﬁrst (k 1) columns of the MINITAB worksheet. Once this is done, the main inferential tools for linear regression analysis are generated using Stat Regression Regression. The Dialog box for the Regression command is shown in Figure 13.21. Select y for the Response variable and x1, x2, . . . , xk for the Predictor variables. You may now choose to generate some residual plots to check the validity of your regression assumptions before you use the model for estimation or prediction. Choose Graphs to display the Dialog box for residual plots, and choose the appropriate diagnostic plot. Once you have veriﬁed the appropriateness of your multiple regression model, you can choose Options and obtain conﬁdence and prediction intervals for either of these cases: • A single set of values x1, x2, . . . , xk (typed in the box marked “Prediction intervals for new observations”) • Several sets of values x1, x2, . . . , xk stored in k columns of the worksheet When you click OK twice, the regression output is generated. 584 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS F IG URE 13 .2 1 ● The only difficulty in performing the multiple regression analysis using MINITAB might be properly entering the data for your particular model. If the model involves polynomial terms or interaction terms, the Calc Calculator command will help you. For example, suppose you want to ﬁt the model E(y) b0 b1x1 b2x2 b3x 2 1 b4x1x2 You will need to enter the observed values of y, x1, and x2 into the ﬁrst three columns of the MINITAB worksheet. Name column C4 “x1-sq” and name C5 “x1x2.” You can now use the calculator Dialog box shown in Figure 13.22 to generate these two columns. In the Expression box, select x1 * x1 or x1 ** 2 and store the results in C4 (x1-sq). Click OK. Similarly, to obtain the data for C5, select x1 * x2 and store the results in C5 (x1x2). Click OK. You are now ready to perform the multiple regression analysis. If you are ﬁtting either a quadratic or a cubic model in one variable x, you can now plot the data points, the polynomial regression curve, and the upper and lower conﬁdence and prediction limits using Stat Regression Fitted line Plot. Select y and x for the Response and Predictor variables, and click “Display conﬁdence interval” and “Display prediction interval” in the Options Dialog box. Make sure that Quadratic or Cubic is selected as the “Type of Regression Model,” so that you will get the proper ﬁt to the data. Recall that in Chapter 12, you used Stat Basic Statistics Correlation to obtain the value of the correlation coefficient r. In multiple regression analysis, the same command will generate a matrix of correlations, one for each pair of variables in the set y, x1, x2, . . . , xk. Make sure that the box marked “Display p-values” is checked. The p-values will provide information on the signiﬁcant correlation between a particular pair, in the presence of all the other variables in the model, and they are identical to the p-values for the individual t-tests of the regression coefficients. FI GUR E 13. 22 ● SUPPLEMENTARY EXERCISES ❍ 585 Supplementary Exercises 13.25 Biotin Intake in Chicks Groups of 10-day-old chicks were randomly assigned to EX1325 seven treatment groups in which a basal diet was supplemented with 0, 50, 100, 150, 200, 250, or 300 micrograms/kilogram (mg/kg) of biotin. The table gives the average biotin intake (x) in micrograms per day and the average weight gain (y) in grams per day.6 Added Biotin Biotin Intake, x Weight Gain, y 0 50 100 150 200 250 300 .14 2.01 6.06 6.34 7.15 9.65 12.50 8.0 17.1 22.3 24.4 26.5 23.4 23.3 In the MINITAB printout, the second-order polynomial model E(y) b0 b1x b2x 2 is ﬁtted to the data. Use the printout to answer the questions. a. What is the ﬁtted least-squares line? b. Find R2 and interpret its value. c. Do the data provide sufficient evidence to conclude that the model contributes signiﬁcant information for predicting y? d. Find the results of the test of H0 : b2 0. Is there sufficient evidence to indicate that the quadratic model provides a better ﬁt to the data than a simple linear model does? e. Do the residual plots indicate that any of the regression assumptions have been violated? Explain. 586 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS MINITAB output for Exercise 13.25 Regression Analysis: y versus x, x-sq The regression equation is y = 8.59 + 3.82 x - 0.217 x-sq Predictor Coef SE Coef T P Constant 8.585 1.641 5.23 0.006 x 3.8208 0.5683 6.72 0.003 x-sq -0.21663 0.04390 -4.93 0.008 S = 1.83318 R-Sq = 94.4% R-Sq(adj) = 91.5% Analysis of Variance Source DF SS MS F P Regression 2 224.75 112.37 33.44 0.003 Residual Error 4 13.44 3.36 Total 6 238.19 Source DF Seq SS x 1 142.92 x-sq 1 81.83 women’s wear departments. Five weeks for observation were randomly selected from each department, and an advertising budget x1 (in hundreds of dollars) was assigned for each. The weekly sales (in thousands of dollars) are shown in the accompanying table for each of the 15 one-week sales periods. If we expect weekly sales E(y) to be linearly related to advertising expenditure x1, and if we expect the slopes of the lines corresponding to the three departments to differ, then an appropriate model for E(y) is E(y) b0 14243 quantitative variable “advertising expenditure” b1x1 b2x2 b3x3 b4x1x2 b5x1x3 1442443 interaction terms that introduce differences in slopes 1442443 dummy variables used to introduce the qualitative variable “department” into the model Residuals versus the Fitted Values (response is y) where x1 Advertising expenditure x2 1 if children’s wear department B 0 if not x3 1 0 if women’s wear department C if not 10 12 14 16 18 20 22 24 26 Fitted Value Advertising Expenditure (hundreds of dollars) Normal Probability Plot of the Residuals (response is y) Department Men’s wear A Children’s wear B Women’s wear C 1 $5.2 8.2 10.0 2 $5.9 9.0 10.3 3 $7.7 9.1 12.1 4 $7.9 10.5 12.7 5 $9.4 10.5 13.6 a. Find the equation of the line relating E( y) to advertising expenditure x1 for the men’s wear department A. [HINT: According to the coding used for the dummy variables, the model represents mean sales E( y) for the men’s wear department A when x2 x3 0. Substitute x2 x3 0 into the equation for E( y) to ﬁnd the equation of this line.] b. Find the equation of the line relating E(y) to x1 for the children’s wear department B. [HINT: According to the coding, the model represents E(y) for the children’s wear department when x2 1 and x3 0.] c. Find the equation of the line relating E(y) to x1 for the women’s wear department C. 4 3 2 1 0 Residual 1 2 3 4 13.26 Advertising and Sales A department store conducted an experiment to investi- EX1326 gate the effects of advertising expenditures on the weekly sales for its men’s wear, children’s wear, and 99 95 90 80 70 60 50 40 30 20 10 5 1 d. Find the difference between the intercepts of the E( y) lines corresponding to the children’s wear B and men’s wear A departments. e. Find the difference in slopes between E(y) lines corresponding to the women’s wear C and men’s wear A departments. f. Refer to part e. Suppose you want to test the null hypothesis that the slopes of the lines corresponding to the three departments are equal. Express this as a test of hypothesis about one or more of the model parameters. 13.27 Advertising and Sales, continued Refer to Exercise 13.26. Use a computer software package to perform the multiple regression analysis and obtain diagnostic plots if possible. a. Comment on the ﬁt of the model, using the analysis of variance F-test, R2, and the diagnostic plots to check the regression assumptions. b. Find the prediction equation, and graph the three department sales lines. c. Examine the graphs in part b. Do the slopes of the lines corresponding to the children’s wear B and men’s wear A departments appear to differ? Test the null hypothesis that the slopes do not differ (H0 : b4 0) versus the alternative hypothesis that the slopes are different. d. Are the interaction terms in the model signiﬁcant? Use the methods described in Section 13.5 to test H0 : b4 b5 0. Do the results of this test suggest that the ﬁtted model should be modiﬁed? e. Write a short explanation of the practical implica- tions of this regression analysis. 13.28 Demand for Utilities Utility companies, which must plan the operation and EX1328 expansion of electricity generation, are vitally interested in predicting customer demand over both short and long periods of time. A short-term study was conducted to investigate the effect of mean monthly daily temperature x1 and cost per kilowatt-hour x2 on the mean daily consumption (in kilowatt-hours, kWh) per household. The company expected the demand for electricity to rise in cold weather (due to heating), fall when the weather was moderate, and rise again when the temperature rose and there was need for air-conditioning. They expected demand to decrease as the cost per kilowatt-hour increased, reﬂecting greater attention to conservation. Data were available for 2 years, a SUPPLEMENTARY EXERCISES ❍ 587 period in which the cost per kilowatt-hour x2 increased owing to the increasing cost of fuel. The company ﬁtted the model E(y) b0 b1x1 b2x2 1 b3x2 b4x1x2 b5x2 1x2 to the data shown in the table. The MINITAB printout for this multiple regression problem is also provided. Price per kWh, x2 Daily Temperature and Consumption Mean Daily Consumption (kWh) per Household 8¢ Mean daily temperature (°F), x1 Mean daily consumption, y 10¢ Mean daily temperature, x1 Mean daily consumption, y MINITAB output for Exercise 13.28 31 62 55 41 32 62 50 39 34 66 49 46 36 66 44 44 39 68 46 44 39 68 42 40 42 71 47 51 42 72 42 44 47 75 40 62 48 75 38 50 56 78 43 73 56 79 40 55 Regression Analysis: y versus x1, x1-sq, x2, x1x2, x1sqx2 The regression equation is y = 326 - 11.4 x1 + 0.113 x1-sq - 21.7 x2 + 0.873 x1x2 - 0.00887 x1sqx2 Predictor Coef SE Coef T P Constant 325.61 83.06 3.92 0.001 x1 -11.383 3.239 -3.51 0.002 x
1-sq 0.11350 0.02945 3.85 0.001 x2 -21.699 9.224 -2.35 0.030 x1x2 0.8730 0.3589 2.43 0.026 x1sqx2 -0.008869 0.003257 -2.72 0.014 S = 2.90763 R-Sq = 89.8% R-Sq(adj) = 87.0% Analysis of Variance Source DF SS MS F P Regression 5 1346.45 269.29 31.85 0.000 Residual Error 18 152.18 8.45 Total 23 1498.63 Source DF Seq SS x1 1 140.71 x1-sq 1 892.78 x2 1 192.44 x1x2 1 57.84 x1sqx2 1 62.68 Unusual Observations Obs x1 y Fit SE Fit Residual St Resid 9 68.0 44.000 49.640 1.104 -5.640 -2.10R 12 78.0 73.000 67.767 2.012 5.233 2.49R R denotes an observation with a large standardized residual. a. Do the data provide sufficient evidence to indicate that the model contributes information for the prediction of mean daily kilowatt-hour consumption per household? Test at the 5% level of signiﬁcance. b. Graph the curve depicting yˆ as a function of temperature x1 when the cost per kilowatt-hour is x2 8¢. Construct a similar graph for the case when x2 10¢ per kilowatt-hour. Are the consumption curves different? 588 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS c. If cost per kilowatt-hour is unimportant in predicting use, then you do not need the terms involving x2 in the model. Therefore, the null hypothesis H0 : x2 does not contribute information for the prediction of y is equivalent to the null hypothesis H0 : b3 b4 b5 0 (if b3 b4 b5 0, the terms involving x2 disappear from the model). The MINITAB printout, obtained by ﬁtting the reduced model E(y) b0 b1x1 b2x 2 1 to the data, is shown here. Use the methods of Section 13.5 to determine whether price per kilowatthour x2 contributes signiﬁcant information for the prediction of y. MINITAB output for Exercise 13.28 Regression Analysis: y versus x1, x1-sq The regression equation is y = 130 - 3.50 x1 + 0.0334 x1-sq Predictor Coef SE Coef T P Constant 130.01 14.88 8.74 0.000 x1 -3.5017 0.5789 -6.05 0.000 x1-sq 0.033371 0.005256 6.35 0.000 S = 4.70630 R-Sq = 69.0% R-Sq(adj) = 66.0% Analysis of Variance Source DF SS MS F P Regression 2 1033.49 516.75 23.33 0.000 Residual Error 21 465.13 22.15 Total 23 1498.63 Source DF Seq SS x1 1 140.71 x1-sq 1 892.78 Unusual Observations Obs x1 y Fit SE Fit Residual St Resid 12 78.0 73.000 59.906 2.243 13.094 3.16R R denotes an observation with a large standardized residual. d. Compare the values of R2(adj) for the two models ﬁt in this exercise. Which of the two models would you recommend? 13.29 Mercury Concentration in Dolphins Because dolphins (and other large marine EX1329 mammals) are considered to be the top predators in the marine food chain, the heavy metal concentrations in striped dolphins were measured as part of a marine pollution study. The concentration of mercury, the heavy metal reported in this study, is expected to differ in males and females because the mercury in a female is apparently transferred to her offspring during gestation and nursing. This study involved 28 males between the ages of .21 and 39.5 years, and 17 females between the ages of .80 and 34.5 years. For the data in the table, x1 Age of the dolphin (in years) x2 y Mercury concentration (in 0 if female 1 if male micrograms/gram) in the liver y 1.70 1.72 8.80 5.90 101.00 85.40 118.00 183.00 168.00 218.00 180.00 264.00 y 241.00 397.00 209.00 314.00 318.00 2.50 9.35 4.01 29.80 45.30 101.00 135.00 x1 .21 .33 2.00 2.20 8.50 11.50 11.50 13.50 16.50 16.50 17.50 20.50 x1 31.50 31.50 36.50 37.50 39.50 .80 1.58 1.75 5.50 7.50 8.05 11.50 x2 x2 481.00 485.00 221.00 406.00 252.00 329.00 316.00 445.00 278.00 286.00 315.00 y 142.00 180.00 174.00 247.00 223.00 167.00 157.00 177.00 475.00 342.00 x1 22.50 24.50 24.50 25.50 26.50 26.50 26.50 26.50 27.50 28.50 29.50 x1 17.50 17.50 18.50 19.50 21.50 21.50 25.50 25.50 32.50 34.50 x2 x2 . Write a second-order model relating y to x1 and x2. Allow for curvature in the relationship between age and mercury concentration, and allow for an interaction between gender and age. Use a computer software package to perform the multiple regression analysis. Refer to the printout to answer these questions. b. Comment on the ﬁt of the model, using relevant statistics from the printout. c. What is the prediction equation for predicting the mercury concentration in a female dolphin as a function of her age? d. What is the prediction equation for predicting the mercury concentration in a male dolphin as a function of his age? e. Does the quadratic term in the prediction equation for females contribute signiﬁcantly to the prediction of the mercury concentration in a female dolphin? f. Are there any other important conclusions that you feel were not considered regarding the ﬁtted prediction equation? EX1330 13.30 The Cost of Flying Does the cost of a plane ﬂight depend on the airline as well as the distance traveled? In Exercise 12.21, you explored the ﬁrst part of this problem. The data shown in this table compare the average cost and distance traveled for two different airlines, measured for 11 heavily traveled air routes in the United States.7 Route Chicago–Detroit Chicago–Denver Chicago–St. Louis Chicago–Seattle Chicago–Cleveland Los Angeles–Chicago Chicago–Atlanta New York–Los Angeles New York–Chicago Los Angeles–Honolulu New York–San Francisco Distance Cost 238 901 262 1736 301 1757 593 2463 714 2556 2574 148 164 256 312 136 152 424 520 129 139 361 473 162 183 444 525 287 334 323 333 513 672 Airline American United American United American United American United American United American United American United American United American United American United American United Use a computer package to analyze the data with a multiple regression analysis. Comment on the ﬁt of the model, the signiﬁcant variables, any interactions that exist, and any regression assumptions that may have been violated. Summarize your results in a report, including printouts and graphs if possible. EX1331 13.31 On the Road Again Until recently, performance tires were ﬁtted mostly on sporty or luxury vehicles. Now they come standard on many everyday sedans. Increased levels of handling and grip have come at the expense of tread wear. The data that follows is abstracted from a report on H-rated performance tires by Consumer Reports8 in which several aspects of performance were evaluated for n = 22 different tires where y overall score x2 wet braking x4 roll resistance x1 dry braking x3 handling x5 tread life SUPPLEMENTARY EXERCISES ❍ 589 Tire Cost y x1 x2 x3 x4 x5 Dunlop® SP Sport 5000 Michelin® Pilot Exalto A/S Falken® Ziex ZE 512 Continental® ContiProContact Michelin Pilot XGT H4 Bridgestone® Potenza RE 950 BFGoodrich® Traction T/A Yokohama® Avid H4s Sumitomo® HTR H4 Bridgestone HP50 Michelin Energy MXV4 Plus Goodyear® Assurance Triple Tred Kumho Solus KH16 Pirelli® P6 Four Seasons Bridgestone Potenza G009 Dayton® Daytona HR Fuzion® HRi Continental ContiPremierContact Cooper® Lifeliner Touring SLE Bridgestone Turanza EL400 Hankook® Optimo H418 General Exclaim 81 78 56 77 98 80 68 62 56 79 103 85 49 71 62 46 49 91 61 91 51 53 85 83 83 81 81 80 78 77 76 75 73 72 72 70 70 69 69 68 65 63 62 50 The variables x1 through x5 are coded using the scale 5 excellent, 4 very good, 3 good, 2 fair, and 1 poor. a. Use a program of your choice to ﬁnd the correlation matrix for the variables under study including cost. Is cost signiﬁcantly correlated with any of the study variables? Which variables appear to be highly correlated with y, the total score? b. Write a model to describe y, total score, as a function of the variables x1 dry braking, x2 wet braking, x3 handling, x4 roll resistance, and x5 tread life. c. Use a regression program of your choice to ﬁt the full model using all of the predictors. What proportion of the variation in y is explained by regression? Does this convey the impression that the model adequately explains the inherent variability in y? d. Which variable or variables appear to be good predictors of y? How might you reﬁne the model in light of these results? Use these variables in reﬁtting the model. What proportion of the variation is explained by this reﬁtted model? Comment on the adequacy of this reduced model in comparison to the full model. EX1332 13.32 Tuna Fish The tuna ﬁsh data from Exercise 11.16 were analyzed as a completely randomized design with four treatments. However, we could also view the experimental design as a 2 2 590 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS factorial experiment with unequal replications. The data are shown below.9 Light Tuna White Tuna .62 .66 .62 .65 .60 .67 Oil 2.56 1.92 1.30 1.79 1.23 1.27 1.22 1.19 1.22 Water .99 1.92 1.23 .85 .65 .53 1.41 1.49 1.29 1.27 1.35 1.12 .63 .67 .69 .60 .60 .66 1.29 1.00 1.27 1.28 Source: Case Study “Tuna Goes Upscale” Copyright 2001 by Consumers Union of U.S., Inc., Yonkers, NY 10703-1057, a nonproﬁt organization. Reprinted with permission from the June 2001 issue of Consumer Reports® for educational purposes only. No commercial use or reproduction permitted. www.ConsumerReports.org. The data can be analyzed using the model y b0 b1x1 b2x2 b3x1x2 e where x1 0 if oil, 1 if water x2 0 if light tuna, 1 if white tuna a. Show how you would enter the data into a computer spreadsheet, entering the data into columns for y, x1, x2, and x1x2. b. The printout generated by MINITAB is shown below. What is the least-squares prediction equation? MINITAB output for Exercise 13.32 Regression Analysis: y versus x1, x2, x1x2 The regression equation is y = 1.15 - 0.251 x1 + 0.078 x2 + 0.306 x1x2 Predictor Coef SE Coef T P Constant 1.1473 0.1370 8.38 0.000 x1 -0.2508 0.1830 -1.37 0.180 x2 0.0777 0.2652 0.29 0.771 x1x2 0.3058 0.3330 0.92 0.365 S = 0.454287 R-Sq = 11.9% R-Sq(adj) = 3.9% Analysis of Variance Source DF SS MS F P Regression 3 0.9223 0.3074 1.49 0.235 Residual Error 33 6.8104 0.2064 Total 36 7.7328 13.33 Tuna, continued Refer to Exercise 13.32. The hypothesis tested in Chapter 11—that the average prices for the four types of tuna are the same—is equivalent to saying that E(y) will not change as x1 and x2 change. This c
an only happen when b1 b2 b3 0. Use the MINITAB printout for the one-way ANOVA shown below to perform the test for equality of treatment means. Verify that this test is identical to the test for signiﬁcant regression in Exercise 13.32. MINITAB output for Exercise 13.33 One-Way ANOVA: Light Water, White Oil, White Water, Light Oil Source DF SS MS F P Factor 3 0.922 0.307 1.49 0.235 Error 33 6.810 0.206 Total 36 7.733 S = 0.4543 R-Sq = 11.93% R-Sq(adj) = 3.92% 13.34 Quality Control A manufacturer recorded the number of defective items (y) EX1334 produced on a given day by each of 10 machine operators and also recorded the average output per hour (x1) for each operator and the time in weeks from the last machine service (x2). y 13 1 11 2 20 15 27 5 26 1 x1 20 15 23 10 30 21 38 18 24 16 x2 3.0 2.0 1.5 4.0 1.0 3.5 0 2.0 5.0 1.5 The printout that follows resulted when these data were analyzed using the MINITAB package using the model: E(y) b0 b1x1 b2x2 Regression Analysis: y versus x1, x2 The regression equation is y = -28.4 + 1.46 x1 + 3.84 x2 Predictor Coef SE Coef T P Constant -28.3906 0.8273 -34.32 0.000 x1 1.46306 0.02699 -54.20 0.000 x2 3.8446 0.1426 26.97 0.000 S = 0.548433 R-Sq = 99.8% R-Sq(adj) = 99.7% c. Is there an interaction between type of tuna and Analysis of Variance type of packing liquid? d. Which, if any, of the main effects (type of tuna and type of packing liquid) contribute signiﬁcant information for the prediction of y? e. How well does the model ﬁt the data? Explain. Source DF SS MS F P Regression 2 884.79 442.40 1470.84 0.000 Residual Error 7 2.11 0.30 Total 9 886.90 Source DF Seq SS x1 1 666.04 x2 1 218.76 a. Interpret R2 and comment on the ﬁt of the model. b. Is there evidence to indicate that the model contributes signiﬁcantly to the prediction of y at the a .01 level of signiﬁcance? c. What is the prediction equation relating yˆ and x1 when x2 4? d. Use the ﬁtted prediction equation to predict the number of defective items produced for an operator whose average output per hour is 25 and whose machine was serviced three weeks ago. e. What do the residual plots tell you about the valid- ity of the regression assumptions? Normal Probability Plot of the Residuals (response is y) t n e c r e P 99 95 90 80 70 60 50 40 30 20 10 .050 0.25 0.00 0.25 0.50 0.75 1.0 0.5 0.0 Residual 0.5 1.0 Residuals versus the Fitted Values (response is y) 0 5 10 15 Fitted Value 20 25 30 EX1335 13.35 Metal Corrosion and Soil Acids In an investigation to determine the relationship between the degree of metal corrosion and the length of time the metal is exposed to the action of soil acids, the percentage of corrosion and exposure time were measured weekly. SUPPLEMENTARY EXERCISES ❍ 591 y x 0.1 1 0.3 2 0.5 3 0.8 4 1.2 5 1.8 6 2.5 7 3.4 8 The data were ﬁtted using the quadratic model, E( y) b0 b1x b2x 2, with the following results. Regression Analysis: y versus x, x-sq The regression equation is y = 0.196 - 0.100 x + 0.0619 x-sq Predictor Coef SE Coef T P Constant 0.19643 0.07395 2.66 0.045 x -0.10000 0.03770 -2.65 0.045 x-sq2 0.061905 0.004089 15.14 0.000 S = 0.0530049 R-Sq = 99.9% R-Sq(adj) = 99.8% Analysis of Variance Source DF SS MS F P Regression 2 9.4210 4.7105 1676.61 0.000 Residual Error 5 0.0140 0.0028 Total 7 9.4350 Source DF Seq SS x 1 8.7771 x-sq 1 0.6438 a. What percentage of the total variation is explained by the quadratic regression of y on x? b. Is the regression on x and x2 signiﬁcant at the a .05 level of signiﬁcance? c. Is the linear regression coefficient signiﬁcant when x2 is in the model? d. Is the quadratic regression coefficient signiﬁcant when x is in the model? e. The data were ﬁtted to a linear model without the quadratic term with the results that follow. What can you say about the contribution of the quadratic term when it is included in the model? Regression Analysis: y versus x The regression equation is y = -0.732 + 0.457 x Predictor Coef SE Coef T P Constant -0.7321 0.2580 -2.84 0.030 x 0.45714 0.05109 8.95 0.000 S = 0.331124 R-Sq = 93.0% R-Sq(adj) = 91.9% Analysis of Variance Source DF SS MS F P Regression 1 8.7771 8.7771 80.05 0.000 Residual Error 6 0.6579 0.1096 Total 7 9.4350 f. The plot of the residuals from the linear regression model in part e shows a speciﬁc pattern. What is the term in the model that seems to be missing? 592 ❍ CHAPTER 13 MULTIPLE REGRESSION ANALYSIS Residuals versus the Fitted Values (response is y) The following computer printout resulted when the data were analyzed using MINITAB. Regression Analysis: y versus x1, x2, x3 The regression equation is y = -3.11 + 0.503 x1 - 1.61 x2 - 1.15 x3 Predictor Coef SE Coef T P Constant -3.112 3.600 -0.86 0.421 x1 0.50314 0.07670 6.56 0.001 x2 -1.6126 0.6579 -2.45 0.050 x3 -1.155 1.791 -0.64 0.543 S = 1.89646 R-Sq = 92.2% R-Sq(adj) = 88.4% Analysis of Variance Source DF SS MS F P Regression 3 256.621 85.540 23.78 0.001 Residual Error 6 21.579 3.597 Total 9 278.200 Source DF Seq SS x1 1 229.113 x2 1 26.012 x3 1 1.496 a. Interpret R2 and comment on the ﬁt of the model. b. Test for a signiﬁcant regression of y on x1, x2, and x3 at the 5% level of signiﬁcance. c. Test the hypothesis H0 : b3 0 against Ha : b3 0 using a .05. Comment on the results of your test. d. What can be said about the utility of x3 as a pre- dictor variable in this problem.5 0.4 0.3 0.2 0.1 0.0 0.1 0.2 0.3 0.4 0.0 0.5 1.0 1.5 Fitted Value 2.0 2.5 3.0 13.36 Managing your Money A particular savings and loan corporation is interested in EX1336 determining how well the amount of money in family savings accounts can be predicted using the three independent variables—annual income, number in the family unit, and area in which the family lives. Suppose that there are two speciﬁc areas of interest to the corporation. The following data were collected, where y Amount in all savings accounts x1 Annual income x2 Number in family unit x3 0 if in Area 1; 1 if not Both y and x1 were recorded in units of $1000. y 0.5 0.3 1.3 0.2 5.4 1.3 12.8 1.5 0.5 15.2 x1 19.2 23.8 28.6 15.4 30.5 20.3 34.7 25.2 18.6 45.8 x2 x3 CASE STUDY Foreign Cars “Made in the U.S.A.”—Another Look The case study in Chapter 12 examined the effect of foreign competition in the automotive industry as the number of imported cars steadily increased during the 1970s and 1980s.10 The U.S. automobile industry has been besieged with complaints about product quality, worker layoffs, and high prices and has spent billions in advertising and research to produce an American-made car that will satisfy consumer demands. Have they been successful in stopping the ﬂood of imported cars purchased by American consumers? The data shown in the table give the number of imported cars (y) sold in the United States (in millions) for the years 1969–2005. To simplify the analysis, we have coded the year using the coded variable x Year 1969. Year 1969, x Number of Imported Cars 10 11 12 13 14 15 16 17 1.1 1.3 1.6 1.6 1.8 1.4 1.6 1.5 2.1 2.0 2.3 2.4 2.3 2.2 2.4 2.4 2.8 3.2 Year 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 CASE STUDY ❍ 593 Year 1969, x Number of Imported Cars, y 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 3.1 3.1 2.8 2.5 2.1 2.0 1.8 1.8 1.6 1.4 1.4 1.4 1.8 2.1 2.2 2.3 2.2 2.2 2.3 Year 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 By examining a scatterplot of these data, you will ﬁnd that the number of imported cars does not appear to follow a linear relationship over time, but rather exhibits a curvilinear response. The question, then, is to decide whether a second-, third-, or higher-order model adequately describes the data. 1. Plot the data and sketch what you consider to be the best-ﬁtting linear, quadratic, and cubic models. 2. Find the residuals using the ﬁtted linear regression model. Does there appear to be any pattern in the residuals when plotted against x? What model do the residuals indicate would produce a better ﬁt? 3. What is the increase in R2 when you ﬁt a quadratic rather than a linear model? Is the coefficient of the quadratic term signiﬁcant? Is the ﬁtted quadratic model signiﬁcantly better than the ﬁtted linear model? Plot the residuals from the ﬁtted quadratic model. Does there seem to be any apparent pattern in the residuals when plotted against x? 4. What is the increase in R2 when you compare the ﬁtted cubic with the ﬁtted quadratic model? Is the ﬁtted cubic model signiﬁcantly better than the ﬁtted quadratic? Are there any patterns in a plot of the residuals versus x? What proportion of the variation in the response y is not accounted for by ﬁtting a cubic model? Should any higher-order polynomial model be considered? Why or why not? 14 Analysis of Categorical Data GENERAL OBJECTIVES Many types of surveys and experiments result in qualitative rather than quantitative response variables, so that the responses can be classiﬁed but not quantiﬁed. Data from these experiments consist of the count or number of observations that fall into each of the response categories included in the experiment. In this chapter, we are concerned with methods for analyzing categorical data. CHAPTER INDEX ● Assumptions for chi-square tests (14.7) ● Comparing several multinomial populations (14.5) ● Contingency tables (14.4) ● The multinomial experiment (14.1) ● Other applications (14.7) ● Pearson’s chi-square statistic (14.2) ● A test of speciﬁed cell probabilities (14.3) How Do I Determine the Appropriate Number of Degrees of Freedom? © Dave Bartruff/CORBIS Can a Marketing Approach Improve Library Services? How do you rate your library? Is the atmosphere friendly, dull, or too quiet? Is the library staff helpful? Are the signs clear and unambiguous? The modern consumer-led approach to marketing, in general, involves the systematic study by organizations of their customers’ wants and needs in order to improve their services or products. In the case study at the end of this chapter, we examine the results
of a study to explore the attitudes of young adults toward the services provided by libraries. 594 14.1 A DESCRIPTION OF THE EXPERIMENT ❍ 595 A DESCRIPTION OF THE EXPERIMENT 14.1 Many experiments result in measurements that are qualitative or categorical rather than quantitative; that is, a quality or characteristic (rather than a numerical value) is measured for each experimental unit. You can summarize this type of data by creating a list of the categories or characteristics and reporting a count of the number of measurements that fall into each category. Here are a few examples: • People can be classiﬁed into ﬁve income brackets. • A mouse can respond in one of three ways to a stimulus. • An M&M’S candy can have one of six colors. • An industrial process manufactures items that can be classiﬁed as “acceptable,” “second quality,” or “defective.” These are some of the many situations in which the data set has characteristics appropriate for the multinomial experiment. THE MULTINOMIAL EXPERIMENT • The experiment consists of n identical trials. • The outcome of each trial falls into one of k categories. • The probability that the outcome of a single trial falls into a particular category—say, category i—is pi and remains constant from trial to trial. This probability must be between 0 and 1, for each of the k categories, and the sum of all k probabilities is Spi 1. • The trials are independent. • The experimenter counts the observed number of outcomes in each category, written as O1, O2, . . . , Ok, with O1 O2 Ok n. You can visualize the multinomial experiment by thinking of k boxes or cells into which n balls are tossed. The n tosses are independent, and on each toss the chance of hitting the ith box is the same. However, this chance can vary from box to box; it might be easier to hit box 1 than box 3 on each toss. Once all n balls have been tossed, the number in each box or cell—O1, O2, . . . , Ok—is counted. You have probably noticed the similarity between the multinomial experiment and the binomial experiment introduced in Chapter 5. In fact, when there are k 2 categories, the two experiments are identical, except for notation. Instead of p and q, we write p1 and p2 to represent the probabilities for the two categories, “success” and “failure.” Instead of x and (n x), we write O1 and O2 to represent the observed number of “successes” and “failures.” When we presented the binomial random variable, we made inferences about the binomial parameter p (and by default, q 1 p) using large-sample methods based on the z statistic. In this chapter, we extend this idea to make inferences about the multinomial parameters, p1, p2, . . . , pk, using a different type of statistic. This statistic, whose approximate sampling distribution was derived by a British statistician named Karl Pearson in 1900, is called the chi-square (or sometimes Pearson’s chi-square) statistic. The multinomial experiment is an extension of the binomial experiment. For a binomial experiment, k 2. 596 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA 14.2 PEARSON’S CHI-SQUARE STATISTIC Suppose that n 100 balls are tossed at the cells (boxes) and you know that the probability of a ball falling into the ﬁrst box is p1 .1. How many balls would you expect to fall into the ﬁrst box? Intuitively, you would expect to see 100(.1) 10 balls in the ﬁrst box. This should remind you of the average or expected number of successes, m np, in the binomial experiment. In general, the expected number of balls that fall into cell i—written as Ei—can be calculated using the formula Ei npi for any of the cells i 1, 2, . . . , k. Now suppose that you hypothesize values for each of the probabilities p1, p2, . . . , pk and calculate the expected number for each category or cell. If your hypothesis is correct, the actual observed cell counts, Oi, should not be too different from the expected cell counts, Ei npi. The larger the differences, the more likely it is that the hypothesis is incorrect. The Pearson chi-square statistic uses the differences (Oi Ei) by ﬁrst squaring these differences to eliminate negative contributions, and then forming a weighted average of the squared differences. The Pearson’s chi-square tests are always uppertailed tests. PEARSON’S CHI-SQUARE TEST STATISTIC X2 S(Oi Ei)2 Ei summed over all k cells, with Ei npi. Although the mathematical proof is beyond the scope of this book, it can be shown that when n is large, X2 has an approximate chi-square probability distribution in repeated sampling. If the hypothesized expected cell counts are correct, the differences (Oi Ei) are small and X2 is close to 0. But, if the hypothesized probabilities are incorrect, large differences (Oi Ei) result in a large value of X2. You should use a right-tailed statistical test and look for an unusually large value of the test statistic. The chi-square distribution was used in Chapter 10 to make inferences about a single population variance s 2. Like the F distribution, its shape is not symmetric and depends on a speciﬁc number of degrees of freedom. Once these degrees of freedom are speciﬁed, you can use Table 5 in Appendix I to ﬁnd critical values or to bound the p-value for a particular chi-square statistic. As an alternative, you can use the Chi-Square Probabilities applet to ﬁnd critical values or exact p-values for the test. The appropriate degrees of freedom for the chi-square statistic vary depending on the particular application you are using. Although we will specify the appropriate degrees of freedom for the applications presented in this chapter, you should use the general rule given next for determining degrees of freedom for the chi-square statistic. How Do I Determine the Appropriate Number of Degrees of Freedom? 1. Start with the number of categories or cells in the experiment. 2. Subtract one degree of freedom for each linear restriction on the cell probabili- ties. You will always lose one df because p1 p2 pk 1. 14.3 TESTING SPECIFIED CELL PROBABILITIES: THE GOODNESS-OF-FIT TEST ❍ 597 3. Sometimes the expected cell counts cannot be calculated directly but must be estimated using the sample data. Subtract one degree of freedom for every independent population parameter that must be estimated to obtain the estimated values of Ei. We begin with the simplest applications of the chi-square test statistic—the goodness- of-ﬁt test. TESTING SPECIFIED CELL PROBABILITIES: THE GOODNESS-OF-FIT TEST 14.3 The simplest hypothesis concerning the cell probabilities speciﬁes a numerical value for each cell. The expected cell counts are easily calculated using the hypothesized probabilities, Ei npi, and are used to calculate the observed value of the X2 test statistic. For a multinomial experiment consisting of k categories or cells, the test statistic has an approximate x 2 distribution with df (k 1). EXAMPLE 14.1 A researcher designs an experiment in which a rat is attracted to the end of a ramp that divides, leading to doors of three different colors. The researcher sends the rat down the ramp n 90 times and observes the choices listed in Table 14.1. Does the rat have (or acquire) a preference for one of the three doors? TABLE 14.1 ● Rat’s Door Choices Door Green Red Blue Observed Count (Oi) 20 39 31 Solution If the rat has no preference in the choice of a door, you would expect in the long run that the rat would choose each door an equal number of times. That is, the null hypothesis is H0 : p1 p2 p3 1 3 versus the alternative hypothesis Ha : At least one pi is different from 1 3 where pi is the probability that the rat chooses door i, for i 1, 2, and 3. The expected cell counts are the same for each of the three categories—namely, npi 90(1/3) 30. The chi-square test statistic can now be calculated as The rejection region and p-value are in the upper tail of the chi-square distribution. X2 S(Oi Ei)2 Ei (31 (39 (20 30)2 30)2 30)2 6.067 30 30 30 598 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA For this example, the test statistic has (k 1) 2 degrees of freedom because the only linear restriction on the cell probabilities is that they must sum to 1. Hence, you can use Table 5 in Appendix I to ﬁnd bounds for the right-tailed p-value. Since the .025 7.38, the p-value observed value, X2 6.067, lies between x 2 is between .025 and .050. The researcher would report the results as signiﬁcant at the 5% level (P .05), meaning that the null hypothesis of no preference is rejected. There is sufficient evidence to indicate that the rat has a preference for one of the three doors. .050 5.99 and x 2 What more can you say about the experiment once you have determined statistically that the rat has a preference? Look at the data to see where the differences lie. The Goodness-of-Fit Test applet, shown in Figure 14.1, will help. F IG URE 14. 1 Goodness-of-Fit applet ● You can see the value of X2 and its exact p-value (.0482) at the bottom of the applet. Just above them, the shaded bar shows the distribution of the observed frequencies. The blue bars represent categories that have an excess of observations relative to expected and red cells (gray in Figure 14.1) indicate a deﬁcit of observations relative to expected. The intensity of the color reﬂects the magnitude of the discrepancy. For this example, that rat chose the red and blue doors more often than expected, and the green door less often. The blue door was chosen only a little more than one-third of the time: 3 1 .344 0 9 However, the sample proportions for the other two doors are quite different from onethird. The rat chooses the green door least often—only 22% of the time: 2 0 .222 0 9 The rat chooses the red door most often—43% of the time: 3 9 .433 0 9 You would summarize the results of the experiment by saying that the rat has a preference for the red door. Can you conclude that the preference is caused by the door color? The answer is no—the cause could be some other physiological or psychological factor that you have not yet explored.
Avoid declaring a causal relationship between color and preference! EXAMPLE 14.2 TABLE 14.2 Degrees of freedom for a simple goodness-of-ﬁt test: df k 1 14.3 TESTING SPECIFIED CELL PROBABILITIES: THE GOODNESS-OF-FIT TEST ❍ 599 The proportions of blood phenotypes A, B, AB, and O in the population of all Caucasians in the United States are .41, .10, .04, and .45, respectively. To determine whether or not the actual population proportions ﬁt this set of reported probabilities, a random sample of 200 Americans were selected and their blood phenotypes were recorded. The observed and expected cell counts are shown in Table 14.2. The expected cell counts are calculated as Ei 200pi. Test the goodness of ﬁt of these blood phenotype proportions. ● Counts of Blood Phenotypes A AB O B Observed (Oi) Expected (Ei) 89 82 18 20 12 8 81 90 Solution The hypothesis to be tested is determined by the model probabilities: H0 : p1 .41; p2 .10; p3 .04; p4 .45 versus Ha : At least one of the four probabilities is different from the speciﬁed value Then X2 S(Oi Ei)2 Ei (81 (89 90)2 82)2 3.70 90 82 From Table 5 in Appendix I, indexing df (k 1) 3, you can ﬁnd that the ob.100 6.25, so that the p-value is greater served value of the test statistic is less than x 2 than .10. You do not have sufficient evidence to reject H0; that is, you cannot declare that the blood phenotypes for American Caucasians are different from those reported earlier. The results are nonsigniﬁcant (NS). You can ﬁnd instructions in the “My MINITAB” section at the end of this chapter that allow you to perform the chi-square goodness-of-ﬁt test and generate the results. This procedure is new to MINITAB 15. If you are using a previous version of MINITAB, you can still generate results using the calculator function. Notice the difference in the goodness-of-ﬁt hypothesis compared to other hypotheses that you have tested. In the goodness-of-ﬁt test, the researcher uses the null hypothesis to specify the model he believes to be true, rather than a model he hopes to prove false! When you could not reject H0 in the blood type example, the results were as expected. Be careful, however, when you report your results for goodness-ofﬁt tests. You cannot declare with conﬁdence that the model is absolutely correct without reporting the value of b for some practical alternatives. 14.3 EXERCISES BASIC TECHNIQUES 14.1 List the characteristics of a multinomial experiment. 14.2 Use Table 5 in Appendix I to ﬁnd the value of x 2 with the following area a to its right: a. a .05, df 3 b. a .01, df 8 600 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA b. k 10, a .01 14.3 Give the rejection region for a chi-square test of speciﬁed probabilities if the experiment involves k categories in these cases: a. k 7, a .05 14.4 Use Table 5 in Appendix I to bound the p-value for a chi-square test: a. X2 4.29, df 5 14.5 Suppose that a response can fall into one of k 5 categories with probabilities p1, p2, . . . , p5 and that n 300 responses produced these category counts: b. X2 20.62, df 6 Category Observed Count 1 47 2 63 3 74 4 51 5 65 a. Are the ﬁve categories equally likely to occur? How would you test this hypothesis? b. If you were to test this hypothesis using the chisquare statistic, how many degrees of freedom would the test have? c. Find the critical value of x 2 that deﬁnes the rejec- tion region with a .05. d. Calculate the observed value of the test statistic. e. Conduct the test and state your conclusions. 14.6 Suppose that a response can fall into one of k 3 categories with probabilities p1 .4, p2 .3, and p3 .3, and n 300 responses produce these category counts: Category Observed Count 1 130 2 98 3 72 Do the data provide sufficient evidence to indicate that the cell probabilities are different from those speciﬁed for the three categories? Find the approximate p-value and use it to make your decision. APPLICATIONS 14.7 Your Favorite Lane A freeway with four lanes in each direction was studied to see whether drivers prefer to drive on the inside lanes. A total of 1000 automobiles were observed during heavy earlymorning traffic, and the number of cars in each lane was recorded: Lane 1 2 3 4 Observed Count 294 276 238 192 Do the data present sufficient evidence to indicate that some lanes are preferred over others? Test using a .05. If there are any differences, discuss the nature of the differences. 14.8 Peonies A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that 75% of the offspring from this cross will have red ﬂowers. To test this claim, 100 seeds from this cross were collected and germinated, and 58 plants had red petals. Use the chi-square goodness-of-ﬁt test to determine whether the sample data conﬁrm the geneticist’s prediction. 14.9 Heart Attacks on Mondays Do you hate Mondays? Researchers from Germany have provided another reason for you: They concluded that the risk of a heart attack for a working person may be as much as 50% greater on Monday than on any other day.1 The researchers kept track of heart attacks and coronary arrests over a period of 5 years among 330,000 people who lived near Augsburg, Germany. In an attempt to verify their claim, you survey 200 working people who had recently had heart attacks and recorded the day on which their heart attacks occurred: Day Observed Count Sunday Monday Tuesday Wednesday Thursday Friday Saturday 24 36 27 26 32 26 29 Do the data present sufficient evidence to indicate that there is a difference in the incidence of heart attacks depending on the day of the week? Test using a .05. 14.10 Mortality Statistics Medical statistics show that deaths due to four major diseases—call them A, B, C, and D—account for 15%, 21%, 18%, and 14%, respectively, of all nonaccidental deaths. A study of the causes of 308 nonaccidental deaths at a hospital gave the following counts: Disease Deaths A 43 B 76 C 85 D 21 Other 83 Do these data provide sufficient evidence to indicate that the proportions of people dying of diseases A, B, C, and D at this hospital differ from the proportions accumulated for the population at large? 14.11 Schizophrenia Research has suggested a link between the prevalence of schizophrenia and birth during particular months of the year in which viral infections are prevalent. Suppose you are working on a similar problem and you suspect a linkage between a disease observed in later life and month of birth. 14.3 TESTING SPECIFIED CELL PROBABILITIES: THE GOODNESS-OF-FIT TEST ❍ 601 You have records of 400 cases of the disease, and you classify them according to month of birth. The data appear in the table. Do the data present sufficient evidence to indicate that the proportion of cases of the disease per month varies from month to month? Test with a .05. Month Births Month Births Jan 38 July 24 Feb Mar Apr May June 31 42 46 28 31 Aug 29 Sept 33 Oct 36 Nov 27 Dec 35 14.12 Snap Peas Suppose you are interested in following two independent traits in snap peas—seed texture (S smooth, s wrinkled) and seed color (Y yellow, y green)—in a second-generation cross of heterozygous parents. Mendelian theory states that the number of peas classiﬁed as smooth and yellow, wrinkled and yellow, smooth and green, and wrinkled and green should be in the ratio 9:3:3:1. Suppose that 100 randomly selected snap peas have 56, 19, 17, and 8 in these respective categories. Do these data indicate that the 9:3:3:1 model is correct? Test using a .01. 14.13 M&M’S The Mars, Incorporated website reports the following percentages of the various colors of its M&M’S candies for the “milk chocolate” variety:2 118 blue, 108 orange, and 85 green candies. Do the data substantiate the percentages reported by Mars, Incorporated? Use the appropriate test and describe the nature of the differences, if there are any. 14.14 Peanut M&M’S The percentage of various colors are different for the “peanut” variety of M&M’S candies, as reported on the Mars, Incorporated website:3 What Colors Come In Your Bag? Brown Yellow Red Blue Orange Green m m m mm mm m 12% 15% 12% 23% 23% 15% What Colors Come In Your Bag? Brown Yellow Red Blue Orange Green m m m mm mm mm 13% 14% 13% 24% 20% 16% A 14-ounce bag of peanut M&M’S is randomly selected and contains 70 brown, 87 yellow, 64 red, 115 blue, 106 orange, and 85 green candies. Do the data substantiate the percentages reported by Mars, Incorporated? Use the appropriate test and describe the nature of the differences, if there are any. 14.15 Admission Standards Previous enrollment records at a large university indicate that of the total number of persons who apply for admission, 60% are admitted unconditionally, 5% are admitted on a trial basis, and the remainder are refused admission. Of 500 applications to date for the coming year, 329 applicants have been admitted unconditionally, 43 have been admitted on a trial basis, and the remainder have been refused admission. Do these data indicate a departure from previous admission rates? Test using a .05. A 14-ounce bag of milk chocolate M&M’S is randomly selected and contains 70 brown, 72 yellow, 61 red, 602 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION 14.4 In some situations, the researcher classiﬁes an experimental unit according to two qualitative variables to generate bivariate data, which we discussed in Chapter 3. • A defective piece of furniture is classiﬁed according to the type of defect and the production shift during which it was made. • A professor is classiﬁed by professional rank and the type of university (public or private) at which she works. • A patient is classiﬁed according to the type of preventive ﬂu treatment he received and whether or not he contracted the ﬂu during the winter. When two categorical variables are recorded, you can summarize the data by counting the observed number of units that fall into each of the various intersections of category levels. The resulting counts are displayed in
an array called a contingency table. A total of n 309 furniture defects were recorded and the defects were classiﬁed into four types: A, B, C, or D. At the same time, each piece of furniture was identiﬁed by the production shift in which it was manufactured. These counts are presented in a contingency table in Table 14.3. EXAMPLE 14.3 TABLE 14.3 ● Contingency Table Type of Defects A B C D Total Shift 3 Total 33 17 49 20 119 74 69 128 38 309 1 15 21 45 13 94 2 26 31 34 5 96 When you study data that involves two variables, one important consideration is the relationship between the two variables. Does the proportion of measurements in the various categories for factor 1 depend on which category of factor 2 is being observed? For the furniture example, do the proportions of the various defects vary from shift to shift, or are these proportions the same, independently of which shift is observed? You may remember a similar phenomenon called interaction in the a b factorial experiment from Chapter 11. In the analysis of a contingency table, the objective is to determine whether or not one method of classiﬁcation is contingent or dependent on the other method of classiﬁcation. If not, the two methods of classiﬁcation are said to be independent. The Chi-Square Test of Independence The question of independence of the two methods of classiﬁcation can be investigated using a test of hypothesis based on the chi-square statistic. These are the hypotheses: H0 : The two methods of classiﬁcation are independent Ha : The two methods of classiﬁcation are dependent With two-way classiﬁcations, we do not test hypotheses about speciﬁc probabilities. We test whether the two methods of classiﬁcation are independent. 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 603 Suppose we denote the observed cell count in row i and column j of the contingency table as Oij. If you knew the expected cell counts (Eij npij ) under the null hypothesis of independence, then you could use the chi-square statistic to compare the observed and expected counts. However, the expected values are not speciﬁed in H0, as they were in previous examples. To explain how to estimate these expected cell counts, we must revisit the concept of independent events from Chapter 4. Consider pij, the probability that an observation falls into row i and column j of the contingency table. If the rows and columns are independent, then pij P(observation falls in row i and column j) P(observation falls in row i) P(observation falls in column j) pi pj where pi and pj are the unconditional or marginal probabilities of falling into row i or column j, respectively. If you could obtain proper estimates of these marginal probabilities, you could use them in place of pij in the formula for the expected cell count. Fortunately, these estimates do exist. In fact, they are exactly what you would in- tuitively choose: • To estimate a row probability, use Degrees of freedom for an r c contingency table: df (r 1)(c 1). pˆi Total observations in row i Total number of observations r i n • To estimate a column probability, use pˆj Total observations in column j Total number of observations c j n The estimate of the expected cell count for row i and column j follows from the independence assumption. ESTIMATED EXPECTED CELL COUNT Eˆij nr ic j ri n n cj n where ri is the total for row i and cj is the total for column j. The chi-square test statistic for a contingency table with r rows and c columns is calculated as Eˆij)2 X2 S(Oij Eˆij and can be shown to have an approximate chi-square distribution with df (r 1)(c 1) If the observed value of X2 is too large, then the null hypothesis of independence is rejected. 604 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA EXAMPLE 14.4 Refer to Example 14.3. Do the data present sufficient evidence to indicate that the type of furniture defect varies with the shift during which the piece of furniture is produced? Solution The estimated expected cell counts are shown in parentheses in Table 14.4. For example, the estimated expected count for a type C defect produced during the second shift is 32 r3 c2 (12 96) 39.77 )( 8 Eˆ n 0 9 3 TABLE 14.4 ● Observed and Estimated Expected Cell Counts Shift Type of Defects 1 2 3 Total A B C D 15 (22.51) 21 (20.99) 45 (38.94) 13 (11.56) 26 (22.99) 31 (21.44) 34 (39.77) 5 (11.81) 33 (28.50) 17 (26.57) 49 (49.29) 20 (14.63) Total 94 96 119 74 69 128 38 309 You can now use the values shown in Table 14.4 to calculate the test statistic as Eˆ X2 S(Oij ij)2 Eˆ ij (20 (26 (15 .63)2 .99)2 .51)2 1 4 2 2 2 2 3 6 14. 9 9 22. 1 5 22. 19.18 When you index the chi-square distribution in Table 5 in Appendix I with df (r 1)(c 1) (4 1)(3 1) 6 the observed test statistic is greater than x 2 .005 18.5476, which indicates that the p-value is less than .005. You can reject H0 and declare the results to be highly signiﬁcant (P .005). There is sufficient evidence to indicate that the proportions of defect types vary from shift to shift. The next obvious question you should ask involves the nature of the relationship between the two classiﬁcations. Which shift produces more of which type of defect? As with the factorial experiment in Chapter 11, once a dependence (or interaction) is found, you must look within the table at the relative or conditional proportions for each level of classiﬁcation. For example, consider shift 1, which produced a total of 94 defects. These defects can be divided into types using the conditional proportions for this sample shown in the ﬁrst column of Table 14.5. If you follow the same procedure for the other two shifts, you can then compare the distributions of defect types for the three shifts, as shown in Table 14.5. Now compare the three sets of proportions (each sums to 1). It appears that shifts 1 and 2 produce defects in the same general order—types C, B, A, and D from most to least—though in differing proportions. Shift 3 shows a different pattern—the most 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 605 type C defects again but followed by types A, D, and B, in that order. Depending on which type of defect is the most important to the manufacturer, each shift should be cautioned separately about the reasons for producing too many defects. TABLE 14.5 ● Conditional Probabilities for Types of Defect Within Three Shifts Types of Defects A B C D Shift 3 2 3 3 2 6 .28 .27 14 .32 41 .35 9 1 1 6 9 2 0 5 .17 .05 9 1 1 9 6 1 1 5 .16 4 9 2 1 .22 4 9 4 5 .48 4 9 1 3 .14 4 9 Total 1.00 1.00 1.00 The Chi-Square Test of Independence applet can help you visualize the distribution of the observed frequencies. In Figure 14.2(a), the blue bars (blue in Figure 14.2(a)) represent categories that have an excess of defectives relative to expected and red cells (gray in Figure 14.2(a)) indicate a deﬁcit of defectives relative to expected. The intensity of the color reﬂects the magnitude of the discrepancy. button to view the expected distribution of In Figure 14.2(b), we used the defectives if the null hypothesis is true. The relative heights of the rectangles in each of the three columns correspond to the conditional distribution of defectives per shift given in Table 14.5. We will use this applet for the MyApplet Exercises at the end of the chapter. (a) (b) FI GUR E 1 4. 2 Chi-Square Test of Independence applet ● 606 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA How Do I Determine the Appropriate Number of Degrees of Freedom? Remember the general procedure for determining degrees of freedom: 1. Start with k rc categories or cells in the contingency table. 2. Subtract one degree of freedom because all of the rc cell probabilities must sum to 1. 3. You had to estimate (r 1) row probabilities and (c 1) column probabilities to calculate the estimated expected cell counts. (The last one of the row and column probabilities is determined because the marginal row and column probabilities must also sum to 1.) Subtract (r 1) and (c 1) df. The total degrees of freedom for the r c contingency table are df rc 1 (r 1) (c 1) rc r c 1 (r 1)(c 1) EXAMPLE 14.5 A survey was conducted to evaluate the effectiveness of a new ﬂu vaccine that had been administered in a small community. The vaccine was provided free of charge in a two-shot sequence over a period of 2 weeks. Some people received the two-shot sequence, some appeared for only the ﬁrst shot, and others received neither. A survey of 1000 local residents the following spring provided the information shown in Table 14.6. Do the data present sufficient evidence to indicate that the vaccine was successful in reducing the number of ﬂu cases in the community? TABLE 14.6 ● 2 3 Contingency Table No Vaccine One Shot Flu No Flu Total 24 289 313 9 100 109 Two Shots Total 13 565 578 46 954 1000 Solution The success of the vaccine in reducing the number of ﬂu cases can be assessed in two parts: • If the vaccine is successful, the proportions of people who get the ﬂu should vary, depending on which of the three treatments they received. • Not only must this dependence exist, but the proportion of people who get the ﬂu should decrease as the amount of ﬂu prevention treatment increases— from zero to one to two shots. The ﬁrst part can be tested using the chi-square test with these hypotheses: H0 : No relationship between treatment and incidence of ﬂu Ha : Incidence of ﬂu depends on amount of ﬂu treatment As usual, computer software packages can eliminate all of the tedious calculations and, if the data are entered correctly, provide the correct output containing the observed 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 607 Use the value of X2 and the p-value from the printout to test the hypothesis of independence. value of the test statistic and its p-value. Such a printout, generated by MINITAB, is shown in Figure 14.3. You can ﬁnd instructions for generating this printout in the section “My MINITAB ” at the end of this chapter. The observed value of the test statistic, X2 17.313, has a p-value of .0
00 and the results are declared highly signiﬁcant. That is, the null hypothesis is rejected. There is sufficient evidence to indicate a relationship between treatment and incidence of ﬂu. FI GUR E 1 4. 3 MINITAB output for Example 14.5 ● Chi-Square Test: No Vaccine, One Shot, Two Shots Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts No Vaccine One Shot Two Shots Total 1 24 9 13 46 14.40 5.01 26.59 6.404 3.169 6.944 2 289 100 565 954 298.60 103.99 551.41 0.309 0.153 0.335 Total 313 109 578 1000 Chi-Sq = 17.313, DF = 2, P-Value = 0.000 What is the nature of this relationship? To answer this question, look at Table 14.7 and Figure 14.4, which give the incidence of ﬂu in the sample for each of the three treatment groups. The answer is obvious. The group that received two shots was less susceptible to the ﬂu; only one ﬂu shot does not seem to decrease the susceptibility! TABLE 14.7 ● Incidence of Flu for Three Treatments No Vaccine One Shot Two Shots 2 4 .08 3 1 3 1 3 9 .02 .08 8 7 5 09 1 FI GUR E 1 4. 4 Chi-Square Test of Independence applet ● 608 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA 14.4 EXERCISES BASIC TECHNIQUES 14.16 Calculate the value and give the number of degrees of freedom for X2 for these contingency tables: a. Columns Rows 1 1 2 3 120 79 31 2 70 108 49 3 55 95 81 4 16 43 140 b. Columns Rows 1 1 2 35 120 2 16 92 3 84 206 14.17 Suppose that a consumer survey summarizes the responses of n 307 people in a contingency table that contains three rows and ﬁve columns. How many degrees of freedom are associated with the chi-square test statistic? 14.18 A survey of 400 respondents produced these cell counts in a 2 3 contingency table: Columns Rows 1 1 2 Total 37 66 103 2 34 57 91 3 93 113 206 Total 164 236 400 a. If you wish to test the null hypothesis of “independence”—that the probability that a response falls in any one row is independent of the column it falls in—and you plan to use a chi-square test, how many degrees of freedom will be associated with the x 2 statistic? b. Find the value of the test statistic. c. Find the rejection region for a .01. d. Conduct the test and state your conclusions. e. Find the approximate p-value for the test and inter- pret its value. 14.19 Gender Differences Male and female respondents to a questionnaire on gender differences were categorized into three groups according to their answers on the ﬁrst question: Group 1 Group 2 Group 3 Men Women 37 7 49 50 72 31 Use the MINITAB printout to determine whether there is a difference in the responses according to gender. Explain the nature of the differences, if any exist. MINITAB output for Exercise 14.19 Chi-Square Test: Group 1, Group 2, Group 3 Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Group 1 Group 2 Group 3 Total 1 37 49 72 158 28.26 63.59 66.15 2.703 3.346 0.517 2 7 50 31 88 15.74 35.41 36.85 4.853 6.007 0.927 Total 44 99 103 246 Chi-Sq = 18.352, DF = 2, P-Value = 0.000 APPLICATIONS 14.20 Mandatory Health Care In 2006, a new law passed in Massachusetts would require all residents to have health insurance. Low-income residents would get state subsidies to help pay insurance premiums, but everyone would pay something for health services. The plan would penalize people without any insurance and charge fees to employers who don’t provide coverage. An ABC News/Washington Post poll4 involving n 1027 adults nationwide asked the question, “Would you support or oppose this plan in your state?” The data that follows is based on the results of this study. Affiliation Support Oppose Unsure Democrats Independents Republicans 256 60 235 163 40 222 22 5 24 a. Are there signiﬁcant differences in the proportions of those surveyed who support, oppose, and are unsure about this plan among Democrats, Independents, and Republicans? Use a .05. b. If signiﬁcant differences exist, describe the nature of the differences by ﬁnding the proportions of those who support, oppose, and are unsure for each of the given affiliations. 14.21 Anxious Infants A study was conducted by Joseph Jacobson and Diane Wille to determine the effect of early child care on infant-mother attachment patterns.5 In the study, 93 infants were classiﬁed as either “secure” or “anxious” using the Ainsworth strange situation paradigm. In addition, the infants were classiﬁed according to the average number of hours per week that they spent in child care. The data are presented in the table. Low (0–3 hours) Moderate (4–19 hours) High (20–54 hours) Secure Anxious 24 11 35 10 5 8 a. Do the data provide sufficient evidence to indicate that there is a difference in attachment pattern for the infants depending on the amount of time spent in child care? Test using a .05. b. What is the approximate p-value for the test in part a? 14.22 Spending Patterns Is there a difference in the spending patterns of high school seniors depending on their gender? A study to investigate this question focused on 196 employed high school seniors. Students were asked to classify the amount of their earnings that they spent on their car during a given month: None or Only a Little Some About Half Most Almost All All or Male 73 Female 57 12 15 6 11 4 9 3 6 A portion of the MINITAB printout is given here. Use the printout to analyze the relationship between spending patterns and gender. Write a short paragraph explaining your statistical conclusions and their practical implications. Partial MINITAB output for Exercise 14.22 Chi-Square Test: None, Some, Half, Most, All Chi-Sq = 6.696, DF = 4, P-Value = 0.153 2 cells with expected counts less than 5. 14.23 Waiting for a Prescription How long do you wait to have your prescriptions ﬁlled? According to USA Today, “about 3 in 10 Americans wait more than 20 minutes to have a prescription ﬁlled.”6 Suppose a comparison of waiting times for pharmacies in HMOs and pharmacies in drugstores produced the following results. Waiting Time 15 minutes 16–20 minutes 20 minutes Don’t know HMO Drugstores 75 44 21 10 119 21 37 23 14.4 CONTINGENCY TABLES: A TWO-WAY CLASSIFICATION ❍ 609 a. Is there sufficient evidence to indicate that there is a difference in waiting times for pharmacies in HMOs and pharmacies in drugstores? Use a .01. b. If we consider only if the waiting time is more than 20 minutes, is there a signiﬁcant difference in waiting times between pharmacies in HMOs and pharmacies in drugstores at the 1% level of signiﬁcance? EX1424 14.24 The JFK Assassination More than 40 years after the assassination of John F. Kennedy, a FOX News poll shows most Americans disagree with the government’s conclusions about the killing. The Warren Commission found that Lee Harvey Oswald acted alone when he shot Kennedy, but many Americans are not so sure. Do you think that we know all the facts about the assassination of President John F. Kennedy or do you think there was a cover-up? Here are the results from a poll of 900 registered voters nationwide:7 We Know All the Facts There Was a Cover-Up (Not Sure) Democrats Republicans Independents 42 64 20 309 246 115 31 46 27 a. Do these data provide sufficient evidence to conclude that there is a difference in voters’ opinions about a possible cover-up depending on the political affiliation of the voter? Test using a .05. b. If there is a signiﬁcant difference in part a, describe the nature of these differences. EX1425 14.25 Telecommuting As an alternative to ﬂextime, many companies allow employees to do some of their work at home. Individuals in a random sample of 300 workers were classiﬁed according to salary and number of workdays per week spent at home. Workdays at Home per Week Salary Less Than One At Least One, but Not All All at Home Under $25,000 $25,000 to $49,999 $50,000 to $74,999 Above $75,000 38 54 35 33 16 26 22 29 14 12 9 12 a. Do the data present sufficient evidence to indicate that salary is dependent on the number of workdays spent at home? Test using a .05. b. Use Table 5 in Appendix I to approximate the p-value for this test of hypothesis. Does the p-value conﬁrm your conclusions from part a? 610 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA 14.26 Telecommuting II An article in American Demographics addressed the same telecommuting issue (Exercise 14.25) in a EX1426A EX1426B slightly different way. They concluded that “people who work exclusively at home tend to be older and better educated than those who have to leave home to report to work.”8 Use the data below based on random samples of 300 workers each to either support or refute their conclusions. Use the appropriate test of hypothesis, and explain why you either agree or disagree with the American Demographics conclusions. Note that “Mixed” workers are those who reported working at home at least one full day in a typical week. Workers Age Non-Home Mixed Home 15–34 35–54 55 and over 73 85 22 23 40 12 12 23 10 Workers Education Non-Home Mixed Home Less than H.S. diploma H.S. graduate Some college/Assoc. degree B.A. or more 23 54 53 41 3 12 24 42 5 11 14 18 14.5 EXAMPLE 14.6 COMPARING SEVERAL MULTINOMIAL POPULATIONS: A TWO-WAY CLASSIFICATION WITH FIXED ROW OR COLUMN TOTALS An r c contingency table results when each of n experimental units is counted as falling into one of the rc cells of a multinomial experiment. Each cell represents a pair of category levels—row level i and column level j. Sometimes, however, it is not advisable to use this type of experimental design—that is, to let the n observations fall where they may. For example, suppose you want to study the opinions of American families about their income levels—say, low, medium, and high. If you randomly select n 1200 families for your survey, you may not ﬁnd any who classify themselves as low-income families! It might be better to decide ahead of time to survey 400 families in each income level. The resulting data will still appear as a two-way classiﬁcation, but the column totals are ﬁxed
in advance. In another ﬂu prevention experiment like Example 14.5, the experimenter decides to search the clinic records for 300 patients in each of the three treatment categories: no vaccine, one shot, and two shots. The n 900 patients will then be surveyed regarding their winter ﬂu history. The experiment results in a 2 3 table with the column totals ﬁxed at 300, shown in Table 14.8. By ﬁxing the column totals, the experimenter no longer has a multinomial experiment with 2 3 6 cells. Instead, there are three separate binomial experiments—call them 1, 2, and 3—each with a given probability pj of contracting the ﬂu and qj of not contracting the ﬂu. (Remember that for a binomial population, pj qj 1.) TABLE 14.8 ● Cases of Flu for Three Treatments No Vaccine One Shot Two Shots Total Flu No Flu Total 300 300 300 r1 r2 n 14.5 COMPARING SEVERAL MULTINOMIAL POPULATIONS: A TWO-WAY CLASSIFICATION WITH FIXED ROW OR COLUMN TOTALS ❍ 611 Suppose you used the chi-square test to test for the independence of row and column classiﬁcations. If a particular treatment (column level) does not affect the incidence of ﬂu, then each of the three binomial populations should have the same incidence of ﬂu so that p1 p2 p3 and q1 q2 q3. The 2 3 classiﬁcation in Example 14.6 describes a situation in which the chisquare test of independence is equivalent to a test of the equality of c 3 binomial proportions. Tests of this type are called tests of homogeneity and are used to compare several binomial populations. If there are more than two row categories with ﬁxed column totals, then the test of independence is equivalent to a test of the equality of c sets of multinomial proportions. You do not need to be concerned about the theoretical equivalence of the chi-square tests for these two experimental designs. Whether the columns (or rows) are ﬁxed or not, the test statistic is calculated as Eˆ X2 S(Oij ij)2 where Eˆ Eˆij icj ij r n which has an approximate chi-square distribution in repeated sampling with df (r 1)(c 1). How Do I Determine the Appropriate Number of Degrees of Freedom? Remember the general procedure for determining degrees of freedom: 1. Start with the rc cells in the two-way table. 2. Subtract one degree of freedom for each of the c multinomial populations, whose column probabilities must add to one—a total of c df. 3. You had to estimate (r 1) row probabilities, but the column probabilities are ﬁxed in advance and did not need to be estimated. Subtract (r 1) df. The total degrees of freedom for the r c (ﬁxed-column) table are rc c (r 1) rc c r 1 (r 1)(c 1) EXAMPLE 14.7 A survey of voter sentiment was conducted in four midcity political wards to compare the fractions of voters who favor candidate A. Random samples of 200 voters were polled in each of the four wards with the results shown in Table 14.9. The values in parentheses in the table are the expected cell counts. Do the data present sufﬁcient evidence to indicate that the fractions of voters who favor candidate A differ in the four wards? TABLE 14.9 ● Voter Opinions in Four Wards Ward 1 2 3 4 Favor A 76 (59) Do Not Favor A 124 (141) Total 200 53 (59) 147 (141) 59 (59) 141 (141) 48 (59) 152 (141) 200 200 200 Total 236 564 800 612 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA Solution Since the column totals are ﬁxed at 200, the design involves four binomial experiments, each containing the responses of 200 voters from each of the four wards. To test the equality of the proportions who favor candidate A in all four wards, the null hypothesis H0 : p1 p2 p3 p4 is equivalent to the null hypothesis H0 : Proportion favoring candidate A is independent of ward and will be rejected if the test statistic X2 is too large. The observed value of the test statistic, X2 10.722, and its associated p-value, .013, are shown in Figure 14.5. The results are signiﬁcant (P .025); that is, H0 is rejected and you can conclude that there is a difference in the proportions of voters who favor candidate A among the four wards. F IG URE 14. 5 MINITAB output for Example 14.7 ● Chi-Square Test: Ward 1, Ward 2, Ward 3, Ward 4, Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Ward 1 Ward 2 Ward 3 Ward 4 Total 1 76 53 59 48 236 59.00 59.00 59.00 59.00 4.898 0.610 0.000 2.051 2 124 147 141 152 564 141.00 141.00 141.00 141.00 2.050 0.255 0.000 0.858 Total 200 200 200 200 800 Chi-Sq = 10.722 DF = 3, P-Value = 0.013 What is the nature of the differences discovered by the chi-square test? To answer this question, look at Table 14.10, which shows the sample proportions who favor candidate A in each of the four wards. It appears that candidate A is doing best in the ﬁrst ward and worst in the fourth ward. Is this of any practical signiﬁcance to the candidate? Possibly a more important observation is that the candidate does not have a plurality of voters in any of the four wards. If this is a two-candidate race, candidate A needs to increase his campaigning! TABLE 14.10 ● Proportions in Favor of Candidate A in Four Wards Ward 3 59/200 .30 Ward 4 48/200 .24 Ward 1 76/200 .38 Ward 2 53/200 .27 14.5 COMPARING SEVERAL MULTINOMIAL POPULATIONS: A TWO-WAY CLASSIFICATION WITH FIXED ROW OR COLUMN TOTALS ❍ 613 14.5 EXERCISES BASIC TECHNIQUES 14.27 Random samples of 200 observations were selected from each of three populations, and each observation was classiﬁed according to whether it fell into one of three mutually exclusive categories: Category Population 1 1 2 3 108 87 112 2 52 51 39 3 40 62 49 Total 200 200 200 You want to know whether the data provide sufficient evidence to indicate that the proportions of observations in the three categories depend on the population from which they were drawn. a. Give the value of X2 for the test. b. Give the rejection region for the test for a .01. c. State your conclusions. d. Find the approximate p-value for the test and inter- pret its value. 14.28 Suppose you wish to test the null hypothesis that three binomial parameters pA, pB, and pC are equal versus the alternative hypothesis that at least two of the parameters differ. Independent random samples of 100 observations were selected from each of the populations. The data are shown in the table. Population A 24 76 B 19 81 C 33 67 Successes Failures Total 100 100 100 Total 76 224 300 a. Write the null and alternative hypotheses for testing the equality of the three binomial proportions. b. Calculate the test statistic and ﬁnd the approximate p-value for the test in part a. c. Use the approximate p-value to determine the statistical signiﬁcance of your results. If the results are statistically signiﬁcant, explore the nature of the differences in the three binomial proportions. APPLICATIONS 14.29 The Sandwich Generation How do Americans in the “sandwich generation” balance the demands of caring for older and younger relatives? In a telephone poll of Americans aged 45 to 55 years conducted by the New York Times,9 the number providing ﬁnancial support for their parents is listed in the next display. Provide Financial Support Yes White Americans African Americans Hispanic Americans Asian Americans 40 56 68 84 No 160 144 132 116 Is there a signiﬁcant difference in the proportion of individuals providing ﬁnancial support for their parents for these subpopulations of Americans? Use a .01. 14.30 Diseased Chickens A particular poultry disease is thought to be noncommunicable. To test this theory, 30,000 chickens were randomly partitioned into three groups of 10,000. One group had no contact with diseased chickens, one had moderate contact, and the third had heavy contact. After a 6-month period, data were collected on the number of diseased chickens in each group of 10,000. Do the data provide sufficient evidence to indicate a dependence between the amount of contact between diseased and nondiseased fowl and the incidence of the disease? Use a .05. No Contact Disease No Disease Total 87 9,913 10,000 Moderate Contact 89 9,911 10,000 Heavy Contact 124 9,876 10,000 14.31 Long-Term Care A study conducted in northwest England made an assessment of EX1431 long-term care facilities that have residents with dementia.10 The homes included those that provided specialized services for elderly people with mental illness/health problems, known as “EMI homes,” as well as others classiﬁed as “non-EMI homes.” It was expected that the EMI homes would score higher on several measures of service quality for people with dementia. One measure included the structure of the home and the services provided, as given in the next table. Home Type Care Type EMI Non-EMI Total Nursing care Residential care Dual-registered Total 54 59 49 162 22 77 26 125 76 136 75 287 614 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA a. Describe the binomial experiments whose proportions are being compared in this experiment. recorded the number of family members in each of the households. The data are shown in the table. b. Do these data indicate that the type of care provided varies by the three types of home? Test at the a .01 level. c. Based upon the results of part b, explain the practical nature of the relationship between home type and care type. 14.32 Deep-Sea Research W.W. Menard has conducted research involving manganese nodules, a mineral-rich concoction found abundantly on the deepsea ﬂoor.11 In one portion of his report, Menard provides data relating the magnetic age of the earth’s crust to the “probability of ﬁnding manganese nodules.” The table gives the number of samples of the earth’s core and the percentage of those that contain manganese nodules for each of a set of magnetic-crust ages. Do the data provide sufficient evidence to indicate that the probability of ﬁnding manganese nodules in the deepsea earth’s crust is dependent on the magnetic-age classiﬁcation? Age Miocene—recent Oligocene Eocene Paleocene Late Cretaceous Early and Middle Cretaceous Jurassic Number of Samples Percentage with Nodules 389 140 214 84
247 1120 99 5.9 17.9 16.4 21.4 21.1 14.2 11.0 EX1433 14.33 How Big Is the Household? A local chamber of commerce surveyed 120 households in their city—40 in each of three types of residence (apartment, duplex, or single residence)—and Type of Residence Family Members 1 2 3 4 or more Apartment Duplex Single Residence 8 16 10 6 20 8 10 2 1 9 14 16 Is there a signiﬁcant difference in the family size distributions for the three types of residence? Test using a .01. If there are signiﬁcant differences, describe the nature of these differences. EX1434 14.34 Churchgoing and Age A snapshot in USA Today indicates that there is a gap in church attendence between 20-year-olds and older Americans.12 Suppose that we randomly select 100 Americans in each of ﬁve age groups and record the numbers who say they attend church in a typical week. Attend Church Regularly? 20s 30s 40s 50s Yes No Source: Barna Research Group 31 69 42 58 47 53 48 52 60 53 47 a. Do the data indicate that the proportion of adults who attend church regularly differs depending on age? Test using a .05. b. If there are signﬁcant differences in part a, describe the nature of these differences by calculating the proportion of churchgoers in each age category. Where do the signiﬁcant differences appear to lie? 14.6 THE EQUIVALENCE OF STATISTICAL TESTS Remember that when there are only k 2 categories in a multinomial experiment, the experiment reduces to a binomial experiment where you record the number of successes x (or O1) in n (or O1 O2) trials. Similarly, the data that result from two binomial experiments can be displayed as a two-way classiﬁcation with r 2 and c 2, so that the chi-square test of homogeneity can be used to compare the two binomial proportions, p1 and p2. For these two situations, we have presented statistical tests for the binomial proportions based on the z-statistic of Chapter 9: 14.7 OTHER APPLICATIONS OF THE CHI-SQUARE TEST ❍ 615 • One sample: z pˆ p0 q0 p0 n • Two samples: z pˆ1 pˆ2 pˆqˆ 1 1 n n 2 1 k 2 Successes Failures r c 2 Sample 1 Sample 2 Successes Successes Failures Failures Why are there two different tests for the same statistical hypothesis? Which one should you use? For these two situations, you can use either the z test or the chisquare test, and you will obtain identical results. For either the one- or two-sample test, we can prove algebraically that z 2 2 so that the test statistic z will be the square root (either positive or negative, depending on the data) of the chi-square statistic. Furthermore, we can show theoretically that the same relationship holds for the critical values in the z and x 2 tables in Appendix I, which produces identical p-values for the two equivalent tests. To test a one-tailed alternative hypothesis such as H0: p1 p2, ﬁrst determine whether pˆ1 pˆ2 0, that is, if the difference in sample proportions has the appropriate sign. If so, the appropriate critical value of x 2 from Table 5 will have one degree of freedom a right-tail area of 2a. For example, the critical x 2 value with 1 df and a .05 will be x 2 .10 2.70554 1.6452. In summary, you are free to choose the test (z or X2 ) that is most convenient. Since most computer packages include the chi-square test, and most do not include the largesample z-tests, the chi-square test may be preferable to you! The one- and two-sample binomial tests from Chapter 9 are equivalent to chi-square tests— z 2 x 2. OTHER APPLICATIONS OF THE CHI-SQUARE TEST 14.7 The application of the chi-square test for analyzing count data is only one of many classiﬁcation problems that result in multinomial data. Some of these applications are quite complex, requiring complicated or calculationally difficult procedures for estimating the expected cell counts. However, several applications are used often enough to make them worth mentioning. • Goodness-of-ﬁt tests: You can design a goodness-of-ﬁt test to determine whether data are consistent with data drawn from a particular probability distribution—possibly the normal, binomial, Poisson, or other distributions. The cells of a sample frequency histogram correspond to the k cells of a multinomial experiment. Expected cell counts are calculated using the probabilities associated with the hypothesized probability distribution. • Time-dependent multinomials: You can use the chi-square statistic to investigate the rate of change of multinomial (or binomial) proportions over time. For example, suppose that the proportion of correct answers on a 100-question exam is recorded for a student, who then repeats the exam in 616 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA each of the next 4 weeks. Does the proportion of correct responses increase over time? Is learning taking place? In a process monitored by a quality control plan, is there a positive trend in the proportion of defective items as a function of time? • Multidimensional contingency tables: Instead of only two methods of classiﬁcation, you can investigate a dependence among three or more classiﬁcations. The two-way contingency table is extended to a table in more than two dimensions. The methodology is similar to that used for the r c contingency table, but the analysis is a bit more complex. • Log-linear models: Complex models can be created in which the logarithm of the cell probability (ln pij) is some linear function of the row and column probabilities. Most of these applications are rather complex and might require that you consult a professional statistician for advice before you conduct your experiment. In all statistical applications that use Pearson’s chi-square statistic, assumptions must be satisﬁed in order that the test statistic have an approximate chi-square probability distribution. ASSUMPTIONS • The cell counts O1, O2, . . . , Ok must satisfy the conditions of a multinomial experiment, or a set of multinomial experiments created by ﬁxing either the row or column totals. • The expected cell counts E1, E2, . . . , Ek should equal or exceed 5. You can usually be fairly certain that you have satisﬁed the ﬁrst assumption by carefully preparing and designing your experiment or sample survey. When you calculate the expected cell counts, if you ﬁnd that one or more is less than 5, these options are available to you: • Choose a larger sample size n. The larger the sample size, the closer the chisquare distribution will approximate the distribution of your test statistic X2. It may be possible to combine one or more of the cells with small expected cell counts, thereby satisfying the assumption. • Finally, make sure that you are calculating the degrees of freedom correctly and that you carefully evaluate the statistical and practical conclusions that can be drawn from your test. CHAPTER REVIEW Key Concepts and Formulas I. The Multinomial Experiment 1. There are n identical trials, and each outcome falls into one of k categories. 2. The probability of falling into category i is pi and remains constant from trial to trial. 3. The trials are independent, Spi 1, and we measure Oi, the number of observations that fall into each of the k categories. MY MINITAB ❍ 617 II. Pearson’s Chi-Square Statistic X2 S(Oi Ei)2 where Ei npi Ei which has an approximate chi-square distribution with degrees of freedom determined by the application. III. The Goodness-of-Fit Test 1. This is a one-way classiﬁcation with cell prob- abilities speciﬁed in H0. 2. Use the chi-square statistic with Ei npi calcu- lated with the hypothesized probabilities. 3. df k 1 (Number of parameters esti- mated in order to ﬁnd Ei) 4. If H0 is rejected, investigate the nature of the differences using the sample proportions. 3. If the null hypothesis of independence of classiﬁcations is rejected, investigate the nature of the dependence using conditional proportions within either the rows or columns of the contingency table. V. Fixing Row or Column Totals 1. When either the row or column totals are ﬁxed, the test of independence of classiﬁcations becomes a test of the homogeneity of cell probabilities for several multinomial experiments. 2. Use the same chi-square statistic as for contin- gency tables. 3. The large-sample z-tests for one and two binomial proportions are special cases of the chisquare statistic. IV. Contingency Tables VI. Assumptions 1. The cell counts satisfy the conditions of a multinomial experiment, or a set of multinomial experiments with ﬁxed sample sizes. 2. All expected cell counts must equal or exceed ﬁve in order that the chi-square approximation is valid. 1. A two-way classiﬁcation with n observations categorized into r c cells of a two-way table using two different methods of classiﬁcation is called a contingency table. 2. The test for independence of classiﬁcation methods uses the chi-square statistic X2 S(Oij Êij)2 Êij with Êij ri cj and df (r 1)(c 1) n The Chi-Square Test Several procedures are available in the MINITAB package for analyzing categorical data. The appropriate procedure depends on whether the data represent a one-way classiﬁcation (a single multinomial experiment) or a two-way classiﬁcation or contingency table. If the raw categorical data have been stored in the MINITAB worksheet rather than the observed cell counts, you may need to tally or cross-classify the data to obtain the cell counts before continuing. For example, suppose you have recorded the gender (M or F) and the college status (Fr, So, Jr, Sr, G) for 100 statistics students. The MINITAB worksheet would contain two columns of 100 observations each. Each row would contain an individual’s gender in column 1 and college status in column 2. To obtain the observed cell counts (Oij) for the 2 5 contingency table, use Stat Tables Cross Tabulation and Chi-Square to generate the Dialog box shown in Figure 14.6. Under “Categorical Variables,” select “Gender” for the row variable and “Status” for the column variable. Leave the boxes marked “For Layers” and “Frequencies 618 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA are
in:” blank. Make sure that the square labeled “Display Counts” is checked. Click the Chi-Square . . . button to display the dialog box in Figure 14.6. Check the boxes for “Chi-Square Analysis” and “Expected Cell Counts.” Click OK twice. This sequence of commands not only tabulates the contingency table, but also performs the chi-square test of independence and displays the results in the Session window shown in Figure 14.7. For the gender/college status data, the large p-value (P .153) indicates a nonsigniﬁcant result. There is insufficient evidence to indicate that a student’s gender is dependent on class status. If the observed cell counts in the contingency table have already been tabulated, simply enter the counts into c columns of the MINITAB worksheet, use Stat Tables Chi-Square Test (Two-Way Table in Worksheet), and select the appropriate columns before clicking OK. For the gender/college status data, you can enter the counts into columns C3–C7 as shown in Figure 14.8. The resulting output will be labeled differently but will look exactly like the output in Figure 14.7. A simple test of a single multinomial experiment can be set up by considering whether the proportions of male and female statistics students are the same—that is, p1 .5 and p2 .5. In MINITAB 15, use Stat Tables Chi-Square Goodness-of-Fit Test (One Variable) to display the dialog box in Figure 14.9. If you have raw categorical data in a column, click the “Categorical data:” button and enter the “Gender” column in the cell. If you have summary values of observed counts for each category, choose “Observed counts.” Then enter the column containing the observed counts or type the observed counts for each category. For this test, we can choose “Equal proportions” to test H0: p1 p2 .5. When you have different proportions for each category, use “Speciﬁc proportions.” You can F IG URE 14. 6 ● FI GUR E 1 4. 7 ● MY MINITAB ❍ 619 FI GUR E 1 4. 8 ● store the proportions for each category in a column, choose “Input column” and enter the column. If you want to type the proportion for each category, choose “Input constants” and type the proportions for the corresponding categories. Click OK. The resulting output will include several graphs along with the values for Oi and Ei for each category, the observed value of the test statistic, X2 1.44, and its p-value 0.230, which is not signiﬁcant. There is insufficient evidence to indicate a difference in the proportion of male and female statistics students. 620 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA If you are using a previous version of MINITAB, you will have to determine the observed and expected cell counts, and enter them into separate columns in the worksheet. Then use Calc Calculator and the expression SUM((‘O’- ‘E’)*2/‘E’) to calculate the observed value of the test statistic. F IG URE 14. 9 ● Supplementary Exercises Starred () exercises are optional. 14.35 Floor Polish A manufacturer of ﬂoor polish conducted a consumer preference experiment to see whether a new ﬂoor polish A was superior to those produced by four competitors, B, C, D, and E. A sample of 100 housekeepers viewed ﬁve patches of ﬂooring that had received the ﬁve polishes, and each indicated the patch that he or she considered superior in appearance. The lighting, background, and so on were approximately the same for all ﬁve patches. The results of the survey are listed here: Polish Frequency A 27 B 17 C 15 D 22 E 19 Do these data present sufficient evidence to indicate a preference for one or more of the polished patches of ﬂoor over the others? If one were to reject the hypothesis of no preference for this experiment, would this imply that polish A is superior to the others? Can you suggest a better way of conducting the experiment? 14.36 Physical Fitness in the U.S. A survey was conducted to investigate the interest of middle-aged adults in physical ﬁtness programs in Rhode Island, Colorado, California, and Florida. The objective of the investigation was to determine whether adult participation in physical ﬁtness programs varies from one region of the United States to another. A random sample of people were interviewed in each state and these data were recorded: Rhode Island Colorado California Florida Participate 46 Do Not Participate 149 63 178 108 192 121 179 Do the data indicate a difference in adult participation in physical ﬁtness programs from one state to another? If so, describe the nature of the differences. 14.37 Fatal Accidents Accident data were analyzed to determine the numbers of fatal accidents for automobiles of three sizes. The data for 346 accidents are as follows: Small Medium Large Fatal Not Fatal 67 128 26 63 16 46 Do the data indicate that the frequency of fatal accidents is dependent on the size of automobiles? Write a short paragraph describing your statistical results and their practical implications. 14.38 Physicians and Medicare Patients An experiment was conducted to investigate the effect of general hospital experience on the attitudes of physicians toward Medicare patients. A random sample of 50 physicians who had just completed 4 weeks of service in a general hospital and 50 physicians who had not were categorized according to their concern for Medicare patients. The data are shown in the table. Do the data provide sufficient evidence to indicate a change in “concern” after the general hospital experience? If so, describe the nature of the change. Hospital Service No Hospital Service Low High High Low Total 27 9 5 9 32 18 Partial MINITAB output for Exercise 14.38 Chi-Square Test: High, Low Chi-Sq = 6.752, DF = 1, P-Value = 0.009 14.39 Discovery-Based Teaching Two biology instructors set out to evaluate the EX1439 effects of discovery-based teaching compared to the standard lecture-based teaching approach in the laboratory.13 The standard lecture-based approach provided a list of instructions to follow at each step of the laboratory exercise, whereas the discovery-based approach SUPPLEMENTARY EXERCISES ❍ 621 asked questions rather than providing directions, and used small group reports to decide the best way to proceed in reaching the laboratory objective. One evaluation of the techniques involved written appraisals of both techniques by students at the end of the course. The comparison of the number of positive and negative responses for both techniques is given in the following table. Group Discovery Control Positive Evaluations Negative Evaluations 37 31 11 17 Total 48 48 a. Is there a signiﬁcant difference in the proportion of positive responses for each of the teaching methods? Use a .05. If so, how would you describe this difference? b. What is the approximate p-value for the test in part a? 14.40 Baby’s Sleeping Position Does a baby’s sleeping position affect the development of motor skills? In one study, 343 full-term infants were examined at their 4-month checkup for various developmental milestones, such as rolling over, grasping a rattle, and reaching for an object.14 The baby’s predominant sleep position—either prone (on the stomach) or supine (on the back) or side—was determined by a telephone interview with the parent. The sample results for 320 of the 343 infants for whom information was received are shown in the table. The researcher reported that infants who slept in the side or supine position were less likely to roll over at the 4-month checkup than infants who slept primarily in the prone position (P .001). Prone Supine or Side Number of Infants Number Who Roll Over 121 93 199 119 a. Use a large-sample z-test to conﬁrm or refute the researcher’s conclusion. b. Rewrite the sample data as a 2 2 contingency table. Use the chi-square test for homogeneity to conﬁrm or refute the researcher’s conclusion. c. Compare the results of parts a and b. Conﬁrm that the two test statistics are related as z2 X2 and that the critical values for rejecting H0 have the same relationship. 14.41 Refer to Exercise 14.40. Find the p-value for the large-sample z test in part a. Compare this p-value with the p-value for the chi-square test, shown in the partial MINITAB printout. 622 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA Partial MINITAB output for Exercise 14.41 Chi-Square Test: Prone, Side Chi-Sq = 9.795, DF = 1, P-Value = 0.002 14.42 Baby’s Sleeping Position II The researchers in Exercise 14.40 also measured several other developmental milestones and their relationship to the infant’s predominant sleep position.14 The results of their research are presented in the table for the 320 infants at their 4-month checkup. Milestone Score Prone Supine or Side P Pulls to sit with no head lag Grasps rattle Reaches for object Pass Fail Pass Fail Pass Fail 79 6 102 3 107 3 144 20 167 1 183 5 .21 .13 .97 Use your knowledge of the analysis of categorical data to explain the experimental design(s) used by the researchers. What hypotheses were of interest to the researchers, and what statistical test would the researchers have used? Explain the conclusions that can be drawn from the three p-values in the last column of the table and the practical implications that can be drawn from the statistical results. Have any statistical assumptions been violated? 14.43 Flower Color and Shape A botanist performs a secondary cross of petunias involving independent factors that control leaf shape and ﬂower color, where the factor A represents red color, a represents white color, B represents round leaves, and b represents long leaves. According to the Mendelian model, the plants should exhibit the characteristics AB, Ab, aB, and ab in the ratio 9:3:3:1. Of 160 experimental plants, the following numbers were observed: AB 95 Ab 30 aB 28 ab 7 Is there sufficient evidence to refute the Mendelian model at the a .01 level? 14.44 Salmonella Is your holiday turkey safe? A “new federal survey found that 13% of turkeys are contaminated with the salmonella bacteria responsible for 1.3 million illnesses and about 500 deaths in a year in the
US.”15 Use the table that follows to determine if there is a signiﬁcant difference in the contamination rate at three processing plants. One hundred turkeys were randomly selected from each of the processing lines at these three plants. Salmonella Present Plant Sample Size 1 2 3 42 23 22 100 100 100 Is there a signiﬁcant difference in the rate of salmonella contamination among these three processing plants? If there is a signiﬁcant difference, describe the nature of these differences. Use a .01. 14.45 An Arthritis Drug A study to determine the effectiveness of a drug (serum) for arthritis resulted in the comparison of two groups, each consisting of 200 arthritic patients. One group was inoculated with the serum; the other received a placebo (an inoculation that appears to contain serum but actually is nonactive). After a period of time, each person in the study was asked to state whether his or her arthritic condition had improved. These are the results: Treated Untreated Improved Not Improved 117 83 74 126 You want to know whether these data present sufficient evidence to indicate that the serum was effective in improving the condition of arthritic patients. a. Use the chi-square test of homogeneity to compare the proportions improved in the populations of treated and untreated subjects. Test at the 5% level of signiﬁcance. b. Test the equality of the two binomial proportions using the two-sample z-test of Section 9.6. Verify that the squared value of the test statistic z2 X2 from part a. Are your conclusions the same as in part a? 14.46 Parking at the University A survey was conducted to determine student, faculty, and administration attitudes about a new university parking policy. The distribution of those favoring or opposing the policy is shown in the table. Do the data provide sufficient evidence to indicate that attitudes about the parking policy are independent of student, faculty, or administration status? Student Faculty Administration Favor Oppose 252 139 107 81 43 40 14.47* The chi-square test used in Exercise 14.45 is equivalent to the two-tailed z-test of Section 9.6 provided a is the same for the two tests. Show algebraically that the chi-square test statistic X2 is the square of the test statistic z for the equivalent test. 14.48 Fitting a Binomial Distribution You can use a goodness-of-ﬁt test to determine whether all of the criteria for a binomial experiment have actually been met in a given application. Suppose that an experiment consisting of four trials was repeated 100 times. The number of repetitions on which a given number of successes was obtained is recorded in the table: Possible Results (number of successes) Number of Times Obtained 0 1 2 3 4 11 17 42 21 9 Estimate p (assuming that the experiment was binomial), obtain estimates of the expected cell frequencies, and test for goodness of ﬁt. To determine the appropriate number of degrees of freedom for X2, note that p was estimated by a linear combination of the observed frequencies. 14.49 Antibiotics and Infection Infections sometimes occur when blood transfusions are given during surgical operations. An experiment was conducted to determine whether the injection of antibodies reduced the probability of infection. An examination of the records of 138 patients produced the data shown in the table. Do the data provide sufficient evidence to indicate that injections of antibodies affect the likelihood of transfusional infections? Test by using a .05. Infection No Infection Antibody No Antibody 4 11 78 45 14.50 German Manufacturing U.S. labor unions have traditionally been content to leave the management of the company to managers and corporate executives. But in Europe, worker participation in management decision making is an accepted idea that is continually spreading. To study the effect of worker participation in managerial decision making, 100 workers were interviewed in each of two separate German manufacturing plants. One plant had active worker participation in managerial decision making; the other did not. Each selected worker was asked whether he or she generally approved of the managerial decisions made within the ﬁrm. The results of the interviews are shown in the table: Participation No Participation Generally Approve Do Not Approve 73 27 51 49 SUPPLEMENTARY EXERCISES ❍ 623 a. Do the data provide sufficient evidence to indicate that approval or disapproval of management’s decisions depends on whether workers participate in decision making? Test by using the X2 test statistic. Use a .05. b. Do these data support the hypothesis that workers in a ﬁrm with participative decision making more generally approve of the ﬁrm’s managerial decisions than those employed by ﬁrms without participative decision making? Test by using the z-test presented in Section 9.6. This problem requires a one-tailed test. Why? 14.51 Three Entrances An occupant-traffic study was conducted to aid in the remodeling of an office building that contains three entrances. The choice of entrance was recorded for a sample of 200 persons who entered the building. Do the data in the table indicate that there is a difference in preference for the three entrances? Find a 95% conﬁdence interval for the proportion of persons favoring entrance 1. Entrance Number Entering 1 83 2 61 3 56 14.52 Homeschool Teachers Parents who are concerned about public school environments and curricula are turning to homeschooling in order to control the content and atmosphere of the learning environments of their children. Although employment as a public school teacher requires a bachelor’s degree in education or a subject area, the educational background of homeschool teachers is quite varied. The educational background of a sample of n 500 parents involved in homeschooling their children in 2003 are provided in the ﬁrst table that follows, along with the corresponding percentages for parents who homeschooled in 1999. The education levels for U.S. citizens in general are given in the second table.16 Parent’s Education 2003 1999 Percentages High school or less Some college/technical Bachelor’s degree Graduate/professional degree 121 153 127 99 18.9 33.7 25.1 22.3 Education Level % U.S. Population, 2003 High school or less Some college Bachelor’s degree or higher 47.5 25.3 27.2 624 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA a. Is there a signiﬁcant change in the educational backgrounds of parents who homeschooled their children in 2003 compared with 1999? Use a .01. b. If there is a signiﬁcant change in the educational backgrounds of these parents, how would you describe that change? c. Using the second table, can we determine if homeschool teachers have the same educational backgrounds as the U.S. population in general? If not, which groups are underrepresented and which are overrepresented? 14.53 Are You a Good Driver? How would you rate yourself as a driver? According to a survey conducted by the Field Institute, most Californians think they are good drivers but have little respect for others’ driving ability. The data show the distribution of opinions according to gender for two different questions, the ﬁrst rating themselves as drivers and the second rating others as drivers.17 Although not stated in the source, we assume that there were 100 men and 100 women in the surveyed group. Rating Self as a Driver Gender Excellent Good Fair Male Female 43 44 48 53 9 3 Rating Others As Drivers Gender Excellent Good Fair Poor/Very Poor Male Female 4 3 42 48 41 35 13 14 a. Is there sufficient evidence to indicate that there is a difference in the self-ratings between male and female drivers? Find the approximate p-value for the test. b. Is there sufficient evidence to indicate that there is a difference in the ratings of other drivers between male and female drivers? Find the approximate p-value for the test. c. Have any of the assumptions necessary for the analysis used in parts a and b been violated? What affect might this have on the validity of your conclusions? 14.54 Vehicle Colors Each model year seems to introduce new colors and different hues for a wide array of vehicles, from luxury cars, to full-size or intermediate models, to compacts and sports cars, to light trucks. However, white and silver/gray continue to make the top ﬁve or six colors across all of these categories of vehicles. The top ﬁve colors and their percentage of the market share for compact/sports cars are shown in the following table.18 Color Percent Silver 20 Gray 17 Blue 16 Black 14 White/ Pearl 10 To verify the ﬁgures, a random sample consisting of 250 compact/sports cars was taken and the color of the vehicles recorded. The sample provided the following counts for the categories given above: 60, 51, 43, 35, and 30, respectively. a. Is any category missing in the classiﬁcation? How many vehicles belong to that category? b. Is there sufficient evidence to indicate that our percentages of compact/sports cars differ from those given? Find the approximate p-value for the test. 14.55 Funny Cards When you choose a greeting card, do you always look for a humorous card, or does it depend on the occasion? A comparison sponsored by two of the nation’s leading manufacturers of greeting cards indicated a slight difference in the proportions of humorous designs made for three different occasions: Father’s Day, Mother’s Day, and Valentine’s Day.19 To test the accuracy of their comparison, random samples of 500 greeting cards purchased at a local card store in the week prior to each holiday were entered into a computer database, and the results in the table were obtained. Do the data indicate that the proportions of humorous greeting cards vary for these three holidays? (HINT: Remember to include a tabulation for all 1500 greeting cards.) Holiday Father’s Day Mother’s Day Valentine’s Day Percent Humorous 20 25 24 EX1456 14.56 Good Tasting Medicine Pﬁzer Canada Inc. is a pharmaceutical company that makes azithromycin, an antibiotic i
n a cherry-ﬂavored suspension used to treat bacterial infections in children. To compare the taste of their product with three competing medications, Pﬁzer tested 50 healthy children and 20 healthy adults. Among other taste-testing measures, they recorded the number of tasters who rated each of the four antibiotic suspensions as the best tasting.20 The results are shown in the table. Is there a difference in the perception of the best taste between adults and children? If so, what is the nature of the difference, and why is it of practical importance to the pharmaceutical company? Flavor of Antibiotic Banana Cherry* Wild Fruit Strawberry-Banana Children Adults 14 4 20 14 7 0 *Azithromycin produced by Pﬁzer Canada Inc. 9 2 EX1457 14.57 Rugby Injuries Knee injuries are a major problem for athletes in many contact sports. However, athletes who play certain positions are more prone to get knee injuries than other players, and their injuries tend to be more severe. The prevalence and patterns of knee injuries among women collegiate rugby players were investigated using a sample questionnaire, to which 42 rugby clubs responded.21 A total of 76 knee injuries were classiﬁed by type as well as the position (forward or back) of the player. Type of Knee Injury Position Forward Back Meniscal MCL Tear Tear 13 12 14 9 ACL Tear 7 14 Patella Dislocation PCL Tear 3 2 1 1 MINITAB output for Exercise 14.57 Chi-Square Test: Men Tear, MCL Tear, ACL Tear, Patella, PCL Tear Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Men Tear MCL Tear ACL Tear Patella PCL Tear Total 1 13 14 7 3 1 38 12.50 11.50 10.50 2.50 1.00 0.020 0.543 1.167 0.100 0.000 2 12 9 14 2 1 38 12.50 11.50 10.50 2.50 1.00 0.020 0.543 1.167 0.100 0.000 Total 25 23 21 5 2 76 Chi-Sq = 3.660, DF = 4, P-Value = 0.454 4 cells with expected counts less than 5.0 a. Use the MINITAB printout to determine whether there is a difference in the distribution of injury types for rugby backs and forwards. Have any of the assumptions necessary for the chi-square test been violated? What effect will this have on the magnitude of the test statistic? b. The investigators report a signiﬁcant difference in the proportion of MCL tears for the two positions (P .05) and a signiﬁcant difference in the SUPPLEMENTARY EXERCISES ❍ 625 proportion of ACL tears (P .05), but indicate that all other injuries occur with equal frequency for the two positions. Do you agree with those conclusions? Explain. EX1458 14.58 Favorite Fast Foods The number of Americans who visit fast-food restaurants regularly has grown steadily over the past decade. For this reason, marketing experts are interested in the demographics of fast-food customers. Is a customer’s preference for a fast-food chain affected by the age of the customer? If so, advertising might need to target a particular age group. Suppose a random sample of 500 fast-food customers aged 16 and older was selected, and their favorite fast-food restaurants along with their age groups were recorded, as shown in the table: Age Group McDonald’s Burger King Wendy’s Other 16–21 21–30 30–49 50 75 89 54 21 34 42 52 25 10 19 28 7 6 10 18 10 Use an appropriate method to determine whether or not a customer’s fast-food preference is dependent on age. Write a short paragraph presenting your statistical conclusions and their practical implications for marketing experts. 14.59 Catching a Cold Is your chance of getting a cold inﬂuenced by the number of social contacts you have? A recent study by Sheldon Cohen, a psychology professor at Carnegie Mellon University, seems to show that the more social relationships you have, the less susceptible you are to colds.22 A group of 276 healthy men and women were grouped according to their number of relationships (such as parent, friend, church member, neighbor). They were then exposed to a virus that causes colds. An adaptation of the results is shown in the table. Number of Relationships Three or Fewer Four or Five Six or More Cold No Cold Total 49 31 80 43 57 100 34 62 96 a. Do the data provide sufficient evidence to indicate that susceptibility to colds is affected by the number of relationships you have? Test at the 5% signiﬁcance level. 626 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA b. Based on the results of part a, describe the nature of the relationship between the two categorical variables: cold incidence and number of social relationships. Do your observations agree with the author’s conclusions? EX1460 14.60 Crime and Educational Achievement A criminologist studying criminal offenders who have a record of one or more arrests is interested in knowing whether the educational achievement level of the offender inﬂuences the frequency of arrests. He has classiﬁed his data using four educational level classiﬁcations: A: completed 6th grade or less B: completed 7th, 8th, or 9th grade C: completed 10th, 11th, or 12th grade D: education beyond 12th grade The contingency table shows the number of offenders in each educational category, along with the number of times they have been arrested. Educational Achievement Number of Arrests A 1 2 3 or more 55 15 7 B 40 25 8 C 43 18 12 D 30 22 10 Do the data present sufficient evidence to indicate that the number of arrests is dependent on the educational achievement of a criminal offender? Test using a .05. 14.61 More Business on the Weekends A department store manager claims that her store has twice as many customers on Fridays and Saturdays than on any other day of the week (the store is closed on Sundays). That is, the probability that a customer visits the store Friday is 2/8, the probability that a customer visits the store Saturday is 2/8, while the probability that a customer visits the store on each of the remaining weekdays is 1/8. During an average week, the following numbers of customers visited the store: Day Number of Customers Monday Tuesday Wednesday Thursday Friday Saturday 95 110 125 75 181 214 Can the manager’s claim be refuted at the a .05 level of signiﬁcance? Exercises 14.62 Use the Chi-Square Probabilities applet to ﬁnd the value of x 2 with the following area a to its right: a. a .05, df 15 b. a .01, df 11 14.63 Use the Chi-Square Probabilities applet to ﬁnd the rejection region for a chi-square test of speciﬁed probabilities for a goodness-of-ﬁt test involving k categories for the following cases: a. k 14, a .005 b. k 3, a .05 14.64 Use the Chi-Square Probabilities applet to calculate the p-value for the following chi-square tests: a. X2 .81, df 3 b. X2 25.40, df 13 14.65 Three hundred people were surveyed, and were asked to select their preferred brand of laptop computer, given that the prices were equivalent. The results are shown in the table. Brand I Brand II Brand III 115 120 65 Use the ﬁrst Goodness-of-Fit applet to determine if consumers have a preference for one of the three brands. If a signiﬁcant difference exists, describe the difference in practical terms. Use a .01. 14.66 In Exercise 14.13, the color distribution of M&M’S milk chocolate candies was given. Use the third Goodness-of-Fit applet to verify the results of Exercise 14.13. Do the data substantiate the percentages reported by Mars, Incorporated? Describe the nature of the differences, if there are any. 14.67 Refer to the color distribution given in Exercise 14.13. Using an individual-sized bag of milk chocolate M&M’S, count the number of M&M’S in MYAPPLET EXERCISES ❍ 627 each of the six colors. Use the third Goodness-of-Fit applet to determine if the percentages reported by Mars, Incorporated can be substantiated. Describe the nature of the differences, if there are any. 14.68 Repeat the instructions in Exercise 14.67 with another individual bag of M&M’S. Are your conclusions the same? 14.69 Opinion and Political Affiliation A group of 306 people were interviewed to determine their opinion concerning a particular current U.S. foreign policy issue. At the same time, their political affiliation was recorded. Do the data in the table present sufficient evidence to indicate a dependence between party affiliation and the opinion expressed for the sampled population? Use the third Chi-Square Test of Independence applet. Approve Do Not Approve No Opinion Republicans Democrats 114 87 53 27 17 8 14.70 A study of the purchase decisions of three stock portfolio managers, A, B, and C, was conducted to compare the numbers of stock purchases that resulted in proﬁts over a time period less than or equal to 1 year. One hundred randomly selected purchases were examined for each of the managers. Do the data provide evidence of differences among the rates of successful purchases for the three managers? Use the third Chi-Square Test of Independence applet. A 63 37 B 71 29 C 55 45 Proﬁt No proﬁt 628 ❍ CHAPTER 14 ANALYSIS OF CATEGORICAL DATA CASE STUDY Libraries Can a Marketing Approach Improve Library Services? Carole Day and Del Lowenthal studied the responses of young adults in their evaluation of library services.23 Of the n 200 young adults involved in the study, n1 152 were students and n2 48 were nonstudents. The table presents the percents and numbers of favorable responses for each group to seven questions in which the atmosphere, staff, and design of the library were examined. Question Student Favorable n1 152 Nonstudent Favorable n2 48 3 4 5 6 7 11 13 77 91.4 Libraries are friendly 79.6% Libraries are dull Library staff are helpful Library staff are less 60.5 helpful to teenagers Libraries are so quiet they feel uncomfortable Libraries should be more brightly decorated Libraries are badly signposted 45.4 75.6 29 121 117 139 92 115 44 69 56.2% 58.3 87.5 45.8 52.05 18.8 43.8 27 28 42 22 25 9 21 P (x 2) .01 .05 NS .01 .01 NS NS Source: Data from C. Day and D. Lowenthal, “The Use of Open Group Discussions in Marketing Library Services to Young Adults,” by C. Day and D. Lowenthal, British Journal of Educational Psychology, 62(1992): 324–340. The entry
in the last column labeled P(x 2) is the p-value for testing the hypothesis of no difference in the proportion of students and nonstudents who answer each question favorably. Hence, each question gives rise to a 2 2 contingency table. 1. Perform a test of homogeneity for each question and verify the reported p-value of the test. 2. Questions 3, 4, and 7 are concerned with the atmosphere of the library; questions 5 and 6 are concerned with the library staff; and questions 11 and 13 are concerned with the library design. How would you summarize the results of your analyses regarding these seven questions concerning the image of the library? 3. With the information given, is it possible to do any further testing concerning the proportion of favorable versus unfavorable responses for two or more questions simultaneously? 15 Nonparametric Statistics GENERAL OBJECTIVE In Chapters 8–10, we presented statistical techniques for comparing two populations by comparing their respective population parameters (usually their population means). The techniques in Chapters 8 and 9 are applicable to data that are at least quantitative, and the techniques in Chapter 10 are applicable to data that have normal distributions. The purpose of this chapter is to present several statistical tests for comparing populations for the many types of data that do not satisfy the assumptions speciﬁed in Chapters 8–10. CHAPTER INDEX ● The Friedman Fr -test (15.7) ● The Kruskal–Wallis H-test (15.6) ● Parametric versus nonparametric tests (15.1) ● The rank correlation coefficient (15.8) ● The sign test for a paired experiment (15.3) ● The Wilcoxon rank sum test: Independent random sam- ples (15.2) ● The Wilcoxon signed-rank test for a paired experiment (15.5) © Don Carstens/Brand X/CORBIS How’s Your Cholesterol Level? What is your cholesterol level? Many of us have become more health conscious in the last few years as we read the nutritional labels on the food products we buy and choose foods that are low in fat and cholesterol and high in ﬁber. The case study at the end of this chapter involves a taste-testing experiment to compare three types of egg substitutes, using nonparametric techniques. 629 630 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS INTRODUCTION 15.1 When sample sizes are small and the original populations are not normal, use nonparametric techniques. Some experiments generate responses that can be ordered or ranked, but the actual value of the response cannot be measured numerically except with an arbitrary scale that you might create. It may be that you are able to tell only whether one observation is larger than another. Perhaps you can rank a whole set of observations without actually knowing the exact numerical values of the measurements. Here are a few examples: • The sales abilities of four sales representatives are ranked from best to worst. • The edibility and taste characteristics of ﬁve brands of raisin bran are rated on an arbitrary scale of 1 to 5. • Five automobile designs are ranked from most appealing to least appealing. How can you analyze these types of data? The small-sample statistical methods presented in Chapters 10–13 are valid only when the sampled population(s) are normal or approximately so. Data that consist of ranks or arbitrary scales from 1 to 5 do not satisfy the normality assumption, even to a reasonable degree. In some applications, the techniques are valid if the samples are randomly drawn from populations whose variances are equal. When data do not appear to satisfy these and similar assumptions, an alternative method of analysis can be used—nonparametric statistical methods. Nonparametric methods generally specify hypotheses in terms of population distributions rather than parameters such as means and standard deviations. Parametric assumptions are often replaced by more general assumptions about the population distributions, and the ranks of the observations are often used in place of the actual measurements. Research has shown that nonparametric statistical tests are almost as capable of detecting differences among populations as the parametric methods of preceding chapters when normality and other assumptions are satisﬁed. They may be, and often are, more powerful in detecting population differences when these assumptions are not satisﬁed. For this reason, some statisticians advocate the use of nonparametric procedures in preference to their parametric counterparts. We will present nonparametric methods appropriate for comparing two or more populations using either independent or paired samples. We will also present a measure of association that is useful in determining whether one variable increases as the other increases or whether one variable decreases as the other increases. THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES 15.2 In comparing the means of two populations based on independent samples, the pivotal statistic was the difference in the sample means. If you are not certain that the assumptions required for a two-sample t-test are satisﬁed, one alternative is to replace the values of the observations by their ranks and proceed as though the ranks were the actual observations. Two different nonparametric tests use a test statistic based on these sample ranks: • Wilcoxon rank sum test • Mann-Whitney U-test 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 631 They are equivalent in that they use the same sample information. The procedure that we will present is the Wilcoxon rank sum test, proposed by Frank Wilcoxon, which is based on the sum of the ranks of the sample that has the smaller sample size. Assume that you have n1 observations from population 1 and n2 observations from population 2. The null hypothesis to be tested is that the two population distributions are identical versus the alternative hypothesis that the population distributions are different. These are the possibilities for the two populations: • • • If H0 is true and the observations have come from the same or identical populations, then the observations from both samples should be randomly mixed when jointly ranked from small to large. The sum of the ranks of the observations from sample 1 should be similar to the sum of the ranks from sample 2. If, on the other hand, the observations from population 1 tend to be smaller than those from population 2, then these observations would have the smaller ranks because most of these observations would be smaller than those from population 2. The sum of the ranks of these observations would be “small.” If the observations from population 1 tend to be larger than those in population 2, these observations would be assigned larger ranks. The sum of the ranks of these observations would tend to be “large.” For example, suppose you have n1 3 observations from population 1—2, 4, and 6—and n2 4 observations from population 2—3, 5, 8, and 9. Table 15.1 shows seven observations ordered from small to large. TABLE 15.1 ● Seven Observations in Order y3 Observation x1 y1 y2 x2 x3 Data Rank y4 9 7 The smallest observation, x1 2, is assigned rank 1; the next smallest observation, y1 3, is assigned rank 2; and so on. The sum of the ranks of the observations from sample 1 is 1 3 5 9, and the rank sum from sample 2 is 2 4 6 7 19. How do you determine whether the rank sum of the observations from sample 1 is signiﬁcantly small or signiﬁcantly large? This depends on the probability distribution of the sum of the ranks of one of the samples. Since the ranks for n1 n2 N observations are the ﬁrst N integers, the sum of these ranks can be shown to be N(N 1)/2. In this simple example, the sum of the N 7 ranks is 1 2 3 4 5 6 7 7(8)/2 or 28. Hence, if you know the rank sum for one of the samples, you can ﬁnd the other by subtraction. In our example, notice that the rank sum for sample 1 is 9, whereas the second rank sum is (28 9) 19. This means that only one of the two rank sums is needed for the test. To simplify the tabulation of critical values for this test, you should use the rank sum from the smaller sample as the test statistic. What happens if two or more observations are equal? Tied observations are assigned the average of the ranks that the observations would have had if they had been slightly different in value. To implement the Wilcoxon rank sum test, suppose that independent random samples of size n1 and n2 are selected from populations 1 and 2, respectively. Let n1 represent the smaller of the two sample sizes, and let T1 represent the sum of the ranks 632 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS of the observations in sample 1. If population 1 lies to the left of population 2, T1 will be “small.” T1 will be “large” if population 1 lies to the right of population 2. FORMULAS FOR THE WILCOXON RANK SUM STATISTIC (FOR INDEPENDENT SAMPLES) Let T1 Sum of the ranks for the ﬁrst sample 1 n1(n1 n2 1) T1 T * T * 1 is the value of the rank sum for n1 if the observations had been ranked from large to small. (It is not the rank sum for the second sample.) Depending on the nature of the alternative hypothesis, one of these two values will be chosen as the test statistic, T. Table 7 in Appendix I can be used to locate critical values for the test statistic for four different values of one-tailed tests with a .05, .025, .01, and .005. To use Table 7 for a two-tailed test, the values of a are doubled—that is, a .10, .05, .02, and .01. The tabled entry gives the value of a such that P(T a) a. To see how to locate a critical value for the Wilcoxon rank sum test, suppose that n1 8 and n2 10 for a one-tailed test with a .05. You can use Table 7(a), a portion of which is reproduced in Table 15.2. Notice that the table is constructed assuming that n1 n2. It is for this reason that we designate the population with the smaller sample size as population 1. Values of n1 are shown across the top of the table, and values of n2 are shown down the left side. The entry—a 56, shaded—is the critical value for
rejection of H0. The null hypothesis of equality of the two distributions should be rejected if the observed value of the test statistic T is less than or equal to 56. TABLE 15.2 ● Table 7 in Appendix I A Portion of the 5% Left-Tailed Critical Values, n2 10 4 9 10 4 10 11 12 13 14 15 16 17 n1 5 19 20 21 23 24 26 6 7 8 28 29 31 33 35 39 41 43 45 51 54 56 THE WILCOXON RANK SUM TEST Let n1 denote the smaller of the two sample sizes. This sample comes from population 1. The hypotheses to be tested are H0 : The distributions for populations 1 and 2 are identical versus one of three alternative hypotheses: 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 633 Ha : The distributions for populations 1 and 2 are different (a two-tailed test) Ha : The distribution for population 1 lies to the left of that for population 2 (a left-tailed test) Ha : The distribution for population 1 lies to the right of that for population 2 (a right-tailed test) 1. Rank all n1 n2 observations from small to large. 2. Find T1, the rank sum for the observations in sample 1. This is the test statistic for a left-tailed test. 3. Find T * 1 n1(n1 n2 1) T1, the sum of the ranks of the observations from population 1 if the assigned ranks had been reversed from large to small. (The value of T * 1 is not the sum of the ranks of the observations in sample 2.) This is the test statistic for a right-tailed test. 4. The test statistic for a two-tailed test is T, the minimum of T1 and T * 1. 5. H0 is rejected if the observed test statistic is less than or equal to the critical value found using Table 7 in Appendix I. We illustrate the use of Table 7 with the next example. EXAMPLE 15.1 The wing stroke frequencies of two species of Euglossine bees were recorded for a sample of n1 4 Euglossa mandibularis Friese (species 1) and n2 6 Euglossa imperialis Cockerell (species 2).1 The frequencies are listed in Table 15.3. Can you conclude that the distributions of wing strokes differ for these two species? Test using a .05. TABLE 15.3 ● Wing Stroke Frequencies for Two Species of Bees Species 1 Species 2 235 225 190 188 180 169 180 185 178 182 Solution You ﬁrst need to rank the observations from small to large, as shown in Table 15.4. TABLE 15.4 ● Wing Stroke Frequencies Ranked from Small to Large Data Species Rank 169 178 180 180 182 185 188 190 225 235 10 634 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS The hypotheses to be tested are H0 : The distributions of the wing stroke frequencies are the same for the two species versus Ha : The distributions of the wing stroke frequencies differ for the two species Since the sample size for individuals from species 1, n1 4, is the smaller of the two sample sizes, you have T1 7 8 9 10 34 and 1 n1(n1 n2 1) T1 4(4 6 1) 34 10 T * For a two-tailed test, the test statistic is T 10, the smaller of T1 34 and T * 1 10. For this two-tailed test with a .05, you can use Table 7(b) in Appendix I with n1 4 and n2 6. The critical value of T such that P(T a) a/2 .025 is 12, and you should reject the null hypothesis if the observed value of T is 12 or less. Since the observed value of the test statistic—T 10—is less than 12, you can reject the hypothesis of equal distributions of wing stroke frequencies at the 5% level of signiﬁcance. A MINITAB printout of the Wilcoxon rank sum test (called Mann–Whitney by MINITAB) for these data is given in Figure 15.1. You will ﬁnd instructions for generating this output in the section “My MINITAB ” at the end of this chapter. Notice that the rank sum of the ﬁrst sample is given as W 34.0, which agrees with our calculations. With a reported p-value of .0142 calculated by MINITAB, you can reject the null hypothesis at the 5% level. F IG URE 15. 1 Printout for Example 15.1 ● Mann-Whitney Test and CI: Species 1, Species 2 N Median Species 1 4 207.50 Species 2 6 180.00 Point estimate for ETA1-ETA2 is 30.50 95.7 Percent CI for ETA1-ETA2 is (5.99,56.01) W = 34.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0142 The test is significant at 0.0139 (adjusted for ties) Normal Approximation for the Wilcoxon Rank Sum Test Table 7 in Appendix I contains critical values for sample sizes of n1 n2 3, 4, . . . , 15. Provided n1 is not too small,† approximations to the probabilities for the Wilcoxon rank sum statistic T can be found using a normal approximation to the distribution of T. It can be shown that the mean and variance of T are n2 1) and s 2 mT n1(n1 2 n2 1) T n1n2(n1 2 1 †Some researchers indicate that the normal approximation is adequate for samples as small as n1 n2 4. 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 635 The distribution of mT z T sT is approximately normal with mean 0 and standard deviation 1 for values of n1 and n2 as small as 10. If you try this approximation for Example 15.1, you get 6 1) 22 n2 1) 4(4 mT n1(n1 2 2 and 6 1) 22 n2 1) 4(6)(4 T n1n2(n1 s 2 12 2 1 The p-value for this test is 2P(T 34). If you use a .5 correction for continuity in calculating the value of z because n1 and n2 are both small,† you have (34 mT z T ) .5 2.45 sT 2 2 22 The p-value for this test is 2P(T 34) 2P(z 2.45) 2(.0071) .0142 the value reported on the MINITAB printout in Figure 15.1. THE WILCOXON RANK SUM TEST FOR LARGE SAMPLES: n1 W 10 and n2 W 10 1. Null hypothesis: H0 : The population distributions are identical 2. Alternative hypothesis: Ha : The two population distributions are not identical (a two-tailed test). Or Ha : The distribution of population 1 is shifted to the right (or left) of the distribution of population 2 (a one-tailed test). 3. Test statistic: z 4. Rejection region: T n1(n1 n2 1)/2 1)/12 n1n2(n 1 n2 a. For a two-tailed test, reject H0 if z za/2 or z za/2. b. For a one-tailed test in the right tail, reject H0 if z za. c. For a one-tailed test in the left tail, reject H0 if z za. Or reject H0 if p-value a. Tabulated values of z are found in Table 3 of Appendix I. EXAMPLE 15.2 An experiment was conducted to compare the strengths of two types of kraft papers: one a standard kraft paper of a speciﬁed weight and the other the same standard kraft paper treated with a chemical substance. Ten pieces of each type of paper, randomly selected from production, produced the strength measurements shown in Table 15.5. Test the null hypothesis of no difference in the distributions of strengths for the two †Since the value of T 34 lies to the right of the mean 22, the subtraction of .5 in using the normal approximation takes into account the lower limit of the bar above the value 34 in the probability distribution of T. 636 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS types of paper versus the alternative hypothesis that the treated paper tends to be stronger (i.e., its distribution of strength measurements is shifted to the right of the corresponding distribution for the untreated paper). Strength Measurements (and Their Ranks) TABLE 15.5 ● for Two Types of Paper Standard 1 Treated 2 1.21 (2) 1.43 (12) 1.35 (6) 1.51 (17) 1.39 (9) 1.17 (1) 1.48 (14) 1.42 (11) 1.29 (3.5) 1.40 (10) Rank sum 1.49 (15) 1.37 (7.5) 1.67 (20) 1.50 (16) 1.31 (5) 1.29 (3.5) 1.52 (18) 1.37 (7.5) 1.44 (13) 1.53 (19) T1 85.5 1 n1(n1 n2 1) T1 210 85.5 124.5 T * Solution Since the sample sizes are equal, you are at liberty to decide which of the two samples should be sample 1. Choosing the standard treatment as the ﬁrst sample, you can rank the 20 strength measurements, and the values of T1 and T * 1 are shown at the bottom of the table. Since you want to detect a shift in the standard (1) measurements to the left of the treated (2) measurements, you conduct a left-tailed test: H0 : No difference in the strength distributions Ha : Standard distribution lies to the left of the treated distribution and use T T1 as the test statistic, looking for an unusually small value of T. To ﬁnd the critical value for a one-tailed test with a .05, index Table 7(a) in Appendix I with n1 n2 10. Using the tabled entry, you can reject H0 when T 82. Since the observed value of the test statistic is T 85.5, you are not able to reject H0. There is insufficient evidence to conclude that the treated kraft paper is stronger than the standard paper. To use the normal approximation to the distribution of T, you can calculate n2 1) 10( mT n1(n1 21) 105 2 2 and n2 1) 10(1 T n1n2(n1 )(21) 175 0 s 2 2 1 1 2 with sT 175 13.23. Then mT 85. z T 105 1.47 5 sT .23 3 1 The one-tailed p-value corresponding to z 1.47 is p-value P(z 1.47) .5 .4292 .0708 which is larger than a .05. The conclusion is the same. You cannot conclude that the treated kraft paper is stronger than the standard paper. 15.2 THE WILCOXON RANK SUM TEST: INDEPENDENT RANDOM SAMPLES ❍ 637 When should the Wilcoxon rank sum test be used in preference to the two-sample unpaired t-test? The two-sample t-test performs well if the data are normally distributed with equal variances. If there is doubt concerning these assumptions, a normal probability plot could be used to assess the degree of nonnormality, and a two-sample F-test of sample variances could be used to check the equality of variances. If these procedures indicate either nonnormality or inequality of variance, then the Wilcoxon rank sum test is appropriate. 15.2 EXERCISES BASIC TECHNIQUES 1 as the test statistic? 15.1 Suppose you want to use the Wilcoxon rank sum test to detect a shift in distribution 1 to the right of distribution 2 based on samples of size n1 6 and n2 8. a. Should you use T1 or T * b. What is the rejection region for the test if a .05? c. What is the rejection region for the test if a .01? 15.2 Refer to Exercise 15.1. Suppose the alternative hypothesis is that distribution 1 is shifted either to the left or to the right of distribution 2. a. Should you use T1 or T * b. What is the rejection region for the test if a .05? c. What is the rejection region for the test if a .01? 15.3 Observations from two random and independent samples, drawn from populations 1 and 2, are given here. Use the Wilcoxon rank su
m test to determine whether population 1 is shifted to the left of population 2. 1 as the test statistic? Sample 1 Sample . State the null and alternative hypotheses to be tested. b. Rank the combined sample from smallest to largest. Calculate T1 and T * 1. c. What is the rejection region for a .05? d. Do the data provide sufficient evidence to indicate that population 1 is shifted to the left of population 2? 15.4 Independent random samples of size n1 20 and n2 25 are drawn from nonnormal populations 1 and 2. The combined sample is ranked and T1 252. Use the large-sample approximation to the Wilcoxon rank sum test to determine whether there is a difference in the two population distributions. Calculate the p-value for the test. 15.5 Suppose you wish to detect a shift in distribution 1 to the right of distribution 2 based on sample sizes n1 12 and n2 14. If T1 193, what do you conclude? Use a .05. APPLICATIONS 15.6 Alzheimer’s Disease In some tests of healthy, elderly men, a new drug has restored their memory almost to that of young people. It will soon be tested on patients with Alzheimer’s disease, the fatal brain disorder that destroys the mind. According to Dr. Gary Lynch of the University of California, Irvine, the drug, called ampakine CX-516, accelerates signals between brain cells and appears to signiﬁcantly sharpen memory.2 In a preliminary test on students in their early 20s and on men aged 65–70, the results were particularly striking. After being given mild doses of this drug, the 65–70-year-old men scored nearly as high as the young people. The accompanying data are the numbers of nonsense syllables recalled after 5 minutes for 10 men in their 20s and 10 men aged 65–70. Use the Wilcoxon rank sum test to determine whether the distributions for the number of nonsense syllables recalled are the same for these two groups. 20s 65–70s 15.7 Alzheimer’s, continued Refer to Exercise 15.6. Suppose that two more groups of 10 men each are tested on the number of nonsense syllables they can remember after 5 minutes. However, this time the 65–70-year-olds are given a mild dose of ampakine CX-516. Do the data provide sufficient evidence to conclude that this drug improves memory in men aged 65–70 compared with that of 20-year-olds? Use an appropriate level of a. 20s 65–70s 11 10 7 3 10 6 3 638 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS 15.8 Dissolved Oxygen Content The observations in the table are dissolved oxygen contents in water. The higher the dissolved oxygen content, the greater the ability of a river, lake, or stream to support aquatic life. In this experiment, a pollution-control inspector suspected that a river community was releasing semitreated sewage into a river. To check this theory, ﬁve randomly selected specimens of river water were selected at a location above the town and another ﬁve below. These are the dissolved oxygen readings (in parts per million): Above Town Below Town 4.8 5.0 5.2 4.7 5.0 4.9 4.9 4.8 5.1 4.9 a. Use a one-tailed Wilcoxon rank sum test with a .05 to conﬁrm or refute the theory. b. Use a Student’s t-test (with a .05) to analyze the data. Compare the conclusion reached in part a. EX1509 15.9 Eye Movement In an investigation of the visual scanning behavior of deaf children, measurements of eye movement were taken on nine deaf and nine hearing children. The table gives the eye-movement rates and their ranks (in parentheses). Does it appear that the distributions of eyemovement rates for deaf children and hearing children differ? Deaf Children Hearing Children 2.75 (15) 2.14 (11) 3.23 (18) 2.07 (10) 2.49 (14) 2.18 (12) 3.16 (17) 2.93 (16) 2.20 (13) Rank Sum 126 .89 (1) 1.43 (7) 1.06 (4) 1.01 (3) .94 (2) 1.79 (8) 1.12 (5.5) 2.01 (9) 1.12 (5.5) 45 15.10 Color TVs The table lists the life (in months) of service before failure of a color television circuit board for 8 television sets manufactured by ﬁrm A and 10 sets manufactured by ﬁrm B. Use the Wilcoxon rank sum test to analyze the data, and test to see whether the life of service before failure of the circuit boards differs for the circuit boards produced by the two manufacturers. Firm A B Life of Circuit Board (months) 32 41 25 39 40 36 31 47 35 45 29 34 37 48 39 44 43 33 15.11 Weights of Turtles The weights of turtles caught in two different lakes were mea- EX1511 sured to compare the effects of the two lake environments on turtle growth. All the turtles were the same age and were tagged before being released into the lakes. The weights for n1 10 tagged turtles caught in lake 1 and n2 8 caught in lake 2 are listed here: Lake Weight (oz) 1 2 14.1 15.2 12.2 13.0 13.9 14.5 14.1 13.6 14.7 12.4 13.8 14.0 16.1 12.7 15.3 11.9 12.5 13.8 Do the data provide sufficient evidence to indicate a difference in the distributions of weights for the tagged turtles exposed to the two lake environments? Use the Wilcoxon rank sum test with a .05 to answer the question. EX1512 15.12 Chemotherapy Cancer treatment by means of chemicals—chemotherapy—kills both cancerous and normal cells. In some instances, the toxicity of the cancer drug—that is, its effect on normal cells—can be reduced by the simultaneous injection of a second drug. A study was conducted to determine whether a particular drug injection reduced the harmful effects of a chemotherapy treatment on the survival time for rats. Two randomly selected groups of 12 rats were used in an experiment in which both groups, call them A and B, received the toxic drug in a dose large enough to cause death, but in addition, group B received the antitoxin, which was to reduce the toxic effect of the chemotherapy on normal cells. The test was terminated at the end of 20 days, or 480 hours. The survival times for the two groups of rats, to the nearest 4 hours, are shown in the table. Do the data provide sufficient evidence to indicate that rats receiving the antitoxin tend to survive longer after chemotherapy than those not receiving the antitoxin? Use the Wilcoxon rank sum test with a .05. Chemotherapy Only A Chemotherapy plus Drug B 84 128 168 92 184 92 76 104 72 180 144 120 140 184 368 96 480 188 480 244 440 380 480 196 15.3 THE SIGN TEST FOR A PAIRED EXPERIMENT ❍ 639 THE SIGN TEST FOR A PAIRED EXPERIMENT 15.3 The sign test is a fairly simple procedure that can be used to compare two populations when the samples consist of paired observations. This type of experimental design is called the paired-difference or matched pairs design, which you used to compare the average wear for two types of tires in Section 10.5. In general, for each pair, you measure whether the ﬁrst response—say, A—exceeds the second response— say, B. The test statistic is x, the number of times that A exceeds B in the n pairs of observations. When the two population distributions are identical, the probability that A exceeds B equals p .5, and x, the number of times that A exceeds B, has a binomial distribution. Only pairs without ties are included in the test. Hence, you can test the hypothesis of identical population distributions by testing H0 : p .5 versus either a one- or two-tailed alternative. Critical values for the rejection region or exact p-values can be found using the cumulative binomial tables in Appendix I. THE SIGN TEST FOR COMPARING TWO POPULATIONS 1. Null hypothesis: H0 : The two population distributions are identical and P(A exceeds B) p .5 2. Alternative hypothesis: a. Ha : The population distributions are not identical and p .5 b. Ha : The population of A measurements is shifted to the right of the popu- lation of B measurements and p .5 c. Ha : The population of A measurements is shifted to the left of the popula- tion of B measurements and p .5 3. Test statistic: For n, the number of pairs with no ties, use x, the number of times that (A B) is positive. 4. Rejection region: a. For the two-tailed test Ha : p .5, reject H0 if x xL or x xU, where P(x xL) a/2 and P(x xU) a/2 for x having a binomial distribution with p .5. b. For Ha : p .5, reject H0 if x xU with P(x xU) a. c. For Ha : p .5, reject H0 if x xL with P(x xL) a. Or calculate the p-value and reject H0 if the p-value a. One problem that may occur when you are conducting a sign test is that the measurements associated with one or more pairs may be equal and therefore result in tied observations. When this happens, delete the tied pairs and reduce n, the total number of pairs. The following example will help you understand how the sign test is constructed and used. EXAMPLE 15.3 The numbers of defective electrical fuses produced by two production lines, A and B, were recorded daily for a period of 10 days, with the results shown in Table 15.6. The response variable, the number of defective fuses, has an exact binomial distribution with a large number of fuses produced per day. Although this variable will have an 640 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS approximately normal distribution, the plant supervisor would prefer a quick-and-easy statistical test to determine whether one production line tends to produce more defectives than the other. Use the sign test to test the appropriate hypothesis. TABLE 15.6 ● Defective Fuses From Two Production Lines Day Line A Line B Sign of Difference 1 2 3 4 5 6 7 8 9 10 170 164 140 184 174 142 191 169 161 200 201 179 159 195 177 170 183 179 170 212 Solution For this paired-difference experiment, x is the number of times that the observation for line A exceeds that for line B in a given day. If there is no difference in the distributions of defectives for the two production lines, then p, the proportion of days on which A exceeds B, is .5, which is the hypothesized value in a test of the binomial parameter p. Very small or very large values of x, the number of times that A exceeds B, are contrary to the null hypothesis. Since n 10 and the hypothesized value of p is .5, Table 1 of Appendix I can be used to ﬁnd the exact p-value for the test of H0 : p .5 versus Ha : p .5 The observed value of the test statistic—which is the n
umber of “plus” signs in the table—is x 1, and the p-value is calculated as p-value 2P(x 1) 2(.011) .022 The fairly small p-value .022 allows you to reject H0 at the 5% level. There is signiﬁcant evidence to indicate that the number of defective fuses is not the same for the two production lines; in fact, line B produces more defectives than line A. In this example, the sign test is an easy-to-calculate rough tool for detecting faulty production lines and works perfectly well to detect a signiﬁcant difference using only a minimum amount of information. Normal Approximation for the Sign Test When the number of pairs n is large, the critical values for rejection of H0 and the approximate p-values can be found using a normal approximation to the distribution of x, which was discussed in Section 6.4. Because the binomial distribution is perfectly symmetric when p .5, this approximation works very well, even for n as small as 10. For n 25, you can conduct the sign test by using the z statistic, n p n x .5 z n q 5 np x . as the test statistic. In using z, you are testing the null hypothesis p .5 versus the alternative p .5 for a two-tailed test or versus the alternative p .5 (or p .5) for a one-tailed test. The tests use the familiar rejection regions of Chapter 9. 15.3 THE SIGN TEST FOR A PAIRED EXPERIMENT ❍ 641 SIGN TEST FOR LARGE SAMPLES: n W 25 1. Null hypothesis: H0 : p .5 (one treatment is not preferred to a second treat- ment) 2. Alternative hypothesis: Ha : p .5, for a two-tailed test (NOTE: We use the two-tailed test as an example. Many analyses might require a one-tailed test.) 5n . 3. Test statistic: z n 5 x . EXAMPLE 15.4 4. Rejection region: Reject H0 if z za/2 or z za/2, where za/2 is the z-value from Table 3 in Appendix I corresponding to an area of a/2 in the upper tail of the normal distribution. A production superintendent claims that there is no difference between the employee accident rates for the day versus the evening shifts in a large manufacturing plant. The number of accidents per day is recorded for both the day and evening shifts for n 100 days. It is found that the number of accidents per day for the evening shift xE exceeded the corresponding number of accidents on the day shift xD on 63 of the 100 days. Do these results provide sufficient evidence to indicate that more accidents tend to occur on one shift than on the other or, equivalently, that P(xE xD) 1/2? Solution This study is a paired-difference experiment, with n 100 pairs of observations corresponding to the 100 days. To test the null hypothesis that the two distributions of accidents are identical, you can use the test statistic n x .5 z n 5 . where x is the number of days in which the number of accidents on the evening shift exceeded the number of accidents on the day shift. Then for a .05, you can reject the null hypothesis if z 1.96 or z 1.96. Substituting into the formula for z, you get n x (.5 100) ( ) .5 3 2.60 1 z 0 n .5 5 5 1 0 63 . Since the calculated value of z exceeds za/2 1.96, you can reject the null hypothesis. The data provide sufficient evidence to indicate a difference in the accident rate distributions for the day versus evening shifts. When should the sign test be used in preference to the paired t-test? When only the direction of the difference in the measurement is given, only the sign test can be used. On the other hand, when the data are quantitative and satisfy the normality and constant variance assumptions, the paired t-test should be used. A normal probability plot can be used to assess normality, while a plot of the residuals (di d) can reveal large deviations that might indicate a variance that varies from pair to pair. When there are doubts about the validity of the assumptions, statisticians often recommend that both tests be performed. If both tests reach the same conclusions, then the parametric test results can be considered to be valid. 642 ❍ CHAPTER 15 NONPARAMETRIC STATISTICS 15.3 EXERCISES BASIC TECHNIQUES 15.13 Suppose you wish to use the sign test to test Ha : p .5 for a paired-difference experiment with n 25 pairs. a. State the practical situation that dictates the alterna- tive hypothesis given. b. Use Table 1 in Appendix I to ﬁnd values of a (a .15) available for the test. 15.14 Repeat the instructions of Exercise 15.13 for Ha : p .5. 15.15 Repeat the instructions of Exercises 15.13 and 15.14 for n 10, 15, and 20. EX1516 15.16 A paired-difference experiment was conducted to compare two populations. The data are shown in the table. Use a sign test to determine whether the population distributions are different. Population 1 1 2 8.9 8.8 2 8.1 7.4 3 9.3 9.0 Pairs 4 7.7 7.8 5 10.4 9.9 6 8.3 8.1 7 7.4 6.9 a. State the null and alternative hypotheses for the test. b. Determine an appropriate rejection region with a .01. c. Calculate the observed value of the test statistic. d. Do the data present sufficient evidence to indicate that populations 1 and 2 are different? APPLICATIONS 15.17 Property Values In Exercise 10.45, you compared the property evaluations of two EX1517 tax assessors, A and B. Their assessments for eight properties are shown in the table: Property Assessor A Assessor B 1 2 3 4 5 6 7 8 76.3 88.4 80.2 94.7 68.7 82.8 76.1 79.0 75.1 86.8 77.3 90.6 69.1 81.0 75.3 79.1 a. Use the sign test to determine whether the data present sufficient evidence to indicate that one of the assessors tends to be consistently more conservative than the other; that is, P(xA xB) 1/2. Test by using a value of a near .05. Find the p-value for the test and interpret its value. b. Exercise 10.45 uses the t statistic to test the null hypothesis that there is no difference in the mean property assessments between assessors A and B. Check the answer (in the answer section) for Exercise 10.45 and compare it with your answer to part a. Do the test results agree? Explain why the answers are (or are not) consistent. EX1518 15.18 Gourmet Cooking Two gourmets, A and B, rated 22 meals on a scale of 1 to 10. The data are shown in the table. Do the data provide sufficient evidence to indicate that one of the gourmets tends to give higher ratings than the other? Test by using the sign test with a value of a near .05. Meal A B Meal 10 11 12 13 14 15 16 17 18 19 20 21 22 10 8 6 3 4 2 3 a. Use the binomial tables in Appendix I to ﬁnd the exact rejection region for the test. b. Use the large-sample z statistic. (NOTE: Although the large-sample approximation is suggested for n 25, it works fairly well for values of n as small as 15.) c. Compare the results of parts a and b. 15.19 Lead Levels in Blood A study reported in the American Journal of Public Health (Science News)—the ﬁrst to follow blood lead levels in lawabiding handgun hobbyists using indoor ﬁring ranges—documents a signiﬁcant risk of lead poisoning.3 Lead exposure measurements were made on 17 members of a law enforcement trainee class before, during, and after a 3-month period of ﬁrearm instruction at a state-owned indoor ﬁring range. No trainee had elevated blood lead levels before the training, but 15 of the 17 ended their training with blood lead levels deemed “elevated” by the Occupational Safety and Health Administration (OSHA). If the use of an indoor 15.4 A COMPARISON OF STATISTICAL TESTS ❍ 643 ﬁring range causes no increase in blood lead levels, then p, the probability that a person’s blood lead level increases, is less than or equal to .5. If, however, use of the indoor ﬁring range causes an increase in a person’s blood lead levels, then p .5. Use the sign test to determine whether using an indoor ﬁring range has the effect of increasing a person’s blood lead level with a .05. (HINT: The normal approximation to binomial probabilities is fairly accurate for n 17.) 15.20 Recovery Rates Clinical data concerning the effectiveness of two drugs in treating EX1520 a particular disease were collected from ten hospitals. The numbers of patients treated with the drugs varied from one hospital to another. You want to know whether the data present sufficient evidence to indicate a higher recovery rate for one of the two drugs. a. Test using the sign test. Choose your rejection region so that a is near .05. b. Why might it be inappropriate to use the Student’s t-test in analyzing the data? Drug A Drug B Hospital Number in Group Number Recovered Percentage Recovered Hospital Number in Group Number Recovered Percentage Recovered 1 2 3 4 5 6 7 8 9 10 84 63 56 77 29 48 61 45 79 62 63 44 48 57 20 40 42 35 57 48 75.0 69.8 85.7 74.0 69.0 83.3 68.9 77.8 72.2 77. 10 96 83 91 47 60 27 69 72 89 46 82 69 73 35 42 22 52 57 76 37 85.4 83.1 80.2 74.5 70.0 81.5 75.4 79.2 85.4 80.4 A COMPARISON OF STATISTICAL TESTS 15.4 The experiment in Example 15.3 is designed as a paired-difference experiment. If the assumptions of normality and constant variance, s 2 d, for the differences were met, would the sign test detect a shift in location for the two populations as efficiently as the paired t-test? Probably not, since the t-test uses much more information than the sign test. It uses not only the sign of the difference, but also the actual values of the differences. In this case, we would say that the sign test is not as efficient as the paired t-test. However, the sign test might be more efficient if the usual assumptions were not met. When two different statistical tests can both be used to test a hypothesis based on the same data, it is natural to ask, Which is better? One way to answer this question would be to hold the sample size n and a constant for both procedures and compare b, the probability of a Type II error. Statisticians, however, prefer to examine the power of a test. Definition Power 1 b P(reject H0 when Ha is true) Since b is the probability of failing to reject the null hypothesis when it is false, the power of the test is the probability of rejecting the null hypothesis when it is false and some speciﬁed alternative is true. It is the probability that the test wil

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