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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 signiﬁc
ant 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. (HINT: 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 ﬂ
uent 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 involves 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 Comp
aring 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 0.0
49 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 degrees 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 27 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 10.41 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 precision 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 spe
ciﬁed 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. Alternative 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 ❍ 4
31 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 both 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 var
iances 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 made 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 Eugloss
a 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 difference 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? (H
INT: 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 signi
ﬁcance? 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 different 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

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