(BQ) Part 2 book Statistics has contents: Nonparametric statistics (available online), categorical data analysis, multiple regression and model building, simple linear regression, analysis of variance - comparing more than two means,... and other contents.
www.downloadslide.net Inferences Based on Two Samples Confidence Intervals and Tests of Hypotheses contents 9.1 9.2 9.3 9.4 9.5 9.6 identifying the Target Parameter Comparing Two Population Means: independent Sampling Comparing Two Population Means: Paired difference experiments Comparing Two Population Proportions: independent Sampling determining the Sample Size Comparing Two Population Variances: independent Sampling (optional) Where We’ve Been • Explored two methods for making statistical inferences: confidence intervals and tests of hypotheses • Studied confidence intervals and tests for a single population mean m, a single population proportion p, and a single population variance s2 • Learned how to select the sample size necessary to estimate a population parameter with a specified margin of error Where We’re Going • Learn how to identify the target parameter for comparing two populations (9.1) • Learn how to compare two means by using confidence intervals and tests of hypotheses (9.2–9.3) • Apply these inferential methods to problems in which we want to compare two population proportions, or two population variances (9.4, 9.6) • Determine the sizes of the samples necessary to estimate the difference between two population parameters with a specified margin of error (9.5) www.downloadslide.net 462 CHAPTER ■ Inferences Based on Two Samples Statistics IN Action Zixit Corp v Visa USA inc.—A Libel Case The National Law Journal (Aug 26–Sept 2, 2002) reported on an interesting court case in volving ZixIt Corp., a start-up Internet credit card clearing center ZixIt claimed that its new online credit card processing system would allow Internet shoppers to make purchases without revealing their credit card numbers This claim violated the established protocols of most major credit card companies, including Visa Without the company’s knowledge, a Visa vice president for technology research and development began writing e-mails and Web site postings on a Yahoo! message board for ZixIt investors, challenging ZixIt’s claim and urging investors to sell their ZixIt stock The Visa executive posted over 400 e-mail and notes before he was caught Once it was discovered that a Visa executive was responsible for the postings, ZixIt filed a lawsuit against Visa Corp., alleging that Visa—using the executive as its agent—had engaged in a “malicious two-part scheme to disparage and interfere with ZixIt” and its efforts to market the new online credit card processing system In the libel case ZixIt asked for $699 million in damages Dallas lawyers Jeff Tillotson and Mike Lynn, of the law firm Lynn Tillotson & Pinker, were hired to defend Visa in the lawsuit The lawyers, in turn, hired Dr James McClave (co-author of this text) as their expert statistician McClave testified in court on an “event study” he did matching the Visa executive’s e-mail postings with movement of ZixIt’s stock price the next business day McClave’s testimony, showing that there was an equal number of days when the stock went up as went down after a posting, helped the lawyers representing Visa to prevail in the case The National Law Journal reported that, after two and a half days of deliberation, “the jurors found [the Visa executive] was not acting in the scope of his employment and that Visa had not defamed ZixIt or interfered with its business.” In this chapter, we demonstrate several of the statistical analyses McClave used to infer that the Visa executive’s postings had no effect on ZixIt’s stock price The daily ZixIt stock prices as well as the timing of the Visa executive’s postings are saved in the ZIXITVISA file.* We apply the statistical methodology presented in this chapter to this data set in two Statistics in Action Revisited examples Statistics IN Action Revisited • Comparing Mean Price Changes (p 473) • Comparing Proportions (p 497) Data Set: ZIXITVISA 9.1 Identifying the Target Parameter Many experiments involve a comparison of two populations For instance, a sociologist may want to estimate the difference in mean life expectancy between inner-city and suburban residents Or a consumer group may want to test whether two major brands of food freezers differ in the average amount of electricity they use Or a political candidate might want to estimate the difference in the proportions of voters in two districts who favor her candidacy Or a professional golfer might be interested in comparing the variability in the distance that two competing brands of golf balls travel when struck with the same club In this chapter, we consider techniques for using two samples to compare the populations from which they were selected The same procedures that are used to estimate and test hypotheses about a single population can be modified to make inferences about two populations As in Chapters and 8, the methodology used will depend on the sizes of the samples and the parameter of interest (i.e., the target parameter) Some key words and the type of data associated with the parameters covered in this chapter are listed in the following box *Data provided (with permission) from Info Tech, Inc., Gainesville, Florida www.downloadslide.net SECTION 9.2 ■ Comparing Two Population Means: Independent Sampling 463 Determining the Target Parameter Parameter Key Words or Phrases Type of Data m1 - m2 p1 - p2 Mean difference; difference in averages Difference between proportions, percentages, fractions, or rates; compare proportions Ratio of variances; difference in variability or spread; compare variation Quantitative Qualitative 1s1 2 > 1s2 2 Quantitative You can see that the key words difference and compare help identify the fact that two populations are to be compared In the previous examples, the words mean in mean life expectancy and average in average amount of electricity imply that the target parameter is the difference in population means, m1 - m2 The word proportions in proportions of voters in two districts indicates that the target parameter is the difference in proportions, p1 - p2 Finally, the key word variability in variability in the distance identifies the ratio of population variances, 1s1 2 > 1s2 2, as the target parameter As with inferences about a single population, the type of data (quantitative or qualitative) collected on the two samples is also indicative of the target parameter With quantitative data, you are likely to be interested in comparing the means or variances of the data With qualitative data with two outcomes (success or failure), a comparison of the proportions of successes is likely to be of interest We consider methods for comparing two population means in Sections 9.2 and 9.3 A comparison of population proportions is presented in Section 9.4 and population variances in optional Section 9.6 We show how to determine the sample sizes necessary for reliable estimates of the target parameters in Section 9.5 9.2 Comparing Two Population Means: Independent Sampling In this section, we develop both large-sample and small-sample methodologies for comparing two population means In the large-sample case, we use the z-statistic; in the small-sample case, we use the t-statistic Large Samples Example 9.1 DIETS A Large-Sample Confidence Interval for M1 − M2 —Comparing Mean Weight Loss for Two Diets Problem A dietitian has developed a diet that is low in fats, carbohydrates, and cholesterol Although the diet was initially intended to be used by people with heart disease, the dietitian wishes to examine the effect this diet has on the weights of obese people Two random samples of 100 obese people each are selected, and one group of 100 is placed on the low-fat diet The other 100 are placed on a diet that contains approximately the same quantity of food, but is not as low in fats, carbohydrates, and cholesterol For each person, the amount of weight lost (or gained) in a three-week period is recorded The data, saved in the DIETS file, are listed in Table 9.1 Form a 95% confidence interval for the difference between the population mean weight losses for the two diets Interpret the result Solution Recall that the general form of a large-sample confidence interval for a single mean m is x { za>2 sx That is, we add and subtract za>2 standard deviations of the sample estimate x to and from the value of the estimate We employ a similar procedure to form the confidence interval for the difference between two population means Let m1 represent the mean of the conceptual population of weight losses for all obese people who could be placed on the low-fat diet Let m2 be similarly defined for the other diet We wish to form a confidence interval for 1m1 - m2 An intuitively appealing estimator for 1m1 - m2 is the difference between the sample means, 1x1 - x2 Thus, we will form the confidence interval of interest with 1x1 - x2 { za>2 s1x1 - x22 www.downloadslide.net 464 CHAPTER ■ Inferences Based on Two Samples Table 9.1 Diet Study Data, Example 9.1 Weight Losses for Low-Fat Diet 21 13 11 16 10 8 12 12 14 16 11 10 10 14 14 11 14 12 12 10 12 13 18 11 11 11 11 11 10 20 19 12 11 11 15 11 14 10 4 13 11 13 12 12 17 15 14 10 18 20 4 12 10 12 12 14 9 9 10 14 10 8 16 6 13 18 10 12 9 12 11 13 13 Weight Losses for Regular Diet 14 13 11 11 14 12 2 10 6 8 13 11 8 16 12 10 Data Set: DIETS Assuming that the two samples are independent, we write the standard deviation of the difference between the sample means (i.e., the standard error of x1 - x2) as s1x1 - x22 = s21 s22 + n2 B n1 Typically (as in this example), the population variances s21 and s22 are unknown Since the samples are both large 1n1 = n2 = 1002, the sample variances s21 and s22 will be good estimators of their respective population variances Thus, the estimated standard error is s1x1 - x22 ≈ s12 s22 + B n1 n2 Summary statistics for the diet data are displayed at the top of the SPSS printout shown in Figure 9.1 Note that x1 = 9.31, x2 = 7.40, s1 = 4.67, and s2 = 4.04 Using these values and observing that a = 05 and z.025 = 1.96, we find that the 95% confidence interval is, approximately, 19.31 - 7.402 { 1.96 14.672 14.042 + = 1.91 { 11.9621.622 = 1.91 { 1.22 100 B 100 or (.69, 3.13) This interval (rounded) is highlighted in Figure 9.1 Figure 9.1 SPSS analysis of diet study data www.downloadslide.net SECTION 9.2 ■ Comparing Two Population Means: Independent Sampling 465 Using this estimation procedure over and over again for different samples, we know that approximately 95% of the confidence intervals formed in this manner will enclose the difference in population means 1m1 - m2 Therefore, we are highly confident that the mean weight loss for the low-fat diet is between 69 and 3.13 pounds more than the mean weight loss for the other diet With this information, the dietitian better understands the potential of the low-fat diet as a weight-reduction diet Look Back If the confidence interval for 1m1 - m2 contains [e.g., -2.5, 1.32], then it is possible for the difference between the population means to be (i.e., m1 - m2 = 0) In this case, we could not conclude that a significant difference exists between the mean weight losses for the two diets ■ Now Work Exercise 9.6a The justification for the procedure used in Example 9.1 to estimate 1m1 - m2 relies on the properties of the sampling distribution of 1x1 - x2 The performance of the estimator in repeated sampling is pictured in Figure 9.2, and its properties are summarized in the following box: Figure 9.2 Sampling distribution of 1x1 - x2 (x – x ) Properties of the Sampling Distribution of (x1 − x2) The mean of the sampling distribution of 1x1 - x2 is 1m1 - m2 2 If the two samples are independent, the standard deviation of the sampling distribution is s1x1 - x22 = s21 s22 + n2 B n1 where s21 and s22 are the variances of the two populations being sampled and n1 and n2 are the respective sample sizes We also refer to s1x1 - x22 as the standard error of the statistic 1x1 - x2 By the Central Limit Theorem, the sampling distribution of 1x1 - x2 is approximately normal for large samples In Example 9.1, we noted the similarity in the procedures for forming a large-sample confidence interval for one population mean and a large-sample confidence interval for the difference between two population means When we are testing hypotheses, the procedures are again similar The general large-sample procedures for forming confidence intervals and testing hypotheses about 1m1 - m2 are summarized in the following boxes: Large, Independent Samples Confidence Interval for (M1 − M2): Normal (z) Statistic s21 and s22 known: 1x1 - x2 { za>2 s1x1 - x22 = 1x1 - x2 { za>2 s22 s21 + n2 B n1 s21 and s22 unknown: 1x1 - x2 { za>2 s1x1 - x22 ≈ 1x1 - x2 { za>2 s21 s22 + B n1 n2 www.downloadslide.net 466 CHAPTER ■ Inferences Based on Two Samples Large, Independent Samples Test of Hypothesis for (M1 − M2): Normal (z) Statistic s21 and s22 known zc = Test statistic: 1x1 - x2 - D0 a s21 C n1 b + a s22 n2 One-Tailed Tests Rejection region: s21 and s22 unknown b H0: 1m1 - m2 = D0 H0: 1m1 - m2 = D0 Ha: 1m1 - m2 D0 Ha: 1m1 - m2 D0 p-value: zc ≈ s21 s22 b + a b C n1 n2 a Two-Tailed Test zc - za zc za P1z zc P1z zc 1x1 - x2 - D0 H0: 1m1 - m2 = D0 Ha: 1m1 - m2 ≠ D0 ͉ zc ͉ za>2 2P1z zc if zc is positive 2P1z zc if zc is negative Decision: Reject H0 if a p@value or if test statistic 1zc falls in rejection region where P1z za = a, P1z za>2 = a>2, and a = P1Type I error2 = P1Reject H0 ͉ H0 true2 [Note: The symbol for the numerical value assigned to the difference 1m1 - m 2 under the null hypothesis is D0 For testing equal population means, D0 = 0.] Conditions Required for Valid Large-Sample Inferences about (M1 − M2) The two samples are randomly selected in an independent manner from the two target populations The sample sizes, n1 and n2, are both large (i.e., n1 Ú 30 and n2 Ú 30) (By the Central Limit Theorem, this condition guarantees that the sampling distribution of 1x1 - x2 will be approximately normal, regardless of the shapes of the underlying probability distributions of the populations Also, s21 and s22 will provide good approximations to s21 and s22 when both samples are large.) Example 9.2 DIETS A Large-Sample Test for M1 − M2 —Comparing Mean Weight Loss for Two Diets Problem Refer to the study of obese people on a low-fat diet and a regular diet presented in Example 9.1 Another way to compare the mean weight losses for the two different diets is to conduct a test of hypothesis Suppose we want to determine if the low-fat diet is more effective than the regular diet Use the information on the SPSS printout shown in Figure 9.1 to conduct the test at a = 05 Solution Again, we let m1 and m2 represent the population mean weight losses of obese people on the low-fat diet and regular diet, respectively If the low-fat diet is more effective in reducing the weights of obese people, then the mean weight loss for the low-fat diet, m1, will exceed the mean weight loss for the regular diet, m2 That is m1 m2 Thus, the elements of the test are as follows: H0: 1m1 - m2 = 1i.e., m1 = m2; note that D0 = for this hypothesis test2 Ha: 1m1 - m2 1i.e., m1 m2 Test statistic: z = 1x1 - x2 - D0 x1 - x2 - = s1x1 - x22 s1x1 - x22 Rejection region: z z.05 = 1.645 (see Figure 9.3) Substituting the summary statistics given in Figure 9.1 into the test statistic, we obtain z = 1x1 - x2 - 9.31 - 7.40 = s1x1 - x22 s21 s22 + n2 B n1 www.downloadslide.net SECTION 9.2 z Rejection 1.645 region z = 3.09 Figure 9.3 Rejection region for Example 9.2 Comparing Two Population Means: Independent Sampling 467 Now, since s21 and s22 are unknown, we approximate the test statistic value as follows: = 05 ■ z ≈ 9.31 - 7.40 s21 s22 + B n1 n2 1.91 = 14.672 14.042 + 100 B 100 = 1.91 = 3.09 617 [Note: The value of the test statistic is highlighted in the SPSS printout of Figure 9.1.] As you can see in Figure 9.3, the calculated z-value clearly falls into the rejection region Therefore, the samples provide sufficient evidence, at a = 05, for the dietitian to conclude that the mean weight losses for the two diets differ Look Back First, note that this conclusion agrees with the inference drawn from the 95% confidence interval in Example 9.1 However, the confidence interval provides more information on the mean weight losses From the hypothesis test, we know only that the two means differ; that is, m1 m2 From the confidence interval in Example 9.1, we found that the mean weight loss m1 of the low-fat diet was between 69 and 3.13 pounds more than the mean weight loss m2 of the regular diet In other words, the test tells us that the means differ, but the confidence interval tells us how large the difference is Second, a one-tailed hypothesis test and a confidence interval (which is two-tailed) may not always agree However, a two-tailed hypothesis test and a confidence interval will always give the same inference about the target parameter for the same value of a ■ Example 9.3 DIETS The p-Value for a Test of M − M 2 Problem Find the observed significance level for the test in Example 9.2 Interpret the result Solution The alternative hypothesis in Example 9.2, Ha: m1 - m2 0, required an upper one-tailed test using z = x1 - x2 s1x1 - x22 as a test statistic Since the z-value calculated from the sample data was 3.09, the observed significance level (p-value) for the one-tailed test is the probability of observing a value of z at least as contradictory to the null hypothesis as z = 3.09; that is, p@value = P1z Ú 3.092 This probability is computed under the assumption that H0 is true and is equal to the highlighted area shown in Figure 9.4 A = 499 p-value = 5–A = 001 Figure 9.4 The observed significance level for Example 9.2 z 3.09 p-value = P(z Ú 3.09) The tabulated area corresponding to z = 3.09 in Table II of Appendix B is 4990 Therefore, the observed significance level for the test is p@value = P1z Ú 3.092 = - 4990 = 0010 Since our selected a value, 05, exceeds this p-value, we have sufficient evidence to reject H0: 1m1 - m2 = in favor of Ha: 1m1 - m2 www.downloadslide.net 468 CHAPTER ■ Inferences Based on Two Samples Look Back The p-value of the test is more easily obtained from a statistical software package A MINITAB printout for the hypothesis test is displayed in Figure 9.5 The onetailed p-value, highlighted on the printout is 001 This value agrees with our calculated p-value Figure 9.5 MINITAB Printout for One-Tailed Test, Example 9.3 ■ Now Work Exercise 9.6b Small Samples Figure 9.6 Assumptions for the two-sample t: (1) normal populations; (2) equal variances In comparing two population means with small samples (say, either n1 30 or n2 30 or both), the methodology of the previous three examples is invalid The reason? When the sample sizes are small, estimates of s21 and s22 are unreliable and the Central Limit Theorem (which guarantees that the z statistic is normal) can no longer be applied But as in the case of a single mean (Section 8.4), we use the familiar Student’s t-distribution described in Chapter To use the t-distribution, both sampled populations must be approximately normally distributed with equal population variances, and the random samples must be selected independently of each other The assumptions of normality and equal variances imply relative frequency distributions for the populations that would appear as shown in Figure 9.6 Since we assume that the two populations have equal variances 1s21 = s22 = s2 2, it is reasonable to use the information contained in both samples to construct a pooled sample estimator S for use in confidence intervals and test statistics Thus, if s21 and s22 are the two sample variances (each estimating the variance s2 common to both populations), the pooled estimator of s2, denoted as s2p, is s2p = or 1n1 - 12s21 + 1n2 - 12s22 1n1 - 12s21 + 1n2 - 12s22 = 1n1 - 12 + 1n2 - 12 n + n2 - From sample e e s2p From sample 2 a 1x1 - x1 + a 1x2 - x2 = n + n2 - where x1 represents a measurement from sample and x2 represents a measurement from sample Recall that the term degrees of freedom was defined in Section 7.2 as less than the sample size Thus, in this case, we have 1n1 - 12 degrees of freedom for sample and 1n2 - 12 degrees of freedom for sample Since we are pooling the information on s2 obtained from both samples, the number of degrees of freedom associated with the pooled variance s2p is equal to the sum of the numbers of degrees of freedom for the two samples, namely, the denominator of s2p; that is, 1n1 - 12 + 1n2 - 12 = n1 + n2 - Note that the second formula given for s2p shows that the pooled variance is simply a weighted average of the two sample variances s21 and s22 The weight given each variance is proportional to its number of degrees of freedom If the two variances have the www.downloadslide.net SECTION 9.2 ■ Comparing Two Population Means: Independent Sampling 469 same number of degrees of freedom (i.e., if the sample sizes are equal), then the pooled variance is a simple average of the two sample variances The result is an average, or “pooled,” variance that is a better estimate of s2 than either s21 or s22 alone BIOGRAPHY BRADLEY EFRON (1938–PRESENT) The Bootstrap Method Bradley Efron was raised in St Paul, Minnesota, the son of a truck driver who was the amateur statistician for his bowling and baseball leagues Efron received a B.S in mathematics from the California Institute of Technology in 1960 but, by his own admission, had no talent for modern abstract math His interest in the science of statistics developed after he read a book by Harold Cramer from cover to cover Efron went to Stanford University to study statistics, and he earned his Ph.D there in 1964 He has been a faculty member in Stanford’s Department of Statistics since 1966 Over his career, Efron has received numerous awards and prizes for his contributions to modern statistics, including the MacArthur Prize Fellow (1983), the American Statistical Association Wilks Medal (1990), and the Parzen Prize for Statistical Innovation (1998) In 1979, Efron invented a method—called the bootstrap—of estimating and testing population parameters in situations in which either the sampling distribution is unknown or the assumptions are violated The method involves repeatedly taking samples of size n (with replacement) from the original sample and calculating the value of the point estimate Efron showed that the sampling distribution of the estimator is simply the frequency distribution of the bootstrap estimates ■ Both the confidence interval and the test-of-hypothesis procedures for comparing two population means with small samples are summarized in the following boxes: Small, Independent Samples Confidence Interval for M1 − M2 : Student’s t-Statistic 1x1 - x2 { ta>2 1n1 - 12s21 + 1n2 - 12s22 n + n2 - where s2p = B s2p a 1 + b n1 n2 and ta>2 is based on 1n1 + n2 - 22 degrees of freedom [Note: s2p = s21 + s22 when n1 = n2] Small, Independent Samples Test of Hypothesis for M1 − M2 : Student’s t-Statistic Test statistic: tc = 1x1 - x2 - D0 C Rejection region: p-value: s2p a where s2p = 1 + b n1 n2 1n1 - 12s21 + 1n2 - 12s22 One-Tailed Tests H0:1m1 - m2 = D0 Ha:1m1 - m2 D0 tc - ta P1t tc H0:1m1 - m2 = D0 Ha:1m1 - m2 D0 tc ta P1t tc 1n1 + n2 - 22 Two-Tailed Test H0:1m1 - m2 = D0 Ha:1m1 - m2 ≠ D0 ͉ tc ͉ ta>2 2P1t tc if tc is positive 2P1t tc if tc is negative Decision: Reject H0 if a p@value or if test statistic (tc) falls in rejection region where P1t ta = a, P1t ta>2 = a>2, the distribution of t is based on 1n1 + n2 - 22df, and a = P1Type I error2 = P1Reject H0 ͉ H0 true2 [Note: The symbol for the numerical value assigned to the difference 1m1 - m2 under the null hypothesis is D0 For testing equal population means, D0 = 0.] www.downloadslide.net 470 CHAPTER ■ Inferences Based on Two Samples Conditions Required for Valid Small-Sample Inferences about (M1 − M2) The two samples are randomly selected in an independent manner from the two target populations Both sampled populations have distributions that are approximately normal The population variances are equal (i.e., s21 = s22) Example 9.4 READING A Small-Sample Confidence Interval for M1 − M2 —Comparing Two Methods of Teaching Problem Suppose you wish to compare a new method of teaching reading to “slow learners” with the current standard method You decide to base your comparison on the results of a reading test given at the end of a learning period of six months Of a random sample of 22 “slow learners,” 10 are taught by the new method and 12 are taught by the standard method All 22 children are taught by qualified instructors under similar conditions for the designated six-month period The results of the reading test at the end of this period are given in Table 9.2 Table 9.2 Reading Test Scores for Slow Learners New Method 80 76 70 80 66 85 79 71 Standard Method 81 76 79 73 72 62 76 68 70 86 75 68 73 66 Data Set: READING a Use the data in the table to estimate the true mean difference between the test scores for the new method and the standard method Use a 95% confidence interval b Interpret the interval you found in part a c What assumptions must be made in order that the estimate be valid? Are they reasonably satisfied? Solution a For this experiment, let m1 and m2 represent the mean reading test scores of “slow learners” taught with the new and standard methods, respectively Then the objective is to obtain a 95% confidence interval for 1m1 - m2 The first step in constructing the confidence interval is to obtain summary statistics (e.g., x and s) on reading test scores for each method The data of Table 9.2 were entered into a computer, and SAS was used to obtain these descriptive statistics The SAS printout appears in Figure 9.7 Note that x1 = 76.4, s1 = 5.8348, x2 = 72.333, and s2 = 6.3437 Figure 9.7 SAS printout for Example 9.4 www.downloadslide.net Credits Chapter p 32 Source: Human Factors, Vol 56, No 31, May 2014; p 32 Source: Academy of Management Journal, Oct 2007; p 49 James T McClave, Terry Sincich; pp 53–56 Screenshots from Minitab Corporation Courtesy of Minitab Corporation Chapter p 59 Based on Li E.C, Williams S.F and Volpe R.D, “The Effects of Topic and Listener Familiarity of Discourse Variables in Procedural and Narrative Discource Tasks”, The Journal of Communication Disorders, Vol: 28, No: 01, Mar 1995, Page: 44 (Table 1) p 60 Summary Table from SPSS Reprint Courtesy of International Business Machines Corporation, © International Business Machines Corporation; p 67 Based on Journal of Moral Education, March 2010; p 67 Source: International Rhino Foundation, 2014; pp 67–68 Based on Sangkuk Lee, Journal of Family History (Jan 2010); p 68 Screenshot from SAS © 1999 SAS In stitute Inc Reprinted with Permission; p. 68 Source: Union of Concerned Scientists, Nov 2012; p 68 Source: Wu J, et al Track Allocation Optimization at a Railway Station: Mean-Variation Model and Case Study, Journal of Transportation Engineering, Vol: 39, No: 5, May 2013 (Extracted from Table 4); p 69 Source: Braga A.A, Hureau D.M & Papachristos A.V, Deterring Gang-Involved Gun Violence: Measuring the Impact of Boston’s Operation Ceasefire on Street Gang Behaviour, Journal of Quantitative Criminology, Vol: 30, No: 01, March 2014 (adapted from Figure 3); p 69 Source: McMunn-Dooley C & Welch M.M, Nature of Interactions Among Young Children and Adult Caregivers in a Children’s Museum, Early Childhood Education Journal, Vol: 42, No: 02, March 2014 (adapted from Figure 2); p 69 Source: Roskes M et al., The Right Side? 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p 834 Based on Gelman “A 55,000 residents desperately need your help!”, Chance, Vol l7 No 2, Spring 2004 (Figures l and 5); p 834 Based on Gangemi.A., and Mancini F “Guilt and focusing in decision-making” Journal of Behavioural Decision Making, Vol 20, Jan 2007 (Table 2); pp 835–836 Screenshots from Minitab Corporation Courtesy of Minitab Corporation Chapter 14 (available online) p 14-7 Source: Patston, T, ‘Teaching stage fright?–Implications for music educators British Joumal ofMusic Education, Vol 31, No 1, March 2014 (adapted from Figure 1); p 14-8 Based on Boyer D.M, Evans A.R and Jernvall J, “Evidence of Dietary Differentiation Among Late Paleocene–Early Eocene Plesiadapids (Mammalia, Primates)”, American Journal of Anthrapology, Vol 142, © 2010, Table A3; p 14-9, Based on Thery.M, et al., “Specific Color Sensitivities of Prey and Predator Explain Camouflage in Different Visual www.downloadslide.net CREDITS 891 Systems”, Behavioural Ecology, Vol.16, No.1, Jan 2005 (Table 1); p 14-9, Based on Shester G.G, “Explaining Catch Variation among Baja California Lobster Fishers Through Spatial Analysis of Trap-Placement Decision”, Bulletin of Marine Science, Vol 86, No 2, April 2010 (Table 1); 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p 14-47, An Illustrated Review of Some Farmer Participatory Research Techniques by Riley J, Fielding W.J from Journal of Agricultural, Biological and Environmental Statistics, Vol 6, No.1, Mar 2001(Table 1) Copyright © 2001 by International Biometric Society Used by permission of International Biometric Society; p 14-48, Based on Hemmings B, Smith M, Graydon J and Dyson R, “Effects of Massage on Psychological Restoration, Perceived Recovery and Repeated Sports Performance”, British Journal of Sports Medicine, Vol 34, No 2, Apr 2000(Adapated from Table 3); p 14-48, Based on Data from Elizabeth Schreiher, Department of Statistics Kansas State University Manhattan, Kansas; p 14-54, Based on Pfeiffer m, et al., “Community Organization and Species Richness of Ants in Mongolia Along an Ecological Gradient from Steppe to Gobi Dessert”, Journal of Biogeography, Vol 30, No 12, Dec 2003 (Tables and 2); p 14-55, Explaining Catch variation Among Baja California Lobster fishers through Spatial Analysis of Trap-Placement Decisions by Shester G.G from Bulletin of Marine Science, Vol 86, No 2, April 2010 (Table 1) Copyright © 2010 by University of Miami, Bulletin of Marine Science Used by permission of University of Miami, Bulletin of Marine Science; p 14-155, Based on Hemmings B, Smith M, Graydon J and Dyson R, “Effects of Massage on Psychological Restoration, Perceived Recovery and Repeated Sports Performance”, British Journal of Sports Medicine, Vol 34, No 2, Apr 2000(Data Adapated from Figure 3); p 14-55, Based on Seal N and Seal J, “Eating Patterns of the Rural Families of Overweight Pre-school Children: A Pilot Study”, Journal of Education and Human Development, Vol 1, No 3, 2009 (Figure 1); p 14-56, Based on Morris P.E and Fritz C.O, “The Name Game: Using Retrieval Practice to Improve the Learning of Names”, Journal of Experimental Psycology-Applied, Vol 6, No 2, June 2000 (Data simulated from figure 2); p 14-57, Based on Singer T, et al., “Empathy for Pain Involves the Affective but not Sensory Components of Pain”, Science, Vol 303, Feb 20,2004 (Adapted from fig 4); p 14-57, Based on Hertwig R, Pachur T and Kurzenhauser S, “Judgements of Risk Frequencies Tests of Possible Cognitive Mechanisms”, Journal of Experimental Psychology Learning, Memory and Cognition, Vol 31, No 4, July 2005 (Table 1); p 14-60, Based on Lichen Radionuclide Baseline Research Project,2003, p 25, Orion, University of Alaska-Fairbanks; p 14-60, Based on C.P Keller, et al., “Planning the Next Generation Regional Atlases: Input from Educators”, Journal of Geograpghy, Vol 94, No 3, May/June 1995, p,413 (Table 1); p 14-60, Based on Shand J, et al., “Variability in the Location of the Retinal Ganglion Cell Area Centralis is Correlated with Ontogenetic Changes in Feeding Behaviour in the Black Bream, Acanthopagrus’ Butcheri’ ” Brain and Behaviour, Vol 55, No 4, Apr 2000 (Figure II); p 14-61, Based on Bonadia P, et al., “Aluminosilicate Refractories for Aluminium Cell Linings”, American Ceramic Society Bulletin, Vol 84, No 2, Feb 2005( Table II); p. 14-62, Based on American Journal of Science, Vol 305, No 1, Jan 2005, p.16 (Table 2); p 14-63, Based on Axelson D, et al., “A Preliminary Study of the Kiddie Schedule for Affective Disorders and Schizophrenia for School-Age Children Mania Rating Scale for Children and www.downloadslide.net 892 CREDITS Adolescents”, Journal of Child and Adolescent Psychopharmacology, Vol 13, No 4, Dec 2003 (Figure 2); p 14-63, Based on Billing B.A, Green B.P and Volz W.H, “Selection of Forum for Litigated Tax Issues”, Journal of Applied Business Research, Vol 12, No.4, Fall 1996, p 38 (Table 2); p 14-63, Based on Yang W.H, et al., “A Freckle Criterion for the Solidification of Super-alloys with a tilted Solidification Front”, JOM: Journal of the Minerals, Metals and Materials Society, Vol 56, No 9, Sept 2004 (Table iv); p 14-64, Based on Uyehara J.C and Narins P.M, “Nest Defense by Willow Fliycatchres to Brood-Parasitic Intruders”, The Condor, Vol 97, No.2, May 1995, p 364 (Figure 1); p 14-65–14-67, Screenshots from Minitab Corporation Courtesy of Minitab Corporation www.downloadslide.net This page intentionally left blank www.downloadslide.net This page intentionally left blank www.downloadslide.net This page intentionally left blank www.downloadslide.net MAKING CONNECTIONS USING DATA ■ Statistics IN Action Social Media network Usage— STATISTICS IN ACTION opens each chapter with a case study showing the connection between statistics and a contemporary, controversial, or high-profile issue The case study is revisited throughout the chapter, showing the relevance of statistical concepts Are You Linked in? The Pew Research Center, a nonpartisan organization funded by a Philadelphia-based charity, has conducted more than 100 surveys on Internet usage in the United States as part of the Pew Internet & American Life Project (PIALP) In a recent report titled Social Media Update, 2013, the PIALP examined adults’ (ages 18 and up) attitudes and behavior toward online social media networks Regarded merely as a fun, online activity for high school and college students just a few years ago, social media now exert tremendous influence over the way people around the world—of all ages—get and share information The five social media sites investigated in this report include Facebook, Twitter, Instagram, Pinterest, and LinkedIn The Pew Research Center contacted 1,445 Internet users via landline telephone or cell phone for the survey Several of the many survey questions asked are provided here as well as the survey results: • Social Networking: When asked if they ever use an online social networking site, adults responded: • Twitter Usage: When asked if they ever use Twitter, adults responded: Yes 18% No 82% • Overall Social Media Usage: When asked about how many of the five social networking sites they use, adults responded: 22% 36% 23% 12% 5% 2% Yes 73% (Average = 1.48 sites) No 27% In the following “Statistics in Action Revisited” sections, we discuss several key statistical concepts covered in this chapter that are relevant to the Pew Internet & American Life Project survey • Facebook Usage: When Facebook users were asked how often they visit the social media site, they responded: Several times a day 40% About once a day 24% 3–5 days a week 10% ■ Statistics IN Action Revisited • Identifying the Population, Sample, and Inference (p 37) 1–2 days a week 13% Every few weeks 6% • Identifying the Data Collection Method and Data Type (p 44) Less often 7% • Critically Assessing the Ethics of a Statistical Study (p 46) Page 30 Example EXAMPLES use real data to show how statistical methods can be applied to real life situations The Example format fosters problem-solving skills by taking a three-step approach: 7.6 Estimating a Population Proportion—Fraction Who Trust the President Problem Public-opinion polls are conducted regularly to estimate the fraction of U.S citizens who trust the president Suppose 1,000 people are randomly chosen and 637 answer that they trust the president How would you estimate the true fraction of all U.S citizens who trust the president? Solution What we have really asked is how you would estimate the probability p of success in a binomial experiment in which p is the probability that a person chosen trusts the president One logical method of estimating p for the population is to use the proportion of successes in the sample That is, we can estimate p by calculating pn = Number of people sampled who trust the president Number of people sampled where pn is read “p hat.” Thus, in this case, • Problem • Solution • Look Back (or Look Ahead) pn = 637 = 637 1,000 Look Back To determine the reliability of the estimator pn , we need to know its sampling distribution That is, if we were to draw samples of 1,000 people over and over again, each time calculating a new estimate pn , what would be the frequency distribution of all the pn -values? The answer lies in viewing pn as the average, or mean, number of successes per trial over the n trials If each success is assigned a value equal to and each failure is assigned a value of 0, then the sum of all n sample observations is x, the total number of successes, and pn = x>n is the average, or mean, number of successes per trial in the n trials The Central Limit Theorem tells us that the relative frequency distribution of the sample mean for any population is approximately normal for sufficiently large samples A NOW WORKEXERCISEsuggestion follows each Example ■ pn Page 365 pn Now Work Exercise 7.59a www.downloadslide.net Connecting facets of everyday life to statistics, McClave and Sincich offer a trusted, comprehensive text that stresses the development of statistical thinking, the assessment of credibility, and the value of inferences made from data USING TECHNOLOGY USING TECHNOLOGY sections, at the end of each chapter, offer statistical software tutorials with step-by-step instructions and screen shots for MINITAB® and, where appropriate, the TI-83/84 Plus Graphing Calculator MINITAB: Simulating a Sampling Distribution Step Select “Calc” on the MINITAB menu bar, and then click on “Random Data” (see Figure 6.M.1) Step Calculate the value of the sample statistic of interest for each sample To this, click on the “Calc” button on the MINITAB menu bar, and then click on “Row Statistics,” as shown in Figure 6.M.3 The resulting dialog box appears in Figure 6.M.4 Figure 6.M.1 MINITAB options for generating random data Step On the resulting menu list, click on the distribution of your choice (e.g., “Uniform”) A dialog box similar to the one (the Uniform Distribution) shown in Figure 6.M.2 will appear Figure 6.M.3 MINITAB selections for generating sample statistics for the simulated data , the arrival www.brighton-webs from the start Figure 6.M.4 MINITAB row statistics dialog box Figure 6.M.2 MINITAB dialog box for simulating the uniform distribution Page 341 Researchers at the University of South Florida Center for Biological Defense , he method has been found to work well powder Consider a powder specimen that has exactly Suppose that the number of anthrax spores in the sample detected by the new method follows MORE THAN 1,800 EXERCISES are available, ind the probability that or fewer anthrax spores are covering a wide range of applications that underscore anthrax the relevance of statistics to everyday life Most of the exercises incorporate real data, and a indicates that Construct the data set is available on the companion website Ethics IN Statistics Purposeful reporting of numerical descriptive statistics in order to mislead the target audience is considered unethical statistical practice Page 129 NEW! ETHICS BOXES appear where appropriate to highlight the importance Maintaining a conof ethical behavior stant temperature in awhen pipe wall in some hot process A new technique that utilizes collecting, analyzing, and the temperature was bolt-on trace elements to maintain (November interpreting statistical data the pipe wall Applying the Concepts—Intermediate 5.14 Social network densities Social networks involve interactions (connections) between members of the network Sociologists define network density as the ratio of actual network connections to the number of possible one-to-one connections For example, a network with 10 members 10 has a b = 45 total possible connections If that network has only connections, the network density is 5/45 = 111 Sociologists at the University of Michigan assumed that the density x of a social network would follow a uniform distribution between and (Social Networks, 2010) a On average, what is the density of a randomly selected social network? b What is the probability that the randomly selected network has a density higher than 7? c Consider a social network with only members Explain why the uniform model would not be a good approximation for the distribution of network density 5.15 Cycle availability of a system In the jargon of system maintenance, “cycle availability” is defined as the probability that a system is functioning at any point in time The U.S Department of Defense developed a series of performance measures for assessing system cycle availability (START, Vol 11, 2004) Under certain assumptions about the failure time and maintenance time of a system, cycle availability is shown to be uniformly distributed between and Find the following parameters for cycle availability: mean, standard deviation, 10th percentile, lower quartile, and upper quartile Interpret the results 5.16 Time delays at a bus stop A bus is scheduled to stop at a certain bus stop every half hour on the hour and the half hour At the end of the day, buses still stop after every 30 minutes, but because delays often occur earlier in the day, the bus is never early and is likely to be late The director of the bus line claims that the length of time a bus is late is uniformly distributed and the maximum time that a bus is late is 20 minutes a If the director’s claim is true, what is the expected number of minutes a bus will be late? b If the director’s claim is true, what is the probability that the last bus on a given day will be more than 19 minutes late? c If you arrive at the bus stop at the end of a day at exactly half-past the hour and must wait more than 19 minutes for the bus, what would you conclude about the director’s claim? Why? Page 266 www.downloadslide.net GLOBAL EDITION For these Global Editions, the editorial team at Pearson has collaborated with educators across the world to address a wide range of subjects and requirements, equipping students with the best possible learning tools This Global Edition preserves the cutting-edge approach and pedagogy of the original, but also features alterations, customization, and adaptation from the North American version This is a special edition of an established title widely used by colleges and universities throughout the world Pearson published this exclusive edition for the benefit of students outside the United States and Canada If you purchased this book within the United States or Canada, you should be aware that it has been imported without the approval of the Publisher or Author Pearson Global Edition ... 1m1 - m2 = 0: 1x1 - x2 { ta >2 21s21 >n1 + 1s 22 >n2 t = 1x1 - x2 > 21 s21 >n1 + 1s 22 >n2 where t is based on degrees of freedom equal to n = 1s21 >n1 + s 22 >n2 2 1s21 >n1 2 n1 - + 1s 22 >n2 2 n2 -... for Exercise 9 .28 Positive Display Rule 3 4 4 5 5 5 5 5 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 5 5 5 5 5 Neutral Display Rule 3 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 5 www.downloadslide.net... s 22 = s2 Calculate the pooled estimator of s2 for each of the following cases: a s21 = 180, s 22 = 20 0, n1 = n2 = 25 b s21 = 25 , s 22 = 40, n1 = 10, n2 = 20 c s21 = 25 , s 22 = 32, n1 = 8, n2 = 12