ISM_chapter15

# ISM_chapter15 - Chapter 15: Nonparametric Statistics 15.1...

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304 Chapter 15: Nonparametric Statistics 15.1 Let Y have a binomial distribution with n = 25 and p = .5. For the two–tailed sign test, the test rejects for extreme values (either too large or too small) of the test statistic whose null distribution is the same as Y . So, Table 1 in Appendix III can be used to define rejection regions that correspond to various significant levels. Thus: Rejection region α Y 6 or Y 19 P ( Y 6) + P ( Y 19) = .014 Y 7 or Y 18 P ( Y 7) + P ( Y 18) = .044 Y 8 or Y 17 P ( Y 8) + P ( Y 17) = .108 15.2 Let p = P(blood levels are elevated after training). We will test H 0 : p = .5 vs H a : p > .5. a. Since m = 15, so p –value = P ( M 15) = ( ) ( ) ( ) 17 17 16 17 17 16 17 17 15 5 . 5 . 5 . + + = 0.0012. b. Reject H 0 . c. P ( M 15) = P ( M > 14.5) P ( Z > 2.91) = .0018, which is very close to part a . 15.3 Let p = P (recovery rate for A exceeds B). We will test H 0 : p = .5 vs H a : p .5. The data are: Hospital A B Sign(A – B) 1 75.0 85.4 2 69.8 83.1 3 85.7 80.2 + 4 74.0 74.5 5 69.0 70.0 6 83.3 81.5 + 7 68.9 75.4 8 77.8 79.2 9 72.2 85.4 10 77.4 80.4 a. From the above, m = 2 so the p –value is given by 2 P ( M 2) = .110. Thus, in order to reject H 0 , it would have been necessary that the significance level α .110. Since this is fairly large, H 0 would probably not be rejected. b. The t –test has a normality assumption that may not be appropriate for these data. Also, since the sample size is relatively small, a large–sample test couldn’t be used either. 15.4 a. Let p = P (school A exceeds school B in test score). For H 0 : p = .5 vs H a : p .5, the test statistic is M = # of times school A exceeds school B in test score. From the table, we find m = 7. So, the p –value = 2 P ( M 7) = 2 P ( M 3) = 2(.172) = .344. With α = .05, we fail to reject H 0 . b. For the one–tailed test, H 0 : p = .5 vs H a : p > .5. Here, the p –value = P ( M 7) = .173 so we would still fail to reject H 0 .

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Chapter 15: Nonparametric Statistics 305 Instructor’s Solutions Manual 15.5 Let p = P (judge favors mixture B). For H 0 : p = .5 vs H a : p .5, the test statistic is M = # of judges favoring mixture B. Since the observed value is m = 2, p –value = 2 P ( M 2) = 2(.055) = .11. Thus, H 0 is not rejected at the α = .05 level. 15.6 a. Let p = P (high elevation exceeds low elevation). For H 0 : p = .5 vs H a : p > .5, the test statistic is M = # of nights where high elevation exceeds low elevation. Since the observed value is m = 9, p –value = P ( M 9) = .011. Thus, the data favors H a . b. Extreme temperatures, such as the minimum temperatures in this example, often have skewed distributions, making the assumptions of the t –test invalid. 15.7 a. Let p = P (response for stimulus 1 is greater that for stimulus 2). The hypotheses are H 0 : p = .5 vs H a : p > .5, and the test statistic is M = # of times response for stimulus 1 exceeds stimulus 2. If it is required that α .05, note that P ( M 1) + P ( M 8) = .04, where M is binomial( n = 9, p = .5) under H 0 . Our rejection region is the set {0, 1, 8, 9}. From the table, m = 2 so we fail to reject H 0 .
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## ISM_chapter15 - Chapter 15: Nonparametric Statistics 15.1...

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