ST102_Solutions_21.pdf

ST102_Solutions_21.pdf - ST102 Outline solutions to...

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ST102 Outline solutions to Exercise 21 1. Visual observation of the table indicates that the ratings for Airline B tend to be higher than for Airline A. We consider the differences B - A. Adding three more rows to the table, we obtain: Passenger 1 2 3 4 5 6 7 8 9 10 Airline A 4 3 5 8 6 2 1 6 7 5 Airline B 5 8 7 6 5 4 4 3 8 9 Sign + + + - - + + - + + Difference 1 5 2 - 2 - 1 2 3 - 3 1 4 Rank 2 10 5 5 2 5 7.5 7.5 2 9 (a) For the sign test, the test statistic is T = the number of negative differences. We have T Bin(10 , π ), where: π = P (rating for Airline B rating for Airline A) . We test H 0 : π = 0 . 5 vs. H 1 : π > 0 . 5. We reject H 0 for small values of T . We observe t = 3, hence the p -value is: p = P 0 . 5 ( T 3) = 3 X i =0 10! i ! (10 - i )! (0 . 5) 10 = (1 + 10 + 45 + 120)(0 . 5) 10 = 0 . 1719 . Hence we do not reject H 0 at the 5% significance level, and conclude that there is no evidence that the rating of Airline B is higher. (b) Let μ B - A be the population mean of the differenced data. We test: H 0 : μ B - A = 0 vs. H 1 : μ B - A > 0 . The Wilcoxon test statistic is T = the sum of the ranks of the negative differences, and we observe t = 14 . 5. The Wilcoxon table for n = 10 identifies a critical value of 11 for a one-sided test at the (approximate) 5% significance level. Hence we do not reject H 0 and conclude that there is no evidence that the rating of Airline B is higher. Noet that the Wilcoxon test is a nonparametric test for means. Table 17 in Murdoch and Barnes’ Statistical Tables lists only the critical values at different significance levels.
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