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Unformatted text preview: &¡¢£¢ ¤¥ ¦§¨©ª«¬­®«¯°± ¡«²§³¨«´µ 8.32. a. The hypothesis of interest is two-tailed: H : μ 1 - μ 2 = 0 H a : μ 1 - μ 2 & 0 The observed value of the test statistic is 807 . 3 66 7100 57 6300 700 , 82 100 , 78 ) ( ) ( 2 2 2 2 2 1 2 1 2 1 2 1- = +- = +--- = n s n s x x z μ μ The p-value is 2 ( ) ) ( 2 807 . 3 = ≈ > Z P Since the p-value is smaller than the significance level 0.1, we can reject H . There is sufficient evidence to indicate that the means differ from April to May. b. In this exercise, H would be rejected in either case. 8.33. The hypothesis of interest is one-tailed: H : μ 1 - μ 2 = 0 H a : μ 1 - μ 2 < 0 since increased scores imply μ 2 > μ 1. The observed value of the test statistic with 60805 . 18 ) 56 (. 9 ) 95 (. 9 2 2 2 = + = s , is 871 . 3 10 1 10 1 ( 60805 . 17 . 8 82 . 6 ) 1 1 ( ) ( 2 1 2 2 1- = +- = +-- = n n s x x t The p-value is ( ) 005 . 871 . 3 < > t P for 9 degrees of freedom. Since the p-value is less than the significance level of 0.05, we can reject H . The training course was effective in increasing customer service scores. 8.44. The hypothesis of interest is H : μ A - μ B = 0 or H : μ d = 0 H a : μ A - μ B > 0 or H a : μ d > 0 where 1 μ is the mean for the current year and 2 μ is the mean for last year. The differences, along with necessary calculations, are given below: i d .54 -.94 -.23 .30 .28 .76 118333 . 6 71 . = = = & n d d i 3780167 . 5 6 ) 71 (. 9741 . 1 1 ) ( 2 2 2 2 =- =-- = & & n n d d s i i d the test statistic is 471 . 6 3780167 . 118333 . =- =- = n s d t d d μ The p-value is 1 . ) 471 . ( > > t P for 5 degrees of freedom. Since the p-value is greater than the significance level of 0.05, we cannot reject Ho....
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