sec_8_3_sol

sec_8_3_sol - APPM 4/5570 Solutions to Problems from...

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APPM 4/5570 Solutions to Problems from Section 8.3 37. ˆ p = 82 / 150 Step One: State the hypotheses: H 0 : p = 0 . 40 H a : p 6 = 0 . 40 Step Two: The test statistic: Z = ˆ p - p 0 q p 0(1 - p 0 ) /n = (82 / 150) - 0 . 40 q (0 . 40)(0 . 60) / 150 3 . 67 Step Three: The rejection region: The rejection region is two-tailed. For α = 0 . 01, it is given by the region on the z -curve above 2 . 575 and below - 2 . 575. Step Four: The conclusion: The test statistic is in the rejection region, so we reject H 0 in favor of H a at 0 . 01 level of signi±cance. These data suggest that the actual percentage of type A blood donation di²ers from 40%. This conclusion would not change for α = 0 . 05 since the new rejection region would then be above 1 . 96 and below - 1 . 96. The test statistic is still in the rejection region. 39. n = 1000 H 0 : p = 0 . 02 H a : p < 0 . 02 a. ˆ p = 15 / 1000 = 0 . 015 The test statistic is Z = ˆ p - p 0 q p 0 (1 - p 0 ) /n = 0 . 015 - 0 . 02 q (0 . 02)(0 . 98) / 1000 ≈ - 1 . 13 . The rejection region is given by the region on the z -curve below - 1 . 645. The test statistic does not fall in the rejection region. There is not signif- icant evidence, given by this data set, to reject H 0 at 0 . 05 level of signif-
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sec_8_3_sol - APPM 4/5570 Solutions to Problems from...

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