Download Statistical Hypothesis Testing Homework Solutions and more Assignments Statistics in PDF only on Docsity! Homework 5 10.2 Note that Y is binomial with parameters n = 20 and p. (a) If the experimenter concludes that less than 80% of insomniacs respond to the drug when actually the drug induces sleep in 80% of insomniacs, a type I error has occured. (b) α = P (rejectH0|H0true) = P (Y ≤ 12|p = .8) = .032 (using Appendix III) (c) If the experimenter does not reject the hypothesis that 80% of insomniacs respond to the drug when actually the drug induces sleep in fewer than 80% of insomniacs, a type II error has occured. (d) β(.6) = P (failtorejectH0|Hαtrue) = P (Y > 12|p = .6) = 1− P (Y ≤ 12|p = .6) = .416. (e) β(.4) = P (failtorejectH0|Hαtrue) = P (Y > 12|p = .4) = .021. 10.20 H0 : µ ≥ 64, Hα : µ < 64. Using the large sample test for a mean, z = −1.77, and with α = .01, −z.01 = −2.326. So, H0 is not rejected: there is not enough evidence to conclude the manufacturer’s claim is false. 10.38 With H0 : µ ≥ 64, this is rejected if z = ȳ−64σ/√n < −2.326, or if ȳ < 64 − 2.326σ√ n = 61.36. If µ = 60, then β = P (Ȳ > 61.36|µ = 60) = P (Z > 61.36−60 8/ √ 50 = P (Z > 1.2) = .1151. 10.42 Using the sample size of formula given in this section, we have n = (zα + zβ)2σ2 (µα − µ0)2 = 607.37, so a sample size of 608 will provide the desired levels. 10.54 (a) The hypotheses are H0 : p = .85, Hα : p > .85, where p=proportion of right-handed executives of large corporations. The computed test statistic is z = 5.34, and with α = .01, z.01 = 2.346. So, we reject H0 and conclude that the proportion of right-handed executives at large corporations is greater than 85%. (b) Since p− value = P (Z > 5.34) < .000001, we can safely reject H0 for any significance level of .000001 or more. This represents strong evidence against H0. 10.70 (a) The hypotheses are H0 : µ1 − µ2 = 0 vs. Hα : µ1 − µ2 > 0. The computed test statistic is t = 2.97 (here, s2p = .0001444). With 21 degrees of freedom, t.05 = 1.721 so we reject H0. (b) For this problem, the hypotheses are H0 : µ1 − µ2 = .01 vs. Hα : µ1 − µ2 > 0.01. Then, t = (.041−.026)−.01√ s2p( 1 9 + 1 12 ) = .989 and p− value > .10. 1
