10.50-10.58

# 10.50-10.58 - 10.50 Let = mean(average occupancy rate To...

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10.50 Let μ = mean (average) occupancy rate. To test H 0 : μ ≥ .6, H a : μ < .6, the computed test statistic is z* = -1.99 . The p–value is given by P(Z < –1.99) = .0233. Since this is less than the significance level of .10, H 0 is rejected. 10.53 a. The hypothesis of interest is H 0 : μ M = 3.8, H a : μ M < 3.8, where μ 1 represents the mean drop in FVC for men on the physical fitness program. With z* = –.996, we have p–value = P(Z < –1) = .1587 . b. With α = .05, p > α . H 0 cannot be rejected (you supply the full conclusion). c. Similarly, we have H 0 : μ W = 3.1, H a : μ W < 3.1. The computed test statistic is z* = –1.826 so that the p–value is P(Z < –1.83) = .0336. d. Since α = .05 is greater than the p–value, we can reject the null hypothesis. There is suff. evidence at the .05 level to conclude that the mean drop in FVC for women is less than 3.1. 10.54 a. The hypotheses are H 0 : p = .85, H a : p ≠ .85, where p = proportion of right–handed executives of large corporations. The computed test statistic is z = 5.34, and with α /2= (.01)/2 = .005, z .005 = 2.567. So, we reject H 0

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