Homework 5
10.2 Note that Yis 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|Hαtr ue) = P(Y > 12|p=.6) = 1 −P(Y≤12|p=.6) = .416.
(e) β(.4) = P(failtorejectH0|Hαtr ue) = 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, H0is 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=¯y−64
σ/√n<−2.326, or if ¯y < 64 −2.326σ
√n= 61.36. If µ= 60, then
β=P(¯
Y > 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 H0and 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 H0for 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, s2
p=.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
√s2
p(1
9+1
12 )=.989 and p−value > .10.
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