Chapter 5

Exercise 1 Fail to reject Ho

Exercise 2 Fail to reject Ho

Exercise 3 CI contains μ, so fail to reject is justified.

Exercise 4 pnorm(-1.265) [1] For a one tailed test, P 80 is

Exercise 5 This is a two tailed test, so P now relects p = 2(p< ) pnorm(-1.265) [1] For a two tailed test, P 80 is

Exercise 6 Reject Ho

Exercise 7 Reject

Exercise 8 CI does not contain contains μ (130), so decision to reject is justified.

Exercise 9 Yes, because the sample mean of 23 is already lower than the value stated by the null hypothesis of μ<25. To reject this hypothesis, we must have a sample mean that is higher than 25.

Exercise 10 Reject Ho

Exercise 11 Reject

Exercise 12 A CI can be a good alternative to a two tailed test. If the CI range does not contain μ, you can reject Ho of μ=546.

Exercise 13 R function: power.t.test(25,4/5,type="one.sample",alternative="one.sided",sig. level=0.01) returns: power = Use case# 2 on P. 188 pnorm(1.67) [1]

Exercise 14 R function power.t.test(36,3/8,type="one.sample",alternative="one.sided",sig.level=0.025) returns power = Apply case #1 P.188 pnorm(-0.29) [1]

Exercise 15 R function: power.t.test(49,3/10,type="one.sample",alternative="two.sided",sig.level=0.05) returns: power = Third Case: pnorm(-0.14) [1] pnorm(-4.06) [1] e-05

Exercise 16 The sample size is two small, so failure to reject can be the result of insufficient power.

Exercise 17 power.t.test(10,2/5,type="one.sample",altern ative="one.sided",sig.level=0.05) power = pnorm(-0.38) [1]

Exercise 18

Exercise 19 Increase α, but this will also increase type I error, which is highly undesirable.

Exercise 20 qt(0.975,24) [1] Only in the last case T>C Fail to reject Reject

Exercise 21 Power depends on the variance; larger variance lower power.

Exercise 22 qt(0.95,15) [1] Fail to reject Reject Only in the last case T>C

Exercise 23 The sample means are consistent with Ho: μ>42, so we fail to reject without testing.

Exercise 24 qt(0.975,9) [1] Fail to reject

Exercise 25 qt(0.025,99) [1] T<C therefore reject.

Exercise 26 C is set by n-2g-1 df=11 qt(0.025,11) [1] Fail to reject in all cases

Exercise 27 qt(0.95,9) [1] Fail to reject in all cases C is set by n-2g-1 df=9

Exercise 28 > qt(0.975,5) [1] Reject C is set by n-2g-1 df=5

Exercise 29 qt(0.995,14) [1] Fail to reject C is set by n-2g-1 df=14