Chapter 5

Exercise 1

Fail to reject Ho

Exercise 2

Fail to reject Ho

80:

05.0,10,25,78 2

Ho

nX

Exercise 3

CI contains μ, so fail to reject is justified.

Exercise 4

pnorm(-1.265)[1] 0.1029357

For a one tailed test, P<0.103, meaning that the probability of obtaining a sample mean of 78 or lower when μ>80 is 0.103.

Exercise 5

This is a two tailed test, so P now relects p<-1.265 + 1-p>1.265 = 2(p<-1.265 )

pnorm(-1.265)[1] 0.1029357

For a two tailed test, P<0.206, meaning that the probability of obtaining a sample mean of 78 or lower or 82 or higher when μ>80 is 0.206.

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 = 0.9254881

Use case# 2 on P. 188

pnorm(1.67)[1] 0.9525403

Exercise 14R function power.t.test(36,3/8,type="one.sample",alternative="one.sided",sig.level=0.025) returns

power = 0.5901872

Apply case #1 P.188

pnorm(-0.29)[1] 0.3859081

Exercise 15

R function: power.t.test(49,3/10,type="one.sample",alternative="two.sided",sig.level=0.05) returns: power = 0.5390021

Third Case:

pnorm(-0.14)[1] 0.44433

pnorm(-4.06)[1] 2.453636e-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",alternative="one.sided",sig.level=0.05)

power = 0.3174914

pnorm(-0.38)[1] 0.3519727

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] 2.063899

Only in the last case T>C

Fail to reject

Fail to reject

Reject

Exercise 21

Power depends on the variance; larger variance lower power.

Exercise 22

qt(0.95,15)[1] 1.75305

Fail to reject

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] 2.262157

Fail to reject

Exercise 25

qt(0.025,99)[1] -1.984217

T<C therefore reject.

Exercise 26

C is set by n-2g-1 df=11

qt(0.025,11)[1] -2.200985

Fail to reject in all cases

Exercise 27

qt(0.95,9)[1] 1.833113

Fail to reject in all cases

C is set by n-2g-1 df=9

Exercise 28

> qt(0.975,5)[1] 2.570582

Reject C is set by n-2g-1 df=5

Exercise 29

qt(0.995,14)[1] 2.976843

Fail to reject

C is set by n-2g-1 df=14