chapter 5. exercise 1 fail to reject ho exercise 2 fail to reject ho
TRANSCRIPT
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Chapter 5
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Exercise 1
Fail to reject Ho
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Exercise 2
Fail to reject Ho
80:
05.0,10,25,78 2
Ho
nX
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Exercise 3
CI contains μ, so fail to reject is justified.
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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.
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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.
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Exercise 6
Reject Ho
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Exercise 7
Reject
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Exercise 8
CI does not contain contains μ (130), so decision to reject is justified.
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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.
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Exercise 10
Reject Ho
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Exercise 11
Reject
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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.
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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
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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
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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
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Exercise 16
The sample size is two small, so failure to reject can be the result of insufficient power.
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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
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Exercise 18
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Exercise 19
Increase α, but this will also increase type I error, which is highly undesirable.
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Exercise 20
qt(0.975,24)[1] 2.063899
Only in the last case T>C
Fail to reject
Fail to reject
Reject
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Exercise 21
Power depends on the variance; larger variance lower power.
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Exercise 22
qt(0.95,15)[1] 1.75305
Fail to reject
Fail to reject
Reject
Only in the last case T>C
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Exercise 23
The sample means are consistent with Ho: μ>42, so we fail to reject without testing.
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Exercise 24
qt(0.975,9)[1] 2.262157
Fail to reject
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Exercise 25
qt(0.025,99)[1] -1.984217
T<C therefore reject.
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Exercise 26
C is set by n-2g-1 df=11
qt(0.025,11)[1] -2.200985
Fail to reject in all cases
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Exercise 27
qt(0.95,9)[1] 1.833113
Fail to reject in all cases
C is set by n-2g-1 df=9
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Exercise 28
> qt(0.975,5)[1] 2.570582
Reject C is set by n-2g-1 df=5
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Exercise 29
qt(0.995,14)[1] 2.976843
Fail to reject
C is set by n-2g-1 df=14