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Consider the hypothesis test Ho: Mi = u2 against Hui < u2 with known variances 01 = 10 and 02 = 6. Suppose that sample sizes

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Answer #1

a)

pop 1 pop 2
sample mean x = 14.30 19.60
std deviation σ= 10.000 6.000
sample size n= 11 14
std error σx1-x2=√(σ21/n122/n2)    = 3.415
test stat z =(x1-x2o)/σx1-x2     = -1.55
p value : = 0.0606 (from excel:1*normsdist(-1.55)

the null hypothesis is not rejected , the p value is 0.0606

b)

Hypothesized mean difference Δo       = 0
true mean difference Δ                                                          = -4
first sample standard deviation =σ1    =   10.000
second sample std deviation =σ2= 6.000
first sample size =n1 = 11.000
second sample size =n2 = 14.000
standard error=(√(σ12/n1+σ12/n1))= 3.4150
for 0.05 level and left tail critival Zα= -1.645
rejecf Ho if x<= Δo +Zα*σx or x<=   -5.6177
P(Type II error) =P(Xbar>-5.618| Δ=-4)=P(Z>(-5.6177--4)/3.415)=P(Z>-0.47)=0.6808
P(Power) =1-type II error =1-0.6808=0.32

c)

Hypothesized mean difference Δo       = 0
true mean difference =Δ = 4
first sample std deviation =σ1   =                10.000
2ndsample std deviation=σ2 = 6.000
for 0.05 level and right tail critical Zα= 1.645
for 0.05 level of type II error critival Zβ= 1.645
n=(Zα/2+Zβ)21222)/(Δo-Δ)2 = 93
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