Consider the hypothesis test H0:μ1=μ2 against H1:μ1<μ2 with known variances σ1=10 and σ2=5. Suppose that sample sizes n1=10 and n2=15 and that x¯1=14.2 and x¯2=19.7. Use α=0.05.
a)
pop 1 | pop 2 | |||
sample mean x = | 14.20 | 19.70 | ||
std deviation σ= | 10.000 | 5.000 | ||
sample size n= | 10 | 15 | ||
std error σ_{x}_{1-}_{x}_{2}=√(σ^{2}1/n_{1}+σ^{2}_{2}/n_{2}) = | 3.416 | |||
test stat z =(x_{1}-x_{2}-Δ_{o})/σ_{x}_{1-}_{x}_{2 } = | -1.61 | |||
p value : = | 0.0537 | (from excel:1*normsdist(-1.61) |
a)
the null hypothesis is not rejected ; the p value is 0.0537
b)
Hypothesized mean difference Δ_{o } = | 0 | ||
true mean difference Δ_{ } = | -4 | ||
first sample standard deviation =σ_{1} = | 10.000 | ||
second sample std deviation =σ_{2}= | 5.000 | ||
first sample size =n_{1} = | 10.000 | ||
second sample size =n_{2} = | 15.000 | ||
standard error=(√(σ_{1}^{2}/n1+σ_{1}^{2}/n_{1}))= | 3.4157 | ||
for 0.05 level and left tail critival Zα= | -1.64 | ||
rejecf Ho if x<= Δo +Zα*σx or x<= | -5.6017 | ||
P(Type II error) =P(Xbar>-5.602| Δ=-4)=P(Z>(-5.6017--4)/3.416)=P(Z>-0.47)=0.6808 | |||
P(Power) =1-type II error =1-0.6808=0.32 |
c)
true mean difference =Δ = | 4 | ||
first sample std deviation =σ1 = | 10.000 | ||
2ndsample std deviation=σ2 = | 5.000 | ||
for 0.05 level and left tail critical Zα= | 1.645 | ||
for 0.05 level of type II error critival Zβ= | 1.645 | ||
n=(Z_{α/2}+Z_{β})^{2}(σ_{1}^{2}+σ_{2}^{2})/(Δ_{o}-Δ)^{2} = | 85 |
Consider the hypothesis test H0:μ1=μ2 against H1:μ1<μ2 with known variances σ1=10 and σ2=5. Suppose that sample...
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