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Consider the hypothesis test H0:μ1=μ2 against H1:μ1<μ2 with known variances σ1=10 and σ2=5. Suppose that sample...

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.

Font Paragraph Styles Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test Ho : = 12 against HI : <H2 with

Font Paragraph Styles Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test Ho : = 12 against HI : <H2 with known variances = 10 and 2 = 5. Suppose that sample sizes nj = 10 and 12 = 15 and that I = 14.2 and 72 = 19.7. Use a = 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) if #1 is 4 units less than 12? (c) Assuming equal sample sizes, what sample size should be used to obtain p = 0.05 if #1 is 4 units less than 12 ? Assume that a = 0.05. (a) The null hypothesis rejected. The p-value is Round your answer to four decimal places (e.g. 98.7654). (b) The power is Round your answer to two decimal places (e.g. 98.76). (c) N = n2 =1 Round your answer up to the nearest integer. Statistical Tables and Charts Question Attempts: 0 of 2 used SAVE FOR LATER SUBMIT ANSWER powered by Maple
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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 σx1-x2=√(σ21/n122/n2)    = 3.416
test stat z =(x1-x2o)/σx1-x2     = -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 Δo       = 0
true mean difference Δ                                                          = -4
first sample standard deviation =σ1    =   10.000
second sample std deviation =σ2= 5.000
first sample size =n1 = 10.000
second sample size =n2 = 15.000
standard error=(√(σ12/n1+σ12/n1))= 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=(Zα/2+Zβ)21222)/(Δo-Δ)2 = 85
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