Question

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 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

Answer 1

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/n1+σ22/n2)    =			3.416
test stat z =(x1-x2-Δo)/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)2(σ12+σ22)/(Δo-Δ)2 =			85