Question

Reserve Problems Chapter 10 Section 1 Problem 1 Your answer is partially correct. Try again. Consider the hypothesis test Ho : 41 – 12 = 0 against H : H1 – Hy 70 samples below: I 37 39 31 33 33 30 31 30 38 38 30 37 36 30 39 31 35 38 II 33 29 34 33 32 29 30 35 31 34 30 30 31 32 34 35 Variances: 6 = 3.8, o2 = 0.1. Use a = 0.05. (a) Test the hypothesis and find the P-value. Find the test statistic. Round your answers to four decimal places (e.g. 98.7654). [2.4103 ZO P-value 0.0159 (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). We are 95% confident that value 1 exceeds value 2 by between 0.404 (c) What is the power of the test for the true difference in means of 2? Round your answer to three decimal places (e.g. 98.765). si-23 3.916 0.8618 The power of the test is 1-=

Answer 1

a)

		pop 1	pop 2
sample mean x =		34.22	32.06
std deviation σ=		3.800	0.100
sample size n=		18	16
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.896
test stat z =(x1-x2-Δo)/σx1-x2     =			2.4104

c)

standard error se=(√(σ12/n1+σ12/n1))=					0.8960
for 0.05 level and two tailed test critival value Zα=					1.9600

rejection region : x1-x2 <=Δ+z*se or x1-x2 >=Δ+z*se or x1-x2 <=-1.7562 or x1-x2>=1.7562
P(Type II error) =P(-1.7562<x1-x2<1.7562 |Δ=2)=P((-1.7562-2)/0.896<z<(1.7562-2)/0.896)=P(-4.19<z<-0.27)=0.393

(please try 0.394 if this comes wrong)