Introduction to Probability and Statistics for Engineers and Scientists

314 Chapter 8:Hypothesis Testing

SOLUTION A simple computation (or the use of Program 8.4.1) shows that the value of
the test statistic is .066. For such a small value of the test statistic (which has a standard
normal distribution whenH 0 is true), it is clear that the null hypothesis is accepted. ■

It follows from Equation 8.4.1 that a test of the hypothesisH 0 :μx=μy(orH 0 :
μx≤μy) against the one-sided alternativeH 1 :μx>μywould be to

accept H 0 if X−Y≤zα

√

σx^2

n

+

σy^2

m

reject H 0 if X−Y>zα

√

σx^2

n

+

σy^2

m

8.4.2 Case of Unknown Variances

Suppose again thatX 1 ,...,XnandY 1 ,...,Ymare independent samples from normal
populations having respective parameters (μx,σx^2 ) and (μy,σy^2 ), but now suppose that all
four parameters are unknown. We will once again consider a test of

H 0 :μx=μy versus H 1 :μx=μy

To determine a significance levelαtest ofH 0 we will need to make the additional
assumption that the unknown variancesσx^2 andσy^2 are equal. Letσ^2 denote their
value — that is,

σ^2 =σx^2 =σy^2

As before, we would like to rejectH 0 whenX−Yis “far” from zero. To determine
how far from zero it need be, let

Sx^2 =

∑n

i= 1

(Xi−X)^2

n− 1

Sy^2 =

∑m

i= 1

(Yi−Y)^2

m− 1

denote the sample variances of the two samples. Then, as was shown in Section 7.3.2,

X−Y−(μx−μy)

√

Sp^2 (1/n+1/m)

∼tn+m− 2