Introduction to Probability and Statistics for Engineers and Scientists

312 Chapter 8:Hypothesis Testing

8.4Testing the Equality of Means of Two Normal Populations

A common situation faced by a practicing engineer is one in which she must decide whether
two different approaches lead to the same solution. Often such a situation can be modeled
as a test of the hypothesis that two normal populations have the same mean value.

8.4.1 Case of Known Variances

Suppose thatX 1 ,...,XnandY 1 ,...,Ymare independent samples from normal populations
having unknown meansμxandμybut known variancesσx^2 andσy^2. Let us consider the
problem of testing the hypothesis

H 0 :μx=μy

versus the alternative

H 1 :μx=μy

SinceXis an estimate ofμxandYofμy, it follows thatX−Ycan be used to estimate
μx−μy. Hence, because the null hypothesis can be written asH 0 :μx−μy=0, it seems
reasonable to reject it whenX−Yis far from zero. That is, the form of the test should
be to

reject H 0 if |X−Y|>c

accept H 0 if |X−Y|≤c

(8.4.1)

for some suitably chosen valuec.
To determine that value ofcthat would result in the test in Equations 8.4.1 having
a significance levelα, we need determine the distribution ofX−Y whenH 0 is true.
However, as was shown in Section 7.3.2,

X−Y∼N

(

μx−μy,

σx^2

n

+

σy^2

m

)

which implies that

X−Y−(μx−μy)

√

σx^2

n

+

σy^2

m

∼N(0, 1) (8.4.2)

Hence, whenH 0 is true (and soμx−μy=0), it follows that

(X−Y)

/√

σx^2

n

+

σy^2

m