Introduction to Probability and Statistics for Engineers and Scientists

322 Chapter 8:Hypothesis Testing

SOLUTION We will test the hypothesis that the machine is effective, since a rejection of
this hypothesis will then enable us to conclude that it is ineffective. Since we are thus
interested in testing

H 0 :σ^2 ≤.0225 versus H 1 :σ^2 >.0225

it follows that we would want to rejectH 0 whenS^2 is large. Hence, thep-value of the
preceding test data is the probability that a chi-square random variable with 19 degrees of
freedom would exceed the observed value of 19S^2 /.0225= 19 ×.025/.0225=21.111.
That is,

p-value=P{χ 192 >21.111}

= 1 −.6693=.3307 from Program 5.8.1.A

Therefore, we must conclude that the observed value ofS^2 =.025 is not large enough
to reasonably preclude the possibility thatσ^2 ≤.0225, and so the null hypothesis is
accepted. ■

8.5.1 Testing for the Equality of Variances of Two

Normal Populations

LetX 1 ,...,XnandY 1 ,...,Ymdenote independent samples from two normal populations
having respective (unknown) parametersμx,σx^2 andμy,σy^2 and consider a test of

H 0 :σx^2 =σy^2 versus H 1 :σx^2 =σy^2

If we 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, then as shown in Section 6.5, (n−1)Sx^2 /σx^2 and (m−1)Sy^2 /σy^2
are independent chi-square random variables withn−1 andm−1 degrees of freedom,
respectively. Therefore, (Sx^2 /σx^2 )/(Sy^2 /σy^2 ) has anF-distribution with parametersn−1 and
m−1. Hence, whenH 0 is true

Sx^2 /Sy^2 ∼Fn−1,m− 1

and so

PH 0 {F 1 −α/2,n−1,m− 1 ≤Sx^2 /Sy^2 ≤Fα/2,n−1,m− 1 }= 1 −α