Mathematical Methods for Physics and Engineering : A Comprehensive Guide

STATISTICS

however, it is difficult to determineaandbfor a given significance levelα,soa

slightly different rejection region, which we now describe, is usually adopted.

The sampling distributionP(u|H 0 ) may be found straightforwardly from the

sampling distribution ofsgiven in (31.35). Let us first determineP(s^2 |H 0 )by

demanding that

P(s|H 0 )ds=P(s^2 |H 0 )d(s^2 ),

from which we find

P(s^2 |H 0 )=

P(s|H 0 )

2 s

=

(

N

2 σ 02

)(N−1)/ 2

(s^2 )(N−3)/^2

Γ

( 1

2 (N−1)

)exp

(

−

Ns^2

2 σ^20

)

. (31.122)

Thus, the sampling distribution ofu=Ns^2 /σ 02 is given by

P(u|H 0 )=

1

2 (N−1)/^2 Γ

( 1

2 (N−1)

)u(N−3)/^2 exp

(

−^12 u

)

.

We note, in passing, that the distribution ofuis precisely that of an (N−1) th-

order chi-squared variable (see subsection 30.9.4), i.e.u∼χ^2 N− 1. Although it does

not give quite the best test, one then takes the rejection region to be

0 <u<a and b<u<∞,

withaandbchosen such that the two tails haveequal areas; the advantage of

this choice is that tabulations of the chi-squared distribution make the size of this

region relatively easy to estimate. Thus, for a given significance levelα, we have

∫a

0

P(u|H 0 )du=α/2and

∫∞

b

P(u|H 0 )du=α/ 2.

Ten independent sample valuesxi,i=1, 2 ,..., 10 , are drawn at random from a Gaussian

distribution with unknown meanμand standard deviationσ. The sample values are as

follows:

2 .22 2.56 1.07 0.24 0.18 0.95 0. 73 − 0 .79 2.09 1. 81

Test the null hypothesisH 0 :σ^2 =2at the10%significance level.

For our null hypothesisσ^20 =2.Sinceforthissamples=1.01 andN= 10, from (31.121)
we haveu=5.10. Forα=0.1 we find, either numerically or using table 31.2, thata=3. 33
andb=16.92. Thus, our rejection region is

0 <u< 3. 33 and 16. 92 <u<∞.

The valueu=5.10 from our sample does not lie in the rejection region, and so we cannot
reject the null hypothesisH 0 :σ^2 =2.