Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.7 HYPOTHESIS TESTING

distributions. In particular, let us consider the case where we have two independent

samples of sizesN 1 andN 2 , drawn respectively from Gaussian distributions with

a common varianceσ^2 but with possibly different meansμ 1 andμ 2. On the basis

of the samples, one wishes to distinguish between the hypotheses

H 0 :μ 1 =μ 2 , 0 <σ^2 <∞ and H 1 :μ 1 =μ 2 , 0 <σ^2 <∞.

In other words, we wish to test the null hypothesis that the samples are drawn

from populations having the same mean. Suppose that the measured sample

means and standard deviations arex ̄ 1 , ̄x 2 ands 1 ,s 2 respectively. In an analogous

way to that presented above, one may show that the generalised likelihood ratio

can be written as

λ=

(

1+

t^2

N 1 +N 2 − 2

)−(N 1 +N 2 )/ 2

.

In this case, the variabletis given by

t=

w ̄−ω

σˆ

(

N 1 N 2

N 1 +N 2

) 1 / 2

, (31.119)

where ̄w=x ̄ 1 − ̄x 2 ,ω=μ 1 −μ 2 and

σˆ=

[

N 1 s^21 +N 2 s^22

N 1 +N 2 − 2

] 1 / 2

.

It is straightforward (albeit with complicated algebra) to show that the variablet

in (31.119) follows Student’st-distribution withN 1 +N 2 −2 degrees of freedom,

and so we may use an appropriate form of Student’st-test to investigate the null

hypothesisH 0 :μ 1 =μ 2 (or equivalentlyH 0 :ω= 0). As above, thet-test can be

used to place a confidence interval onω=μ 1 −μ 2.

Suppose that two classes of students take the same mathematics examination and the

following percentage marks are obtained:

Class1:6662345577805560694750

Class2:64907656817270

Assuming that the two sets of examinations marksare drawn from Gaussian distributions

with a common variance, test the hypothesisH 0 :μ 1 =μ 2 at the5%significance level. Use

your result to obtain the95%classical central confidence interval onω=μ 1 −μ 2.

We begin by calculating the mean and standard deviation of each sample. The number of
values in each sample isN 1 =11andN 2 = 7 respectively, and we find

̄x 1 =59. 5 ,s 1 =12.8and ̄x 2 =72. 7 ,s 2 =10. 3 ,

leading tow ̄= ̄x 1 −x ̄ 2 =− 13 .2andσˆ=12.6. Settingω= 0 in (31.119), we thus find
t=− 2 .17.
The rejection region forH 0 is given by (31.114), wheretcritsatisfies

CN 1 +N 2 − 2 (tcrit)=1−α/ 2 , (31.120)