Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.7 HYPOTHESIS TESTING

0

0

0. 1

0. 2

0. 3

0. 4

0. 5

− 4 − 3 − 2 − 1123 4

t

P(t|H 0 )

N=2

N=3

N=5

N=10

Figure 31.11 Student’st-distribution for various values ofN. The broken

curve shows the standard Gaussian distribution for comparison.

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

distribution with unknown meanμand unknown 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 :μ=0at the10%significance level.

For our null hypothesis,μ 0 = 0. Since for this sample ̄x=1.11,s=1.01 andN= 10, it
follows from (31.113) that

t=

̄x

s/

√

N− 1

=3. 33.

The rejection region fortis given by (31.114) wheretcritis such that

CN− 1 (tcrit)=1−α/ 2 ,

andαis the required significance of the test. In our caseα=0.1andN= 10, and from
table 31.3 we findtcrit=1.83. Thus our rejection region forH 0 at the 10% significance
level is

t<− 1. 83 and t> 1. 83.

For our samplet=3.30 and so we can clearly reject the null hypothesisH 0 :μ=0atthis
level.

It is worth noting the connection between thet-test and the classical confidence

interval on the meanμ. The central confidence interval onμat the confidence

level 1−αis the set of values for which

−tcrit<

̄x−μ

s/

√

N− 1

<tcrit,