Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.7 HYPOTHESIS TESTING

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

distribution with standard deviationσ=1. 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.

We must test the (simple) null hypothesisH 0 :μ= 0 against the (composite) alternative
hypothesisH 1 :μ= 0. Thus, the subspaceSis the single pointμ=0,whereasAis the
entireμ-axis. The likelihood function is

L(x;μ)=

1

(2π)N/^2

exp

[

−^12

∑

i(xi−μ)

2 ],

which has its global maximum atμ= ̄x. The test statistictis then given by

t(x)=

L(x;0)

L(x; ̄x)

=

exp

[

−^12

∑

ix

(^2) i]
exp

[

−^12

∑

i(xi− ̄x)

2 ]=exp

(

−^12 Nx ̄^2

)

.

It is in fact more convenient to consider the test statistic

v=−2lnt=N ̄x^2.

Since−2lntis a monotonically decreasing function oft, the rejection region now becomes
v>vcrit,where
∫∞

vcrit

P(v|H 0 )dv=α, (31.111)

αbeing the significance level of the test. Thus it only remains to determine the sampling
distributionP(v|H 0 ). Under the null hypothesisH 0 ,weexpect ̄xto be Gaussian distributed,
with mean zero and variance 1/N. Thus, from subsection 30.9.4,vwill follow achi-squared
distribution of order 1. Substituting the appropriate form forP(v|H 0 ) in (31.111) and setting
α=0.1, we find by numerical integration (or from table 31.2) thatvcrit=N ̄x^2 crit=2.71.
SinceN= 10, the rejection region on ̄xat the 10% significance level is thus

̄x<− 0. 52 and ̄x> 0. 52.

As noted before, for this samplex ̄=1.11, and so we may reject the null hypothesis
H 0 :μ= 0 at the 10% significance level.

The above example illustrates the general situation that if the maximum-

likelihood estimatesaˆof the parameters fall in or near the subspaceSthen the

sample will be considered consistent withH 0 and the value oftwill be near

unity. Ifaˆis distant fromSthen the sample will not be in accord withH 0 and

ordinarilytwill have a small (positive) value.

It is clear that in order to prescribe the rejection region fort,orforarelated

statisticuorv, it is necessary to know the sampling distributionP(t|H 0 ). IfH 0

is simple then one can in principle determineP(t|H 0 ), although this may prove

difficult in practice. Moreover, ifH 0 is composite, then it may not be possible

to obtainP(t|H 0 ), even in principle. Nevertheless, a useful approximate form for

P(t|H 0 ) exists in the large-sample limit. Consider the null hypothesis

H 0 :(a 1 =a^01 ,a 2 =a^02 ,...,aR=a^0 R), whereR≤M

and thea^0 i are fixed numbers. (In fact, we may fix the values of any subset