378 Maximum Likelihood Methods
Note that under the null hypothesis,H 0 ,thestatistic(2/θ 0 )
∑n
i=1Xihas aχ
2
distribution with 2ndegrees of freedom. Based on this, the following decision rule
results in a levelαtest:
RejectH 0 if (2/θ 0 )
∑n
i=1Xi≤χ
2
1 −α/ 2 (2n)or(2/θ^0 )
∑n
i=1Xi≥χ
2
α/ 2 (2n),(6.3.5)
whereχ^21 −α/ 2 (2n)isthelowerα/2 quantile of aχ^2 distribution with 2ndegrees
of freedom andχ^2 α/ 2 (2n) is the upperα/2 quantile of aχ^2 distribution with 2n
degrees of freedom. Other choices ofc 1 andc 2 can be made, but these are usually
the choices used in practice. Exercise 6.3.2 investigates the power curve for this
test.
g(t)
t
c 1 c 2
c
Figure 6.3.1:Plot for Example 6.3.1, showing that the functiong(t)≤cif and
only ift≤c 1 ort≥c 2.
Example 6.3.2(Likelihood Ratio Test for the Mean of a Normal pdf).Consider
a random sampleX 1 ,X 2 ,...,Xnfrom aN(θ, σ^2 ) distribution where−∞<θ<∞
andσ^2 >0 is known. Consider the hypotheses
H 0 :θ=θ 0 versusH 1 : θ =θ 0 ,
whereθ 0 is specified. The likelihood function is
L(θ)=
(
1
2 πσ^2
)n/ 2
exp
{
−(2σ^2 )−^1
∑n
i=1
(xi−θ)^2
}
=
(
1
2 πσ^2
)n/ 2
exp
{
−(2σ^2 )−^1
∑n
i=1
(xi−x)^2
}
exp{−(2σ^2 )−^1 n(x−θ)^2 }.
Of course, in Ω ={θ:−∞<θ<∞},themleisθ̂=Xand thus
Λ=
L(θ 0 )
L(̂θ)
=exp{−(2σ^2 )−^1 n(X−θ 0 )^2 }.