Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

8.1. Most Powerful Tests 475

Example 8.1.2.LetX′=(X 1 ,...,Xn) denote a random sample from the distri-
bution that has the pdf

f(x;θ)=

1

√

2 π

exp

(

−

(x−θ)^2

2

)

, −∞<x<∞.

It is desired to test the simple hypothesisH 0 :θ=θ′= 0 against the alternative
simple hypothesisH 1 :θ=θ′′=1. Now

L(θ′;x)

L(θ′′;x)

=

(1/

√

2 π)nexp

[

−

∑n

1

x^2 i/2

]

(1/

√

2 π)nexp

[

−

∑n

1

(xi−1)^2

/

2

]

=exp

(

−

∑n

1

xi+

n

2

)

.

Ifk>0, the set of all points (x 1 ,x 2 ,...,xn) such that

exp

(

−

∑n

1

xi+

n

2

)

≤k

is a best critical region. This inequality holds if and only if

−

∑n

1

xi+

n

2

≤logk

or, equivalently,

∑n

1

xi≥

n

2

−logk=c.

In this case, a best critical region is the setC={(x 1 ,x 2 ,...,xn):

∑n
1 xi≥c},
wherecis a constant that can be determined so that the size of the critical region
is a desired number α. The event

∑n
1 Xi≥c is equivalent to the eventX ≥
c/n=c 1 , for example, so the test may be based upon the statisticX.IfH 0 is
true, that is,θ=θ′=0,thenXhas a distribution that isN(0, 1 /n). Given the
significance levelα,thenumberc 1 is computed in R asc 1 =qnorm(1−α, 0 , 1 /

√
n);
hence,PH 0 (X≥c 1 )=α. So, if the experimental values ofX 1 ,X 2 ,...,Xnwere,
respectively,x 1 ,x 2 ,...,xn, we would computex=

∑n
1 xi/n.Ifx≥c^1 ,thesimple
hypothesisH 0 :θ=θ′= 0 would be rejected at the significance levelα;ifx<c 1 ,
the hypothesisH 0 would be accepted. The probability of rejectingH 0 whenH 0 is
true isαthe level of significance. The probability of rejectingH 0 ,whenH 0 is false,
is the value of the power of the test atθ=θ′′=1,whichis,

PH 1 (X≥c 1 )=

∫∞

c 1

1

√

2 π

√

1 /n

exp

[

−

(x−1)^2

2(1/n)

]

dx. (8.1.7)