Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

504 Optimal Tests of Hypotheses

provided thatβis not equal to 0 or 1.

Now letαaandβabe preassigned proper fractions; some typical values in the

applications are 0.01, 0.05, and 0.10. If we take

k 0 =

αa

1 −βa

,k 1 =

1 −αa

βa

,

then inequalities (8.4.5) become

α

1 −β

≤

αa

1 −βa

,

1 −αa

βa

≤

1 −α

β

; (8.4.6)

or, equivalently,

α(1−βa)≤(1−β)αa,β(1−αa)≤(1−α)βa.

If we add corresponding members of the immediately preceding inequalities, we find
that
α+β−αβa−βαa≤αa+βa−βαa−αβa

and hence
α+β≤αa+βa;
that is, the sumα+βof the probabilities of the two kinds of errors is bounded
above by the sumαa+βaof the preassigned numbers. Moreover, sinceαandβare
positive proper fractions, inequalities (8.4.6) imply that

α≤

αa

1 −βa

,β≤

βa

1 −αa

;

consequently, we have an upper bound on each ofαandβ. Various investigations
of the sequential probability ratio test seem to indicate that in most practical cases,
the values ofαandβare quite close toαaandβa. This prompts us to approximate
the power function at the pointsθ=θ′andθ=θ′′byαaand 1−βa, respectively.

Example 8.4.2.LetX beN(θ,100). To find the sequential probability ratio
test for testingH 0 :θ= 75 againstH 1 :θ= 78 such that each ofαandβis
approximately equal to 0.10, take

k 0 =

0. 10

1 − 0. 10

=

1

9

,k 1 =

1 − 0. 10

0. 10

=9.

Since

L(75,n)

L(78,n)

=

exp

[

−

∑

(xi−75)^2 /2(100)

]

exp [−

∑

(xi−78)^2 /2(100)]

=exp

(

−

6

∑

xi− 459 n

200

)

,

the inequality

k 0 =

1

9

<

L(75,n)

L(78,n)

<9=k 1