510 Optimal Tests of Hypothesesor, equivalently,
R(θ′′,C)≤R(θ′′,A).That is,
R(θ′,C)=R(θ′′,C)≤R(θ′′,A).
This means that
max[R(θ′,C),R(θ′′,C)]≤R(θ′′,A).
Then certainly,
max[R(θ′,C),R(θ′′,C)]≤max[R(θ′,A),R(θ′′,A)],and the critical regionCprovides a minimax solution, as we wanted to show.
Example 8.5.1. LetX 1 ,X 2 ,...,X 100 denote a random sample of size 100 from
a distribution that isN(θ,100). We again consider the problem of testingH 0 :
θ= 75 againstH 1 :θ= 78. We seek a minimax solution withL(75,78) = 3 and
L(78,75) = 1. SinceL(75)/L(78)≤kis equivalent tox≥c, we want to determine
c, and thusk,sothat
3 P(X≥c;θ= 75) =P(X<c;θ= 78). (8.5.2)BecauseXisN(θ,1), the preceding equation can be rewritten as3[1−Φ(c−75)] = Φ(c−78).As requested in Exercise 8.5.4, the reader can show by using Newton’s algorithm
that the solution to one place isc=76.8. The significance level of the test is
1 −Φ(1.8) = 0.036, approximately, and the power of the test whenH 1 is true is
1 −Φ(− 1 .2) = 0.885, approximately.
8.5.2 Classification
The summary above has an interesting application to the problem ofclassification,
which can be described as follows. An investigator makes a number of measurements
on an item and wants to place it into one of several categories (or classify it).
For convenience in our discussion, we assume that only two measurements, say
XandY, are made on the item to be classified. Moreover, letX andY have
ajointpdff(x, y;θ), where the parameterθrepresents one or more parameters.
In our simplification, suppose that there are only two possible joint distributions
(categories) forXandY, which are indexed by the parameter valuesθ′andθ′′,
respectively. In this case, the problem then reduces to one of observingX=xand
Y =yand then testing the hypothesisθ=θ′against the hypothesisθ=θ′′,with
the classification ofX andY being in accord with which hypothesis is accepted.
From the Neyman–Pearson theorem, we know that a best decision of this sort is of
the following form: If
f(x, y;θ′)
f(x, y;θ′′)
≤k,