Page 441 - Probability and Statistical Inference

418 8. Tests of Hypotheses

α test for H versus H .
0 1
How can one prove this claim? One may argue as follows.
Suppose that Q(θ) is the power function of the MP level α test between θ 0
and some fixed θ (> θ ). Suppose that there exists a level θ test with its power
1
0
function Q*(θ) for choosing between θ and some fixed θ* (> θ ) such that
0
0
Q*(θ*) > Q(θ*). But, we are working under the assumption that the MP level
α test between θ and θ is also the MP level α test between θ and θ*, that is
1
0
0
Q(θ*) is the maximum among all level α tests between θ and θ*. This leads
0
to a contradiction.
The case of a lower-sided composite alternative is handled analogously.
Some examples follow.
Example 8.4.1 (Example 8.3.1 Continued) Suppose that X , ..., X are iid
1
n
2
+
N(µ, σ ) with unknown µ ∈ ℜ, but assume that σ ∈ ℜ is known. With
preassigned α ∈ (0, 1), we wish to obtain the UMP level a test for H : µ = µ 0
0
versus H : µ > µ where µ is a real number. Now, fix a value µ (> µ ) and
0
0
0
1
1
then from (8.3.8) recall that the MP level a test between µ and arbitrarily
0
chosen µ will have the following form:
1
where z is the upper 100α% point of the standard normal distribution. See
α
the Figure 8.3.1. Obviously this test does not depend on the specific choice of
µ (> µ ). Hence, the test given by (8.4.1) is UMP level α for testing H : µ =
1 0 0
µ versus H : µ > µ . !
0 1 0

Figure 8.4.1. Chi-Square Lower 100a% or Upper 100(1 - α)%
Point with Degrees of Freedom n

Example 8.4.2 Let X , ..., X be iid N(0, σ ) with unknown σ ∈ ℜ .
2
+
n
1
With preassigned α ∈ (0, 1), we wish to obtain the UMP level α test for
H : σ = σ versus H : σ < σ where σ is a positive number. Now, fix a
0 0 1 0 0