Page 428 - Probability and Statistical Inference

8. Tests of Hypotheses 405

Now since µ > µ , the condition in (8.3.6) can be rephrased as:
1 0

Since E [X] = µ, it does make sense to reject H when is large (> k)
0
µ
because the alternative hypothesis postulates a value µ which is larger than
1
µ . But the MP test given in (8.3.7) is not yet in the implementable form and
0
the test must also be size a. Let us equivalently rewrite the same test as fol-
lows:

where z is the upper 100α% point of the standard normal distribution. See
α
the Figure 8.3.1. The form of the test given in (8.3.7) asks us to reject H for
0
large enough values of while (8.3.8) equivalently asks us to reject H for
0
large enough values of Under H , observe that
0
is a statistic, referred to as the test statistic, which is distributed as a standard
normal random variable.
Here, the critical region .
Now, we have:

Thus, we have the MP level a test by the Neyman-Pearson Lemma. !
Example 8.3.2 (Example 8.3.1 Continued) Suppose that X , ..., X are iid
n
1
N(µ, σ ) with unknown µ ∈ ℜ, but σ ∈ ℜ is known. With preassigned α ∈
2
+
(0, 1) we wish to derive the MP level α test for H : µ = µ versus H : µ = µ 1
1
0
0
where µ < µ and µ , µ are two real numbers.
1 0 0 1

Figure 8.3.2. Standard Normal PDF: Lower 100α% Point