Section 8.2 (p.452)
4. (a) Here we want to minimize a(d) given b(d) £
0.05. As argued in the proof of the Neyman-Pearson lemma, the test procedure is
also one such that reject H0 if f0(x) < k f1(x).
That is, if 
(see Example 1 for
details).
Since we want b(d)
= 0.05, it follows that
Therefore k = 5 - 1.645 n-1/2.
That is, we reject H0 if 
(b) When n=4, we reject H0 if 
. Therefore,
6. (a) By N-P lemma, the test procedure rejects H0
if f0 < k f1. That is
Therefore, we have S (Xi-m)2 > c for a constant c.
(b) 
where Y has a c2
distribution with 8 d.f. Therefore c/2 = 15.51, c=31.02.
7. (a) Let the test be: Reject H0 if X > 1.5.
Then a(d)
= Pr(X > 1.5|q = 1) = 0 and b(d) = Pr(X £1.5
|q = 2) = 1.5/2 < 1.
(b) According to N-P lemma, the test rejects H0
if f0(x) < k f1(x)
If x < 0, f0(x)
= 0 and f1(x) = 0
If 0 £ x £
1, f0(x) = 1 and f1(x) = 1/2
If 1 < x £ 2, f0(x) = 0 and f1(x) =
1/2
If x > 2, f0(x)
= 0 and f1(x) = 0
Therefore, we have the
following 3 regions:
(1) X < 0 or X > 2:
f0(x) = k f1(x) for all k > 0. Therefore it can be
chosen to be either critical region or not and doesn’t matter.
(2) 1 < x £ 2: f0(x) < k f1(x) for
all k > 0. Therefore it is always the critical region.
(3) 0£ x £
1: f0(x) < k f1(x) for k > 2. Therefore it may or
may not be the critical region depend on the choice of k.
Now we want a = 0 = Pr(reject H0|q=1). Note that we cannot let k > 2 because in
that case the region (3) is part of the rejection region and therefore a = Pr(0 < X £ 1|q=1) = 1 ¹
0. That is, region (3) can not be in the critical region. In summary, the desired
test is one such that reject H0 if 1 < X £ 2, or simpler if X > 1.
9. (a) and (b) By Theorem 1, the test rejects H0 if
f0(x) < f1(x). That is,
Therefore, the test
rejects H0 if 
where 
(c) The test rejects H0 if 
where SXi has a poison distribution with mean
20l. Also that 
. Therefore
a(d)
= Pr(Poisson(l=20*1/4) ³
8) = .1333
b(d)
= Pr(Poisson(l=20*1/2) £
7) = .2203
Ž a(d) + b(d) = .3536