Section 8.4 (p.473)

1. Let l₁ < l₂, increases in S X_i.

2. Let Y = S (X_i-m)² and s₁² < s₂². increases in Y.

3. Let a₁ < a₂, increases in P X_i.

5. Let q₁ < q₂, increases in S d(X_i)

6. Since the probability of rejecting H₀ is always 0.05, the power function is 0.05 for every q.

7. The joint pdf of X₁,…, X_n has a MLR in SX_i² (Exercise 2). Therefore by Theorem 1 the test that rejects H₀ if SX_i² ³ c is a UMP test. It is possible for any a₀ because SX_i² is continuous and the constant c con be derived from the equation Pr(SX_i² ³ c | s²=2) = a₀.

9. The joint pdf of X₁,…, X_n has a MLR in SX_i and the UMP test rejects H₀ when SX_i ³ c (Example 3). Note that SX_i has a Binomial(n, p) distribution. When a₀ = 0.0577 = Pr(SX_i ³ c | p=1/2), it can be shown that c = 14 (use table on p.684). When a₀ = 0.0207, c = 15.

10. The joint pdf of X₁,…, X_n has a MLR in SX_i (Exercise 1). Therefore by Theorem 1 the test that rejects H₀ if SX_i² ³ c is a UMP test. To determine c, we have a₀ = 0.0143 = Pr(SX_i ³ c | l=1). Since SX_i has a Poisson distribution with mean 10, we have c = 18 (use table on p. 688).

14. The joint pdf of X₁,…, X_n has a MLR in because

increases in if b₂>b₁.

Now we apply the results from Exercise 11, which can be proved in the same manner as the textbook proves Theorem 1. Therefore, the test that rejects H₀ if is a UMP test. Or equivalently, rejects H₀ if .

17. The test that rejects H₀ if is a UMP test (Example 4). Given a₀=0.025, we have

(a) . Since p(m|d^*) is increasing in m, we want to find the smallest n for which p(m=0.5|d^*) ³ 0.9. That is

(b)