Section 8.5 (p.483)

 

1.         Since p(m|d) is symmetric with respect to m0, we let c1 = m0 - c and c2 = m0+c.

 

5.        

p(0.1|d) = 0.07  Þ  Pr(Z £ 0.25-5c) + Pr(Z ³ 0.25+5c) = 0.07

p(0.2|d) = 0.07  Þ  Pr(Z £ -0.25-5c) + Pr(Z ³ -0.25+5c) = 0.07

Using the normal table, we have 5c=1.87 (try and error), or c1=-0.224, c2=0.524

 

7.         (a) Let q1 < q2, we have

Therefore, fn(x|q) has a MLR in max(X1,…,Xn) and the UMP test is the one that rejects H0 if max(X1,…,Xn) ³ c.

 

            (b) a0 = Pr(max(X1,…,Xn) ³ c | q=3) = 1 – (c/3)n   Þ   c = 3(1-a0)1/n

 

8.         The power function is p(q|d) = Pr(max(X1,…,Xn) ³ c|q), and

where c = q (1-a0)1/n as shown in 7(b).

 

11.       p(q|d) = Pr(max(X1,…,Xn) £ c1|q) + Pr(max(X1,…,Xn) ³ c2|q) = 1 + (c1/q)n - (c2/q)n. We want d to be unbiased, so p(q|d) is minimized when q = 3. Therefore we let c2=3. Also, p(q=3|d) = 0.05 = 1 + (c1/3)n - (3/3)n   Þ   c1 = 3 (0.05)1/n.