Section 7.5 (p.401)

 

1.         Based on the discussion on p. 399, the confidence interval is , where c is a constant such that Pr (-c < U < c) = g, U has a t distribution with n-1 = 7 degrees of freedom. = 3.0625, = 0.5125.

 

            (a) g = 0.90 ̃ c = 1.895 ̃ CI = (2.719, 3.406)

            (b) g = 0.95 ̃ c = 2.365 ̃ CI = (2.634, 3.491)

            (c) g = 0.99 ̃ c = 3.499 ̃ CI = (2.428, 3.687)

 

2.         L = 2cs¢/n1/2 ̃ L2 = 4c2s¢2/n ̃ E(L2) = 4c2/n E(s¢2).

Since  has a c2 distribution with n-1 degrees of freedom, E(Y) = n-1. Therefore . So we get E(L2) = 4c2s2/n.

 

            (a) c = 2.776, E(L2) = 6.16 s2.

            (b) c = 2.262, E(L2) = 2.05 s2.

            (c) c = 2.045, E(L2) = 0.56 s2.

            (d) c = 1.895, E(L2) = 1.80 s2.

            (e) c = 2.365, E(L2) = 2.80 s2.

            (f) c = 3.449, E(L2) = 6.12 s2.

 

3.         Since s2 is known,  has a standard normal distribution. We want , or . Now c = 1.96, so the length of confidence interval is .  £ 0.01s ̃ n ³ 153664.

 

5.         Since an exponential distribution with mean m is the same as a gamma distribution with a = 1 and b = 1/m. Therefore, S Xi has a gamma distribution with a = n and b = 1/m (Theorem 3 of Sec.5.9). It can be shown that (1/m) S Xi has a gamma distribution with a = n and b = 1 (See Exercise 1 in Sec. 5.9. If you don’t know how to solve it, review Sec. 3.8). Now the distribution of (1/m) S Xi is known (both a and b are known), we can get c1 and c2 such that Pr (c1 < (1/m) S Xi < c2) = g for a desired confidence coefficient. Rewrite the expression of the probability, we have Pr ((1/c2) S Xi < m < (1/c1) S Xi) = g. That is, the confidence interval is ((1/c2) S Xi,  (1/c1) S Xi).