Let X1,..., Xn be iid exponential A).
a. Find a UMP size a hypothesis test of H0: λ = λ0 versus H1: λ b. Find a UMA 1 - α confidence interval based on inverting the test in part (a). Show that the interval can be expressed as

c. Find the expected length of C*(x1,... ,xn).
d. Madansky (1962) exhibited a 1 - α interval whose expected length is shorter than that of the UMA interval. In general, Madansky's interval is difficult to calculate, but in the following situation calculation is relatively simple. Let 1 - α = .3 and n = 120. Madansky's interval is

which is a 30% confidence interval. Use the fact that X2240,.7 = 251.046 to show that the 30% UMA interval satisfies
E [Length (C*(x1, ...,xn))] = .956λ > E [Length (CM(x1,..., xn))] = .829A.

log(99)