Note for 9.28: the question doesn't ask for it, but you should also know how to find the value of c when the size of the test is specified.
Note for 9.34: just simplify the form of the test; you don't need to find an explicit value or function for c.

Here are the problems written out, in case you don't have the text:

• 8.17: Let X_1, ..., X_n be a random sample from each of the following distributions. In each case, find the MLE theta-hat, var(theta-hat), 1/[ n*I(theta) ], where I(theta) is the Fisher information of a single observation X, and compare var(theta-hat) and 1/[ n*I(theta) ].
  (b) N(theta, 1), -infinity < theta < infinity.
  (d) Gamma(alpha=5, beta=theta), 0 < theta < infinity.
• 8.18: Referring to exercise 8.17 and using the fact that theta-hat has an approximate N( theta, 1/[ n*I(theta) ] ) distribution, in each case construct an approximate 95 percent confidence interval for theta.
• 9.28: Let X_1, ..., X_n be a random sample from the normal distribution N(theta, 1). Show that the likelihood ratio principle for testing H_0: theta=theta' where theta' is specified, against H_1: theta <> theta' (that's a "not equal to") leads to the inequality |x-bar - theta'| >= c. Is this a uniformly most powerful test of H_0 against H_1? (Note: the question doesn't ask for it, but you should also know how to find the value of c when the size of the test is specified.)
• 9.34: A random sample X_1, ..., X_n arises from a distribution given by:
  H_0: f(x|theta) = 1/theta, 0 < x < theta, zero elsewhere,
  or
  H_1: f(x|theta) = (1/theta) * exp(-x/theta), 0 < x < infinity, zero elsewhere.
  Determine the likelihood ratio test associated with the test of H_0 against H_1.
  (Note: just simplify the form of the test; you don't need to find an explicit value or function for c.)