Let \(X\) be a random sample of size \(n\) from a \(N\left(\mu, \sigma^{2}ight)\) population distribution representing the weights, in ounces, of cereal placed in cereal boxes for a certain brand and type of breakfast cereal. Define \(\hat{\sigma}\) as in Theorem 6.19.

(a) Show that the random variable \(T=n^{1 / 2}(\bar{X}-\mu) / \hat{\sigma}\) has the \(t\)-distribution with \(n\) - 1 degrees of freedom.

(b) Let \(n=25\). What is the probability that the random interval \(\left(\bar{X}-2.06 \hat{\sigma} / n^{1 / 2}, \bar{X}+2.06 \hat{\sigma} / n^{1 / 2}ight.\) ) will have an outcome that contains the value of \(\mu\) ? (This random interval is an example of a confidence interval in this case for the population mean \(\mu\). See Section 10.6.)

(c) Suppose that \(\bar{x}=16.3\) and \(s^{2}=.01\). Define a confidence interval that is designed to have a .90 probability of generating an outcome that contains the value of the population mean weight of cereal placed in the cereal boxes. Generate a confidence interval outcome for the mean weight.