Let \(X\) and \(Y\) be two independent random samples of sizes \(n_{X}\) and \(n_{y}\), respectively, from two normal population distributions that do not necessarily have the same means or variances. The two distributions refer to the miles per gallon achieved by two \(1 / 2\)-ton pickup trucks produced by two rival Detroit manufacturers. Define \(\hat{\sigma}\) as in Theorem 6.19.

(a) Show that the random variable \(F=\left(\hat{\sigma}_{X}^{2} / \sigma_{X}^{2}ight) /\left(\hat{\sigma}_{Y}^{2} / \sigma_{Y}^{2}ight)\) has the \(F\)-distribution with \(\left(n_{X}-1ight)\) numerator and \(\left(n_{y}-1ight)\) denominator degrees of freedom.

(b) Let \(n_{x}=21\) and \(n_{y}=31\). What is the probability that the random interval \(\left(.49\left(\hat{\sigma}_{Y}^{2} / \hat{\sigma}_{X}^{2}ight), 1.93\left(\hat{\sigma}_{Y}^{2} / \hat{\sigma}_{X}^{2}ight)ight)\) will have an outcome that will contain the value of the ratio of the variances \(\sigma_{Y}{ }^{2} / \sigma_{X}{ }^{2}\) ? (This random interval is another example of a confidence interval - in this case for the ratio of the population variances \(\sigma_{Y}{ }^{2} / \sigma_{X}{ }^{2}\).)

(c) Suppose that \(s_{X}{ }^{2}=.25\) and \(s_{y}{ }^{2}=.04\). Define a confidence interval that is designed to have a .98 probability of generating an outcome that contains the value of the ratio of population variances associated with the miles per gallon achieved by the two pickup trucks. Generate a confidence interval outcome for the ratio of variances.