Chapter 4

In Exercise 4.2.11, the sampling was from the \(N(0,1)\) distribution. Show, however, that setting \(\mu=0\) and \(\sigma=1\) is without loss of generality. Hint: First, \(X_{1}, \ldots, X_{n}\) is a random sample from the \(N\left(\mu, \sigma^{2}\right)\) if and only if \(Z_{1}, \ldots, Z_{n}\) is a random sample from the \(N(0,1)\), where \(Z_{i}=\left(X_{i}-\mu\right) / \sigma\). Then show the confidence interval based on the \(Z_{i}\) 's contains 0 if and only if the confidence interval based on the \(X_{i}\) 's contains \(\mu\).

Let \(p\) denote the probability that, for a particular tennis player, the first serve is good. Since \(p=0.40\), this player decided to take lessons in order to increase \(p\). When the lessons are completed, the hypothesis \(H_{0}: p=0.40\) is tested against \(H_{1}: p>0.40\) based on \(n=25\) trials. Let \(Y\) equal the number of first serves that are good, and let the critical region be defined by \(C=\\{Y: Y \geq 13\\}\). (a) Show that \(\alpha\) is computed by \(\alpha=1\) -pbinom \((12,25, .4)\). (b) Find \(\beta=P(Y<13)\) when \(p=0.60\); that is, \(\beta=P(Y \leq 12 ; p=0.60)\) so that \(1-\beta\) is the power at \(p=0.60\).

Suppose a random sample of size 2 is obtained from a distribution that has pdf \(f(x)=2(1-x), 0

For the proof of Theorem 4.8.1, we assumed that the cdf was strictly increasing over its support. Consider a random variable \(X\) with cdf \(F(x)\) that is not strictly increasing. Define as the inverse of \(F(x)\) the function $$ F^{-1}(u)=\inf \\{x: F(x) \geq u\\}, \quad 0

Let \(Y_{1}

Let \(\bar{x}\) be the observed mean of a random sample of size \(n\) from a distribution having mean \(\mu\) and known variance \(\sigma^{2}\). Find \(n\) so that \(\bar{x}-\sigma / 4\) to \(\bar{x}+\sigma / 4\) is an approximate \(95 \%\) confidence interval for \(\mu\).

Let \(Y_{1}

Assume that \(Y_{1}\) has a \(\Gamma(\alpha+1,1)\) -distribution, \(Y_{2}\) has a uniform \((0,1)\) distribution, and \(Y_{1}\) and \(Y_{2}\) are independent. Consider the transformation \(X_{1}=\) \(Y_{1} Y_{2}^{1 / \alpha}\) and \(X_{2}=Y_{2}\) (a) Show that the inverse transformation is: \(y_{1}=x_{1} / x_{2}^{1 / \alpha}\) and \(y_{2}=x_{2}\) with support \(0

Let \(Y_{1}0\), provided that \(x \geq 0\), and \(f(x)=0\) elsewhere. Show that the independence of \(Z_{1}=Y_{1}\) and \(Z_{2}=Y_{2}-Y_{1}\) characterizes the gamma pdf \(f(x)\), which has parameters \(\alpha=1\) and \(\beta>0 .\) That is, show that \(Y_{1}\) and \(Y_{2}\) are independent if and only if \(f(x)\) is the pdf of a \(\Gamma(1, \beta)\) distribution. Hint: Use the change-of-variable technique to find the joint pdf of \(Z_{1}\) and \(Z_{2}\) from that of \(Y_{1}\) and \(Y_{2}\). Accept the fact that the functional equation \(h(0) h(x+y) \equiv\) \(h(x) h(y)\) has the solution \(h(x)=c_{1} e^{c_{2} x}\), where \(c_{1}\) and \(c_{2}\) are constants.

Assume a binomial model for a certain random variable. If we desire a \(90 \%\) confidence interval for \(p\) that is at most \(0.02\) in length, find \(n\). Hint: Note that \(\sqrt{(y / n)(1-y / n)} \leq \sqrt{\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)}\).