Chapter 5: Problem 11
Let \(Y_{1}
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Chapter 5: Problem 11
Let \(Y_{1}
These are the key concepts you need to understand to accurately answer the question.
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Let a random sample of size 17 from the normal distribution \(N\left(\mu, \sigma^{2}\right)\) yield \(\bar{x}=4.7\) and \(s^{2}=5.76 .\) Determine a 90 percent confidence interval for \(\mu .\)
Let \(\bar{X}\) be the mean of a random sample from the exponential
distribution, \(\operatorname{Exp}(\theta)\)
(a) Show that \(\bar{X}\) is an unbiased point estimator of \(\theta\).
(b) Using the mgf technique determine the distribution of \(\bar{X}\).
(c) Use (b) to show that \(Y=2 n \bar{X} / \theta\) has a \(\chi^{2}\)
distribution with \(2 n\) degrees of freedom.
(d) Based on Part (c), find a \(95 \%\) confidence interval for \(\theta\) if
\(n=10 .\) Hint: Find \(c\) and \(d\) such that \(P\left(c<\frac{2 n
\bar{X}}{\theta}
Let the observed value of the mean \(\bar{X}\) of a random sample of size 20 from a distribution that is \(N(\mu, 80)\) be \(81.2 .\) Find a 95 percent confidence interval for \(\mu .\)
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(1, \beta)\) distribution. (a) Show that the confidence interval \(\left(2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(1-(\alpha / 2))}, 2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(\alpha / 2)}\right)\) is an exact \((1-\alpha) 100 \%\) confidence interval for \(\beta\). (b) Show that value of a \(90 \%\) confidence interval for the data of the example is \((64.99,136.69)\)
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 \(H_{0}: p=0.40\) will be 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) Determine \(\alpha=P(Y \geq 13 ; ; p=0.40)\). (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 \(p=0.60\).
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