Chapter 5: Problem 22
Let \(Y_{1}
Chapter 5: Problem 22
Let \(Y_{1}
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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 \(f(x)=\frac{1}{6}, x=1,2,3,4,5,6\), zero elsewhere, be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is $$ g_{1}\left(y_{1}\right)=\left(\frac{7-y_{1}}{6}\right)^{5}-\left(\frac{6-y_{1}}{6}\right)^{5}, \quad y_{1}=1,2, \ldots, 6 $$ zero elsewhere. Note that in this exercise the random sample is from a distribution of the discrete type. All formulas in the text were derived under the assumption that the random sample is from a distribution of the continuous type and are not applicable. Why?
A Weibull distribution with pdf
$$
f(x)=\left\\{\begin{array}{ll}
\frac{1}{\theta^{3}} 3 x^{2} e^{-x^{3} / \theta^{3}} & 0
Consider the sample of data: \(\begin{array}{rrrrrrrrrrr}13 & 5 & 202 & 15 & 99 & 4 & 67 & 83 & 36 & 11 & 301 \\ 23 & 213 & 40 & 66 & 106 & 78 & 69 & 166 & 84 & 64 & \end{array}\) (a) Obtain the five number summary of these data. (b) Determine if there are any outliers. (c) Boxplot the data. Comment on the plot. (d) Obtain a \(92 \%\) confidence interval for the median \(\xi_{1 / 2}\).
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