應用數理統計第三次作業

18大數據 LD 180*******511

23.

$\begin{alignedat}{4} \bar{\xi} \sim N(\alpha,\frac{\sigma^2}{n}) \\ (\xi_{n+1}-\bar{\xi})\sim N(0,\frac{(n-1)\sigma^2}{n})\\ \sqrt{\frac{n}{(n-1)\sigma}}(\xi_{n+1}-\bar{\xi}) \sim N(0,1)\\ \frac{(n+1)S^2}{\sigma^2} \sim \chi^2(n)\\ \end{alignedat}$
$化簡後得：\frac{(\xi_{n+1}-\bar{\xi})}{S} \sqrt{\frac{n-1}{n+1}} \sim t(n-1)$

24.

$\begin{alignedat}{7} \xi_{i} \sim N(a,\sigma_i^2) \\ \sum_{i=1}^n\frac{\xi_i}{\sigma_i}/\sum_i^n\frac{1}{\sigma_i} \sim N(a,\frac{n}{\left(\sum \frac{1}{\sigma_i}\right)^2})\\ \eta \sim N(a,\frac{n}{\left(\sum \frac{1}{\sigma_i}\right)^2})\\ \zeta_i=\frac{\xi_i-a}{\sigma_i} \sim N(0,1)\\ \frac{\eta -a}{n}\sum\frac{1}{\sigma} \sim N(0,1/n)\\ \bar{\zeta} \sim N(0,1/n)\\ \zeta \sim \sum_{i=1}^n(\zeta_i - \bar{\zeta})^2/1 \sim\chi^2(n-1)\\ \end{alignedat}$

33.

$\begin{alignedat}{3} & P \{\sum_{i=1}^{10} (\xi_i -\bar{\xi})^2 < 0.18\}\\ &=P \{\frac{\sum_{i=1}^{10} (\xi_i -\bar{\xi})^2}{0.3^2} < 2\}\\ &=P\{\chi^2(9)<2\}\\ &=0.01\\ \end{alignedat}$

$\begin{alignedat}{} &P\{\frac{\bar{\xi}}{\tilde{S}}<0.9\}\\ &=P\{\frac{\bar{\xi} \times \sqrt{10}/0.3}{\sqrt{{\frac{n-1}{n}}\times \tilde{S}^2}}\times \sqrt{\frac{n-1}{n}}<0.9\}\\ &=P\{t(9)<0.9\sqrt{9}\}\\ &=0.9932 \end{alignedat}$
筆者親自證明，血的教訓：千萬別用Katex寫數學作業，
時間複雜度O(t^1000000)
附贈作業答案：https://wenku.baidu.com/view/e9bc536366ec102de2bd960590c69ec3d4bbdb67.html?rec_flag=default&sxts=1584862071463