231 lines
12 KiB
TeX
231 lines
12 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\setcounter{chapter}{7}
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\section{假设检验的基本思想与概念}
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\begin{enumerate}
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\questionandanswer[1]{
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设$x_1,x_2, \cdots ,x_n$是来自$N(\mu,1)$的样本，考虑如下假设检验问题
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$$
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H_0: \mu=2 \quad \mathrm{vs}\quad H_1:\mu=3,
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$$
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若检验由拒绝域为$W=\{ \bar{x}\geqslant 2.6 \}$确定。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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当$n=20$时求检验犯两类错误的概率；
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}{
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第一类错误：$\alpha=P(\bar{x}\geqslant 2.6|H_0)$，当$H_0$成立即$\mu=2$时$\bar{x}\sim N\left(2,\frac{1}{20}\right)$，所以
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$$
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\alpha=P\left( \frac{\bar{x}-2}{\sqrt{\frac{1}{20}}}\geqslant \frac{2.6-2}{\sqrt{\frac{1}{20}}} \right) = 1-\Phi\left( \frac{2.6-2}{\sqrt{\frac{1}{20}}} \right) = 0.0036452
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$$
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\begin{center}
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\includegraphics[width=0.3\linewidth]{imgs/2024-05-27-16-38-24.png}
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\end{center}
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第二类错误：$\beta=P(\bar{x}<2.6|H_1)$，当$H_1$成立即$\mu=3$时$\bar{x}\sim N\left( 3,\frac{1}{20} \right) $，所以
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$$
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\beta=P\left( \frac{\bar{x}-3}{\sqrt{\frac{1}{20}}}<\frac{2.6-3}{\sqrt{\frac{1}{20}}} \right) =\Phi\left( \frac{2.6-3}{\sqrt{\frac{1}{20}}} \right) =0.036819
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$$
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\begin{center}
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\includegraphics[width=0.3\linewidth]{imgs/2024-05-27-16-41-54.png}
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\end{center}
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}
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\questionandanswerSolution[]{
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如果要使得检验犯第二类错误的概率$\beta\leqslant 0.01$，$n$最小应取多少？
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}{
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$$
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\beta=P(\bar{x}<2.6|H_1)=P\left( \frac{\bar{x}-3}{\sqrt{\frac{1}{n}}}<\frac{2.6-3}{\sqrt{\frac{1}{n}}} \right) \leqslant 0.01
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$$
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即$\Phi\left( \frac{-0.4}{\sqrt{\frac{1}{n}}} \right) \leqslant 0.01$，即$\Phi\left( \frac{0.4}{\sqrt{\frac{1}{n}}} \right) \geqslant 0.99$，即$\frac{0.4}{\sqrt{\frac{1}{n}}}\geqslant 2.33 $，解得$ n \geqslant 33.930625$，所以$n$最小应取34.
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}
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\questionandanswerProof[]{
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证明：当$n \to \infty$时，$\alpha \to 0, \beta \to 0$。
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}{
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$$
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\alpha=P\left( \frac{\bar{x}-2}{\sqrt{\frac{1}{n}}}\geqslant \frac{2.6-2}{\sqrt{\frac{1}{n}}} \right) =1-\Phi\left( \frac{0.6}{\sqrt{\frac{1}{n}}} \right) \xrightarrow{n \to \infty} 0
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$$
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$$
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\beta=P\left( \frac{\bar{x}-3}{\sqrt{\frac{1}{n}}}<\frac{2.6-3}{\sqrt{\frac{1}{n}}} \right) =\Phi \left( \frac{-0.4}{\sqrt{\frac{1}{n}}} \right) \xrightarrow{n \to \infty}0
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$$
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}
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\end{enumerate}
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\questionandanswerSolution[3]{
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设$x_1,x_2, \cdots ,x_{16}$是来自正态总体$N(\mu,4)$的样本，考虑检验问题
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$$
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H_0:\mu=6\quad \mathrm{vs}\quad H_1:\mu\neq 6
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$$
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拒绝域取为$W=\{ \left\vert \bar{x}-6 \right\vert \geqslant c \}$，试求$c$使得检验的显著性水平为$0.05$，并求该检验在$\mu=6.5$处犯第二类错误的概率。
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}{
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当$H_0$成立即$\mu=6$时，$\bar{x}\sim N(\mu, \frac{4}{16})=N(6, \frac{1}{4})$，所以
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$$
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p = P\left( \left\vert \bar{x}-6 \right\vert \geqslant c | \mu=6 \right) =P\left( \frac{\left\vert \bar{x}-6 \right\vert }{\frac{1}{2}} \geqslant 2c \middle| \mu=6\right) =2(1-\Phi(2c)) = 0.05
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$$
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则$\Phi(2c)=0.975$，所以$2c=1.96$，从而$\bm{c=0.98}$。
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当$\mu=6.5$时，$\bar{x} \sim N(6.5, 0.25)$，所以该检验在$\mu=6.5$处犯第二类错误的概率为
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$$
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\begin{aligned}
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\bm{\beta} &= P\left( \left\vert \bar{x}-6 \right\vert < c \ \middle|\ \mu=6.5 \right) = P\left( 5.02<\bar{x}<6.98 \right) \\
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&=P\left( \frac{5.02-6.5}{0.5}<\bar{x}<\frac{6.98-6.5}{0.5} \right) =\Phi\left( \frac{6.98-6.5}{0.5} \right) - \Phi\left( \frac{5.02-6.5}{0.5} \right) \bm{ = 0.8299317} \\
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\end{aligned}
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$$
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\begin{center}
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\includegraphics[width=0.3\linewidth]{imgs/2024-05-29-14-19-20.png}
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\end{center}
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}
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\questionandanswerSolution[4]{
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设总体为均匀分布$U(0, \theta)$，$x_1,x_2, \cdots ,x_n$是样本，考虑检验问题
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$$
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H_0: \theta\geqslant 3 \quad\mathrm{vs}\quad H_1:\theta<3,
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$$
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拒绝域取为$W=\{ x_{(n)}\leqslant 2.5 \}$，求检验犯第一类错误的最大值$\alpha$，若要使得该最大值$\alpha$不超过$0.05$，$n$至少应取多大？
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}{
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$x_{(n)}$的密度函数为$f_n(x)=\frac{nx^{n-1}}{\theta^{n}} 1_{(0,\theta)}(x)$，所以检验犯第一类错误的概率为
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$$
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\alpha' = P(x_{(n)}\leqslant 2.5|H_0)=P(x_{(n)}\leqslant 2.5|\theta\geqslant 3)=\int_{0}^{2.5} \frac{n x^{n-1}}{\theta^{n}} \mathrm{d}x = \left(\frac{5}{2}\right)^{n} \theta^{- n}
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$$
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当$\theta$取$3$时$\alpha'$取到最大值$\bm{\alpha} = \left( \frac{5}{2} \right) ^{n} 3^{-n} \bm{= \left(\frac{6}{5}\right)^{- n}}$，而$\alpha = \left( \frac{6}{5} \right) ^{-n}= 0.05 $解得$ n = - \frac{\ln{(20)}}{- \ln{(6)} + \ln{(5)}} =16.4310371534373$，所以$n$至少应取$\bm{17}$。
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}
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\questionandanswer[8]{
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设$x_1,x_2, \cdots ,x_{30}$为取自泊松分布$P(\lambda)$的随机样本。
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}{}
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\begin{enumerate}
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\questionandanswer[]{
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试给出单侧假设检验问题$H_0:\lambda\leqslant 0.1\ \ \mathrm{vs}\ \ H_1:\lambda>0.1$的显著性水平$\alpha=0.05$的检验；
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}{
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由于泊松分布关于参数$\lambda$具有可加性，所以$\sum_{k=1}^{n} x_k\sim P(30\lambda)$，所以选取$\sum_{k=1}^{n} x_k$作为统计量，设拒绝域为$W$，则
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$$
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P(W|H_0)=P(W|\lambda\leqslant 0.1)= \sum_{k=c}^{\infty} \frac{(30\lambda)^{k}}{k!} e^{-30\lambda} \leqslant 0.05
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$$
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当$\lambda$越大则犯第一类错误的概率越大，所以此时$\lambda$可以取$0.1$，则
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$$
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\sum_{k=c}^{\infty} \frac{(30\lambda)^{k}}{k!} e^{-30\lambda} = \sum_{k=c}^{\infty} \frac{3^{k}}{k!}e^{-3}
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$$
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\begin{center}
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\includegraphics[width=0.3\linewidth]{imgs/2024-05-29-15-56-01.png}
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\includegraphics[width=0.3\linewidth]{imgs/2024-05-29-15-55-51.png}
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\end{center}
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（图中的$50$可以为任何较大的自然数）可以看到当$c$取$6$时上式大于$0.05$，当$c$取$7$时上式小于$0.05$，所以所求检验的拒绝域为$\displaystyle W= \left\{ \sum_{k=1}^{30} x_k\geqslant 7 \right\} $。
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}
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\questionandanswer[]{
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求此检验的势函数$\beta(\lambda)$在$\lambda=0.05,0.2,0.3, \cdots ,0.9$时的值，并据此画出$\beta(\lambda)$的图像。
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}{
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$$
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\begin{aligned}
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\beta(\lambda)&= P_{\lambda} \left( \sum_{k=1}^{30} x_i \geqslant 7 \right) = \sum_{k=7}^{\infty} \frac{(30\lambda)^{k}}{k!} e^{-30 \lambda} \\
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& = (- 1012500 \lambda^{6} - 202500 \lambda^{5} - 33750 \lambda^{4} - 4500 \lambda^{3} - 450 \lambda^{2} - 30 \lambda + e^{30 \lambda} - 1) e^{- 30 \lambda} \\
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\end{aligned}
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$$
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使用\LaTeX 的 pgfplots 宏包画图如下：
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\begin{center}
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\noindent\hspace{-6em} % ylabel会导致图片偏右，需要向左移动回来
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\begin{tikzpicture}
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\begin{axis}[
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xlabel={$\lambda$},
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ylabel={$\beta(\lambda)$}
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]
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\addplot[domain=0:1] {(- 1012500*x^6 - 202500*x^5 - 33750 *x^4 - 4500 *x^3 - 450 *x^2 - 30 *x + e^(30*x) - 1)* e^(-30*x)};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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}
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\end{enumerate}
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\end{enumerate}
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\section{正态总体参数假设检验}
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说明：本节习题均采用拒绝域的形式完成，在可以计算检验的$p$值时要求计算出$p$值。
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\begin{enumerate}
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\questionandanswerSolution[]{
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有一批枪弹，出厂时，其初速率$v \sim N(950, 100)$（单位：m/s）。经过较长时间储存，取9发进行测试，得样本值（单位：m/s）如下：
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$$
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914\quad 920\quad 910\quad 934\quad 953\quad 945\quad 912\quad 924\quad 940
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$$
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据经验，枪弹经储存后其初速率仍服从正态分布，且标准差保持不变，问是否可以认为这批枪弹的初速率有显著降低（$\alpha=0.05$）？
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}{
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设总体的均值为$\mu$，则待检验的原假设$H_0$和备选假设$H_1$分别为
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$$
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H_0:\mu=950 \quad\mathrm{vs}\quad H_1:\mu<950
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$$
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拒绝域为$\{ u\leqslant u_{\alpha} \}$，即$\left\{ \frac{\bar{x}-950}{10/3}\leqslant u_{0.05} \right\} $即 $\left\{ \bar{x}\leqslant -1.645\times \frac{10}{3}+950 \approx 944.5167 \right\} $。
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根据样本计算得出$\bar{x}=928$，在拒绝域内，因此可以认为这批枪弹的初速率有显著降低。
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再计算$p$值，
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$$
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p=\Phi\left( \frac{928-950}{10/3} \right) = \bm{2.0665\times 10^{-11}} < 0.05
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$$
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}
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\questionandanswerSolution[5]{
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设需要对某正态总体的均值进行假设检验
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$$
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H_0:\mu=15 \quad\mathrm{vs}\quad H_1:\mu<15
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$$
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已知$\sigma^{2}=2.5$，取$\alpha=0.05$，若要求当$H_1$中的$\mu\leqslant 13$时犯第二类错误的概率不超过$0.05$，求所需的样本容量。
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}{
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由于已知$\sigma^{2}=2.5$，所以拒绝域为$\left\{ \frac{\bar{x}-15}{\sqrt{2.5/n}}\leqslant u_{0.05} \right\} $
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$$
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\beta=P\left( \frac{\bar{x}-15}{\sqrt{2.5/n}} >u_{0.05} \middle| \mu\leqslant 13 \right) \leqslant 0.05
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$$
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其中
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$$
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\begin{aligned}
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P\left( \frac{\bar{x}-15}{\sqrt{2.5/n}}>u_{0.05} \right) &=P\left( \frac{\bar{x}-\mu+\mu-15}{\sqrt{2.5 /n}} >u_{0.05} \right) =P\left( \frac{\bar{x}-\mu}{\sqrt{2.5 /n}}>u_{0.05}+\frac{15-\mu}{\sqrt{2.5 /n}} \right) \\
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&=1-\Phi\left( -1.645+\frac{15-\mu}{\sqrt{2.5 /n}} \right) \leqslant 0.05 \\
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\end{aligned}
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$$
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所以
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$\displaystyle
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\Phi\left( -1.645+\frac{15-\mu}{\sqrt{2.5 /n}} \right) \geqslant 0.95
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$
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，从而
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$\displaystyle
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-1.645+\frac{15-\mu}{\sqrt{2.5 /n}} \geqslant 1.645
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$
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需要在$\mu\leqslant 13$时成立，由于左侧关于$\mu$递减，所以当$\mu=13$时，解$-1.645+\frac{15-13}{\sqrt{2.5 /n} }=1.645 $可得$ n = 6.7650625$，所以所需的样本容量至少为$\bm{7}$。
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}
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\questionandanswer[6]{
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从一批钢管中抽取10根，测得其内径（单位：mm）为
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$$
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100.36\quad 100.31\quad 99.99\quad 100.11\quad 100.64\quad 100.85\quad 99.42\quad 99.91\quad 99.35\quad 100.10
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$$
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设这批钢管内径服从正态分布$N(\mu,\sigma^{2})$，试分别在下列条件下检验假设（$\alpha=0.05$）：
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$$
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H_0:\mu=100 \quad\mathrm{vs}\quad H_1:\mu>100
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$$
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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已知$\sigma=0.5$；
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}{
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拒绝域为
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$\displaystyle
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\left\{ \frac{\bar{x}-100}{0.5/\sqrt{10}}\geqslant u_{1-\alpha} \right\} = \left\{ \bar{x} \geqslant u_{0.95} \times 0.5 \sqrt{10} + 100 \right\} =\left\{ \bar{x}\geqslant 102.60 \right\}
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$
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根据样本计算得出$\bar{x}=100.104$，不在拒绝域中，所以不能拒绝原假设。
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再计算$p$值，
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$$
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p=1-\Phi\left( \frac{100.104-100}{0.5} \right) = \bm{0.082385} >0.05
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$$
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}
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\questionandanswerSolution[]{
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$\sigma$未知。
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}{
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拒绝域为
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$$
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\left\{ \frac{\bar{x}-100}{s /\sqrt{10}} \geqslant t_{0.95}(9) \right\} = \left\{ \frac{\bar{x}-100}{s /\sqrt{10}}\geqslant 1.8331 \right\}
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$$
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根据样本计算得出$\bar{x}=100.104, s=0.4759598489$，所以 $\frac{\bar{x}-100}{s /\sqrt{10}}=0.690976092663247$不在拒绝域内，所以不能拒绝原假设。
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再计算$p$值，
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$$
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p=P_{t\sim t(9)}\left( t \geqslant 0.690976092663247 \right) > 1-0.7027 = \bm{0.2973} > 0.05
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$$
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}
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\end{enumerate}
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\end{enumerate}
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\end{document}
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