300 lines
16 KiB
TeX
300 lines
16 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}{6}
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\setcounter{section}{5}
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\section{区间估计}
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\begin{enumerate}
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\questionandanswer[3]{
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$0.50, 1.25, 0.80, 2.00$是取自总体$X$的样本，已知$Y=\ln X$服从正态分布$N(\mu,1)$。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求$\mu$的置信水平为$95\%$的置信区间；
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}{
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$$
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\frac{1-0.95}{2} = 0.025
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,\quad
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u_{0.025} = -1.96
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,\quad
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\frac{u_{0.025}}{\sqrt{n}}=\frac{-1.96}{\sqrt{4}}=-0.98
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$$
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$$
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\overline{\ln x} = \frac{1}{4}(\ln 0.50+ \ln 1.25+\ln 0.80+\ln 2.00) = 0
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$$
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所以$\mu$的置信水平为$95\%$的置信区间为$[-0.98, 0.98]$。
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}
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\questionandanswerSolution[]{
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求$X$的数学期望的置信水平为$95\%$的置信区间。
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}{
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$$
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EX = E e^{Y} = e^{EY} = e^{\mu}
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$$
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所以$EX$的置信水平为$95\%$的置信区间为$[e^{-0.98}, e^{0.98}]，即[0.3753110988514, 2.66445624192942]$。
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}
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\end{enumerate}
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\questionandanswer[5]{
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已知某种材料的抗压强度$X\sim N(\mu,\sigma^{2})$，现随机地抽取10个试件进行抗压试验，测得数据如下：
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$$
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482\quad 493\quad 457\quad 471\quad 510\quad 446\quad 435\quad 418\quad 394\quad 469
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$$
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求平均抗压强度$\mu$的置信水平为$95\%$的置信区间；
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}{
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由于$\sigma$未知，所以$\mu$的置信区间为$\left[ \bar{x}-t_{1-0.025}(10-1)s/ \sqrt{10}, \bar{x}+t_{1-0.025}(10-1)s /\sqrt{10} \right] $
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之后计算得
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$$
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\bar{x}=457.5,\quad s\approx 35.21757768
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$$
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所以$\mu$的置信水平为$95\%$的置信区间为
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$$
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[457.5- 2.2622\times 35.21757768/ \sqrt{10}, 457.5-2.2622\times 35.21757768 /\sqrt{10}]
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$$
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即
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$$
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[432.306385526736, 482.693614473264]
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$$
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}
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\questionandanswerSolution[]{
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若已知$\sigma=30$，求平均抗压程度$\mu$的置信水平为$95\%$的置信区间；
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}{
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由于$\sigma$已知，所以$\mu$的$95\%$置信区间为$\left[ \bar{x}-u_{0.975}\sigma/\sqrt{n}, \bar{x}+u_{0.975}\sigma/\sqrt{n} \right] $，代入得
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$$
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\left[ 457.5-1.96\times 30 / \sqrt{10}, 457.5 + 1.96\times 30 /\sqrt{10} \right]
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$$
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即
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$$
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\left[ 438.90580735821, 476.09419264179 \right]
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$$
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}
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\questionandanswerSolution[]{
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求$\sigma$的置信水平为$95\%$的置信区间。
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}{
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$\mu$未知时$\sigma$的置信水平为$95\%$的置信区间为$\left[ \frac{s\sqrt{n-1}}{\sqrt{\chi^{2}_{0.975}(n-1)}}, \frac{s\sqrt{n-1}}{\sqrt{\chi^{2}_{0.025}(n-1)}} \right] $，代入得
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$$
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\left[ \frac{35.21757768\times \sqrt{10-1}}{\sqrt{19.0228}}, \frac{35.21757768\times \sqrt{10-1}}{\sqrt{2.7004}} \right]
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$$
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即
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$$
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\left[ 24.2238693218913, 64.2934434191729 \right]
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$$
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}
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\end{enumerate}
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\questionandanswerSolution[6]{
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在一批货物中随机抽取80件，发现有11件不合格品，试求这批货物的不合格品率的置信水平为$0.90$的置信区间。
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}{
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样本的分布为$b(1,p)$。由于样本量较大，可以使用近似置信区间，即\\
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$\left[ \bar{x}-u_{0.95}\sqrt{\frac{\bar{x}(1-\bar{x})}{n}}, \bar{x}+u_{0.95}\sqrt{\frac{\bar{x}(1-\bar{x})}{n}} \right] $
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，其中$\bar{x}=\frac{11}{80} = 0.1375$，$n=80$，$u_{0.95}=1.645$，代入得
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$$
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\left[ 0.1375-1.645 \times \sqrt{\frac{0.1375\times (1-0.1375)}{80}}, 0.1375+1.645\times \sqrt{\frac{0.1375\times (1-0.1375)}{80}} \right]
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$$
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即
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$$
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\left[ 0.0741638282314373, 0.200836171768563 \right]
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$$
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}
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\questionandanswer[9]{
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设从总体$X\sim N(\mu_1,\sigma_1^{2})$和总体$Y\sim N(\mu_2,\sigma_2^{2})$中分别抽取容量为$n_1=10,n_2=15$的独立样本，可计算得$\bar{x}=82, s_x^{2}=56.5, \bar{y}=76, s_y^{2}=52.4$。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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若已知$\sigma_1^{2}=64, \sigma_2^{2}=49$，求$\mu_1-\mu_2$的置信水平为$95\%$的置信区间；
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}{
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$\sigma_1^{2}$和$\sigma_2^{2}$均已知，则$\mu_1-\mu_2$的置信水平为$95\%$的置信区间为\\
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$\left[ \bar{x}-\bar{y}-u_{0.975}\sqrt{\frac{\sigma_1^{2}}{n_1} +\frac{\sigma_2^{2}}{n_2}}, \bar{x}-\bar{y}+u_{0.975}\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}} \right] $，代入得
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$$
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\left[ 82-76-1.96\times \sqrt{\frac{64}{10}+\frac{49}{15}},\ 82-76+1.96\times \sqrt{\frac{64}{10}+\frac{49}{15}} \right]
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$$
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即
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$$
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\left[ -0.0938876480180258,\ 12.093887648018 \right]
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$$
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}\questionandanswerSolution[]{
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若已知$\sigma_1^{2}=\sigma_2^{2}$，求$\mu_1-\mu_2$的置信水平为$95\%$的置信区间；
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}{
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$\sigma_1^{2}=\sigma_2^{2}$未知，则$\mu_1-\mu_2$的置信水平为$95\%$的置信区间为\\
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$$
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\left[\bar{x}-\bar{y} - \sqrt{\frac{n_1+n_2}{n_1n_2}}s_w t_{0.975}(n_1+n_2-2),\ \bar{x}-\bar{y}+\sqrt{\frac{n_1+n_2}{n_1n_2}}s_w t_{0.975}(n_1+n_2-2)\right]
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$$
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其中$\displaystyle s_w^{2} = \frac{(n_1-1)s_{x}^{2}+(n_2-1)s_{y}^{2}}{n_1+n_2-2}$，$t_{0.975}(23)=2.0687$，代入得
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$$
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s_w^{2}=\frac{(10-1)\times 56.5+(15-1)\times 52.4}{10+15-2} = \frac{12421}{230}
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$$
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置信区间为
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$$
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\left[ 82-76-\sqrt{\frac{10+15}{10\times 15}} \sqrt{\frac{12421}{230}}\times 2.0687,\ 82-76+\sqrt{\frac{10+15}{10\times 15}}\sqrt{\frac{12421}{230}}\times 2.0687 \right]
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$$
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即
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$$
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\left[ -0.206349837966326,\ 12.2063498379663 \right]
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$$
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}
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\questionandanswerSolution[]{
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若对$\sigma_1^{2},\sigma_2^{2}$一无所知，求$\mu_1-\mu_2$的置信水平为$95\%$的置信区间；
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}{
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此时为一般场合下，$\mu_1-\mu_2$的置信水平为$95\%$的近似置信区间为\\
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$[\bar{x}-\bar{y}-s_0 t_{0.975}(l),\ \bar{x}-\bar{y}+s_0 t_{0.975}(l)]$，其中$s_0^{2} = \frac{s_{x}^{2}}{n_1}+\frac{s_y^{2}}{n_2}=\frac{56.5}{10}+\frac{52.4}{15} = \frac{2743}{300}$, \\
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$\displaystyle l=\frac{s_0^{4}}{\frac{s_{x}^{4}}{n_1^{2}(n_1-1)}+\frac{s_y^{4}}{n_2^{2}(n_2-1)}} = \frac{\left( \frac{2743}{300} \right) ^{2}}{\frac{56.5^{2}}{10^{2}(10-1)}+\frac{52.4^{2}}{15^{2}(15-1)}} = \frac{52668343}{2783727} \approx 18.92008\approx 19$,\\
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$t_{0.975}(19)=2.0930$。
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所以置信区间为
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$$
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\left[ 82-76-\sqrt{\frac{2743}{300}}\times 2.0930, \ 82-76+\sqrt{\frac{2743}{300}}\times 2.0930 \right]
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$$
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即
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$$
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\left[ -0.328801942179367, \ 12.3288019421794 \right]
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$$
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}
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\questionandanswerSolution[]{
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求$\sigma_1^{2}/\sigma_2^{2}$的置信水平为$95\%$的置信区间。
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}{
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置信区间为
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$$
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\left[\frac{s_{x}^{2}}{s_y^{2}}\cdot \frac{1}{F_{0.975}(9, 14)}，\ \frac{s_{x}^{2}}{s_y^{2}}\cdot \frac{1}{F_{0.025}(9,14)} \right]
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$$
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由于$F_{\frac{\alpha}{2}}(n_1,n_2) = {1}/{F_{1-\frac{\alpha}{2}}(n_2,n_1)}$，所以可以代入得
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$$
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\left[ \frac{56.5}{52.4}\cdot \frac{1}{3.21},\ \frac{56.5}{52.4}\cdot {3.80} \right]
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$$
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即
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$$
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\left[0.335901643242729,\ 4.09732824427481 \right]
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$$
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}
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\end{enumerate}
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\questionandanswerSolution[12]{
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设某电子产品的寿命服从指数分布，其密度函数为$\lambda e^{-\lambda x}I_{\{ x>0 \}}$，现从此批产品中抽取容量为9的样本，测得寿命为（单位：千小时）
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$$
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15\quad 45\quad 50\quad 53\quad 60\quad 65\quad 70\quad 83\quad 90
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$$
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求平均寿命$1/\lambda$的置信水平为0.9的置信区间和置信上、下限。
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}{
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首先尝试构造枢轴量，设样本为$x_1,x_2, \cdots ,x_9$，则$x_1,x_2, \cdots x_9\overset{\text{i.i.d.}}{\sim}\operatorname{Exp}(\lambda)$，则$\sum_{i=1}^{9} x_i \sim \operatorname{Ga}(9, \lambda)$，所以${2\lambda}\sum_{i=1}^{9} x_i \sim \operatorname{Ga}(9, \frac{1}{2})=\chi^{2}(18)$，分布不依赖于$\lambda$，所以$G=2\lambda\sum_{i=1}^{9} x_i$为枢轴量，所以
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$$
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P\left( \chi^{2}_{0.05}(18)\leqslant G\leqslant \chi^{2}_{0.95}(18) \right) = 0.9
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$$
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$$
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P\left( \frac{\chi^{2}_{0.05}(18)}{2\sum_{i=1}^{9} x_i}\leqslant \lambda\leqslant \frac{\chi^{2}_{0.95}(18)}{2\sum_{i=1}^{9} x_i} \right) =0.9
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$$
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所以$\lambda$的置信水平为0.9的双侧置信区间为
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$$
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\left[ \frac{\chi^{2}_{0.05}(18)}{2\sum_{i=1}^{9} x_i}, \ \frac{\chi^{2}_{0.95}(18)}{2\sum_{i=1}^{9} x_i} \right] = \left[ \frac{9.3905}{2\times 531},\ \frac{28.8693}{2\times 531} \right] = \left[ 0.00884227871939736,\ 0.0271838983050847 \right]
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$$
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同理，单侧置信上限为
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$$
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\frac{\chi^{2}_{0.9}(18)}{2\sum_{i=1}^{9} x_i}=\frac{25.9894}{2\times 531} = 0.0244721280602637
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$$
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单侧置信下限为
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$$
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\frac{\chi^{2}_{0.1}(18)}{2\sum_{i=1}^{9} x_i}=\frac{10.8649}{2\times 531} = 0.0102306026365348
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$$
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所以$\frac{1}{\lambda}$的置信水平为0.9的置信区间为
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$$
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\left[ \frac{2\times 531}{28.8693}, \ \frac{2\times 531}{9.3905} \right] = \mathbf{\left[ 36.7864825264208, \ 113.093019541025 \right]}
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$$
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单侧置信上限为
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$$
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\frac{2\times 531}{10.8649} = \mathbf{97.74595256284}
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$$
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单侧置信下限为
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$$
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\frac{2\times 531}{25.9894} = \mathbf{40.8628133008073}
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$$
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}
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\questionandanswerSolution[13]{
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设总体$X$的密度函数为
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$$
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p(x;\theta)=\frac{1}{\pi\left[ 1+(x-\theta)^{2} \right] }, \quad -\infty<x<\infty, \quad -\infty<\theta<\infty
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$$
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$x_1,x_2, \cdots ,x_n$为抽自此总体的简单随机样本，求位置参数$\theta$的置信水平近似为$1-\alpha$的置信区间。
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}{
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设$m_{0.5}$表示样本中位数，则根据例5.3.10，当然$n$较大时，$m_{0.5}\sim N\left( \theta, \frac{\pi^{2}}{4n} \right) $，将$m_{0.5}$看作样本容量为1的样本，则$m_{0.5}$服从方差已知，期望未知的正态分布，所以$\theta$的置信水平近似为$1-\alpha$的置信区间为
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$$
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\left[ m_{0.5}-u_{1-\frac{\alpha}{2}} \sqrt{\frac{\pi^{2}}{4n}} / \sqrt{1} , \ m_{0.5}+u_{1-\frac{\alpha}{2}}\sqrt{\frac{\pi^{2}}{4n}} / \sqrt{1}\right]
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$$
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即
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$$
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\left[ m_{0.5}-u_{1-\frac{\alpha}{2}} \frac{\pi}{2 \sqrt{n}} , \ m_{0.5}+u_{1-\frac{\alpha}{2}}\frac{\pi}{2\sqrt{n}}\right]
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$$
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}
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\questionandanswerSolution[14]{
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设$x_1,x_2, \cdots ,x_n$为抽自正态总体$N(\mu,16)$的简单随机样本，为使得$\mu$的置信水平为$1-\alpha$的置信区间的长度不大于给定的$L$，试问样本容量$n$至少要多少？
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}{
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$\sigma^{2}=16$已知，则置信区间为$\left[ \bar{x}-u_{1-\frac{\alpha}{2}}\sigma / \sqrt{n}, \bar{x}+u_{1-\frac{\alpha}{2}}\sigma / \sqrt{n} \right] $，区间长度为$2 u_{1-\frac{\alpha}{2}}\times 4 / \sqrt{n}\leqslant L$，则$\displaystyle n \geqslant \left( \frac{8 u_{1-\frac{\alpha}{2}}}{L} \right) ^{2}=\frac{64 u_{1-\frac{\alpha}{2}}^{2}}{L^{2}}$。
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}
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\questionandanswerSolution[16]{
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设$x_1,x_2, \cdots ,x_n$是来自$U(\theta-\frac{1}{2}, \theta+\frac{1}{2})$的样本，求$\theta$的置信水平为$1-\alpha$的置信区间（提示：证明$\displaystyle \frac{x_{(n)}+x_{(1)}}{2}-\theta$为枢轴量，并求出对应的密度函数）。
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}{
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设总体为$X$，则$X\sim U(\theta-\frac{1}{2}, \theta+\frac{1}{2})$，则$X-\theta+\frac{1}{2}\sim U(0,1)$，则根据例5.3.9，$(Y,Z)=\left( x_{(1)}-\theta+\frac{1}{2}, x_{(n)}-\theta+\frac{1}{2} \right) $的联合密度函数为
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$$
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p(y,z)=n(n-1)(z-y)^{n-2}
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$$
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再根据卷积公式，$y+z=x_{(1)}+x_{(n)}-2\theta+1$的概率密度函数为
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$$
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p(x)=\int_{0}^{1} n(n-1)(x-2t)^{n-2} \mathrm{d}t = \frac{n}{2} x^{n-1} - \frac{n}{2} (x-2)^{n-1}
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$$
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所以$\frac{(y+z)-1}{2}=\frac{x_{(1)}+x_{(n)}}{2}-\theta$的概率密度函数为
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$$
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p'(x) = \frac{n}{2}(2x+1)^{n-1}-\frac{n}{2}(2x-1)^{n-1}
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$$
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显然与$\theta$无关，所以令$G=\frac{x_{(1)}+x_{(n)}}{2}-\theta$即为枢轴量，则可知
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$$
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\int_{- \frac{1-\alpha^{\frac{1}{n}}}{2}}^{\frac{1-\alpha ^{\frac{1}{n}}}{2}} \frac{n}{2}(2x+1)^{n-1}-\frac{n}{2}(2x-1)^{n-1} \mathrm{d}x = 1-\alpha
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$$
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所以
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$
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-\frac{1-\alpha ^{\frac{1}{n}}}{2} \leqslant G=\frac{x_{(1)}+x_{(n)}}{2}-\theta\leqslant \frac{1-\alpha ^{\frac{1}{n}}}{2}
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$
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，所以
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$
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\frac{x_{(1)}+x_{(n)}}{2}-\frac{1-\alpha ^{\frac{1}{n}}}{2} \leqslant \theta \leqslant \frac{x_{(1)}+x_{(n)}}{2}+\frac{1-\alpha ^{\frac{1}{n}}}{2}
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$
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所以$\theta$的置信水平为$1-\alpha$的置信区间为
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$$
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\left[ \frac{x_{(1)}+x_{(n)}}{2}-\frac{1-\alpha ^{\frac{1}{n}}}{2},\ \frac{x_{(1)}+x_{(n)}}{2}+\frac{1-\alpha ^{\frac{1}{n}}}{2} \right]
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$$
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}
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\questionandanswer[19]{
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设总体$X$的密度函数为
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$$
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p(x,\theta)=e^{-(x-\theta)} I_{\{ x>\theta \}}, \quad -\infty<\theta<\infty,
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$$
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$x_1,x_2, \cdots ,x_n$为抽自此总体的简单随机样本。
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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证明：$x_{(1)}-\theta$的分布与$\theta$无关，并求出此分布；
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}{
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令$y = x-\theta$，则$p(y) = e^{-y}I_{\{ y>0 \}}$。
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由于$y=x-\theta$是单调增函数，所以$y_{(1)}=x_{(1)}-\theta$。
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$F(y)=\int_{0}^{y} e^{-t} \mathrm{d}t=1-e^{-y}$，从而次序统计量$y_{(1)}=x_{(1)}-\theta$的概率密度函数为
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$$
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p_{(1)}(y) = \frac{n!}{0!(n-1)!} [F(y)]^{0} \left[ 1-F(y) \right] ^{n-1} p(y) = n \left( e^{-y} \right) ^{n-1} e^{-y} I_{\{ y>0 \}} = ne^{-ny} I_{\{ y>0 \}}
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$$
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所以$x_{(1)}-\theta \sim \operatorname{Exp}(n)$，与$\theta$无关。
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}
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\questionandanswerSolution[]{
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求$\theta$的置信水平为$1-\alpha$的置信区间。
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}{
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$$
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P(c\leqslant x_{(1)}-\theta\leqslant d)=\int_{c}^{d} ne^{-ny} \mathrm{d}y
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$$
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因为被积函数在$[0,+\infty)$上单调递减，所以区间长度最短则$c=0$，所以
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$$
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\int_{0}^{d} n e^{-ny} \mathrm{d}y = \left. -e^{-ny} \right|_{0}^{d} = 1-e^{-nd} = 1-\alpha
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$$
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所以 $d = \dfrac{-\ln \alpha}{n}$。
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而$c \leqslant x_{(1)}-\theta\leqslant d \implies x_{(1)}-d\leqslant \theta\leqslant x_{(1)}-c\implies x_{(1)}+\frac{\ln \alpha}{n}\leqslant \theta\leqslant x_{(1)}$，所以$\theta$的置信水平为$1-\alpha$的置信区间为
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$$
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\left[ x_{(1)}+\frac{\ln \alpha}{n},\ x_{(1)} \right]
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$$
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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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