diff --git a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf index b06ed08..b884520 100644 Binary files a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf and b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf differ diff --git a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex index 4564501..5eedd14 100644 --- a/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex +++ b/advanced-math/knowledge/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex @@ -291,6 +291,8 @@ $=\int_0^\frac{\pi}{2}\textrm{d}\theta\int_0^{+\infty}e^{-r^2}r\,\textrm{d}r=\di \subsection{计算} +基本思想还是三重积分化为一重积分。 + \subsubsection{基础方法} \paragraph{直角坐标系} \leavevmode \medskip @@ -307,7 +309,7 @@ $=\int_0^\frac{\pi}{2}\textrm{d}\theta\int_0^{+\infty}e^{-r^2}r\,\textrm{d}r=\di $\iiint\limits_\Omega f(x,y,z)\,\textrm{d}v=\iint\limits_{D_{xy}}\textrm{d}\sigma\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)\,\textrm{d}z$。 -\textbf{例题：}计算三重积分$I=\displaystyle{\iiint\limits_D\dfrac{\textrm{d}x\textrm{d}y\textrm{d}z}{(1+x+y+z)^3}}$，其中$\Omega$是由平面$x=0,y=0,z=0$及$x+y+z=1$所围成的四面体。 +\textbf{例题：}计算三重积分$I=\displaystyle{\iiint\limits_\Omega\dfrac{\textrm{d}x\textrm{d}y\textrm{d}z}{(1+x+y+z)^3}}$，其中$\Omega$是由平面$x=0,y=0,z=0$及$x+y+z=1$所围成的四面体。 解：根据图形，已知是一个四面体，所以下底面是一个1×1的等腰直角三角形$D_{xy}$，上曲面为一个等边三角形$z=-1-x-y$，有两个侧柱面。 diff --git a/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.pdf b/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.pdf index 07ce470..b270e98 100644 Binary files a/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.pdf and b/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.pdf differ diff --git a/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.tex b/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.tex index 9c2fc64..25ae257 100644 --- a/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.tex +++ b/probability-theory-and-mathematical-statistics/exercise/5-mathematical-statistics/mathematical-statistics.tex @@ -208,6 +208,14 @@ $\ln L(\lambda)=-n\ln2-n\ln\lambda-\dfrac{1}{\lambda}\sum\limits_{i=1}^n\vert x_ 即可以认为平均水平为70分。 +\textbf{例题：}设$X_1,X_2,\cdots,X_{36}$是取自正态总体$N(\mu,0.04)$的简单随机样本，其中$\mu$为未知参数，即$\overline{X}=\dfrac{1}{36}\sum\limits_{i=1}^{36}X_i$，若对于检验问题$H_0:\mu\leqslant0.5$，$H_1:\mu>0.5$在显著性水平$\alpha=0.05$，取得检验拒绝域$D=\{(x_1,x_2,\cdots,x_{36}):\overline{x}>C\}$，求$C$。 + +解：当$H_0$成立，则$X\sim N(0.5,0.04)$，$\overline{X}\sim N(0.5,0.04\div36)=N\left(\dfrac{1}{2},\dfrac{1}{900}\right)$。 + +$\alpha=0.05=P\{$拒绝$H_0|H_0$成立$\}=P\{\overline{X}>C\}=1-P\{\overline{X}\leqslant C\}=1-\varPhi((C-0.5)\times30)=1-\varPhi(30C-15)$。 + +$\therefore\varPhi(30C-15)=0.95=\varPhi(1.645)$，即$30C-15=1.645$，$C=0.5548$。 + \textbf{例题：}已知某机器生产出来的零件长度$X$（单位：$cm$）服从正态分布$N(\mu,\delta^2)$，现从中随意抽取容量为16的一个样本，测得样本均值$\overline{x}=10$，样本方差$s^2=0.16$，$t_{0.025}(15)=2.132$。 (1)求总体均值$\mu$置信水平为0.95的置信区间。 diff --git a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf index e298830..3ebcb7c 100644 Binary files a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf and b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.pdf differ diff --git a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex index 8039ab3..03341a3 100644 --- a/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex +++ b/probability-theory-and-mathematical-statistics/knowledge/5-mathematical-statistics/mathematical-statistics.tex @@ -451,6 +451,27 @@ $\therefore P\left(\vert Z\vert<\dfrac{\Delta}{\sigma/\sqrt{n}}\right)=1-\alpha$ \subsubsection{正态总体下的六大检验与拒绝域} +\begin{center} + \scalebox{0.77}{ + \begin{tabular}{|c|c|c|c|c|c|} + \hline + 检验参数 & 条件 & 原假设$H_0$ & 备择假设$H_1$ & 检验法与统计量 & 拒绝域 \\ \hline + \multirow{6}{*}{$\mu$} & \multirow{3}{*}{$\sigma=\sigma_0$} & $\mu=\mu_0$ & $\mu\neq\mu_0$ & \multirow{3}{*}{\tabincell{c}{$U$检验\\$U=\dfrac{\overline{X}-\mu_0}{\sigma_0/\sqrt{n}}\sim N(0,1)$}} & $\vert u\vert\geqslant u_{\frac{\alpha}{2}}$ \\ \cline{3-4} \cline{6-6} + & & $\mu\leqslant\mu_0$ & $\mu>\mu_0$ & & $u\geqslant u_\alpha$ \\ \cline{3-4} \cline{6-6} + & & $\mu\geqslant\mu_0$ & $\mu<\mu_0$ & & $u\leqslant u_\alpha$ \\ \cline{2-6} + & \multirow{3}{*}{$\sigma$未知} & $\mu=\mu_0$ & $\mu\neq\mu_0$ & \multirow{3}{*}{\tabincell{c}{$T$检验\\$U=\dfrac{\overline{X}-\mu_0}{S/\sqrt{n}}\sim t(n-1)$}} & $\vert t\vert\geqslant t_{\frac{\alpha}{2}}(n-1)$ \\ \cline{3-4} \cline{6-6} + & & $\mu\leqslant\mu_0$ & $\mu>\mu_0$ & & $t\geqslant t_\alpha(n-1)$ \\ \cline{3-4} \cline{6-6} + & & $\mu\geqslant\mu_0$ & $\mu<\mu_0$ & & $t\leqslant t_\alpha(n-1)$ \\ \hline + \multirow{6}{*}{$\sigma^2$} & \multirow{3}{*}{$\mu$已知} & $\sigma^2=\sigma^2_0$ & $\sigma^2\neq\sigma^2_0$ & \multirow{3}{*}{\tabincell{c}{$\chi^2$检验\\$\chi^2=\dfrac{\sum\limits_{i=1}^n(X_i-\mu)^2}{\sigma^2_0}\sim\chi^2(n)$}} & \tabincell{c}{$\chi^2\leqslant\chi^2_{1-\frac{\alpha}{2}}(n)$\\$\chi^2\geqslant\chi^2_{\frac{\alpha}{2}}(n)$} \\ \cline{3-4} \cline{6-6} + & & $\sigma^2\leqslant\sigma^2_0$ & $\sigma^2>\sigma^2_0$ & & $\chi^2\geqslant\chi^2_\alpha(n)$ \\ \cline{3-4} \cline{6-6} + & & $\sigma^2\geqslant\sigma^2_0$ & $\sigma^2<\sigma^2_0$ & & $\chi^2\leqslant\chi^2_{1-\alpha}(n)$ \\ \cline{2-6} + & \multirow{3}{*}{$\mu$未知} & $\sigma^2=\sigma^2_0$ & $\sigma^2\neq\sigma^2_0$ & \multirow{3}{*}{\tabincell{c}{$\chi^2$检验\\$\chi^2=\dfrac{(n-1)S^2}{\sigma_0^2}\sim\chi^2(n-1)$}} & \tabincell{c}{$\chi^2\leqslant\chi^2_{1-\frac{\alpha}{2}}(n-1)$\\$\chi^2\geqslant\chi^2_{\frac{\alpha}{2}}(n-1)$} \\ \cline{3-4} \cline{6-6} + & & $\sigma^2\leqslant\sigma^2_0$ & $\sigma^2>\sigma^2_0$ & & $\chi^2\geqslant\chi^2_\alpha(n-1)$ \\ \cline{3-4} \cline{6-6} + & & $\sigma^2\geqslant\sigma^2_0$ & $\sigma^2<\sigma^2_0$ & & $\chi^2\leqslant\chi^2_{1-\alpha}(n-1)$ \\ \hline + \end{tabular} + } +\end{center} + \subsection{两类错误} 显著性水平$\alpha$实际上是犯第一类错误的概率的上界。\medskip @@ -465,25 +486,4 @@ $\therefore P\left(\vert Z\vert<\dfrac{\Delta}{\sigma/\sqrt{n}}\right)=1-\alpha$ \end{tabular} }\end{center} \medskip -\begin{center} - \scalebox{0.8}{ - \begin{tabular}{|c|c|c|c|c|c|} - \hline - 检验参数 & 条件 & 原假设$H_0$ & 备择假设$H_1$ & 检验法与统计量 & 拒绝域 \\ \hline - \multirow{6}{*}{$\mu$} & \multirow{3}{*}{$\sigma=\sigma_0$} & $\mu=\mu_0$ & $\mu\neq\mu_0$ & \multirow{3}{*}{\tabincell{c}{$U$检验\\$U=\dfrac{\overline{X}-\mu_0}{\sigma_0/\sqrt{n}}\sim N(0,1)$}} & $\vert u\vert\geqslant u_{\frac{\alpha}{2}}$ \\ \cline{3-4} \cline{6-6} - & & $\mu\leqslant\mu_0$ & $\mu>\mu_0$ & & $u\geqslant u_\alpha$ \\ \cline{3-4} \cline{6-6} - & & $\mu\geqslant\mu_0$ & $\mu<\mu_0$ & & $u\leqslant u_\alpha$ \\ \cline{2-6} - & \multirow{3}{*}{$\sigma$未知} & $\mu=\mu_0$ & $\mu\neq\mu_0$ & \multirow{3}{*}{\tabincell{c}{$T$检验\\$U=\dfrac{\overline{X}-\mu_0}{S/\sqrt{n}}\sim t(n-1)$}} & $\vert t\vert\geqslant t_{\frac{\alpha}{2}}(n-1)$ \\ \cline{3-4} \cline{6-6} - & & $\mu\leqslant\mu_0$ & $\mu>\mu_0$ & & $t\geqslant t_\alpha(n-1)$ \\ \cline{3-4} \cline{6-6} - & & $\mu\geqslant\mu_0$ & $\mu<\mu_0$ & & $t\leqslant t_\alpha(n-1)$ \\ \hline - \multirow{6}{*}{$\mu$} & \multirow{3}{*}{$\mu$已知} & $\sigma^2=\sigma^2_0$ & $\sigma^2\neq\sigma^2_0$ & \multirow{3}{*}{\tabincell{c}{$\chi^2$检验\\$\chi^2=\dfrac{1}{\sigma^2_0}\sum\limits_{i=1}^n(X_i-\mu)^2$\\$\sim\chi^2(n)$}} & \tabincell{c}{$\chi^2\leqslant\chi^2_{1-\frac{\alpha}{2}}(n)$\\$\chi^2\geqslant\chi^2_{\frac{\alpha}{2}}(n)$} \\ \cline{3-4} \cline{6-6} - & & $\sigma^2\leqslant\sigma^2_0$ & $\sigma^2>\sigma^2_0$ & & $\chi^2\geqslant\chi^2_\alpha(n)$ \\ \cline{3-4} \cline{6-6} - & & $\sigma^2\geqslant\sigma^2_0$ & $\sigma^2<\sigma^2_0$ & & $\chi^2\leqslant\chi^2_{1-\alpha}(n)$ \\ \cline{2-6} - & \multirow{3}{*}{$\mu$未知} & $\sigma^2=\sigma^2_0$ & $\sigma^2\neq\sigma^2_0$ & \multirow{3}{*}{\tabincell{c}{$\chi^2$检验\\$\chi^2=\dfrac{(n-1)S^2}{\sigma_0^2}$\\$\sim\chi^2(n-1)$}} & \tabincell{c}{$\chi^2\leqslant\chi^2_{1-\frac{\alpha}{2}}(n-1)$\\$\chi^2\geqslant\chi^2_{\frac{\alpha}{2}}(n-1)$} \\ \cline{3-4} \cline{6-6} - & & $\sigma^2\leqslant\sigma^2_0$ & $\sigma^2>\sigma^2_0$ & & $\chi^2\geqslant\chi^2_\alpha(n-1)$ \\ \cline{3-4} \cline{6-6} - & & $\sigma^2\geqslant\sigma^2_0$ & $\sigma^2<\sigma^2_0$ & & $\chi^2\leqslant\chi^2_{1-\alpha}(n-1)$ \\ \hline - \end{tabular} - } -\end{center} - \end{document}