⚖️假设检验

\( \text{总体的分布类型是已知的，仅涉 及总体未知参数的检验称为参数假设检验} \)

\( \begin{aligned} &\text{设总体 }X\text{ 的分布函数为 }F(x),\;F(x)\text{ 的形式一般未知}.\; (X_{1},X_{2},\dots,X_{n})\text{ 为来自总体 }X\text{ 的简单随机样本，}\\ &\text{相应样本观测值为 }(x_{1},x_{2},\dots,x_{n}).\;\text{参数假设检验的主要步骤可概括如下：}\\[6pt] &\textbf{1.}\;\text{将实际问题转化为假设检验问题，提出原假设 }H_{0}\text{ 及备择假设 }H_{1}.\\[4pt] &\textbf{2.}\;\text{在 }H_{0}\text{ 成立的前提下，构造合适的检验统计量 }\\ &\qquad U\;=\;g\bigl(X_{1},X_{2},\dots,X_{n}\bigr),\\ &\text{要求 }U\text{ 的分布完全已知（不含未知参数）}.\\[4pt] &\textbf{3.}\;\text{给定显著性水平 }\alpha\;(0<\alpha<1)，\text{利用 }U\text{ 构造拒绝域 }\mathcal W,\text{使当 }H_{0}\text{ 成立时}\\ &\qquad P\bigl(\,(X_{1},X_{2},\dots,X_{n})\in\mathcal W\,\bigr)\;=\;P\bigl(U\in\mathcal W\bigr)\;\le\;\alpha.\\ &\text{此即“构造小概率事件” }(X_{1},X_{2},\dots,X_{n})\in\mathcal W.\\[4pt] &\textbf{4.}\;\text{代入样本数据，计算 }U\text{ 的观测值 }\\ &\qquad \hat U\;=\;g\bigl(x_{1},x_{2},\dots,x_{n}\bigr).\\ &\text{若 }\hat U\in\mathcal W,\;\text{则拒绝 }H_{0};\;\text{若 }\hat U\notin\mathcal W,\;\text{则接受 }H_{0}.\\ \end{aligned} \)

\[ \boxed{\text{方差已知，均值的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \mu=\mu_{0} & U=\dfrac{\bar X-\mu_{0}}{\sigma/\sqrt n}\sim N(0,1) & \mu\neq\mu_{0} & |U|\gt u_{\alpha/2} \\ \\ \mu\ge\mu_{0} & U=\dfrac{\bar X-\mu_{0}}{\sigma/\sqrt n}\sim N(0,1) & \mu\lt \mu_{0} & U\lt -u_{\alpha} \\ \\ \mu\le\mu_{0} & U=\dfrac{\bar X-\mu_{0}}{\sigma/\sqrt n}\sim N(0,1) & \mu\gt \mu_{0} & U\gt u_{\alpha} \end{array} \]

\[ \boxed{\text{方差未知，均值的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \mu=\mu_{0} & T=\dfrac{\bar X-\mu_{0}}{S/\sqrt n}\sim t(n-1) & \mu\neq\mu_{0} & |T|\gt t_{\alpha/2}(n-1) \\ \\ \mu\ge\mu_{0} & T=\dfrac{\bar X-\mu_{0}}{S/\sqrt n}\sim t(n-1) & \mu\lt \mu_{0} & T\lt -t_{\alpha}(n-1) \\ \\ \mu\le\mu_{0} & T=\dfrac{\bar X-\mu_{0}}{S/\sqrt n}\sim t(n-1) & \mu \gt \mu_{0} & T\gt t_{\alpha}(n-1) \end{array} \]

\[ \boxed{\text{均值已知，方差的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \sigma^{2}\le\sigma_{0}^{2} & \chi^{2}=\dfrac{\displaystyle\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\sigma_{0}^{2}}\sim\chi^{2}(n) & \sigma^{2}\gt \sigma_{0}^{2} & \chi^{2}\gt \chi^{2}_{\alpha}(n) \\ \\ \sigma^{2}\ge\sigma_{0}^{2} & \chi^{2}=\dfrac{\displaystyle\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\sigma_{0}^{2}}\sim\chi^{2}(n) & \sigma^{2}\lt \sigma_{0}^{2} & \chi^{2}\lt \chi^{2}_{1-\alpha}(n) \\ \\ \sigma^{2}=\sigma_{0}^{2} & \chi^{2}=\dfrac{\displaystyle\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\sigma_{0}^{2}}\sim\chi^{2}(n) & \sigma^{2}\neq\sigma_{0}^{2} & \chi^{2}\lt \chi^{2}_{1-\alpha/2}(n)\quad\text{或}\quad \chi^{2}\gt \chi^{2}_{\alpha/2}(n) \end{array} \]

\[ \boxed{\text{均值未知，方差的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \sigma^{2}\le\sigma_{0}^{2} & \chi^{2}=\dfrac{(n-1)S^{2}}{\sigma_{0}^{2}}\sim\chi^{2}(n-1) & \sigma^{2}\gt \sigma_{0}^{2} & \chi^{2}\gt \chi^{2}_{\alpha}(n-1) \\ \\ \sigma^{2}\ge\sigma_{0}^{2} & \chi^{2}=\dfrac{(n-1)S^{2}}{\sigma_{0}^{2}}\sim\chi^{2}(n-1) & \sigma^{2}\lt \sigma_{0}^{2} & \chi^{2}\lt \chi^{2}_{1-\alpha}(n-1) \\ \\ \sigma^{2}=\sigma_{0}^{2} & \chi^{2}=\dfrac{(n-1)S^{2}}{\sigma_{0}^{2}}\sim\chi^{2}(n-1) & \sigma^{2}\neq\sigma_{0}^{2} & \chi^{2}\lt \chi^{2}_{1-\alpha/2}(n-1)\quad\text{或}\quad \chi^{2}\gt \chi^{2}_{\alpha/2}(n-1) \end{array} \]

\[ \boxed{\text{方差已知，均值差的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \mu_{1}-\mu_{2}=\delta & U=\dfrac{\bar X-\bar Y-\delta}{\sqrt{\dfrac{\sigma_{1}^{2}}{n}+\dfrac{\sigma_{2}^{2}}{m}}}\sim N(0,1) & \mu_{1}-\mu_{2}\neq\delta & |U|\gt u_{\alpha/2} \\ \\ \mu_{1}-\mu_{2}\le\delta & U=\dfrac{\bar X-\bar Y-\delta}{\sqrt{\dfrac{\sigma_{1}^{2}}{n}+\dfrac{\sigma_{2}^{2}}{m}}}\sim N(0,1) & \mu_{1}-\mu_{2}\gt \delta & U\gt u_{\alpha} \\ \\ \mu_{1}-\mu_{2}\ge\delta & U=\dfrac{\bar X-\bar Y-\delta}{\sqrt{\dfrac{\sigma_{1}^{2}}{n}+\dfrac{\sigma_{2}^{2}}{m}}}\sim N(0,1) & \mu_{1}-\mu_{2}\lt \delta & U\lt -u_{\alpha} \end{array} \]

\[ \boxed{\text{两总体方差相等，配对检验}t\text{ 检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \mu_{1}-\mu_{2}=\delta & T=\dfrac{\bar X-\bar Y-\delta}{S_{w}\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}}\sim t\bigl(n+m-2\bigr),\; S_{w}^{2}=\dfrac{(n-1)S_{1}^{2}+(m-1)S_{2}^{2}}{n+m-2} & \mu_{1}-\mu_{2}\neq\delta & |T|\gt t_{\alpha/2}(n+m-2) \\ \\ \mu_{1}-\mu_{2}\le\delta & T\text{ 同上} & \mu_{1}-\mu_{2}\gt \delta & T\gt t_{\alpha}(n+m-2) \\ \\ \mu_{1}-\mu_{2}\ge\delta & T\text{ 同上} & \mu_{1}-\mu_{2}\lt \delta & T\lt -t_{\alpha}(n+m-2) \end{array} \]

\[ \boxed{\text{两总体方差不相等，大容量}t\text{ 检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \mu_{1}-\mu_{2}=\delta & T=\dfrac{\bar Z-\delta}{S_{Z}/\sqrt n}\sim t(n-1),\; \bar Z=\dfrac{1}{n}\sum_{i=1}^{n}Z_{i},\; Z_{i}=X_{i}-Y_{i},\; S_{Z}^{2}=\dfrac{1}{n-1}\sum_{i=1}^{n}(Z_{i}-\bar Z)^{2} & \mu_{1}-\mu_{2}\neq\delta & |T|\gt t_{\alpha/2}(n-1) \\ \\ \mu_{1}-\mu_{2}\le\delta & T\text{ 同上} & \mu_{1}-\mu_{2}\gt \delta & T\gt t_{\alpha}(n-1) \\ \\ \mu_{1}-\mu_{2}\ge\delta & T\text{ 同上} & \mu_{1}-\mu_{2}\lt \delta & T\lt -t_{\alpha}(n-1) \end{array} \]

\[ \boxed{\text{均值已知，方差比的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \sigma_{1}^{2}\le\sigma_{2}^{2} & F=\dfrac{\displaystyle \frac{m}{n}\sum_{i=1}^{m}(X_{i}-\mu_{1})^{2}} {\displaystyle \sum_{j=1}^{n}(Y_{j}-\mu_{2})^{2}} \sim F(n,m) & \sigma_{1}^{2}\lt \sigma_{2}^{2} & F\gt F_{\alpha}(n,m) \\ \\ \sigma_{1}^{2}\ge\sigma_{2}^{2} & F\text{ 同上} & \sigma_{1}^{2}\lt \sigma_{2}^{2} & F\lt F_{1-\alpha}(n,m) \\ \\ \sigma_{1}^{2}=\sigma_{2}^{2} & F\text{ 同上} & \sigma_{1}^{2}\neq\sigma_{2}^{2} & F\lt F_{1-\alpha/2}(n,m)\quad\text{或}\quad F\gt F_{\alpha/2}(n,m) \end{array} \]

\[ \boxed{\text{均值未知，方差比的检验}} \] \[ \begin{array}{c|c|c|c} \text{原假设 }H_{0} & \text{检验统计量在 }H_{0}\text{ 下的分布} & \text{备择假设 }H_{1} & \text{拒绝域} \\ \hline \sigma_{1}^{2}\le\sigma_{2}^{2} & F=\dfrac{S_{1}^{2}}{S_{2}^{2}}\sim F(n-1,m-1) & \sigma_{1}^{2}\gt \sigma_{2}^{2} & F\gt F_{\alpha}(n-1,m-1) \\ \\ \sigma_{1}^{2}\ge\sigma_{2}^{2} & F\text{ 同上} & \sigma_{1}^{2}\lt \sigma_{2}^{2} & F\lt F_{1-\alpha}(n-1,m-1) \\ \\ \sigma_{1}^{2}=\sigma_{2}^{2} & F\text{ 同上} & \sigma_{1}^{2}\neq\sigma_{2}^{2} & F\lt F_{1-\alpha/2}(n-1,m-1)\quad\text{或}\quad F\gt F_{\alpha/2}(n-1,m-1) \end{array} \]

\[ \boxed{\text{二项分布参数 }p\text{ 的 }U\text{ 检验（正态近似）}} \] \[ \begin{array}{c|c} \text{原假设 }H_{0}\;\&\;\text{备择假设 }H_{1} & \text{显著性水平 }\alpha\text{ 下的拒绝域} \\ \hline H_{0}:p=p_{0},\;H_{1}:p\neq p_{0} & \displaystyle \left|\frac{\bar X-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}\right|\gt u_{\alpha/2} \\ \\ H_{0}:p\ge p_{0},\;H_{1}:p\lt p_{0} & \displaystyle \frac{\bar X-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}\lt -u_{\alpha} \\ \\ H_{0}:p\le p_{0},\;H_{1}:p\gt p_{0} & \displaystyle \frac{\bar X-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}\gt u_{\alpha} \end{array} \] \[ \text{其中 }\; U=\frac{\bar X-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}} \;\stackrel{\text{近似}}{\sim}\;N(0,1). \]

\[ \begin{aligned} &\textbf{Pearson 定理}\\ &\text{设总体按 }k\text{ 个区间分组，记原假设 }H_{0}:X\sim F(x)。\\ &\text{若样本容量 }n\text{ 充分大，则无论 }F(x)\text{ 为何分布，统计量}\\[4pt] &\qquad \chi^{2}=\sum_{i=1}^{k} \frac{\bigl(n_{i}-np_{i}\bigr)^{2}}{np_{i}} \;\stackrel{\text{近似}}{\sim}\; \chi^{2}\bigl(k-r-1\bigr),\\[6pt] &\text{其中 }p_{i}=P\bigl(X\text{ 落入第 }i\text{ 组}\bigr),\; r\text{ 为 }F(x)\text{ 中未知参数的个数。}\\[8pt] &\text{因此在显著性水平 }\alpha\text{ 下的拒绝域为}\\[4pt] &\qquad \chi^{2}=\sum_{i=1}^{k} \frac{\bigl(n_{i}-np_{i}\bigr)^{2}}{np_{i}} \gt \chi^{2}_{\alpha}\bigl(k-r-1\bigr). \end{aligned} \]