概率笔记 P08：假设检验
1 定义
假设检验
根据样本，按照一定的规则判断所做的假设真伪，并作出接受还是拒绝此假设的决定。
两类错误
对于假设\(H_0\)，
- 第一类错误：拒绝实际真的假设。
- 第二类错误：接受实际不真的假设。
显著性
- 显著性水平：在假设检验中允许犯第一类错误的概率。记作\(\alpha\)，它表现了对弃真程度的控制。即发生\(H_0\)为真而拒绝了\(H_0\)这种事件的概率。
- 显著性检验：只控制第一类错误的检验。
显著性检验的一般步骤
- 根据问题提出原假设\(H_0\)，（由此也可以得出对立的假设\(H_1）\)；
- 给出显著性水平\(\alpha\)，可以取\(0.1, 0.05, 0.01, 0.001\)等；
- 确定检验的统计量、拒绝域\(W\)形式；
- 按照犯第一类错误的概率等于\(\alpha\)求出拒绝域\(W\)的具体值；
- 根据样本值计算检验统计量\(T\)的观测值\(t\)，若\(t \in W\)，则拒绝原假设\(H_0\)，否则，接受原假设\(H_0\)。
2 正态总体参数的假设检验
设显著性水平为\(\alpha\)，单个正态总体为\(N(\mu, \sigma^2)\)的参数的假设检验以及两个正态总体\(N(\mu_1, \sigma_1^2), N(\mu_2, \sigma_2^2)\)的\(\mu_1 - \mu_2\)和\(\sigma_1^2 = \sigma_2^2\)的假设检验。
表中\(S_\omega = \sqrt{\cfrac {(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}}\)
\[\begin{array}{c|c|c|c|c|c} \hline \\ \text{parameter test} & H_0 & H_1 & \text{static test} & {\text {distribution of} \\ \text{static when} \\ H_0 \text{ is true}} & \text{reject region} \\ \hline \\ \mu \ (\sigma^2 \text{ known}) & \begin{array}{c} \mu = \mu_0 \\ \mu \le \mu_0 \\ \mu \ge \mu_0 \\ \end{array} & \begin{array}{c} \mu \neq \mu_0 \\ \mu \gt \mu_0 \\ \mu \lt \mu_0 \\ \end{array} & U = \cfrac {\overline{X} - \mu_0}{\sigma / \sqrt{n}} & N(0, 1) & \begin{array}{c} |U| \ge u_{\frac \alpha 2} \\ U \ge u_\alpha \\ U \le -u_\alpha \\ \end{array} \\ \hline \\ \mu \ (\sigma^2 \text{ unknown}) & \begin{array}{c} \mu = \mu_0 \\ \mu \le \mu_0 \\ \mu \ge \mu_0 \\ \end{array} & \begin{array}{c} \mu \neq \mu_0 \\ \mu \gt \mu_0 \\ \mu \lt \mu_0 \\ \end{array} & T = \cfrac {\overline{X} - \mu_0}{S / \sqrt{n}} & t(n-1) & \begin{array}{c} |T| \ge t_{\frac \alpha 2}(n-1) \\ T \ge t_\alpha(n-1) \\ T \le -t_\alpha(n-1) \\ \end{array} \\ \hline \\ \sigma^2 \ (\mu \text{ known}) & \begin{array}{c} \sigma^2 = \sigma_0^2 \\ \sigma^2 \le \sigma_0^2 \\ \sigma^2 \ge \sigma_0^2 \\ \end{array} & \begin{array}{c} \sigma^2 \neq \sigma_0^2 \\ \sigma^2 \gt \sigma_0^2 \\ \sigma^2 \lt \sigma_0^2 \\ \end{array} & \chi^2 = \cfrac 1{\sigma_0^2} \displaystyle \sum_{i=1}^n (X_i - \mu)^2 & \chi^2(n) & \begin{array}{c} \chi^2 \le \chi_{1-\frac{\alpha}2}^2(n) \ or \\ \chi^2 \ge \chi_{\frac \alpha 2}^2(n) \\ \chi^2 \ge \chi_\alpha^2(n) \\ \chi^2 \le \chi_{1 - \alpha}^2(n) \\ \end{array} \\ \hline \\ \sigma^2 \ (\mu \text{ unknown}) & \begin{array}{c} \sigma^2 = \sigma_0^2 \\ \sigma^2 \le \sigma_0^2 \\ \sigma^2 \ge \sigma_0^2 \\ \end{array} & \begin{array}{c} \sigma^2 \neq \sigma_0^2 \\ \sigma^2 \gt \sigma_0^2 \\ \sigma^2 \lt \sigma_0^2 \\ \end{array} & \chi^2 = \cfrac {(n-1)S^2}{\sigma_0^2} & \chi^2(n-1) & \begin{array}{c} \chi^2 \le \chi_{1-\frac{\alpha}2}^2(n-1) \ or \\ \chi^2 \ge \chi_{\frac \alpha 2}^2(n-1) \\ \chi^2 \ge \chi_\alpha^2(n-1) \\ \chi^2 \le \chi_{1 - \alpha}^2(n-1) \\ \end{array} \\ \hline \\ \mu_1 - \mu_2 \\ (\sigma_1^2, \sigma_2^2 \text{ known}) & \begin{array}{c} \mu_1 - \mu_2 = \mu_0 \\ \mu_1 - \mu_2 \le \mu_0 \\ \mu_1 - \mu_2 \ge \mu_0 \end{array} & \begin{array}{c} \mu_1 - \mu_2 \neq \mu_0 \\ \mu_1 - \mu_2 \gt \mu_0 \\ \mu_1 - \mu_2 \lt \mu_0 \end{array} & U = \cfrac {\overline{X} - \overline{Y} - \mu_0}{\sqrt{\cfrac {\sigma_1^2}{n_1} + \cfrac {\sigma_2^2}{n_2}}} & N(0, 1) & \begin{array}{c} |U| \ge u_{\frac \alpha 2} \\ U \ge u_\alpha \\ U \le -u_\alpha \\ \end{array} \\ \hline \\ \mu_1 - \mu_2 \\ (\sigma_1^2, \sigma_2^2 \text{ unknown} \\ \text{but } \sigma_1^2 = \sigma_2^2) & \begin{array}{c} \mu_1 - \mu_2 = \mu_0 \\ \mu_1 - \mu_2 \le \mu_0 \\ \mu_1 - \mu_2 \ge \mu_0 \end{array} & \begin{array}{c} \mu_1 - \mu_2 \neq \mu_0 \\ \mu_1 - \mu_2 \gt \mu_0 \\ \mu_1 - \mu_2 \lt \mu_0 \end{array} & T = \cfrac {\overline{X} - \overline{Y} - \mu_0}{S_\omega \sqrt{\cfrac 1{n_1} + \cfrac 1{n_2}}} & t(n_1 + n_2 - 2) & \begin{array}{c} |T| \ge t_{\frac \alpha 2}(n_1 + n_2 - 2) \\ T \ge t_\alpha(n_1 + n_2 - 2) \\ T \le -t_\alpha(n_1 + n_2 - 2) \\ \end{array} \\ \hline \\ \sigma_1^2 = \sigma_2^2 \\ (\mu_1, \mu_2 \text{ known}) & \begin{array}{c} \sigma_1^2 = \sigma_2^2 \\ \sigma_1^2 \le \sigma_2^2 \\ \sigma_1^2 \ge \sigma_2 \\ \end{array} & \begin{array}{c} \sigma_1^2 \neq \sigma_2^2 \\ \sigma_1^2 \gt \sigma_2^2 \\ \sigma_1^2 \lt \sigma_2 \\ \end{array} & F = \cfrac {n_2 \displaystyle \sum_{i=1}^{n_1} (X_i - \mu_1)^2}{n_1 \displaystyle \sum_{j = 1}^{n_2} (Y_j - \mu_2)^2} & F(n_1, n_2) & \begin{array}{c} F \le F_{1-\frac \alpha 2} (n_1, n_2) \ or \\ F \ge F_{\frac \alpha 2} (n_1, n_2) \\ F \ge F_\alpha (n_1, n_2) \\ F \le F_{1-\alpha} (n_1, n_2) \\ \end{array} \\ \hline \\ \sigma_1^2 = \sigma_2^2 \\ (\mu_1, \mu_2 \text{ unknown}) & \begin{array}{c} \sigma_1^2 = \sigma_2^2 \\ \sigma_1^2 \le \sigma_2^2 \\ \sigma_1^2 \ge \sigma_2 \\ \end{array} & \begin{array}{c} \sigma_1^2 \neq \sigma_2^2 \\ \sigma_1^2 \gt \sigma_2^2 \\ \sigma_1^2 \lt \sigma_2 \\ \end{array} & F = \cfrac {S_1^2}{S_2^2} & F(n_1 - 1, n_2 - 2) & \begin{array}{c} F \le F_{1-\frac \alpha 2} (n_1 - 1, n_2 - 1) \ or \\ F \ge F_{\frac \alpha 2} (n_1 - 1, n_2 - 1) \\ F \ge F_\alpha (n_1 - 1, n_2 - 1) \\ F \le F_{1-\alpha} (n_1 - 1, n_2 - 1) \\ \end{array} \\ \hline \end{array}
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