第4講統計的検定

正規母集団の母平均の検定と推定

母分散が既知の場合

統計量

T = \frac{\bar{X} - μ_{0}}{\sqrt{\frac{{σ_{0}}^{2}}{n}}} \sim N (0, 1)

$\begin{equation} T = \frac{\bar{X}-\mu_{0}}{\ds\sqrt{\frac{{\sigma_0}^{2}}{n}}} \sim N(0, 1) \end{equation}$

検定

$H_{0}:~\mu = \mu_{0}~~~~H_{1}:~\mu \ne \mu_0$

| T | > z_{\frac{α}{2}}

$\begin{equation} |T| > z_{\frac{\alpha}{2}} \end{equation}$

$H_{0}:~\mu = \mu_{0}~~~~H_{1}:~\mu > \mu_0$

T > z_{α}

$\begin{equation} T > z_{\alpha} \end{equation}$

$H_{0}:~\mu = \mu_{0}~~~~H_{1}:~\mu < \mu_0$

T < - z_{α}

$\begin{equation} T < -z_{\alpha} \end{equation}$

区間推定

\bar{X} - z_{\frac{α}{2}} \sqrt{\frac{{σ_{0}}^{2}}{n}} \leq μ \leq \bar{X} + z_{\frac{α}{2}} \sqrt{\frac{{σ_{0}}^{2}}{n}}

$\begin{equation} \bar{X} - z_{\frac{\alpha}{2}}\sqrt{\frac{{\sigma_0}^2}{n}} \le \mu \le \bar{X} + z_{\frac{\alpha}{2}}\sqrt{\frac{{\sigma_0}^2}{n}} \end{equation}$

母分散が未知の場合

統計量

T = \frac{\bar{X} - μ_{0}}{\sqrt{\frac{U^{2}}{n}}} \sim t (n - 1)

$\begin{equation} T = \frac{\bar{X}-\mu_{0}}{\ds\sqrt{\frac{{U}^{2}}{n}}} \sim t(n-1) \end{equation}$

U^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (X_{i} - \bar{X})^{2}

$\begin{equation} {U}^{2}= \frac{1}{n-1}\sum_{i=1}^{n}(X_{i} - \bar{X})^2 \end{equation}$

検定

$H_{0}:~~\mu = \mu_{0}~~~~H_{1}:~\mu \ne \mu_0$

| T | > t_{\frac{α}{2}} (n - 1)

$\begin{equation} |T| > t_{\frac{\alpha}{2}}(n-1) \end{equation}$

$H_{0}:~~\mu = \mu_{0}~~~~H_{1}:~\mu > \mu_0$

T > t_{α} (n - 1)

$\begin{equation} T > t_{\alpha}(n-1) \end{equation}$

$H_{0}:~~\mu = \mu_{0}~~~~H_{1}:~\mu < \mu_0$

T < - t_{α} (n - 1)

$\begin{equation} T < -t_{\alpha}(n-1) \end{equation}$

区間推定

\bar{X} - t_{\frac{α}{2}} (n - 1) \sqrt{\frac{U^{2}}{n}} \leq μ \leq \bar{X} + t_{\frac{α}{2}} (n - 1) \sqrt{\frac{U^{2}}{n}}

$\begin{equation} \bar{X} - t_{\frac{\alpha}{2}}(n-1)\sqrt{\frac{{U}^2}{n}} \le \mu \le \bar{X} + t_{\frac{\alpha}{2}}(n-1)\sqrt{\frac{{U}^2}{n}} \end{equation}$

第4講 統計的検定

正規母集団の母平均の検定と推定

母分散が既知の場合

統計量

検定

区間推定

母分散が未知の場合

統計量

検定

区間推定

第4講統計的検定