第4講統計的検定

二項母集団の母比率の差の検定と推定

統計量

T = \frac{{\hat{p}}_{1} - {\hat{p}}_{2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_{1}} + \frac{1}{n_{2}})}} \sim N (0, 1)

$\begin{equation} T = \frac{\hat{p}_{1}-\hat{p}_{2}}{\ds\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \sim N(0, 1) \end{equation}$

\hat{p} = \frac{n_{1} {\hat{p}}_{1} + n_{2} {\hat{p}}_{2}}{n_{1} + n_{2}}

$\begin{equation} \hat{p} = \frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2} \end{equation}$

検定

$H_{0}:~p_{1} = p_{2}~~~~H_{1}:~\mu \ne \mu_0$

| T | > z_{\frac{α}{2}}

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

$H_{0}:~p_{1} = p_{2}~~~~H_{1}:~\mu > \mu_0$

T > z_{α}

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

$H_{0}:~p_{1} = p_{2}~~~~H_{1}:~\mu < \mu_0$

T < - z_{α}

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

区間推定

{\hat{p}}_{1} - {\hat{p}}_{2} - z_{\frac{α}{2}} \sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_{1}} + \frac{1}{n_{2}})} \leq p_{1} - p_{2} \leq {\hat{p}}_{1} - {\hat{p}}_{2} + z_{\frac{α}{2}} \sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_{1}} + \frac{1}{n_{2}})}

$\begin{equation} \hat{p}_{1}-\hat{p}_{2} - z_{\frac{\alpha}{2}} \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)} \le p_{1} - p_{2} \le \hat{p}_{1}-\hat{p}_{2} + z_{\frac{\alpha}{2}} \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)} \end{equation}$

第4講 統計的検定

二項母集団の母比率の差の検定と推定

統計量

検定

区間推定

第4講統計的検定