第4講統計的検定

正規母集団の等分散の検定と推定

統計量

{U_{1}}^{2} = \frac{1}{n_{1} - 1} \sum_{i = 1}^{n_{1}} (X_{1 i} - {\bar{X}}_{1})^{2}

$\begin{equation} {U_1}^{2}= \frac{1}{n_1 - 1}\sum_{i=1}^{n_1}(X_{1i} - \bar{X}_1)^2 \end{equation}$

{U_{2}}^{2} = \frac{1}{n_{2} - 1} \sum_{i = 1}^{n_{2}} (X_{2 i} - {\bar{X}}_{2})^{2}

$\begin{equation} {U_2}^{2}= \frac{1}{n_2 - 1}\sum_{i=1}^{n_2}(X_{2i} - \bar{X}_2)^2 \end{equation}$

F = \frac{{U_{1}}^{2}}{{U_{2}}^{2}} \sim F (n_{1} - 1, n_{2} - 1)

$\begin{equation} F = \frac{{U_1}^{2}}{{U_2}^{2}} \sim F(n_1 - 1,~n_2 - 1) \end{equation}$

検定

$H_{0}:~~{\sigma_{1}}^{2} = {\sigma_{2}}^{2}~~~~H_{1}:~\sigma^{2} \ne {\sigma_{0}}^{2}$

T = \frac{{U_{1}}^{2}}{{U_{2}}^{2}} (\geq 1) > F_{\frac{α}{2}} (n_{1} - 1, n_{2} - 1)

$\begin{equation} T = \frac{{U_1}^2}{{U_2}^2}~(\ge 1)~ > F_{\frac{\alpha}{2}}(n_1-1,n_2-1) \end{equation}$

T = \frac{{U_{2}}^{2}}{{U_{1}}^{2}} (\geq 1) > F_{\frac{α}{2}} (n_{2} - 1, n_{1} - 1)

$\begin{equation} T = \frac{{U_2}^2}{{U_1}^2}~(\ge 1)~ > F_{\frac{\alpha}{2}}(n_2-1,n_1-1) \end{equation}$

$H_{0}:~~{\sigma_{1}}^{2} = {\sigma_{2}}^{2}~~~~H_{1}:~{\sigma_{1}}^{2} > {\sigma_{2}}^{2}$

T = \frac{{U_{1}}^{2}}{{U_{2}}^{2}} > F_{α} (n_{1} - 1, n_{2} - 1)

$\begin{equation} T = \frac{{U_1}^2}{{U_2}^2} > F_{\alpha}(n_1-1,n_2-1) \end{equation}$

$H_{0}:~~{\sigma_{1}}^{2} = {\sigma_{2}}^{2}~~~~H_{1}:~{\sigma_{1}}^{2} < {\sigma_{2}}^{2}$

T = \frac{{U_{2}}^{2}}{{U_{1}}^{2}} > F_{α} (n_{2} - 1, n_{1} - 1)

$\begin{equation} T = \frac{{U_2}^2}{{U_1}^2} > F_{\alpha}(n_2-1,n_1-1) \end{equation}$

区間推定

F_{1 - \frac{α}{2}} (n_{1} - 1, n_{2} - 1) \frac{{U_{2}}^{2}}{{U_{1}}^{2}} \leq \frac{{σ_{2}}^{2}}{{σ_{1}}^{2}} \leq F_{\frac{α}{2}} (n_{1} - 1, n_{2} - 1) \frac{{U_{2}}^{2}}{{U_{1}}^{2}}

$\begin{equation} F_{1-\frac{\alpha}{2}}(n_1-1,n_2-1)\frac{{U_{2}}^{2}}{{U_{1}}^{2}} \le \frac{{\sigma_{2}}^{2}}{{\sigma_{1}}^{2}} \le F_{\frac{\alpha}{2}}(n_1-1,n_2-1)\frac{{U_{2}}^{2}}{{U_{1}}^{2}} \end{equation}$

第4講 統計的検定

正規母集団の等分散の検定と推定

統計量

検定

区間推定

第4講統計的検定