統計検定 1級過去問解答/解答例と解説
2015年11月29日 (日) 試験

統計数理問5

n_{2} = n - n_{1}

$\begin{equation} n_2 = n - n_1 \end{equation}$

{\bar{y}}_{1} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} y_{i}

$\begin{equation} \bar{y}_1 = \frac{1}{n_1}\sum_{i=1}^{n_1}y_i \end{equation}$

{s_{1}}^{2} = \frac{1}{n_{1} - 1} \sum_{i = 1}^{n_{1}} (y_{i} - {\bar{y}}_{1})^{2}

$\begin{equation} {s_1}^2 = \frac{1}{n_1 - 1}\sum_{i=1}^{n_1}(y_i - \bar{y}_1)^2 \end{equation}$

{\bar{y}}_{2} = \frac{1}{n_{2}} \sum_{i = n_{1} + 1}^{n} y_{i}

$\begin{equation} \bar{y}_2 = \frac{1}{n_2}\sum_{i=n_1 + 1}^{n}y_i \end{equation}$

{s_{2}}^{2} = \frac{1}{n_{2} - 1} \sum_{i = n_{1} + 1}^{n} (y_{i} - {\bar{y}}_{1})^{2}

$\begin{equation} {s_2}^2 = \frac{1}{n_2 - 1}\sum_{i=n_1 + 1}^{n}(y_i - \bar{y}_1)^2 \end{equation}$

\bar{y} = \frac{1}{n} \sum_{i = 1}^{n_{1}} y_{i}

$\begin{equation} \bar{y} = \frac{1}{n}\sum_{i=1}^{n_1}y_i \end{equation}$

s^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (y_{i} - \bar{y})^{2}

$\begin{equation} {s}^2 = \frac{1}{n - 1}\sum_{i=1}^{n}(y_i - \bar{y})^2 \end{equation}$