第5講回帰分析

残差分析と決定係数

残差分析

H_{0} : β_{1} = β_{2} = \dots = β_{p} = 0

$\begin{equation} H_{0}~:~~\beta_{1} = \beta_{2} = \cdots = \beta_{p} = 0 \end{equation}$

要因	平方和	自由度	平均変動	分散比
$R$	$S_{R}$	$\phi_{R} = p$	$\ds V_{R}=\frac{S_{R}}{\phi_{R}}$	$\ds F=\frac{V_{R}}{V_{e}}$
$e$	$S_{e}$	$\phi_{e} = n-p-1$	$\ds V_{e}=\frac{S_{e}}{\phi_{e}}$
合計	$S_{yy}$	$\phi_{yy} = n-1$

重相関係数

\begin{aligned} R & = \frac{\sum_{i = 1}^{n} (y_{i} - \bar{y}) ({\hat{y}}_{i} - \bar{y})}{\sqrt{\sum_{i = 1}^{n} (y_{i} - \bar{y})^{2}} \sqrt{\sum_{i = 1}^{n} ({\hat{y}}_{i} - \bar{y})^{2}}} \\ = \frac{S_{y \hat{y}}}{\sqrt{S_{y y}} \sqrt{S_{\hat{y} \hat{y}}}} \\ = \frac{S_{\hat{y} \hat{y}}}{\sqrt{S_{y y}} \sqrt{S_{\hat{y} \hat{y}}}} \\ = \frac{\sqrt{S_{\hat{y} \hat{y}}}}{\sqrt{S_{y y}}} \\ = \sqrt{\frac{S_{\hat{y} \hat{y}}}{S_{y y}}} \\ = \sqrt{\frac{S_{R}}{S_{y y}}} \end{aligned}

$\begin{align} R &= \dfrac{\sum_{i=1}^{n}(y_i - \bar{y})(\hat{y}_i - \bar{y})}{\sqrt{\sum_{i=1}^{n}(y_i - \bar{y})^2}\sqrt{\sum_{i=1}^{n}(\hat{y}_i - \bar{y})^2}} \\ &=\frac{S_{y\hat{y}}}{\sqrt{S_{yy}}\sqrt{S_{\hat{y}\hat{y}}}} \\ &= \frac{S_{\hat{y}\hat{y}}}{\sqrt{S_{yy}}\sqrt{S_{\hat{y}\hat{y}}}} \\ &= \frac{\sqrt{S_{\hat{y}\hat{y}}}}{\sqrt{S_{yy}}} \\ &= \sqrt{\frac{S_{\hat{y}\hat{y}}}{S_{yy}}} \\ &= \sqrt{\frac{S_{R}}{S_{yy}}} \end{align}$

決定係数

R^{2} = \frac{S_{R}}{S_{y y}}

$\begin{equation} R^2 = \dfrac{S_{R}}{S_{yy}} \end{equation}$

自由度調整済み決定係数

R^{* 2} = 1 - \frac{\frac{S_{e}}{ϕ_{e}}}{\frac{S_{y y}}{ϕ_{y y}}}

$\begin{equation} R^{*2} = 1 - \dfrac{\dfrac{S_{e}}{\phi_e}}{\dfrac{S_{yy}}{\phi_{yy}}} \end{equation}$