【数学I】例題5.1.13：相関係数の計算（One More）★★

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% 例題I5.1.13：相関係数の計算（One More）★★
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次の表は，5名の生徒A，B，C，D，Eの読書時間$x$（時間）とレポートの評価$y$（点）を測定した結果である．このとき，$x$と$y$の相関係数$r$を求めよ．

\begin{tabular}{|c||c|c|c|c|c|}
\hline&A&B&C&D&E\\
\hline 読書時間$x$（時間）&8&8&10&6&8\\
\hline 評価$y$（点）&5&6&10&4&10\\
\hline
\end{tabular}

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% 例題I5.1.13の解答（One More）★★
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$x,y$のデータの平均値をそれぞれ$\overline{x},\overline{y}$とすると，

\begin{align*}
\overline{x}&=\frac{1}{5}(8+8+10+6+8)=8(\text{時間}),\\
\overline{y}&=\frac{1}{5}(5+6+10+4+10)=7(\text{点})
\end{align*}

したがって，下のような表が得られる．

\begin{tabular}{|c||c|c|c|c|c|c|c|}
\hline&$x$&$y$&$x-\overline{x}$&$y-\overline{y}$&$(x-\overline{x})^2$&$(y-\overline{y})^2$&$(x-\overline{x})(y-\overline{y})$\\
\hline A&$8$&$5$&$0$&$-2$&$0$&$4$&$0$\\
\hline B&$8$&$6$&$0$&$-1$&$0$&$1$&$0$\\
\hline C&$10$&$10$&$2$&$3$&$4$&$9$&$6$\\
\hline D&$6$&$4$&$-2$&$-3$&$4$&$9$&$6$\\
\hline E&$8$&$10$&$0$&$3$&$0$&$9$&$0$\\
\hline 計&$40$&$35$&$0$&$0$&$8$&$32$&$12$\\
\hline
\end{tabular}

よって，相関係数$r$は，

\[
r=\frac{12}{\sqrt{8\times 32}}=\frac{12}{\sqrt{256}}=\frac{12}{16}=0.75
\]

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% 問題I5.1.13：相関係数の計算（One More）★★
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次の表は，5名の生徒A，B，C，D，Eの運動時間$x$（時間）と体力テストの点数$y$（点）を測定した結果である．このとき，$x$と$y$の相関係数$r$を求めよ．

\begin{tabular}{|c||c|c|c|c|c|}
\hline&A&B&C&D&E\\
\hline 運動時間$x$（時間）&8&6&8&5&8\\
\hline 点数$y$（点）&10&12&15&10&13\\
\hline
\end{tabular}

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% 問題I5.1.13の解答（One More）★★
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$x,y$のデータの平均値をそれぞれ$\overline{x},\overline{y}$とすると，

\begin{align*}
\overline{x}&=\frac{1}{5}(8+6+8+5+8)=7(\text{時間}),\\
\overline{y}&=\frac{1}{5}(10+12+15+10+13)=12(\text{点})
\end{align*}

したがって，下のような表が得られる．

\begin{tabular}{|c||c|c|c|c|c|c|c|}
\hline&$x$&$y$&$x-\overline{x}$&$y-\overline{y}$&$(x-\overline{x})^2$&$(y-\overline{y})^2$&$(x-\overline{x})(y-\overline{y})$\\
\hline A&$8$&$10$&$1$&$-2$&$1$&$4$&$-2$\\
\hline B&$6$&$12$&$-1$&$0$&$1$&$0$&$0$\\
\hline C&$8$&$15$&$1$&$3$&$1$&$9$&$3$\\
\hline D&$5$&$10$&$-2$&$-2$&$4$&$4$&$4$\\
\hline E&$8$&$13$&$1$&$1$&$1$&$1$&$1$\\
\hline 計&$35$&$60$&$0$&$0$&$8$&$18$&$6$\\
\hline
\end{tabular}

よって，相関係数$r$は，$r={\frac{6}{\sqrt{8\times 18}}}=\frac{6}{\sqrt{144}}=\frac{6}{12}=0.5$

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% 例題I5.1.13の側注：相関係数の計算（One More）★★
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% 側注1
表を用いるとよい．

% 側注2
$r=\frac{s_{xy}}{s_xs_y}$

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% 問題I5.1.13の側注：相関係数の計算（One More）★★
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% 側注1
表を用いるとよい．

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% 例題I5.1.13の考え方・補足など：相関係数の計算（One More）★★
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% 考え方
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$x,y$のデータの標準偏差をそれぞれ$s_x,s_y$とし，$x$と$y$の共分散を$s_{xy}$とするとき，次のように{相関係数$r$}は共分散を$x$と$y$の標準偏差で割った値で表される．

\[
{r}=\frac{s_{xy}}{s_xs_y}=\frac{\frac{1}{n}\{(x-\overline{x})(y-\overline{y})\text{の和}\}}{\sqrt{\frac{1}{n}\{(x-\overline{x})^2\text{の和}\}}\sqrt{\frac{1}{n}\{(y-\overline{y})^2\text{の和}\}}}=\frac{\{(x-\overline{x})(y-\overline{y})\text{の和}\}}{\sqrt{\{(x-\overline{x})^2\text{の和}\}\times\{(y-\overline{y})^2\text{の和}\}}}
\]

このとき，下の解答のように表を作成して計算するとよい．