<p>相関係数の基本的な計算方法を学ぶ問題です。</p><h4>相関係数の計算公式</h4><p>ピアソンの積率相関係数は以下の公式で計算されます：</p><p class='formula'>$r = \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum(x_i - \bar{x})^2 \sum(y_i - \bar{y})^2}}
lt;/p><p class='step'>1. 平均値の計算</p><p>まず、xとyの平均値を求めます：</p><p class='formula'>$\bar{x} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3
lt;/p><p class='formula'>$\bar{y} = \frac{3 + 5 + 7 + 9 + 11}{5} = \frac{35}{5} = 7
lt;/p><p class='step'>2. 偏差の計算</p><p>各データ点の平均からの偏差を計算します：</p><ul><li>$(x_1, y_1) = (1, 3)$：$x_1 - \bar{x} = 1 - 3 = -2$, $y_1 - \bar{y} = 3 - 7 = -4
lt;/li><li>$(x_2, y_2) = (2, 5)$：$x_2 - \bar{x} = 2 - 3 = -1$, $y_2 - \bar{y} = 5 - 7 = -2
lt;/li><li>$(x_3, y_3) = (3, 7)$：$x_3 - \bar{x} = 3 - 3 = 0$, $y_3 - \bar{y} = 7 - 7 = 0
lt;/li><li>$(x_4, y_4) = (4, 9)$：$x_4 - \bar{x} = 4 - 3 = 1$, $y_4 - \bar{y} = 9 - 7 = 2
lt;/li><li>$(x_5, y_5) = (5, 11)$：$x_5 - \bar{x} = 5 - 3 = 2$, $y_5 - \bar{y} = 11 - 7 = 4
lt;/li></ul><p class='step'>3. 各項の計算</p><p><strong>分子の計算：</strong></p><p class='formula'>$\begin{align}\sum(x_i - \bar{x})(y_i - \bar{y}) &= (-2)(-4) + (-1)(-2) + 0 + (1)(2) + (2)(4) \\&= 8 + 2 + 0 + 2 + 8 = 20\end{align}
lt;/p><p><strong>分母の計算：</strong></p><p class='formula'>$\sum(x_i - \bar{x})^2 = (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 4 + 1 + 0 + 1 + 4 = 10
lt;/p><p class='formula'>$\sum(y_i - \bar{y})^2 = (-4)^2 + (-2)^2 + 0^2 + 2^2 + 4^2 = 16 + 4 + 0 + 4 + 16 = 40
lt;/p><p class='step'>4. 相関係数の算出</p><p class='formula'>$r = \frac{20}{\sqrt{10 \times 40}} = \frac{20}{\sqrt{400}} = \frac{20}{20} = 1.00
lt;/p><div class='key-point'><div class='key-point-title'>結果の解釈</div><ul><li><strong>r = 1.00：</strong> 完全な正の相関</li><li><strong>データの関係：</strong> y = 2x + 1（完全な直線関係）</li><li><strong>実用的意味：</strong> xの値から yの値を100%正確に予測可能</li></ul></div><p>したがって、相関係数は<strong>1.00</strong>です。</p>