前言
相关公式
(1) . 如果 \(y_1=x_1+b\), \(y_2=x_2+b\), \(\cdots\), \(y_n=x_n+b\), 那么 \(s_y^2\)\(=\)\(s_x^2\);
证明:由于 \(s_x^2=\cfrac{1}{n}\sum\limits_{i}^{n}(x_i-\bar{x})^2\), 且 \(\bar{y}=\bar{x}+b\)
\(s_y^2=\cfrac{1}{n}\sum\limits_{i=1}^{n}(y_i-\bar{y})^2=\cfrac{1}{n}\sum\limits_{i=1}^{n}[(x_i+b)-(\bar{x}+b)]^2=\cfrac{1}{n}\sum\limits_{i}^{n}(x_i-\bar{x})^2=s_x^2\)
故有 \(s_y^2\)\(=\)\(s_x^2\);
(2) . 如果 \(y_1=ax_1\), \(y_2=ax_2\), \(\cdots\), \(y_n=a x_n\), 那么 \(s_y^2\)\(=\)\(a^2\)\(\cdot s_x^2\).
证明:由于 \(s_x^2=\cfrac{1}{n}\sum\limits_{i}^{n}(x_i-\bar{x})^2\), 且 \(\bar{y}=a\cdot\bar{x}\)
\(s_y^2=\cfrac{1}{n}\sum\limits_{i=1}^{n}(y_i-\bar{y})^2\)\(=\)\(\cfrac{1}{n}\sum\limits_{i=1}^{n}(a\cdot x_i-a\cdot\bar{x})^2\)
\(=\)\(\cfrac{1}{n}\sum\limits_{i=1}^{n}a^2(x_i-\bar{x})^2\)\(=\)\(a^2\cdot\cfrac{1}{n}\sum\limits_{i=1}^{n}(x_i-\bar{x})^2\)\(=\)\(a^2\cdot s_x^2\)
仿上题证明;
证明:由于 \(s_x^2=\cfrac{1}{n}\sum\limits_{i}^{n}(x_i-\bar{x})^2=0\),故 \(x_-\bar{x}=0\),即 \(x_i=\bar{x}\), \(x_i\)(\(i\)\(=\)\(1\),\(2\),\(\cdots\),\(n\)) ,得证 .
(1). \(\bar{w}\)\(=\)\(\cfrac{l}{l+m+n}\cdot\bar{x}\)\(+\)\(\cfrac{m}{l+m+n}\cdot\bar{y}\)\(+\)\(\cfrac{n}{l+m+n}\cdot\bar{z}\);
(2). \(s^2\)\(=\)\(\cfrac{1}{l+m+n}\)\(\bigg\{l\cdot\left[s_1^2+(\bar{x}-\bar{w})^2\right]\)\(+\)\(m\cdot\left[s_2^2+(\bar{y}-\bar{w})^2\right]\)\(+\)\(n\cdot\left[s_3^2+(\bar{z}-\bar{w})^2\right]\bigg\}\).