[Statistics] a群體與b群體各別標準差求整體標準差
Given \(\sigma_a\) and \(\sigma_b\), Ask for \(\sigma\) 1. 簡化 \(\sigma\) 將 \(\sigma\) 乘開 \(\sigma = \sqrt{\frac{\sum{(x_i-\bar x)^2}}{n}}\) \(\sigma = \sqrt{\frac{\sum{x_i^2-2\bar x\sum{x_i}+n\bar x^2}}{n}}\) 平均等於總和除以個數 \(\frac{\sum x_i}{n}=\bar x\)，故 \(\sigma = \sqrt{\frac{\sum x_i^2}{n}-\frac{2\bar x\sum x_i}{n}+\frac{n\bar x^2}{n}}\) \(\sigma = \sqrt{\frac{\sum x_i^2}{n}-2\bar x^2+\bar x^2}\) 得 \(\boxed{\sigma = \sqrt{\frac{\sum x_i^2}{n}-\bar x^2}}-(1)\) 2. 求個別平方和 由\((1)\)式可推得各別的標準差為 \(\boxed{\sigma_a = \sqrt{\frac{\sum x_{ai}^2}{n_a}-\bar x_a^2}}-(2)\) 且 \(\boxed{n = n_a+n_b}-(3)\) \(\boxed{\sum x_i^2=\sum x_{ai}^2+\sum x_{bi}^2}-(4)\) 欲求 \(\sum x_{ai}^2\)，我們將\((2)\)式展開 \(\sigma_a^2 = \frac{\sum x_{ai}^2}{n_a}-\bar x_a^2\) \(\sigma_a^2+\bar x_a^2= \frac{\sum x_{ai}^2}{n_a}\) 得\(\boxed{\sum x_{ai}^2=n_a(\sigma_a^2+\bar x_a^2)}-(5)\) 3....