E 方差分析的前提假定

1. \(\sigma_1^2=\sigma_2^2=...=\sigma_m^2\)；b) \(n_1=n_2=...=n_m\)

方差分析的统计量

\[F=\frac{MSB}{MSW}\sim F(m-1;n-m)\]

其中，

\[MSB=\frac{SSB}{m-1},SSB=\sum_{i=1}^mn_i(\overline{y_i}-\overline y)^2\] \[MSW=\frac{SSW}{n-m},SSW=\sum_{i=1}^m\sum^{n_i}_{j=1}(y_{ij}-\overline {y_i})^2\]

我们知道，对满足正态分布的总体而言，修正的样本方差\(S_d^2\)满足：

\[\frac{(n-1)S_d^2}{\sigma^2}\sim χ^2 (n-1),S_d^2=\frac{1}{n-1}\sum_{i=1}^n(y_i-\overline y)^2\]

观察\(SSB\)和\(SSW\)：

\[SSB=\sum_{i=1}^mn_i(\overline{y_i}-\overline y)^2=\sum_{i=1}^m\frac{\sigma_i^2 (\overline{y_i}-\overline y)^2}{\sigma_i^2⁄n_i }=\sum_{i=1}^m\sigma_i^2χ_{(1-\frac{1}{m})}^2\]

\[SSW=\sum_{i=1}^m\sum^{n_i}_{j=1}(y_{ij}-\overline {y_i})^2=\sum_{i=1}^m\sigma_i^2[\sum_{j=1}^{n_i}\frac{(y_{ij}-\overline{y_i})^2}{\sigma_i^2}]=\sum_{i=1}^m\sigma_i^2\chi_{(n_i-1)}^2\] 由\(\chi^2\)的可加性可知：

\[SSB=\sum_{i=1}^m\sigma_i^2χ_{(1-\frac{1}{m})}^2=\chi_{\sum_{i=1}^m\sigma_i^2(1-\frac{1}{m})}^2\]

\[SSW=\sum_{i=1}^m\sigma_i^2\chi_{(n_i-1)}^2=\chi_{\sum_{i=1}^m\sigma_i^2 (n_i-1)}^2\] 由于

\[n\sigma^2=\sum_{i=1}^mn_i\sigma_i^2=\frac{n}{m}\sum_{i=1}^m\sigma_i^2\rightarrow\sum_{i=1}^m\sigma_i^2=m\sigma^2\]

因此

\[SSB=\chi_{\sum_{i=1}^m\sigma_i^2(1-\frac{1}{m})}^2=\chi_{m\sigma^2 (1-\frac{1}{m})}^2=\sigma^2\chi_{(m-1)}^2\] 当\(n_1=n_2=...=n_m=n⁄m\)时

\[SSW=\chi_{\sum_{i=1}^m\sigma_i^2 (n_i-1)}^2=χ_{m\sigma^2 (\frac{n}{m-1})}^2=\sigma^2\chi_{(n-m)}^2\]

或当\(\sigma_1^2=\sigma_2^2=...=\sigma_m^2=\sigma^2\)时

\[SSW=\chi_{\sum_{i=1}^m\sigma_i^2 (n_i-1)}^2=χ_{\sigma^2 \sum_{i=1}^m(n_i-1)}^2=\sigma^2\chi_{(n-m)}^2\]

从而

\[F=\frac{MSB}{MSW}=\frac{SSB⁄(m-1)}{SSW⁄(n-m)}=\frac{\chi_{(m-1)}^2/(m-1)}{\chi_{(n-m)}^2)⁄(n-m)}\sim F(m-1;n-m)\]