水準と繰返しが有る標本平均周りの平方和の式展開(分解あり)
L:水準, R:繰返し
$$\begin{eqnarray}
\sum_{n=1}^{N}{(x_{n}-\bar{x})^2}
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x})^2}&&\dots L:水準, R:繰返し, N=LR
\\&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(\bar{x_{l}}-\bar{x})^2}
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x_{l}})^2}
&\dots \sum_{l=1}^{L}\sum_{r=1}^{R}{(\bar{x_{l}}-\bar{x})^2}:各水準平均と全体平均の 差の平方和, \sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x_{l}})^2}:各値と各水準との差の平方和\\
&=&R\sum_{l=1}^{L}{(\bar{x_{l}}^2-2\bar{x_{l}}\bar{x}+\bar{x}^2)}
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}^2-2x_{lr}\bar{x_{l}}+\bar{x_{l}}^2)}
&\dots (a-b)^2=a^2-2ab+b^2\\
&=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-R\sum_{l=1}^{L}{2\bar{x_{l}}\bar{x}}+R\sum_{l=1}^{L}{\bar{x}^2}
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-\sum_{l=1}^{L}\sum_{r=1}^{R}{2x_{lr}\bar{x_{l}}}+\sum_{l=1}^{L}\sum_{r=1}^{R}{\bar{x_{l}}^2}
&\dots \sum_{k=1}^{K}{(x_k+y_k)}=\sum_{k=1}^{K}{x_k}+\sum_{k=1}^{K}{y_k}\\
&=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-2R\bar{x}\sum_{l=1}^{L}{\bar{x_{l}}}+R\bar{x}^2\sum_{l=1}^{L}{1}
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\sum_{l=1}^{L}{(\bar{x_{l}}\sum_{r=1}^{R}{x_{lr}})}+\sum_{l=1}^{L}{(\bar{x_{l}}^2\sum_{r=1}^{R}{1})}
&\dots \sum_{k=1}^{K}{Cx_k}=C\sum_{k=1}^{K}{x_k}\\
&=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-2R\bar{x}L\bar{x}+R\bar{x}^2L
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\sum_{l=1}^{L}{(\bar{x_{l}}R\bar{x_{l}})}+\sum_{l=1}^{L}{\bar{x_{l}}^2R}
&\dots \sum_{k=1}^{K}{1}=K, \bar{x}:=\frac{\sum_{k=1}^{K}{x_k}}{K} \rightarrow \sum_{k=1}^{K}{x_k}=K\bar{x}\\
&=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-2LR\bar{x}^2+LR\bar{x}^2
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2R\sum_{l=1}^{L}{\bar{x_{l}}^2}+R\sum_{l=1}^{L}{\bar{x_{l}}^2}\\
&=&\color{red}{R\sum_{l=1}^{L}{\bar{x_{l}}^2}}\color{black}{-LR\bar{x}^2}
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}\color{red}{-R\sum_{l=1}^{L}{\bar{x_{l}}^2}}\\
&=&-LR\bar{x}^2
&+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-LR\bar{x}^2\\
&=&\sum_{n=1}^{N}{x_{n}^2}-N\bar{x}^2&\dots LR=N\\
\end{eqnarray}$$
水準と繰返しが有る標本平均周りの平方和の式展開(分解なし)
$$\begin{eqnarray}
\sum_{n=1}^{N}{(x_{n}-\bar{x})^2}
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x})^2}\dots L:水準, R:繰返し, N=LR\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}^2-2x_{lr}\bar{x}+\bar{x}^2)}\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-\sum_{l=1}^{L}\sum_{r=1}^{R}{2x_{lr}\bar{x}}+\sum_{l=1}^{L}\sum_{r=1}^{R}{\bar{x}^2}\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\bar{x}\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}}+\bar{x}^2\sum_{l=1}^{L}\sum_{r=1}^{R}{1}\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\bar{x}LR\bar{x}+\bar{x}^2LR\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2LR\bar{x}^2+LR\bar{x}^2\\
&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-LR\bar{x}^2\\
&=&\sum_{n=1}^{N}{x_{n}^2}-N\bar{x}^2&\dots LR=N\\
&=&\sum_{n=1}^{N}{x_{n}^2}-N(\frac{\sum_{n=1}^{N}{x_{n}}}{N})^2\\
&=&\sum_{n=1}^{N}{x_{n}^2}-\frac{1}{N}(\sum_{n=1}^{N}{x_{n}})^2\\
&=&\sum_{n=1}^{N}{x_{n}^2}-\frac{1}{N}(N\bar{x})^2\\
&=&\sum_{n=1}^{N}{x_{n}^2}-\frac{1}{N}N^2\bar{x}^2\\
&=&\sum_{n=1}^{N}{x_{n}^2}-N\bar{x}^2\\
\end{eqnarray}$$
各水準平均と全体平均の差の平方和と各値と各水準との差の平方和に分解できること
以上より分解有りと無しで同じ値になることから，\(\sum_{n=1}^{N}{(x_{n}-\bar{x})^2}=\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x})^2}\)は\(\sum_{l=1}^{L}\sum_{r=1}^{R}{(\bar{x_{l}}-\bar{x})^2}\)と\( \sum_{l=1}^{L}\sum_{r=1}^{R}{ (x_{lr}-\bar{x_{l}})^2 } \)に分解できることがわかる．
