Let \(X_{i j k}, i=1, \ldots, a ; j=1, \ldots, b, k=1, \ldots, c\), be a random
sample of size \(n=a b c\) from a normal distribution \(N\left(\mu,
\sigma^{2}\right) .\) Let \(\bar{X}_{\ldots}=\sum_{k=1}^{c} \sum_{j=1}^{b}
\sum_{i=1}^{a} X_{i j k} / n\) and
$$
\begin{aligned}
\bar{X}_{i_{r}} &=\sum_{k=1}^{c} \sum_{j=1}^{b} X_{i j k} / b c . \text {
Prove that } \\
& \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j
k}-\bar{X}_{\ldots}\right)^{2}=\sum_{i=1}^{a} \sum_{j=1}^{b}
\sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i . .}\right)^{2}+b c
\sum_{i=1}^{a}\left(\bar{X}_{i .}-\bar{X}_{\cdots}\right)^{2}
\end{aligned}
$$
Show that \(\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j
k}-\bar{X}_{i . .}\right)^{2} / \sigma^{2}\) has a chi-square distribution with
\(a(b c-1)\)
degrees of freedom. Prove that the two terms in the right-hand member are
independent. What, then, is the distribution of \(b c
\sum_{i=1}^{a}\left(\bar{X}_{i .}-\bar{X}_{\ldots}\right)^{2} / \sigma^{2} ?\)
Furthermore, let \(X_{. j .}=\sum_{k=1}^{c} \sum_{i=1}^{a} X_{i j k} / a c\) and
\(\bar{X}_{i j .}=\sum_{k=1}^{c} X_{i j k} / c .\) Show that
$$
\begin{aligned}
\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j
k}-\bar{X}_{\cdots}\right)^{2}=& \sum_{i=1}^{a} \sum_{j=1}^{b}
\sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i j .}\right)^{2} \\
&+b c \sum_{i=1}^{a}\left(\bar{X}_{i_{n}}-\bar{X}_{\ldots}\right)^{2}+a c
\sum_{j=1}^{b}\left(\bar{X}_{. j}-\bar{X}_{\ldots}\right)^{2} \\
&+c \sum_{i=1}^{a} \sum_{j=1}^{b}\left(\bar{X}_{i j .}-\bar{X}_{i .}-X_{. j
.}+X_{\ldots}\right)
\end{aligned}
$$
Prove that the four terms in the right-hand member, when divided by
\(\sigma^{2}\), are independent chi-square variables with \(a b(c-1), a-1, b-1\),
and \((a-1)(b-1)\) degrees of freedom, respectively.
