(In some of the exercises that follow, we must
Chapter 9, Problem 4E(choose chapter or problem)
Show that the cross-product terms formed from \(\left(\bar{X}_{i \cdot}-\bar{X}_{. .}\right),\left(\bar{X}_{. j}-\bar{X}_{. .}\right)\), and \(\left(X_{i j}-\bar{X}_{i .}-\bar{X}_{. j}+\bar{X}_{. .}\right)\) sum to zero, \(i=1,2, \ldots a\) and \(j=1,2, \ldots, b\). Hint: For example, write
\(\begin{aligned}&\sum_{i=1}^{a} \sum_{j=1}^{b}\left(\bar{X}_{\cdot j}-\bar{X}_{. .}\right)\left(X_{i j}-\bar{X}_{i .}-\bar{X}_{. j}+\bar{X}_{. .}\right) \\&\quad=\sum_{j=1}^{b}\left(\bar{X}_{. j}-\bar{X}_{. .}\right) \sum_{i=1}^{a}\left[\left(X_{i j}-\bar{X}_{. j}\right)-\left(\bar{X}_{i .}-\bar{X}_{. .}\right)\right] \end{aligned}\)
and sum each term in the inner summation, as grouped here, to get zero.
Equation Transcription:
Text Transcription:
(Bar X_i-Bar X..),(Bar X_.j-Bar X..)
(X_ij-Bar X_i.-Bar X_.j+Bar X..)
i=1,2,…a
j=1,2,…,b
sum_i=1^a sum_ j=1^b ( Bar X_j-Bar X.) (X_ij-Bar X_i.-Bar X_.j+Bar X..) =sum_j=1^b (Bar X_.j-Bar X_..) sum_i=1^a[( X_ij-Bar X_.j)-(Bar X_i.-Bar X..)]
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