An interaction sum of squares
Four cell means, with one observation each, are
| B1 | B2 | |
|---|---|---|
| A1 | 3 | 5 |
| A2 | 4 | 10 |
Compute the interaction residual of each cell, \overline Y_{jk}-\overline Y_{j\cdot}-\overline Y_{\cdot k}+\overline Y_{\cdot\cdot}, and the interaction sum of squares. With one observation per cell, why can this sum of squares not be compared with an error mean square from the same model?
The row means are 4 and 7, the column means are 3.5 and 7.5, and the grand mean is 5.5. The interaction residuals are
\begin{aligned} 3-4-3.5+5.5&=1,\\ 5-4-7.5+5.5&=-1,\\ 4-7-3.5+5.5&=-1,\\ 10-7-7.5+5.5&=1. \end{aligned}
With one observation per cell the interaction sum of squares is 1^2+1^2+1^2+1^2=4. Fitting the interaction uses up every degree of freedom in a 2\times 2 table of single observations: the error sum of squares is identically zero, so there is no error mean square left to divide by. The interaction can be described. It cannot be tested until there is a replicate inside the cells.