In the balanced case with no missing cells, the standard analysis first reduces the data to 24
Question:
In the balanced case with no missing cells, the standard analysis first reduces the data to 24 rat averages \(\bar{Y}_{i .}\), the treatment and control averages \(\bar{Y}_{T}, \bar{Y}_{C}\), and the overall average \(\bar{Y}_{. .}\). The sum of squares for treatment effects is
\[
\mathrm{SS}_{T}=70 \bar{Y}_{T}^{2}+50 \bar{Y}_{C}^{2}-120 \bar{Y}_{. .}^{2}=\left(\bar{Y}_{T}-\bar{Y}_{C}ight)^{2} \frac{5 \times 10 \times 14}{24}
\]
The total sum of squares for rats splits into two orthogonal parts
\[
5 \sum_{i}\left(\bar{Y}_{i .}-\bar{Y}_{. .}ight)^{2}=\mathrm{SS}_{T}+\mathrm{SS}_{R}
\]
which are independent on one and 22 degrees of freedom respectively. If treatment effects are null, the mean squares \(\mathrm{SS}_{T} / 1\) and \(\mathrm{SS}_{R} / 22\) have the same expected value, and the mean-square ratio
\[
F=\frac{\mathrm{SS}_{T} / 1}{\mathrm{SS}_{R} / 22}
\]
is distributed as \(F_{1,22}\). Simulate a complete design with additive effects, and check that the two terms shown above agree with parts of the decomposition reported by anova (Im (y ~ site+treat+rat)) .
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