Matrix algebra derivation of Sums of Squares
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$$Z^{T}$$ = |
1,...,1 |
$$z_text{1}, … ,z_text{N}$$ |
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$$Z^{T}Z =$$ |
N |
$$\sum_text{i}z_text{i}$$ |
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$$\sum_text{i}z_text{i}$$ |
$$\sum_text{i}z_text{i}^text{2}$$ |
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$$(Z{T}Z){-1} =$$ |
$$\frac{\sum_{i}z_{i}{2}}{N\sum_text{i}z_{i}{2}} $$ |
0 |
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$$\frac{N}{N\sum_text{i}z_{i}^{2}}$$ |
which may be more simply as expressed as
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$$(Z{T}Z){-1} =$$ |
$$\frac{1}{N} $$ |
0 |
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0 |
$$\frac{1}{\sum_text{i}z_{i}^{2}}$$ |
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$$Z^{T}Y =$$ |
$$\sum_text{i} y_text{i}$$ |
$$\sum_text{i}z_text{i}y_text{i}$$ |
Then the regression terms are obtained using the least squares estimate B = $$(Ztext{T}Z)text{-1}Z^text{T}Y$$. Two terms are required: in the regression of the standardised covariate on the difference in a pair of response level means we require the regression estimate of the intercept (W1) and the covariate (W1 x covariate).
For the intercept using the above B = average difference between levels of W1, call this $$\bar{y}$$
For the W1 x covariate interaction B = $$\frac{\sum_text{i}z_text{i}y_text{i}}{\sum_text{i}z_{i}^{2}}$$
Taking SS explained by the regression equal to B $$Ztext{T}Z$$ B (see e.g., Rao, Toutenburg, Shalabh and Heumann(2007)) and using the appropriate Bs above and the diagonal entries in $$Z\text{T}Z$$ this gives W1 SS of $$\bar{y}\mbox{ x }\bar{y}$$ N and $$(\frac{\sum_text{i}z_text{i}y_text{i}}{\sum_text{i}z_{i}{2}})text{2}$$ x $$\sum_text{i}z_text{i}text{2}$$ = $$\frac{(\sum_text{i}z_text{i}y_text{i})text{2}}{\sum_text{i}z_{i}^{2}}$$ for the W1 x covariate SS.
Reference
Rao CR, Toutenberg H, Shalabh, and Heumann C (2007). Linear models and generalizations: least squares and alternatives. Third Edition. Springer-Verlag:Berlin.