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Could anybody demonstrate or direct me to a readily available proof of the following:

For the cell means model:

$$ y_{ij} = \mu_{i} + \epsilon_{ij},\ \text{ for }\ i = 1, \ldots, r\ \text{ and }\ j = 1, \ldots, n_{i}.$$

Show that:

$$ \sum_{i=1}^{r}\sum_{j=1}^{n_{i}}(y_{ij}-\bar{y}_{\cdot\cdot})^{2} = \sum_{i=1}^{r}\sum_{j=1}^{n_{i}}(y_{ij}-\bar{y}_{i\cdot})^{2} + \sum_{i=1}^{r}n_{i}(\bar{y}_{i\cdot}-\bar{y}_{\cdot\cdot})^{2} $$

where the first term is $\text{SS}_{\text{TO}}$, the second $\text{SS}_{\text{E}}$ and the third $\text{SS}_{\text{TR}}$.

(I have an upcoming examination in linear models and this proof was required for one of the previous year's examinations, but, so far, I haven't had much success finding the proof.)

share|improve this question
Hint: this is the Pythagorean Theorem in disguise. – whuber Dec 14 '12 at 21:21
Let's say $A = \sqrt{\text{SS}_{\text{E}}}$, $B = \sqrt{\text{SS}_{\text{TR}}}$, $C = \sqrt{\text{SS}_{\text{TO}}}$; Is it correct to assume that $2AB = 0$? – user9171 Dec 14 '12 at 22:25
Of course not: all three values you name, being lengths, are (typically) positive numbers. You need to view $SS_E$ etc. as squared norms of vectors: that reduces this problem to showing that two vectors are orthogonal (their dot product is zero). – whuber Dec 14 '12 at 22:35

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