where
$$SS_{\text{total}}=\sum_{j=1}^a\sum_{i=1}^b (x_{ij}-\bar{x})^2$$
$$SST=SS_{\text{treatment}}=b\sum_{j=1}^a (\bar{x_j}-\bar{x})^2$$
$$SSB=SS_{\text{block}}=a\sum_{i=1}^b (\bar{x_i}-\bar{x})^2$$
$$SSE=SS_{\text{error}}=\sum_{j=1}^a\sum_{i=1}^b (x_{ij}-\bar{x_i}-\bar{x_j}+\bar{x})^2$$
$$\bar{x}=\frac{1}{ab} \sum_{j=1}^a\sum_{i=1}^b x_{ij}$$
$$\bar{x_i}=\frac{1}{a}\sum_{j=1}^a x_{ij}$$
$$\bar{x_j}=\frac{1}{b}\sum_{i=1}^b x_{ij}$$
I think it had to start by adding and substarcting terms: $$SS_{\text{total}}=\sum_{j=1}^a\sum_{i=1}^b (x_{ij}-\bar{x_i}+\bar{x_i}-\bar{x_j}+\bar{x_j}-\bar{x})^2$$
Then square it to $$\sum_{j=1}^a\sum_{i=1}^b [(x_{ij}-\bar{x_i})^2+(\bar{x_i}-\bar{x_j})^2+(\bar{x_j}-\bar{x})^2+2(x_{ij}-\bar{x_i})(\bar{x_i}-\bar{x_j})+2(x_{ij}-\bar{x_i})(\bar{x_j}-\bar{x})+2(\bar{x_i}-\bar{x_j})(\bar{x_j}-\bar{x})]$$
After that how should I continue?
Best Answer
For clarity, use instead the notation
$$\overline x_{i\bullet}=\frac1a\sum_{j=1}^a x_{ij}\quad,\quad\overline x_{\bullet j}=\frac1b\sum_{i=1}^b x_{ij}$$ Rewrite $x_{ij}$ as
$$x_{ij}=\overline x+(\overline x_{i\bullet}-\overline x)+(\overline x_{\bullet j}-\overline x)+(x_{ij}-\overline x_{i\bullet}-\overline x_{\bullet j}+\overline x)$$
Square both sides and sum over $i$ and $j$ to get
$$\sum_{i,j}x_{ij}^2=ab\,\overline x^2+a\sum_i(\overline x_{i\bullet}-\overline x)^2+b\sum_j(\overline x_{\bullet j}-\overline x)^2+\sum_{i,j}(x_{ij}-\overline x_{i\bullet}-\overline x_{\bullet j}+\overline x)^2$$
Convince yourself that the product terms vanish.
For instance, $$\sum_{i, j}(\overline x_{i\bullet}-\overline x)(\overline x_{\bullet j}-\overline x)=\sum_{i}(\overline x_{i\bullet}-\overline x)\sum_j(\overline x_{\bullet j}-\overline x)=0$$
because $$\sum_{i}(\overline x_{i\bullet}-\overline x)=\sum_i\overline x_{i\bullet}-b\overline x=b\overline x-b\overline x=0$$
So,
$$\sum_{i,j}(x_{ij}-\overline x)^2=a\sum_i(\overline x_{i\bullet}-\overline x)^2+b\sum_j(\overline x_{\bullet j}-\overline x)^2+\sum_{i,j}(x_{ij}-\overline x_{i\bullet}-\overline x_{\bullet j}+\overline x)^2$$