Consider the two factor additive ANOVA model
$$\begin{align} X_{ij} &=\mu_{ij}+e_{ij} \\ \mu_{ij}&=\mu+\alpha_i+\beta_j \end{align}$$
where as usual $\sum_{i=1}^a \alpha_i=0$ and $\sum_{j=1}^b \beta_j=0$ and $e_{ij}\sim ^{iid} N(0,\sigma^2)$.
We want to test
$H_0:\beta_1=\beta_2=\ldots=\beta_n=0$ vs $H_1: \beta_j \neq 0$ for some $j$
and so an F test is used
$$F=\frac{\sum_{i=1}^a \sum_{j=1}^b \left( \bar{X}_{.j}-\bar{X}.. \right)^2 /(b-1)}{\sum_{i=1}^a \sum_{j=1}^b \left( X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}.. \right)^2/(b-1)(a-1)}\sim F \left(b-1,(b-1)(a-1) \right)$$
where $$\bar{X}_{i_.}=\frac{\sum_{j=1}^b X_{ij}}{b}, \bar{X}_{.j}=\frac{\sum_{i=1}^a X_{ij}}{a}$$ and $$\bar{X}_{..}=\frac{\sum_{i=1}^a \sum_{j=1}^b X_{ij}}{ab}$$
Standard notation for the ANOVA model.
My question now is how is it that the numerator, over $\sigma^2$ of course, follows a central chi-squared distribution under the null? Even if the $\beta$'s are zero there is still the $\alpha$i's to consider and thus I think there should be a non centrality parameter in the $F$ distribution, no? (This was answered by @gung below)
My second question is, where does the independence of the quadratic forms in question come from? I do not think Cochran's theorem applies since the random variables come from different distributions even under the null (there is still the $\alpha$i's to consider).
I would appreciate some help here.
