# In a two-way ANOVA, how can the F-statistic for one factor have a central distribution if the null is false for the other factor?

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.

I think the $F$-test you are trying to use is the global test of the ANOVA model. It is a simultaneous test that all parameters (i.e., $\alpha_i$'s and $\beta_j$'s) equal $0$. So yes, if all $\beta_j$'s were $0$, but some $\alpha_i$'s were not, then the numerator would follow a non-central $\chi^2$ distribution and the quotient would follow a non-central $F$ distribution.

To test $\sum\beta_j = 0$ alone, you want the following $F$ statistic:
$$F=\frac{\frac{J\sum_{j=1}^J \left( \bar X_{.j}-\bar{X}_{..} \right)^2}{(J-1)}}{\frac{\sum_{i=1}^I \sum_{j=1}^J \left( X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}.. \right)^2}{(I-1)(J-1)}}\sim F \left(b-1,(b-1)(a-1) \right)$$

• I have edited my post since I had made a mistake in the numerator of my fraction. It is your F-statistic that I care about. And I believe that answers my question. Could you please tell me additionally where does the independence of those two quadratic forms of the statistic comes from? As far as I know Cochran's theorem does not apply. – JohnK Nov 15 '14 at 19:01
• This version assumes an additive model, @JohnK. If there is an interaction, the F-stat won't follow the assumed distribution, but if there isn't an interaction, the effect of A has been partialled out, so it doesn't matter if there is an effect of A or not. – gung Nov 15 '14 at 19:05
• Does this mean that we can use Cochran's theorem to verify the independence of those two under the null? We simple disregard the $\alpha$ 's? – JohnK Nov 15 '14 at 19:09
• @JohnK, I'm not sure--someone else will have to answer that for you--but I think so. – gung Nov 15 '14 at 19:13