# Thread: Regression question regarding F-ratio

1. ## Regression question regarding F-ratio

Why does the F-ratio follow an F-distribution when the null hypothesis of equal grouped means is satisfied?

I'm really stuck on this.

I know that an F distribution is:

$\displaystyle \displaystyle \frac{\frac{\chi^{2}_1}{df_1}}{\frac{\chi^{2}_2}{d f_2}}$

And the F-Ratio is:

MST= mean squares for treatment
MSE= mean squares for error

$\displaystyle \frac{MST}{MSE}$

From here I'm totally stuck.

I think it all boils down to me not understanding why

$\displaystyle \display \frac{SST}{\sigma^2}~{\chi^2}_{a-1}$

Only when all of the group means equal each other.

It seems that

$\displaystyle \frac{SSE}{\sigma^2}~\chi^{2}_{N-a}$

is always a chi squared distribution.

2. The short answer is that you can rewrite $\displaystyle SST / \sigma^2$ as a sum of squares of $\displaystyle a - 1$ normal random variables and that, under the null hypothesis, each of the terms is $\displaystyle N(0, 1)$ and hence you get the chi-square distribution you asked for. The crucial role the null hypothesis plays is that, if the null hypothesis is false, then the terms you are summing up will not have mean 0 and so you don't get a chi-square when you add them up (you get a noncentral chi-square instead).

The long answer is very long unless you know a good bit of matrix algebra.

3. Thank you. If you could elaborate more on the ole of the null hypothesis, that'd be awesome

4. Take the expectation of SST and you'll get a feel for what happens when the null hypothesis is true. It's quite a bit of algebra, but doable. You should get $\displaystyle (a - 1) \sigma^2 + junk$ where the junk term goes away when all the treatment means are the same, and what's leftover is the expectation of chi-square with a-1 degrees of freedom after division by $\displaystyle \sigma^2$.

5. Thanks a lot. I see what you're getting at now.