F distribution: Why is Fnm =/= Fmn^-1

#1
(not sure if this is the correct forum, sorry if so)

Hello everyone

I'm trying to understand the F-distribution. I realize that the F-distribution is not symmetric in the degrees of freedom, and understand why. However I would think that it should be Fnm = 1/Fmn, but it isn't. Here is my reasoning:

When testing whether the variances of two sets are equal, one uses the F-statistic: F = S1^2/S2^2. This is then compared with the f-value at the desired significance level: Fnm (with degrees of freedom n and m respectively).
However the zero hypothesis is symmetric: H0: sigma1 = sigma2. So I should get the same result if I switch set1 and set2, and get the zero hypothesis H0': sigma2 = sigma1. The new F-statistic is
F' = S2^2/S1^2 = 1/F.
Now the degrees of freedom are also switched and so F' should be compared with the f-value Fmn. But if we want to reach the same conclusion for both hypotheses H0 and H0', we should have Fnm = 1/Fmn... I would think.

What is wrong with my reasoning? Because, in general, Fmn is not equal to 1/Fnm.

Thanks in advance
Lukas
 

Dason

Ambassador to the humans
#2
Notice that if we denote \(F_{\alpha, n, m}\) to be the \(\alpha\) quantile of a F distribution with n numerator df and m denominator df that it is true that

\(F_{\alpha, n, m} = 1/F_{1 - \alpha, m, n}\)

This is more useful and if you think about your problem for a little while longer you might realize that this is what you should have been looking at in the first place.
 
#3
Notice that if we denote \(F_{\alpha, n, m}\) to be the \(\alpha\) quantile of a F distribution with n numerator df and m denominator df that it is true that

\(F_{\alpha, n, m} = 1/F_{1 - \alpha, m, n}\)

This is more useful and if you think about your problem for a little while longer you might realize that this is what you should have been looking at in the first place.
Yes, of course, you're right.
Because if we switch the variances, we go from a problem where \(F = \frac{S_1}{S_2}>1 \) to a problem where \(F' = \frac{S_2}{S_1}<1 \) (or vice versa). So we obviously have to look at the other side of the distribution.

Thanks, Dason.