First, I'm sorry for the long post, but I needed a second opinion from you, the experts, about this problem!
I was reading a paper by Gallagher (2006), where he puts an example on "how one may correctly perform an F-test and how some softwares perform poorly on this task" (which could have been a pretty interesting papers for students, but I think I found an error, which maybe it'll make it even more interesting!). He poses the following example:
Consider the following two artificial data sets:
Sample 1: 1,2,3,3.3 $\Rightarrow$ $s_1^2=1.08917$
Sample 2: 1,2,3 $\Rightarrow$ $s_2^2=1$
(where $s_1^2$ and $s_2^2$ are the estimated sample variances).
This is ok (thanks to @Procastinator!). However, here's what bothers me:
(...) it may be erroneously assumed that $F=1.08917$ lies in the upper tail of the F distribution on 3 and 2 degrees of freedom, as $F>1$. Clearly, from (1) (where the equation shows that $P(F_{3,2}>1.08917)=0.5114$) this is not the case (...)"
Correct me if I'm wrong, but the mode for the F-distribution is $$\hat{f}=\frac{\nu_2(\nu_1-2)}{\nu_1(\nu_2+2)}$$ which for $\nu_1=3$ and $\nu_2=2$, gives $\hat{f}=1/6=0.17$. Because $F>\hat{f}$, then actually the value resides in the upper tail of the distribution. There's nothing strange with having a probability greather than $0.5$ because the PDF of the F-distribution for $3$ and $2$ degrees of freedom is actually pretty skewed.
Did I miss something here?