# Am I getting something wrong? A question about a paper on the F-test

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?

• You made a mistake while typing sample 1, this should be $(1, 2, 3, 3.3)$. This is, there are 4 observations, not 5. With this you get the result in the paper. I guess the rest follows by taking this into account. – user10525 May 2 '12 at 9:59
• Oooooh! Right! IT WAS the lack of sleep! Thanks a lot. However, doesn't the last part of my post still counts? (the part in which I say that the $F$ value is actually in the upper tail of the distribution?). – Néstor May 2 '12 at 10:02
• Perhaps 'tails' here are identified with respect to the median, not w.r.t. the mode. If ${\mathbb P}(X>a)>0.5$, then this is interpreted as: $a$ lies in the left-tail of the distribution of $X$. – user10525 May 2 '12 at 10:12
• That makes a lot of sense, thanks! Can you post that as an answer, so I can accept it? :-). – Néstor May 2 '12 at 10:22

You made a mistake while typing Sample 1, this should be $(1,2,3,3.3)$. This is, there are 4 observations, not 5. With this you get the result in the paper. I guess the rest follows by taking this into account.
Perhaps 'tails' here are identified with respect to the median, not w.r.t. the mode. If ${\mathbb P}(X>a)>0.5$, then this is interpreted as: $a$ lies in the left-tail of the distribution of $X$.