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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?

[1]: The F test for Comparing Two Normal Variances: Correct and Incorrect Calculation of the Two-Sided p-value?

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  • $\begingroup$ 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. $\endgroup$ – user10525 May 2 '12 at 9:59
  • $\begingroup$ 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?). $\endgroup$ – Néstor May 2 '12 at 10:02
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    $\begingroup$ 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 $X$. $\endgroup$ – user10525 May 2 '12 at 10:12
  • $\begingroup$ That makes a lot of sense, thanks! Can you post that as an answer, so I can accept it? :-). $\endgroup$ – Néstor May 2 '12 at 10:22
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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 $X$.

Thanks.

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    $\begingroup$ I'd like to add that it actually makes more sense to identify the median when calculating two-sided p-values for asymmetric distributions than to identify the mode, as I stated on my original post. This is because of the very definition of a two-sided p-value: you actually search the probability of your data or more extreme values given the null hypothesis on both extremes, which you can define arbitrarly from the median. However, this isn't by any means the panacea for the problem, as it's still arbitrarily chosen (e.g., it wouldn't make sense for bimodal distributions). $\endgroup$ – Néstor May 2 '12 at 10:36

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