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Stats newbie here.

Is there any meaning to the critical F value if the two samples have the same degrees of freedom?

For example, I have 2 sets of data, each with the same number of degrees of freedom. If I calculate the chi-square value for each data set, then the F-test is defined as:

$$F(\nu_1,\nu_2) = \frac{\chi_1/\nu_1}{\chi_2/\nu_2}.$$

If $\nu_1 = \nu_2$, then it seems like they would just cancel and have no impact on the F distribution resulting in

$$F = \frac{\chi_1}{\chi_2}.$$

Is this correct?

However, when I try to do an inverse F-test to determine the critical F-value (e.g. 5% significance), the result is clearly dependent on both $\nu_1$ and $\nu_2$, even if they are identical.

In MATLAB:

>> finv(0.95,1,1)

ans =

  161.4476

>> finv(0.95,2,2)

ans =

   19.0000

>> finv(0.95,4,4)

ans =

    6.3882

Please explain as if you are talking to someone who knows very little about stats (a.k.a. me).

Thanks

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  • $\begingroup$ What do you mean by "an inverse F-test"? Do you actually just mean "compute an F critical value"? Note that "95% significance level" isn't really a thing (well, it's conceivable but certainly not what you mean). Presumably you mean "5% significance level" $\endgroup$ – Glen_b Aug 26 '17 at 5:33
  • $\begingroup$ @Glen_b Yes, I think I might mean the "critical F value with 5% significance level". I called it "inverse F-test" because the command in MATLAB is "finv". I've edited my question for clarity $\endgroup$ – Darcy Aug 28 '17 at 16:02
  • $\begingroup$ What do you mean by "cancel"? The wording of this phrase suggests the misconception that the ratio of two identically distributed (but independent) random variables would be a constant. Is that what you are asking about? $\endgroup$ – whuber Aug 28 '17 at 16:05
  • $\begingroup$ @whuber I have edited the answer for clarity. What I mean is that the two $\nu$ values should cancel leaving you with $F = \chi_1/\chi_2$. Is that correct? $\endgroup$ – Darcy Aug 28 '17 at 16:17
  • $\begingroup$ Sure it's correct-but I don't understand the motivation behind your post. You seem to acknowledge the distribution depends on the ordered pair $(\nu_1, \nu_2)$. When $\nu_1=\nu_2$, it still depends on what that common value is. It would be surprising if it didn't-. What is your actual question? $\endgroup$ – whuber Aug 28 '17 at 18:56
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The $\nu$ values cancel but the numerator and denominator in an F test are still scaled $\chi^2$ variates, and the shape of those -- and hence of their ratio -- is dependent on the degrees of freedom.

For example, when $\nu=2$ the distribution of both numerator and denominator is negative exponential in shape. When $\nu=6$ they have a more humped shape (but still fairly skews). When $\nu=200$ they both look fairly close to normal. This changes the distribution of the ratio.

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