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I don't understand why in the F test we calculate the ratio between MSE between subject and MSE within subject. As far as I know, this is due because we want to use the F distribution, which is a rate between two $\chi^2$ distribution divided their degree of freedom.

My question is: Why is not used a simple version of this method? Couldn't we just take the ratio of the sum of squares between subjects with the sum of squares within subjects? Then we could compare this outcome with a distribution that is the result of the ratio between two $\chi^2$ distribution, instead of dividing the numerator and denominator for the degrees of freedom.

Does it makes sense? Why they use the other way?

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  • $\begingroup$ With the F, under the null hypothesis, the means squares are two different estimates of $\sigma^2$. The only difference for what you propose is the scaling by the ratio of degrees of freedom. So instead of the present F tables, you'd have tables that did exactly the same job but contained different numbers (the same as F but scaled by $\nu_1/\nu_2$). This would seem to make no difference, apart from the fact that when you work with tables, the F-tables "converge" at high degrees of freedom in a way that your tables would not. ...(ctd) $\endgroup$
    – Glen_b
    Jul 27, 2013 at 1:21
  • $\begingroup$ (ctd)... That convergence confers some calculation advantages that make appropriate interpolation work very well, and at sufficiently large denominator d.f. let you just use the asymptotic values. $\endgroup$
    – Glen_b
    Jul 27, 2013 at 1:22

1 Answer 1

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  • We have that (ref) $$ \frac{\text{RSS}}{\sigma²} \sim \chi²_{n - p} $$
  • Under the null hypothesis that all parameters are zero, we also have that $$ \frac{\text{ESS}}{\sigma²} \stackrel{\small \text{H}_0}{\sim} \chi²_p $$
  • RSS and ESS are independent.

You are proposing to consider the ratio of $\frac{\text{ESS}}{\sigma²}$ and of $\frac{\text{RSS}}{\sigma²}$, $$ \cfrac{\frac{\text{ESS}}{\sigma²}}{\frac{\text{RSS}}{\sigma²}} = \frac{\text{ESS}}{\text{RSS}} = \frac{p}{n-p} \times \cfrac{\frac{\text{ESS}}{n-p}}{\frac{\text{RSS}}{p}} $$ which, under the null, is distributed as a constant multiple of an F distribution (as noted by @Whuber in his comment to the previous version of my answer).

In summary, what you propose is not really different from what is done.


RSS: residual sum of squares

ESS : explained sum of squares

n : number of observations

p: number of parameters

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  • $\begingroup$ The only change proposed by the OP is multiplication by a constant (achieved by not doing the divisions to obtain the means from the sums). Therefore the ratio of the question is a constant multiple of an $F$ distribution. This answer only seems to make this simple relationship appear complicated. $\endgroup$
    – whuber
    Jul 26, 2013 at 19:35
  • $\begingroup$ Thank you @whuber. I have edited to put it more clearly. $\endgroup$
    – ocram
    Jul 26, 2013 at 20:04
  • $\begingroup$ OK: Given that the two distributions are so closely related, we should search elsewhere for an answer. I think if you were to review the relationships between F and other distributions you could come up with some good reasons why we use the convention we do. Or, just maybe, you might conclude that the O.P.'s suggested convention is superior! (Related question: why is the PDF of the Standard Normal distribution proportional to $\exp(-\frac{1}{2}x^2)$ and not the apparently simpler $\exp(-x^2)$?) $\endgroup$
    – whuber
    Jul 26, 2013 at 20:05
  • $\begingroup$ @whuber: mmmm, interesting. At first glance, I would simply say that we sometimes add superficial constants at some places to get rid of them at other places. $\endgroup$
    – ocram
    Jul 26, 2013 at 20:12
  • $\begingroup$ @whuber I believe the present way of doing it has advantages (though not really critical these days) over the OP's suggestion. These are outlined in my comment under the question. In respect of the standard normal, I've also seen the suggestion for taking the standard normal as being the one which makes the normalizing constant 1; it has a couple of nice features. Like so many things, once you have a convention, even if some other way might be a little better, there's a lot of inertia to overcome; in the short term it's usually better not to change, so conventions may stick around for ever. $\endgroup$
    – Glen_b
    Jul 27, 2013 at 1:18

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