# Analysis of variance

1. Aug 13, 2011

### Rasalhague

Well, just as I thought I'd got the hang of this...

Koosis: Statistics: A Self-Teaching Guide, 4th ed., §§ 6.29-43.

The "degreeses of freedom" are 3 and 36. This critical value, 4.38, is found by looking up the score for 1% in the table at the back of the book, or in Excel with F.INV.RT(0.01,3,36) = 4.38.

But this is a two-tailed test, right? So why do we not divide the desired significance level by 2 before inputting it into this function, as was explained by Koosis in §§ 6.19-22, and on this page? Thus: F.INV.RT(0.01/2,3,36) = 5.06. This latter method makes sense because, in a two-tailed test a result could achieve significance by being far enough from the mean of the F distribution on either side, so extreme results on one side of the mean will only constitute half of that 1%.

Koosis concludes that the result, which is 4.42, is significant, since it's greater than 4.38. But I'd have concluded that it's not significance, since it's less than 5.06.

Last edited: Aug 13, 2011
2. Aug 13, 2011

### Rasalhague

Ah, here we go:

http://www.le.ac.uk/bl/gat/virtualfc/Stats/variance.html [Broken]

If I've (half) understood this right, an F-test for ANOVA is always 1-tailed because what the F-test is directly testing is the alternative hypothesis that the variance of population means is greater than the mean of population variances.

If it is greater, one concludes that the population means are not equal. But this is an indirect consequence, an inference from a test of variances.

The objects that play the same role as the two population variances in the other kind of F-test - the kind of F-test used simply to test variances - are being used in the ANOVA test in the manner of a 1-tailed test; we're asking specifically whether the numerator is greater than the denominator, not merely whether they're unequal.

Last edited by a moderator: May 5, 2017
3. Aug 14, 2011

### Rasalhague

Hello, Yesterday Rasalhague! There's a problem with this rationalization. When you get to § 6.73, you'll find it says ANOVA and the t test for the difference of two means depend on the same assumptions, namely that the population distributions are normal, and that the population variances are equal.

What's being tested is whether there's a difference between the variance of the group means (approximated by the numerator: the between-groups variance estimate) and the mean of the variances of the groups (approxmated by the denominator: the within-groups variance estimate). And one only considers the result significant if the numerator is greater than the denominator.

Last edited: Aug 14, 2011