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It seems that when running an ANOVA, the degrees of freedom of the mean sum of squares between groups (SS_b) will always be less than or equal to the degrees of freedom of the (summed for all groups) mean sum of squares within group (SS_w). So, since the first parameter (d1) for the F distribution corresponds to the ANOVA's numerator degrees of freedom (i.e. the degrees of freedom for SS_b), and the second parameter (d2) corresponds to the ANOVA's denominator degrees of freedom (i.e. the degrees of freedom for SS_w), it seems to me that it should always be d1 <= d2, yet on wikipedia they show the following figure, which shows F distributions where d1 > d2. Could someone explain how this is possible?

F distributions with different parameter values

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  • $\begingroup$ Leaving aside the case discussed in the answer, it is not always the case that numerator df exceeds denominator df in an ANOVA F-test; in some of the more complicated designed experiments there may be many more df in the model than are left in the error term. $\endgroup$
    – Glen_b
    Commented May 3, 2020 at 3:10
  • $\begingroup$ oh interesting, can you point me to a concrete example of this? Could a general example of this be when you are comparing the fit of two models to the data, and one model has many many more parameters than the number of samples, or am I barking up the wrong tree? $\endgroup$
    – jaib1
    Commented May 3, 2020 at 11:45
  • $\begingroup$ A simple example off the top of my head would be a Latin square with t=3; the overall F test has 6 df for the model and 2 df for error. It's possible to have more extreme examples of experimental designs (it's easy to come up with factorial models where you'd end up with no error df unless you either omit the highest order interactions or you start confounding high-order interactions with main effects to claw some df back for the error). Books on experimental design have relevant information of the sort of thing I was talking about $\endgroup$
    – Glen_b
    Commented May 4, 2020 at 0:36
  • $\begingroup$ For a book example, take Cochran and Cox, Experimental Designs (1957). p290 has a fractional factorial design with 4 factors A,B,C,D (each at 3 levels) and 3 blocks and a total of 27 observations (26 df total), where the D-interactions (pairwise) - as well as any higher order interactions - are omitted (/assumed to be absent - to constitute error). Block effects have 2df, Main effects 8df, Interaction (AB,AC,BC) 12df and error 4df. An overall F test (if you cared to do one) would have 22 and 4 df $\endgroup$
    – Glen_b
    Commented May 4, 2020 at 0:53

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It’s possible because it follows the rules of a density about being non-negative and integrating to one, and someone chooses to make the degrees of freedom what they are, with $d_1>d_2$.

One place where this could come up is comparing the variances of two groups of unequal size. Note that this is not ANOVA, which is a test of means. This situation would be comparing the variance of a treatment group to a control group.

For the ANOVA you mention, you’re right: you will always have more total observations than groups, so $d_1\le d_2$. However, that is not the only use of the F-test.

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  • $\begingroup$ ah I see. so is the case you describe a Welch's t-test? Is this because an F-distribution generalizes a t-distribution? And would it be the case that d1 > d2 because the number of samples in the treatment group > number of samples in control group? $\endgroup$
    – jaib1
    Commented May 3, 2020 at 11:42
  • $\begingroup$ @jaib1 Welch’s t-test is a test of means. I mean testing the equality of the variances of two groups, regardless of what their means are. The F-distribution is not a generalization of the t-distribution in any way I know. As for why $d_1>d_2$, it will depend on which group’s variance you choose to put in the numerator of your F-stat. If that group has the larger sample size, then $d_1>d_2$. I suggest watching JBStatistics on testing variances, if you’re not familiar with it: youtube.com/…. (The video has $d_1>d_2$!) $\endgroup$
    – Dave
    Commented May 3, 2020 at 14:13

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