Re: Covariate in ANCOVA question



On Oct 25, 7:31 pm, David Winsemius <doe_s...@xxxxxxxxxxx> wrote:
Bruce Weaver <bwea...@xxxxxxxxxxxx> wrote in news:1193318943.008583.179600@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

On Oct 24, 8:57 pm, David Winsemius <doe_s...@xxxxxxxxxxx> wrote:
Bruce Weaver <bwea...@xxxxxxxxxxxx> wrote
innews:1193226596.924562.47100@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

--- snip ---

So because there are 3 groups we should not think about type I erors
in post-hoc testing? Why, then, do the texts I have talk about
controlling for alpha and use three group examples?

--
David Winsemius

Good question. I can only speculate about the reasons. Fisher's LSD
fell out of favour generally because it does not control the family-
wise alpha well when there are 4 or more groups. I think that over
time, we (collectively) have forgotten that it *does* control family-
wise alpha when there are 3 groups. Howell's book is the only one I'm
aware of that says this.

The only one, eh. Could that be because the logic is questionable?

I think you should go back to the argument you offered from Howell and
think more deeply. When you say he says "assume none of the mu's are
equal" and as a result says type I error is not possible...that is a
tautology rather than a statistical inference. When you _know_ there is a
difference then the question is what is power of the test to support
that. The risk is a type II error. Type I errors occur when there is no
difference and one risks (falsely) concluding that there is a difference.

--
David Winsemius

Whatever its faults may be when there are more than 3 groups, the
Fisher LSD procedure does keep the actual familywise type I error
rate at its nominal level when there are 3 groups.

.



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