Re: Holm's method for multiple comparison??

I'm pretty sure i=j which is the index number of the contrast. So in
effect you use a per test alpha of familywise alpha/(n-i+1). So for 20
tests you start with per test alpha/20 for the largest difference and
alpha/19 for the next largest and so forth.

Note that this looks like the Lazerlere (spelling) & Mulaik procedure
which is equivalent to Holm, but more generic.

Thom

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