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# bottstrapping evolutionary trees

David Skibinski D.O.F.Skibinski at Swansea.AC.UK
Fri May 5 09:53:38 EST 1995

evolutionary trees. With three species, A, B,  and C there
are three rooted trees I (AB) C II (AC) B and III A (BC). My a priori
expectation is that these trees can be equally probable,
thus p=0.33 for tree I. I get some data, make a tree and do
bootstrapping and find that tree I is supported in a
proportion of 0.99 of a large sample of bootstrap reps.

My question is, can I use a likelihood ratio test to compare
0.99 with 0.33 ? If so what is the test ? Is it
G = 2ln((1-0.33)/(1-0.99) with 1 (or 2 df) or something else ?

Then if I get two other data sets and find bootstrap probabilities
of 0.95 and 0.75 for tree I, how can these be combined with 0.33
in a single test, and what would be the degrees of freedom ? I
am puzzled by the fact that in the `combining probabilities test`
(as described in Biometry by Sokal and Rohlf) which appears to be a
likelihood ratio test in which the alternative hypotheses are p=1,
the df = 2k where k is number of probabilities. Yet in my example
there are only three data sets, so should sum to 3 df.

Any help would be very greatly appreciated.

David Skibinski