Consider the following simple example using R-2.9.0 and 'perm' 2.9.1:
> require('perm')
> p<- c(15,21,26,32,39,45,52,60,70,82)
> g<- c('y','n','y','y', rep('n',6)) #Patients ranked 1,3,4 receive treatment
> permTS(p ~ g, alternative = 'two.sided', method='exact.ce') #find
p-value by complete enumeration
Exact Permutation Test (complete enumeration)
data: p by g
p-value = 0.05
alternative hypothesis: true mean of g=n minus mean of g=y is not equal to 0
sample estimates:
mean of g=n minus mean of g=y
28.38095
The permutation observed is '134', which has a rank sum of 8. Other
permutations with rank sums of 8 or less are '123', '124' and '125'.
So there are a total of 4 out of 4! = 120 possible, or a one-tail
p-value of 4/120 = 0.0333, or a 2-tail p-value of 2*4/120 = 0.067.
This is not, however, what permTS() returns. The permTS() value of
0.05 appears to correspond to 3 patterns, not 4.
I am misunderstanding how to solve this simple problem, or is
something going on with permTS() that I'm missing.
Thanks.
================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
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