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: ral at 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 "Vere scire est per causas scire"