I'm re-running Kaplan-Meier Survival Curves from previously published data, using the exact data set used in the publication (Charpentier et al. 2008 - Inbreeding depression in ring-tailed lemurs (Lemur catta): genetic diversity predicts parasitism, immunocompetence, and survivorship). This publication ran the curves in SAS Version 9, using LIFETEST, to analyze the age at death structured by genetic heterozygosity and sex of the animal (n=64). She reports a Chi square value of 6.31 and a p value of 0.012; however, when I run the curves in R, I get a Chi square value of 0.9 and a p value of 0.821. Can anyone explain this??
R Code used: Age is the time to death, mort is the censorship code, sex is the stratum of gender, and ho2 is the factor delineating the two groups to be compared.
> survdiff(Surv(age, mort1)~ho2+sex,data=mariekmsurv1)
Call:
survdiff(formula = Surv(age, mort1) ~ ho2 + sex, data = mariekmsurv1)
N Observed Expected (O-E)^2/E (O-E)^2/V
ho2=1, sex=F 18 3 3.23 0.0166 0.0215
ho2=1, sex=M 12 3 2.35 0.1776 0.2140
ho2=2, sex=F 17 5 3.92 0.3004 0.4189
ho2=2, sex=M 17 4 5.50 0.4088 0.6621
Chisq= 0.9 on 3 degrees of freedom, p= 0.821
> str(mariekmsurv1)
'data.frame': 64 obs. of 6 variables:
$ id : Factor w/ 65 levels "","aeschylus",..: 14 31 33 30 47 57 51 39 36 3 ...
$ sex : Factor w/ 3 levels "","F","M": 3 2 3 2 2 2 2 2 2 2 ...
$ mort1: int 0 0 0 0 0 0 0 0 0 0 ...
$ age : num 0.12 0.192 0.2 0.23 1.024 ...
$ sex.1: Factor w/ 3 levels "","F","M": 3 2 3 2 2 2 2 2 2 2 ...
$ ho2 : int 1 1 1 2 1 1 1 1 1 2 ...
- attr(*, "na.action")=Class 'omit' Named int [1:141] 65 66 67 68 69 70 71 72 73 74 ...
.. ..- attr(*, "names")= chr [1:141] "65" "66" "67" "68" ...
survdiff
withrho=0
) is log-rank. – Heldentenor