Hi,
1/ I wonder why a anova.lme on a single lme object does not print the sum of
squares (as expected from the help: "a data frame with the sums of squares,
numerator degrees of freedom, denominator
degrees of freedom, F-values, and P-values").
Example:
> fm2 <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1)
> anova(fm2)
numDF denDF F-value p-value
(Intercept) 1 80 4123.156 <.0001
age 1 80 114.838 <.0001
Sex 1 25 9.292 0.0054
How can one get the sum of squares easily?
2/ I have tried a home-made permutation test for comparison with results
obtained above. I am a quite surprised that the intercept has a probability
close to 1:
> lme.perm.test(fm2)
p.value
(Intercept) 0.998
age 0.000
Sex 0.001
The home made function is grounded on 1000 permutations of the response variable
(distance) and the comparison of the F-values obtained with the F-values
observed with the original (non permuted)data.
Thanks in advance for any hint,
Patrick Giraudoux
PS: below the current script of the "draft" permutation test:
lme.perm.test<-function(lme.model,pn=1000){
# Giraudoux 17.4.2004 permutation test for linear mixed effect models
# lme.model = a lme object
# pn = permutation number (default = 1000)
# value: a data.frame of the p.values of each independent variable F
# the permutation number is kept in the ouput attributes and thus
# can be called from there
an<-anova(lme.model)
Fobs<-an[[3]]
n<-rep(0,length(an[[3]]))
for (i in 1:pn){
lm1<-update(lme.model, sample(.)~.)
an<-anova(lm1)
Frdm<-an[[3]]
for (j in 1:(length(an[[3]]))){
if (Frdm[j]>=Fobs[j]) n[j]<-n[j]+1
}
}
n<-n/pn
names(n)<-row.names(an)
n<-as.data.frame(n)
names(n)<-"p.value"
attributes(n)<-c(attributes(n),permutations=pn)
return(n)
}
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