Using R 1.8.1, and the negative binomial glm implemented in MASS,
the default when using anova and a chi-square test is to divide the
deviance by the estimated dispersion. Using my UNIX version of S-plus (v
3.4), and the same MASS functions, the deviances are *not* divided by the
estimated dispersion.
Firstly, I'm wondering if anyone can enlighten about the correct procedure
(I thought the F-test was more appropriate when dispersion is estimated)?
Secondly, after a bit of muddling with the negative binomial pdf, I
concluded that, like for the Poisson, phi is actually 1. This result is
borne out by simulations. Is this correct?
# an example in R 1.81 with library(MASS)
y<-rnegbin(n=100,mu=1,theta=1)
x<-1:length(y)
model<-glm(y~x,family=neg.bin(1))
summary(model)$dispersion
[1] 1.288926
anova(model,test='Chisq")
#...
Df Deviance Resid. Df Resid. Dev P(>|Chi|)
NULL 99 102.038
x 1 0.185 98 101.853 0.705
# But the "real" chi-square probability is
1-pchisq(0.185,1)
[1] 0.6671111
Thanks in advance,
Dan
____________________________________
Daniel Kehler
Dept. of Biology
Dalhousie University, B3H 4J1
Halifax, Nova Scotia, Canada
Office: LSC 800
email: kehler at mscs.dal.ca
phone: 902 494 3910
