A poisson family has dispersion 1 by definition of 'Poisson'.
You need a quasipoisson family to use a moment estimate of dispersion.

This difference is described on the help page for anova.glm.
Note the estimated dispersion *is* used and *is* one:

summary(glm.D93)$dispersion
[1] 1

as it ought to be (and is in S-PLUS too). That it is not used in S-PLUS is not documented (and it is not used in summary.glm either) and is at least an inconsistency.

Your assumption that R is wrong and S-PLUS (sic) is correct is not appreciated. Why did you file this as an R bug and not an S-PLUS one?


On Wed, 2 Feb 2005 [EMAIL PROTECTED] wrote:

There may be a bug in the anova.glm function.

But not in the one in R!

deathstar[32] R

R : Copyright 2004, The R Foundation for Statistical Computing
Version 2.0.1  (2004-11-15), ISBN 3-900051-07-0

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for a HTML browser interface to help.
Type 'q()' to quit R.

counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
glm.D93 <- glm(counts ~ outcome + treatment, family=poisson())
anova(glm.D93,test="Chisq")
Analysis of Deviance Table

Model: poisson, link: log

Response: counts

Terms added sequentially (first to last)


Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 8 10.5814 outcome 2 5.4523 6 5.1291 0.0655 treatment 2 0.0000 4 5.1291 1.0000

anova(glm.D93,test="F")
Analysis of Deviance Table

Model: poisson, link: log

Response: counts

Terms added sequentially (first to last)


Df Deviance Resid. Df Resid. Dev F Pr(>F) NULL 8 10.5814 outcome 2 5.4523 6 5.1291 2.7262 0.06547 . treatment 2 0.0000 4 5.1291 0.0000 1.00000 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1



----------------------------------------------------------------------
The F test should use the estimated dispersion parameter, so should be
different from the Chisq test. The following is what I got from Splus:

deathstar[31] Splus
S-PLUS : Copyright (c) 1988, 2000 MathSoft, Inc.
S : Copyright Lucent Technologies, Inc.
Version 6.0 Release 1 for Linux 2.2.12 : 2000
Working data will be in /data/home/faculty/yuedong/MySwork
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- factor(rep(1:3,3))
treatment <- factor(rep(1:3,c(3,3,3)))
glm.D93 <- glm(counts ~ outcome + treatment, family=poisson())
anova(glm.D93, test="Chisq")
Analysis of Deviance Table

Poisson model

Response: counts

Terms added sequentially (first to last)
          Df  Deviance Resid. Df Resid. Dev   Pr(Chi)
     NULL                      8   10.58145
  outcome  2  5.452305         6    5.12914 0.0654707
treatment  2  0.000000         4    5.12914 1.0000000
anova(glm.D93, test="F")
Analysis of Deviance Table

Poisson model

Response: counts

Terms added sequentially (first to last)
          Df  Deviance Resid. Df Resid. Dev  F Value     Pr(F)
     NULL                      8   10.58145
  outcome  2  5.452305         6    5.12914 2.108359 0.2369863
treatment  2  0.000000         4    5.12914 0.000000 1.0000000


-------------------------------------------------------------------- Yuedong Wang Phone: (805) 893-4870 Dept of Statistics & Applied Probability Fax: (805) 893-2334 Univ of California [EMAIL PROTECTED] Santa Barbara, CA 93106 http://www.pstat.ucsb.edu/faculty/yuedong

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-- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595

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