[R-sig-ME] Questions on the results from glmmPQL(MASS)
David.Duffy at qimr.edu.au
Mon Dec 8 08:29:06 CET 2008
On Mon, 8 Dec 2008, zhijie zhang wrote:
> Do u mean the following method?
> > model0<-glmmML(y ~ trt + I(week > 2), cluster=ID, family=binomial,
>> model1<-glm(y ~ trt + I(week > 2), family=binomial, data=bacteria)
> Error message occurred.
anova does not have a method for glmmML, but the deviances seem to be
calculated the same (see model0$cluster.null.deviance etc):.
model0 Residual deviance: 192.3 on 215 degrees of freedom AIC: 202.3
model1 Residual deviance: 199.18 on 216 degrees of freedom
LRTS = 6.88. We will assume that the test statistic is distributed 1/2
X2(0) and 1/2 X2(1), so P ~ 0.004. Comparing to a Wald test using the SE
on the random effect SD, I get:
Z = 1.242/0.4024 = 3.08, P=0.001
| David Duffy (MBBS PhD) ,-_|\
| email: davidD at qimr.edu.au ph: INT+61+7+3362-0217 fax: -0101 / *
| Epidemiology Unit, Queensland Institute of Medical Research \_,-._/
| 300 Herston Rd, Brisbane, Queensland 4029, Australia GPG 4D0B994A v
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