[R-sig-ME] Questions on the results from glmmPQL(MASS)

[David Duffy]

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,
> data=bacteria)
>> model1<-glm(y ~ trt + I(week > 2), family=binomial, data=bacteria)
>> anova(model0,model1)
> 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