Questions on the results from glmmPQL(MASS)
On 08/12/2008, at 6:42 PM, zhijie zhang wrote:
On Mon, Dec 8, 2008 at 3:29 PM, David Duffy <David.Duffy at qimr.edu.au> wrote:
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.
1/2 X2(0) and 1/2 X2(1): ?? what do they mean?
X2 is chi-square. Because the test is on the boundary the test statistic is distributed as the weighted sum of chi-square rather than the usual chi-square. Verbeke and Molenberghs cover this in their books. Simulation (parametric bootstrap) seems a better way of doing this. Information criteria (AIC or BIC) can also be used. Most usual justification for a random effect is that there is expected to be one, so provided it can be estimated it is included.
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
Is this a "clerical error"? Based on your hints, it seems that p should be 0.003475077 .
dnorm(3.08)
[1] 0.003475077 Thanks.
P value for z test is 2*dnorm(3.08) which is close enough to 0.001 given that the test is only an approximation. Ken
-- | 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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