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Question on significancy of terms
3 messages · Leonel Lopez, Emmanuel Curis, Geoff Brookshire
1 day later
Hello Leonel, I think your problem is not specific of GLMM. The (fixed-part) models a*b = a+b+a:b and b+a:b will lead to similar results, but expressed with a different set of coefficients. Assume a and b are two-levels factors (a1,a2 and b1,b2) for sake of simplicity : the first model will have coefficients - associated to a : for a2 - associated to b : for b2 - associated to a:b : for a2:b2 The second model will have coefficients - associated to b : for b2 - associated to a:b : for a1:b2 and a2:b2 (you can check this also with the simple lm function, by the way). So they have the same number of coefficients and, I think, express the same linear relationship but in a different basis. So the likelyhood will be the same... I never tested that with GLMM, but I guess it is the same problem as with lm and lmer... Best regards,
On Tue, Sep 11, 2012 at 08:29:03PM +0100, Leonel Lopez wrote:
? Hi lme4 users: ? I am new to mixed models in R and started using lme4 and apart of all I?m not an statistician. ? I am developing a GLMM model like the one below which contains the independent effect and interaction of two factors, plus the repeated measurement effect on individuals. ? Y~a*b+(1/Ind)I am using theglmer( function, family=poisson.?This model has three terms (a+b+a:ab). If I want to test the significance of the three terms I am using the?likelihood ratio test (anova) comparing the model?with the term and without the term ? Let say for example: ? m1<-glmer(y~a*b+(1/ind), family=poisson) ? m2<-glmer(y~a+b+(1/ind),?family=poisson) ? anova(m1,m2) ? but, how to test the significancy of only "a" or only "b" ? For example to test a: ? ? m1<-glmer(y~a*b+(1/ind),?family=poisson) ? m3<-glmer(y~b+a:b+(1/ind),?family=poisson) ? anova(m1,m3) ? ? This seems to be a incorrect model development as the LRT gave exacttly the same values for the two models. ? ? ? I am only considering transform my variable, get a normal distribution and conduct the analysis with the lmer() function. ? ? I?ll be very grateful for your help!! ? ? Cheers ? ? Leo ? [[alternative HTML version deleted]] ? ? _______________________________________________ ? R-sig-mixed-models at r-project.org mailing list ? https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
Emmanuel CURIS
emmanuel.curis at univ-paris5.fr
Page WWW: http://emmanuel.curis.online.fr/index.html
Hi there, I asked a similar question about a year ago: https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q3/006690.html To test for the significance of a, you could either use pvals.mcmc (in the languageR package) to get an MCMC p-value, or do Wald chi-square tests of model fits without the interaction term like so: m1 <- glmer(y ~ a + b + (1|ind), family=poisson) m2 <- glmer(y ~ a + (1|ind), family=poisson) anova(m1, m2) (Make sure you use the pipe character | for random effects.) cheers, Geoff
On Tue, Sep 11, 2012 at 3:29 PM, Leonel Lopez <llopezt2004 at yahoo.co.uk> wrote:
Hi lme4 users:
I am new to mixed models in R and started using lme4 and apart of all I?m not an statistician.
I am developing a GLMM model like the one below which contains the independent effect and interaction of two factors, plus the repeated measurement effect on individuals.
Y~a*b+(1/Ind)I am using theglmer( function, family=poisson. This model has three terms (a+b+a:ab). If I want to test the significance of the three terms I am using the likelihood ratio test (anova) comparing the model with the term and without the term
Let say for example:
m1<-glmer(y~a*b+(1/ind), family=poisson)
m2<-glmer(y~a+b+(1/ind), family=poisson)
anova(m1,m2)
but, how to test the significancy of only "a" or only "b"
For example to test a:
m1<-glmer(y~a*b+(1/ind), family=poisson)
m3<-glmer(y~b+a:b+(1/ind), family=poisson)
anova(m1,m3)
This seems to be a incorrect model development as the LRT gave exacttly the same values for the two models.
I am only considering transform my variable, get a normal distribution and conduct the analysis with the lmer() function.
I?ll be very grateful for your help!!
Cheers
Leo
[[alternative HTML version deleted]]
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