seatales <ssphadke <at> uh.edu> writes:
> > Hello, > I am using the following model > > model1=lmer(PairFrequency~MatingPair+(1|DrugPair)+(1|DrugPair:MatingPair), > data=MateChoice, REML=F) > > 1. After reading around through the R help, I have learned that the above > code is the right way to analyze a mixed model with the MatingPair as the > fixed effect, DrugPair as the random effect and the interaction between > these two as the random effect as well. Please confirm if that seems > correct. You should probably send this sort of question to the r-sig-mixed-models mailing list ... You probably want (MatingPair|DrugPair) rather than (1|DrugPair:MatingPair). Whether REML=FALSE or REML=TRUE depends what you want to do next. > > 2. Assuming the above code is correct, I have model2 in which I remove the > interaction term, model3 in which I remove the DrugPair term and model4 in > which I only keep the fixed effect of MatingPair. > > 3. I want to perform the log likelihood ratio test to compare these models > and that's why I have REML=F. However the code anova(model1, model2, model3, > model4) gives me a chisq estimate and a p-value, not the LRT values. How do > I get LRT (L.Ratio) while using lmer? The chi-squared values are the differences in deviance (-2 log likelihood) between the respective papers of models, which under the null hypothesis of the LRT will be chi-squared distributed. In other words, these *are* the LRT test statistics. > > 4. I am under the impression after reading a few posts that LRT is not > usually obtained with lmer but it is given if I use lme (the old model). I don't know what you mean by this. The main difference between lmer and lme in the testing/inference context is that lme is willing to guess at "denominator degrees of freedom" to perform conditional F-tests. > > 5. I could not find how to input the random interaction term while using > lme? Is it the following way? Would someone please guide me to some already > existing posts or help here? = ran > > lme(PairFrequency~MatingPair, random=~(1|DrugPair)+(1|DrugPair:MatingPair), > data=MateChoice, method="ML")...is this the right way? would lme give me > loglikelihood ratio test values (L.ratio)? > See above. > Thanks a lot. I hope someone can help. Most posts I have found deal with > nesting but there is absolutely no nesting in my data. > > Sujal P. > p.s: If it matters how data is arranged, then I have one vector called > MatingPair which has 3 levels and another vector DrugPair which also has 3 > levels. The PairFrequency data is a count data and is normally distributed. > The data are huge, hence I am not able to post it here. It is probably unwise to estimate DrugPair as a random effect if it only has three levels. > > Also, here is what I mean by getting chisq value rather than L.Ratio: See above. > Data: MateChoice > Models: > model2: PairFrequency ~ MatingPair + (1 | DrugPair) > model3: PairFrequency ~ MatingPair + (1 | DrugPair:MatingPair) > model1: PairFrequency ~ MatingPair + (1 | DrugPair) + (1 | > DrugPair:MatingPair) > > Df AIC BIC logLik Chisq Chi Df Pr(>Chisq) > model2 5 274.90 282.82 -132.45 > model3 5 282.44 290.36 -136.22 0.0000 0 1.00000 > model1 6 276.90 286.40 -132.45 7.5443 1 0.00602 ** > --- > Signif. codes: 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 > â â 1 > > -- > View this message in context: http://r.789695.n4.nabble.com/mixed-model-random-interaction-term-log-likelihood-ratio-test-tp3448718p3448718.html > Sent from the R help mailing list archive at Nabble.com. > [[alternative HTML version deleted]] > > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
