mixed model random interaction term log likelihood ratio test
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 ???***??? 0.001 ???**??? 0.01 ???*??? 0.05 ???.??? 0.1
??? ??? 1
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