[R] mixed model random interaction term log likelihood ratio test
Ben Bolker
bbolker at gmail.com
Fri Apr 15 00:14:05 CEST 2011
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
>
> --
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http://r.789695.n4.nabble.com/mixed-model-random-interaction-term-log-likelihood-ratio-test-tp3448718p3448718.html
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