Thanks.
Chris

On Wed, Oct 6, 2010 at 5:05 PM, Ned Dochtermann
<ned.dochtermann@gmail.com>wrote:

> Hi Chris,
>
> You're not going to be able to do that test using lmer. To conduct the test
> you want you'll need to know the likelihood estimates for two models, one
> with the random factor and another without it. You can't run a model
> without
> the random factor in lme4 and you can't use the likelihood from lm because
> they aren't "commensurate" between lme4 and lm (this issue is discussed at:
> http://glmm.wikidot.com/random-effects-testing). I've run the same sorts
> of
> tests for lmer and lm, as I'm sure many other people have and they aren't
> compatible.
>
> You can, however, get what you want using nlme:
> m.rand<-lme(Y~1,random=~1|Group,data=data)
> m.null<-gls(Y~1,data=data)
> (I don't use nlme much so you may want to double check the code syntax)
>
> Then you just run the likelihood ratio test from there, I think with nlme
> LRT is built in as anova(m.rand,m.null).
>
> This issue has been discussed a lot so you may find more detailed info by
> searching the archives.
>
>
> Good luck,
> Ned
>
> --
> Ned Dochtermann
> Department of Biology
> University of Nevada, Reno
>
> ned.dochtermann@gmail.com
> http://wolfweb.unr.edu/homepage/mpeacock/Dochter/
> --
>
> Hi,
> I originally ran a model in HLM 6 that I am now in lme4. In lme4 the model
> would look like the following:
>
> lmer(Y ~ 1 + (1 | Group), data= data)
>
> So I only have a random intercept for Group.
>
> I noticed that HLM 6 gives a chi-square test statistic associated with this
> random variable. Does anyone know how I can calculate this chi-square
> statistic in R or what formula the HLM authors are using?
>
> Thanks!
> Chris
>
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>
>
>


-- 
Christopher David Desjardins
Ph.D. student, Quantitative Methods in Education
M.S. student, Statistics
University of Minnesota

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