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 at 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
Chi-square test on random effects
3 messages · Ned Dochtermann, Christopher David Desjardins, Robert Kushler
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I believe the RLRsim package provides a better solution. Regards, Rob Kushler
On 10/6/2010 7:01 PM, Christopher Desjardins wrote:
Thanks. Chris On Wed, Oct 6, 2010 at 5:05 PM, Ned Dochtermann <ned.dochtermann at 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 at 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 [[alternative HTML version deleted]]