[R] Random effects ANOVA
Hi, I used the command lmer, in package lme4, to fit a random effects ANOVA. But i didn't get the p-values of significance tests of variance components. Does anyone know how to do it? Thanks, Luis Ernesto - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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.
Re: [R] Random effects ANOVA
Dear Luis, I believe that lmer() doesn't print out standard errors or Wald tests for variance-covariance components because these tests aren't reliable. You can omit each random effect from the model in turn, and use anova() to perform a likelihood-ratio test for the corresponding variance and covariance components, comparing each null model with the full model. I hope this helps, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Luis Salasar Sent: Friday, November 24, 2006 1:57 PM To: r-help@stat.math.ethz.ch Subject: [R] Random effects ANOVA Hi, I used the command lmer, in package lme4, to fit a random effects ANOVA. But i didn't get the p-values of significance tests of variance components. Does anyone know how to do it? Thanks, Luis Ernesto - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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. __ R-help@stat.math.ethz.ch 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.
Re: [R] Random effects ANOVA?
On 3/30/06, Prof Brian Ripley [EMAIL PROTECTED] wrote: I think you want print or summary rather than anova. anova() is not very useful for aov() models even without error strata. That's sort of better. summary(aov(time ~ drink + Error(video), data = df)); gives me: Error: video Df Sum Sq Mean Sq F value Pr(F) Residuals 2160 80 Error: Within Df Sum Sq Mean Sq F valuePr(F) drink 1 240.000 240.000 44.211 1.313e-08 *** Residuals 56 304.000 5.429 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ...what I'm really looking for is something akin to the output from SAS, which is: Source DF Anova SSMean Square F ValuePr F DRINK 1 240.000240.000 12.000.0742 VIDEO 2 160.000 80.000 4.000.2000 I didn't follow how the videos were chosen. Random effects apply when the 'treatments' were chosen from a large population (which might apply if each subject watched (on separate occasions) three videos chosen randomly from a larger pool), and if the interest is in the variability of the response over videos in the pool. If subjects were observed more than once then I suspect you most likely want a random effect for subjects. This problem comes directly from the final for my Experimental Stats class, which is why it feels a little odd. The videos were randomly selected from a library. Subjects watched one of the three videos, drank one of the two drinks, and completed the tasks. There were no repeated measures, so we can't block on subjects. The hypothesis test, according to SAS, treats DRINK*VIDEO as an error term. Setting aside whether this is the right analysis, how can I replicate this analysis in R? -- Chris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Random effects ANOVA?
Chris Bergstresser [EMAIL PROTECTED] writes: On 3/30/06, Prof Brian Ripley [EMAIL PROTECTED] wrote: I think you want print or summary rather than anova. anova() is not very useful for aov() models even without error strata. That's sort of better. summary(aov(time ~ drink + Error(video), data = df)); gives me: Error: video Df Sum Sq Mean Sq F value Pr(F) Residuals 2160 80 Error: Within Df Sum Sq Mean Sq F valuePr(F) drink 1 240.000 240.000 44.211 1.313e-08 *** Residuals 56 304.000 5.429 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ...what I'm really looking for is something akin to the output from SAS, which is: Source DF Anova SSMean Square F ValuePr F DRINK 1 240.000240.000 12.000.0742 VIDEO 2 160.000 80.000 4.000.2000 I didn't follow how the videos were chosen. Random effects apply when the 'treatments' were chosen from a large population (which might apply if each subject watched (on separate occasions) three videos chosen randomly from a larger pool), and if the interest is in the variability of the response over videos in the pool. If subjects were observed more than once then I suspect you most likely want a random effect for subjects. This problem comes directly from the final for my Experimental Stats class, which is why it feels a little odd. The videos were randomly selected from a library. Subjects watched one of the three videos, drank one of the two drinks, and completed the tasks. There were no repeated measures, so we can't block on subjects. The hypothesis test, according to SAS, treats DRINK*VIDEO as an error term. Setting aside whether this is the right analysis, how can I replicate this analysis in R? Possibly, time ~ drink + Error(video/drink) It won't give you the F test for video though. Not quite sure whether time ~ drink + video + Error(video:drink) works. It might, although it is a bit unnatural to have a random interaction between to systematic effects. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Random effects ANOVA?
On 30 Mar 2006 22:41:51 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote: Not quite sure whether time ~ drink + video + Error(video:drink) works. It might, although it is a bit unnatural to have a random interaction between to systematic effects. This exactly reproduces the given SAS output. Whether it's actually the right model to use, given the problem, is a different question. And luckily not one I have to answer. -- Chris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Random effects ANOVA?
Hi all -- So I have a very simple dataset, which consists of 60 subjects, who watched one of three videos, drank one of two drinks, and completed a task. The response variable is the time to complete the task. The ANOVA command is simple enough: anova(aov(time ~ drink * video, data = df)); However, the videos were randomly selected; I need to use the random effects model for them. So I tried anova(aov(time ~ drink + Error(video), data = df)); This gives me a no applicable method for 'anova' error. The command aov works, but doesn't give me anything I can interpret effectively. Is there a simpler command I should be using? Am I doing something wrong? -- Chris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Random effects ANOVA?
I think you want print or summary rather than anova. anova() is not very useful for aov() models even without error strata. The point of the aov classes is that they present the results of lm() fits in ways which are more conventional for designed experiments, including giving conventional ANOVA tables without recourse to anova(). I didn't follow how the videos were chosen. Random effects apply when the 'treatments' were chosen from a large population (which might apply if each subject watched (on separate occasions) three videos chosen randomly from a larger pool), and if the interest is in the variability of the response over videos in the pool. If subjects were observed more than once then I suspect you most likely want a random effect for subjects. What I guess is more likely is that you have 60 measurements on 60 subjects, each of whom was assigned one video and one drink. Then there appear to be no random effects, and the problem caused by randomly selecting videos is that the experiment is likely to be unbalanced (since each video is not going to be selected 20 times by chance). aov() is primarily intended for balanced designs, and interpretation is tricky without balance (see the warnings on its help page). There are many loose ends hindering offering help by email, and the best advice is to consult a statistician face-to-face about an appropriate analysis. On Wed, 29 Mar 2006, Chris Bergstresser wrote: Hi all -- So I have a very simple dataset, which consists of 60 subjects, who watched one of three videos, drank one of two drinks, and completed a task. The response variable is the time to complete the task. The ANOVA command is simple enough: anova(aov(time ~ drink * video, data = df)); However, the videos were randomly selected; I need to use the random effects model for them. So I tried anova(aov(time ~ drink + Error(video), data = df)); This gives me a no applicable method for 'anova' error. The command aov works, but doesn't give me anything I can interpret effectively. Is there a simpler command I should be using? Am I doing something wrong? -- Chris -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
