I'm certainly no expert, but I believe that aov( ) is simply a front-end to the lm( ) function, which fits fixed-effects linear models. As you may know, it is possible to write any ANOVA model as a regression model with appropriate use of indicator variables for the various levels of treatments. So, aov( ) does not, in general, yield results that are comparable to those generated by lme( ). You have specified the same _fixed_ effect ( ~ Age ) in both aov( ) and lme( ) which is why they give you the same parameter estimates for this covariate. When you perform anova( ) to compare the two lme( ) models, you are performing a likelihood ratio test. This is not generally the same as the F-test/t-test used to evaluate the marginal significance of a fixed-effect term in a linear model, which explains the difference between the p-values. Best, david paul -----Original Message----- From: Martin Hoyle [mailto:plxmh at nottingham.ac.uk] Sent: Tuesday, April 08, 2003 10:09 AM To: r-help at stat.math.ethz.ch Subject: [R] Basic LME Hello R Users, I am investigating the basic use of the LME function, using the following example; Response is Weight, covariate is Age, random factor is Genotype model.lme <- lme (Weight~Age, random=~ 1|Genotype) After summary(model.lme), I find that the estimate of Age is 0.098 with p=0.758. I am comparing the above model with the AOV function; model.aov <- aov (Weight~Age + Genotype) I find that the estimate of Age is also 0.098, and p=0.758 as in the LME model above. So, my questions are; 1: I expected that the LME model would be a better way to analyse this data compared to the AOV model, since Genotype is a random factor. However, I obtain the same parameter estimate and p value for Age. Please can someone tell me why? 2: When using LME, when I am after a p value for the covariate Age, is it better to do the following; Model.lme2 <- lme (Weight~Age, random=~ 1|Genotype, method="ML") Model.lme3 <- lme (Weight~1, random=~ 1|Genotype, method="ML") Anova(Model.lme2, Model.lme3) Giving likelihood ratio=0.102, with p=0.749, which is slightly different to the p values of 0.758 above. Thanks for your attention, Martin. Martin Hoyle, School of Life and Environmental Sciences, University of Nottingham, University Park, Nottingham, NG7 2RD, UK Webpage: http://myprofile.cos.com/martinhoyle ______________________________________________ R-help at stat.math.ethz.ch mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Basic LME
1 message · Paul, David A