Hi, I'm trying to convert the following SAS code in R to get the same result that I get from SAS. Here is the SAS code: DATA plants; INPUT sample $ treatmt $ y ; cards;
1 trt1 6.426264755 1 trt1 6.95419631 1 trt1 6.64385619 1 trt2 7.348728154 1 trt2 6.247927513 1 trt2 6.491853096 2 trt1 2.807354922 2 trt1 2.584962501 2 trt1 3.584962501 2 trt2 3.906890596 2 trt2 3 2 trt2 3.459431619 3 trt1 2 3 trt1 4.321928095 3 trt1 3.459431619 3 trt2 3.807354922 3 trt2 3 3 trt2 2.807354922 4 trt1 0 4 trt1 0 4 trt1 0 4 trt2 0 4 trt2 0 4 trt2 0 ; RUN; PROC MIXED ASYCOV NOBOUND DATA=plants ALPHA=0.05 method=ML; CLASS sample treatmt; MODEL y = treatmt ; RANDOM int treatmt/ subject=sample ; RUN; I get the following covariance estimates from SAS:Intercept sample ==> 5.5795treatmt sample ==> -0.08455Residual ==> 0.3181I tried the following in R, but I get different results. options(contrasts = c(factor = "contr.SAS", ordered = "contr.poly")) df$sample=as.factor(df$sample) lmer(y~ 1+treatmt+(1+treatmt|sample),REML=FALSE, data = df) Since the results from R are standard deviations, I have to square all results to get the variances. sample==> 2.357412^2 = 5.557391 sample*treatmt==>0.004977^2 = 2.477053e-05 residual==>0.517094^2 = 0.2673862As shown above, the results from SAS and R are different. Do you know how to get the exact values in R?I appreciate any help.Thanks,Gram [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.