Dear Gram, A few things first: Please don't post in HTML, it mangles your text. R-sig-mixed model is a better list for questions on mixed models. Send further replies only to that list and not to r-help.
You are probably not fitting the same model in R as the one in SAS. Please provide the equations of the SAS model and then you can help you translate that into R code. You are assuming that we all speak SAS, but this is an R mailing list. The lingua franca among statistical software is mathematics. Best regards, ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey 2015-06-17 19:52 GMT+02:00 Grams Robins <grams_rob...@yahoo.com>: > 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. [[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.