[R-sig-ME] [R] Proc Mixed variance of random effects in R

[Ken Beath]

Actually what you want is the R results in SAS. The default in SAS is for
uncorrelated random effects, which is definitely not what is needed for
random intercept/slope models. Add TYPE=UN after the random statement.

On 18 June 2015 at 17:54, Thierry Onkelinx <thierry.onkelinx at inbo.be> wrote:

> 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_robins at 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]]
> >
> > ______________________________________________
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> > and provide commented, minimal, self-contained, reproducible code.
>
>         [[alternative HTML version deleted]]
>
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-- 

*Ken Beath*
Lecturer
Statistics Department
MACQUARIE UNIVERSITY NSW 2109, Australia

Phone: +61 (0)2 9850 8516

Level 2, AHH
http://stat.mq.edu.au/our_staff/staff_-_alphabetical/staff/beath,_ken/

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