R help - Jun 2015 - Proc Mixed variance of random effects in R

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

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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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