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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