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
following in R, but I get different results. options(contrasts =
= "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
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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