On Tue, Sep 15, 2009 at 10:34 AM, Peng Yu <pengyu...@gmail.com> wrote:
> Hi,
>
> I run the following commands. 'A' has 3 levels and 'B' has 4 levels.
> Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3,
> B4)?
If you were to define a model matrix for those coefficients it would
be rank deficient. (the sum of the first three columns is 1 and the
sum of the last four columns is 1) You only have 6 degrees of freedom
in total, not 7. Because of the -1 in the formula the first factor is
expanded to the indicator columns but the second is represented with a
set of 3 "contrasts" which are the contr.treatment version, by
default.
>> a=3
>> b=4
>> n=1000
>> A = rep(sapply(1:a,function(x){rep(x,n)}),b)
>> B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}),
>> function(x){rep(x,a)}))
>> Y = A + B + rnorm(a*b*n)
>>
>> fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B))
>> afit=aov(Y ~ A + B - 1,fr)
>> afit$coefficients
> A1 A2 A3 B2 B3 B4
> 1.993067 2.969252 3.965888 1.005445 2.020717 3.023940
>
> Regards,
> Peng
>
> ______________________________________________
> R-help@r-project.org mailing list
> 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.
>
______________________________________________
R-help@r-project.org mailing list
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.
