Hi Dennis,
The cell mean mu_12 from the model involves the intercept and factor 2:
Coefficients:
(Intercept) factor(treat)2
factor(treat)3
0.429244
0.499982 0.352971
factor(treat)4 factor(treat)5
factor(treat)6
-0.204752
0.142042 0.044155
factor(treat)7 factor(group)2
factor(treat)2:factor(group)2
-0.007775
-0.337907 -0.208734
factor(treat)3:factor(group)2 factor(treat)4:factor(group)2
factor(treat)5:factor(group)2
-0.195138
0.800029 0.227514
factor(treat)6:factor(group)2 factor(treat)7:factor(group)2
0.331548 NA
So mu_12 = 0.429244-0.337907 = 0.091337. This can be verified by:
> predict(fit,data.frame(list(treat=1,group=2)))
1
0.09133691
Warning message:
In predict.lm(fit, data.frame(list(treat = 1, group = 2))) :
prediction from a rank-deficient fit may be misleading
But as you can see, it gave a warning about rank-deficient fit... why this is a
rank-deficient fit?
Because "treat 1_group 2" has no cases, so why it is still estimable
while on the contrary, "treat 7_group 2" which has 2 cases is not?
Thanks
John
________________________________
From: Dennis Murphy <djmuser@gmail.com>
Sent: Monday, November 7, 2011 9:29 PM
Subject: Re: [R] why NA coefficients
Hi John:
What is the estimate of the cell mean \mu_{12}? Which model effects
involve that cell mean? With this data arrangement, the expected
population marginal means of treatment 1 and group 2 are not estimable
either, unless you're willing to assume a no-interaction model.
Chapters 13 and 14 of Milliken and Johnson's Analysis of Messy Data
(vol. 1) cover this topic in some detail, but it assumes you're
familiar with the matrix form of a linear statistical model. Both
chapters cover the two-way model with interaction - Ch.13 from the
cell means model approach and Ch. 14 from the model effects approach.
Because this was written in the mid 80s and republished in the early
90s, all the code used is in SAS.
HTH,
Dennis
> Thanks David. The only category that has no cases is "treat 1-group
2":
>
>> with(test,table(treat,group))
> group
> treat 1 2
> 1 8 0
> 2 1 5
> 3 5 5
> 4 7 3
> 5 7 4
> 6 3 3
> 7 8 2
>
> But why the coefficient for "treat 7-group 2" is not estimable?
>
> Thanks
>
> John
>
>
>
>
> ________________________________
> From: David Winsemius <dwinsemius@comcast.net>
>
> Cc: "r-help@r-project.org" <r-help@r-project.org>
> Sent: Monday, November 7, 2011 5:13 PM
> Subject: Re: [R] why NA coefficients
>
>
> On Nov 7, 2011, at 7:33 PM, array chip wrote:
>
>> Hi, I am trying to run ANOVA with an interaction term on 2 factors
(treat has 7 levels, group has 2 levels). I found the coefficient for the last
interaction term is always 0, see attached dataset and the code below:
>>
>>>
test<-read.table("test.txt",sep='\t',header=T,row.names=NULL)
>>> lm(y~factor(treat)*factor(group),test)
>>
>> Call:
>> lm(formula = y ~ factor(treat) * factor(group), data = test)
>>
>> Coefficients:
>> (Intercept) factor(treat)2
factor(treat)3
>> 0.429244 0.499982
0.352971
>> factor(treat)4 factor(treat)5
factor(treat)6
>> -0.204752 0.142042
0.044155
>> factor(treat)7 factor(group)2
factor(treat)2:factor(group)2
>> -0.007775 -0.337907
-0.208734
>> factor(treat)3:factor(group)2 factor(treat)4:factor(group)2
factor(treat)5:factor(group)2
>> -0.195138 0.800029
0.227514
>> factor(treat)6:factor(group)2 factor(treat)7:factor(group)2
>> 0.331548 NA
>>
>>
>> I guess this is due to model matrix being singular or collinearity
among the matrix columns? But I can't figure out how the matrix is singular
in this case? Can someone show me why this is the case?
>
> Because you have no cases in one of the crossed categories.
>
> --David Winsemius, MD
> West Hartford, CT
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>
>
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