As Peter Dalgaard has pointed out, the model you fit has redundant parameters, as you can see from its summary (or from the coefficients):
> summary(example.lm)
. . .
Coefficients: (13 not defined because of singularities)
. . .
> coef(example.lm)
(Intercept) treatmentS period2 sequence2 sequence1:subject2
305.35714 -46.60714 15.89286 -135.00000 NA
sequence2:subject2 sequence1:subject3 sequence2:subject3 sequence1:subject4 sequence2:subject4
222.50000 NA 200.00000 -5.00000 NA
sequence1:subject5 sequence2:subject5 sequence1:subject6 sequence2:subject6 sequence1:subject7
NA 240.00000 45.00000 NA 110.00000
sequence2:subject7 sequence1:subject9 sequence2:subject9 sequence1:subject10 sequence2:subject10
NA NA 150.00000 -60.00000 NA
sequence1:subject11 sequence2:subject11 sequence1:subject12 sequence2:subject12 sequence1:subject13
75.00000 NA NA 145.00000 NA
sequence2:subject13 sequence1:subject14 sequence2:subject14
NA 57.50000 NA
The computational procedure used by Anova (in the car package) assumes a full-rank parametrization of the model. The difference between the Anova methods for lm and glm is that the former uses the coefficient estimates and their covariance matrix to compute sums of squares while the latter refits the model. You got a result from SAS because it works with a deficient-rank parametrization. Finally (and more generally), if you really want "type-III" sums of squares in R, be careful with the kind of contrast-coding that you use. The default "treatment" contrasts won't give you what you want.
I'm sorry that you encountered this problem.
John
At 11:00 AM 1/29/2003 +0100, martin wolfsegger wrote:
I am looking for help to analyze an unbalanced AB/BA cross-over design by
requesting the type III SS !
# Example 3.1 from S. Senn (1993). Cross-over Trials in Clinical
Research
outcome<-c(310,310,370,410,250,380,330,270,260,300,390,210,350,365,370,310,380,290,260,90,385,400,410,320,340,220)
subject<-as.factor(c(1,4,6,7,10,11,14,1,4,6,7,10,11,14,2,3,5,9,12,13,2,3,5,9,12,13))
period<-as.factor(c(1,1,1,1,1,1,1,2,2,2,2,2,2,2,1,1,1,1,1,1,2,2,2,2,2,2))
treatment<-as.factor(c('F','F','F','F','F','F','F','S','S','S','S','S','S','S','S','S','S','S','S','S','F','F','F','F','F','F'))
sequence<-as.factor(c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2))
example<-data.frame(outcome,subject,period,treatment,sequence)
The recommended SAS code equals
PROC GLM DATA=example;
CLASS subject period treatment sequence;
MODEL outcome = treatment sequence period subject(sequence);
RANDOM subject(sequence);
RUN;
For PROC GLM, the random effects are treated in a post hoc fashion after the
complete fixed effect model is fit. This distinction affects other features
in the GLM procedure, such as the results of the LSMEANS and ESTIMATE
statements. Looking only on treatment, period, sequence and subject effects, the
random statement can be omitted.
The R code for type I SS equals
example.lm<-lm(outcome~treatment+period+sequence+subject%in%sequence,
data=example)
anova(example.lm)
Response: outcome
Df Sum Sq Mean Sq F value Pr(>F)
treatment 1 13388 13388 17.8416 0.001427 **
period 1 1632 1632 2.1749 0.168314
sequence 1 335 335 0.4467 0.517697
sequence:subject 11 114878 10443 13.9171 6.495e-05 ***
Residuals 11 8254 750
According to the unbalanced design, I requested the type III SS which
resulted in an error statement
library(car)
Anova(example.lm, type="III")
Error in linear.hypothesis.lm(mod, hyp.matrix, summary.model = sumry, :
One or more terms aliased in model.
by using glm I got results with 0 df for the sequence effect !!!!
example.glm<-glm(outcome~treatment+period+sequence+subject%in%sequence,
data=example, family=gaussian)
library(car)
Anova(example.glm,type="III",test.statistic="F")
Anova Table (Type III tests)
Response: outcome
SS Df F Pr(>F)
treatment 14036 1 18.7044 0.001205 **
period 1632 1 2.1749 0.168314
sequence 0 0
sequence:subject 114878 11 13.9171 6.495e-05 ***
Residuals 8254 11
The questions based on this output are
1) Why was there an error statement requesting type III SS based on lm ?
2) Why I got a result by using glm with 0 df for the period effect ?
3) How can I get the estimate, the StdError for the constrast (-1,1) of the
treatment effect ?
Thanks
--
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----------------------------------------------------- John Fox Department of Sociology McMaster University Hamilton, Ontario, Canada L8S 4M4 email: [EMAIL PROTECTED] phone: 905-525-9140x23604 web: www.socsci.mcmaster.ca/jfox -----------------------------------------------------
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