Dear advanced statisticians, *******Objectif********
I try to set up linear models with mean as intercept: Answer: y Variable: x, as factor of two modalities: x(1), x(2). I would like to have a model as: y = mean(y)+A(i)+residuals, with i in (1,2) and A(1) coefficient for x(1) and A(2) coefficient for x(2). *******Trials in R******* ## Firstly: I write in R: >Model<-lm(y~x,Data) >summary(Model) ... I've got the coefficients for: - the intercept (x(1) as been choosen) that we can call B(1) - the second modality: x(2) that we can call B(2) If I have well understood we have for the model and predictions: if x(1): y=B(1) if x(2): y=B(1)+B(2) which is quite different as y=mean(y)+A(i) ## Secondly I tried to skip the intercept >Model2<-lm(y~0+x,Data) >summary(Model2) ... I've got the coefficients for: - the first modality: x(1) that we can call C(1) - the second modality: x(2) that we can call C(2) And the model and predictions, if I'm right, are: if x(1): y=C(1) if x(2): y=C(2) ******* Questions *********** How can I obtain a predictive model y=mean(y)+A(i) ? Is it possible to settle mean(y) as intercept? Thanks for your help. Ghislain V., retarded statistician. ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
