search for: intecept

R-gurus, I am plotting data against each other using t-tests. In this scenario, I am plotting coinage / total money. When there is no total money, then there is no coinage, but the plots I generate do not reflect this. Instead the y-axis always has an intecept that is not zero in these instances. Is there any way to force the y-intercept to begin at zero? Thank you in advance for any input. Tristan _________________________________________________________________ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r...

...SI) lower upper Intercept -4.0778905 -3.1366836 Treatment2 3.4455972 4.3196598 Treatment 3 0.399302 1.287747 Treatment 4 -1.7898933 -0.2980325 log(Treatment*Site.(in)) -22.2198233 -19.7342530 log(Site.(In)) -28.7857601 -23.0952939 mcmcpvalue(as.Matrix(fitSI[,1])) etc Intecept 0 Treatment 2 0 Treatment 3 0.0075 Treatment 4 0.00758 log(Treatment*Site.(in)) 0 log(Site.(In)) 0 Any help in explaining why these two results are completely different would be much appreciated. I have tried the example and read all of the posts. Cheers Beth

....comb(vec[-1]) tmp2=sapply(tmp,function(x)c(vec[1],x)) c(tmp,tmp2) } } #################### ## function: CV1glm--CV1 using K fold CV ## input: y -- response vector; x -- predictors without intercept ## output: the vector whether each predictor should be selected. E.g. (0,1,0,1) means no intecept while var1 and var3 should be selected #################### CV1glm=function(y,x){ n.var=ncol(x) n=nrow(x) comb=rec.comb(1: (n.var)) n.comb=length(comb) pe=c() data=data.frame(y=y) glm=glm(y~1,data=data) pe[1]=cv.glm(data,glm)$delta[1] for(i in 2:n.comb){ data=data.frame(y=y,x=x[,comb[[i...

...) lower upper Intercept -4.0778905 -3.1366836 Treatment2 3.4455972 4.3196598 Treatment 3 0.399302 1.287747 Treatment 4 -1.7898933 -0.2980325 log(Treatment*Site.(in)) -22.2198233 -19.7342530 log(Site.(In)) -28.7857601 -23.0952939 mcmcpvalue(as.Matrix(fitSI[,1])) etc Intecept 0 Treatment 2 0 Treatment 3 0.0075 Treatment 4 0.00758 log(Treatment*Site.(in)) 0 log(Site.(In)) 0 Any help in explaining why these two results are completely different would be much appreciated. I have tried the example and found the same results as given. Cheers Beth

Hello everyone, Recently, I faced a problem on explanatory of *Interaction variable* in Linear Regression, could anyone give me some help on how to explain that? the response variable Y is significantly correlated with *Interaction variable X* which is consisted of Continuous predictor A and Categorical predictor B. The Categorical predictor B has two factors B1 (value=1) and B2 (value=0). The

I am having troubles in fitting functions of the form y~a*x^b+c to data, for example x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23) y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86) I tried for example with nls, which did only work with really good initial guessed values. Any suggestion, what I should use? Thanks a lot Thomas [[alternative HTML version deleted]]

Hello, I wanted to create a histogram, but somehow I got stuck... The interval limits are: x = 1, 2, 3, 3.5, 4.5, 5, 5.5 The interval widths are therefore: 1, 1, 0.5, 1, 0.5, 0.5 Nothing I tried worked... Can anyone help me please? Thanks Tobias -- View this message in context: http://www.nabble.com/Histogram-tf4315900.html#a12288850 Sent from the R help mailing list archive at Nabble.com.

Hello, I have started in "Ruby on Rails: Talk" a discussion about my problems with current ActiveRecord implementation. Please take a look here: http://groups.google.com/group/rubyonrails-talk/browse_thread/thread/d40b754fc5872a99/47cff815df123b8a?lnk=gst&rnum=2 Regards Oliver Paulus --~--~---------~--~----~------------~-------~--~----~ You received this message because you are