R help - Nov 2011 - compare linear regressions

Hi,

        I'm trying to compare two linear regressions. I'm using the
following approach:
##################
xx<-1:100
df1 <- data.frame(x = xx, y = xx * 2 + 30 + rnorm(n=length(xx),sd=10), g = 1)
df2 <- data.frame(x = xx, y = xx * 4 + 9  + rnorm(n=length(xx),sd=10), g = 2)
dta <- rbind(df1, df2)
dta$g <- factor(dta$g)
plot(df2$x,df2$y,type="l",col="red")
lines(df1$x,df1$y,col="blue")
summary(lm(formula = y ~ x + g + x:g, dta))
##################
I learned that the coefficients (g2 and x:g2) tell me about the
differences in intercept and slope and the corresponding p-values.

Now I'm trying to do the same except that there should be no intercept
term:
##################
xx<-1:100
df1 <- data.frame(x = xx, y = xx * 2 + rnorm(n=length(xx),sd=10), g = 1)
df2 <- data.frame(x = xx, y = xx * 4 + rnorm(n=length(xx),sd=10), g = 2)
dta <- rbind(df1, df2)
dta$g <- factor(dta$g)
plot(df2$x,df2$y,type="l",col="red")
lines(df1$x,df1$y,col="blue")
summary(lm(formula = y ~ x - 1 + x:g, dta))
##################
I assume that the last line is the correct way to specify a linear model
without intercept. But I'm not certain about that. Can someone please
confirm?

Thanks a lot,
        Holger

Yes, adding "-1" or "+0" both return a lm without an
intercept term.

Michael

On Tue, Nov 8, 2011 at 8:52 AM, Holger Taschenberger
<Holger.Taschenberger at mpi-bpc.mpg.de> wrote:
> Hi,
>
> ? ? ? ?I'm trying to compare two linear regressions. I'm using the
> following approach:
> ##################
> xx<-1:100
> df1 <- data.frame(x = xx, y = xx * 2 + 30 + rnorm(n=length(xx),sd=10), g
= 1)
> df2 <- data.frame(x = xx, y = xx * 4 + 9 ?+ rnorm(n=length(xx),sd=10), g
= 2)
> dta <- rbind(df1, df2)
> dta$g <- factor(dta$g)
> plot(df2$x,df2$y,type="l",col="red")
> lines(df1$x,df1$y,col="blue")
> summary(lm(formula = y ~ x + g + x:g, dta))
> ##################
> I learned that the coefficients (g2 and x:g2) tell me about the
> differences in intercept and slope and the corresponding p-values.
>
> Now I'm trying to do the same except that there should be no intercept
> term:
> ##################
> xx<-1:100
> df1 <- data.frame(x = xx, y = xx * 2 + rnorm(n=length(xx),sd=10), g = 1)
> df2 <- data.frame(x = xx, y = xx * 4 + rnorm(n=length(xx),sd=10), g = 2)
> dta <- rbind(df1, df2)
> dta$g <- factor(dta$g)
> plot(df2$x,df2$y,type="l",col="red")
> lines(df1$x,df1$y,col="blue")
> summary(lm(formula = y ~ x - 1 + x:g, dta))
> ##################
> I assume that the last line is the correct way to specify a linear model
> without intercept. But I'm not certain about that. Can someone please
> confirm?
>
> Thanks a lot,
> ? ? ? ?Holger
>
> ______________________________________________
> R-help at 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.
>

Thank you. I was confused because the output of the line
"summary(lm(formula
=..." reads:

"Coefficients: (1 not defined because of singularities)",

which did not look like a normal message (which can safely be ignored)
to me.

--Holger

On Tue, 8 Nov 2011 15:06:28 +0100
Peter Konings <peter.l.e.konings at gmail.com> wrote:
> On Tue, Nov 8, 2011 at 2:52 PM, Holger Taschenberger <
> Holger.Taschenberger at mpi-bpc.mpg.de> wrote:
> <snip>
> 
> > summary(lm(formula = y ~ x - 1 + x:g, dta))
> > ##################
> > I assume that the last line is the correct way to specify a linear
model
> > without intercept. But I'm not certain about that. Can someone
please
> > confirm?
> 
> 
> Yes, that's true. See chapter 11 of the "Introduction To R"
manual that was
> installed with R for an overview of model specification in R.
> 
> HTH
> Peter.