[R] Comparing two regression slopes

[Spencer Graves]

      I missed the earlier thread, but if I had the data and not just 
the table of coefficients and standard errors, then I'd try combining 
the data sets as follows: 

 > set.seed(1)
 > df1 <- data.frame(x=1:3, y=1:3+rnorm(3))
 > df2 <- data.frame(x=1:3, y=1:3+rnorm(3))
 > DF. <- data.frame(source=rep(c(-1,1), each=3), rbind(df1, df2))
 >
 > anova(lm(y~x+I(source*x), DF.))
Analysis of Variance Table

Response: y
              Df  Sum Sq Mean Sq F value Pr(>F)
x              1 0.47271 0.47271  0.5696 0.5053
I(source * x)  1 0.23386 0.23386  0.2818 0.6323
Residuals      3 2.48964 0.82988              
 >
      hope this helps.  spencer graves

Joerg Schaber wrote:

> I would suggest the method of Sokal and Rholf (1995) S. 498, using the 
> F test.
> Below I repeat the analysis by Spencer Graves:
>
> Spencer:
> > df1 <- data.frame(x=1:3, y=1:3+rnorm(3))
> > df2 <- data.frame(x=1:3, y=1:3+rnorm(3))
> > fit1 <- lm(y~x, df1)
> > s1 <- summary(fit1)$coefficients
> > fit2 <- lm(y~x, df2)
> > s2 <- summary(fit2)$coefficients
> > db <- (s2[2,1]-s1[2,1])
> > sd <- sqrt(s2[2,2]^2+s1[2,2]^2)
> > df <- (fit1$df.residual+fit2$df.residual)
> >  td <- db/sd
> > 2*pt(-abs(td), df)
> [1] 0.8757552
>
> Sokal & Rholf
> > n <- length(df1$x)      > ssx1 <- var(df1$x)*(n-1)      # sums of 
> squares
> > ssx2 <- var(df2$x)*(n-1)
> > ssy1 <- var(df1$y)*(n-1)
> > ssy2 <- var(df2$y)*(n-1)
> > sxy1 <- cor(df1$x,df1$y)*(n-1)
> > sxy2 <- cor(df2$x,df2$y)*(n-1)
> > d2xy1 <- ssy1 - sxy1^2/ssx1         # unexplained
> > d2xy2 <- ssy2 - sxy2^2/ssx2
> > Fs <- db^2/((ssx1+ssx2)/(ssx1*ssx2)*(d2xy1+ d2xy2)/(2*n-4))  # F 
> statistic
> > 1-pf(Fs,1,2*n-4)
> [1] 0.8827102
>
> slight differences, but can be MUCH larger with more data!
>
> greetings,
>
> joerg
>
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