Hi Benedikt
in principle this is correct. A p-value less than a prespecified alpha-level leads to rejection of the null hypothesis H0. I'm a little bit puzzled about your given p-value, since actually your slopes are the same, so H0 is valid but rejected. This might be a matter of chance, but it is so to say against all odds (~2 out of 10 000 000) having such a special data set.

Alternatively - and using well established estimators - you can incorporate your factor in a joint model.
df1 <- data.frame(x=1:3, y=1:3+rnorm(3),g=0)
df2 <- data.frame(x=1:3, y=1:3+rnorm(3),g=1)
dta<-rbind(df1,df2)
dta$g<-factor(dta$g) # not really necessary for a two-group factor
ll<-lm(y~x*g,dta)
summary(ll)

This model has a varying slope and intercept term (which is the same as fitting two separate regression lines as you did), the test for different slopes is to test significance of x:g. If you are in fact asking for a pure varying slope model you have to restrict your fit1 and fit2 to have the same intercept or you use the joint model with
ll<-lm(y~x+x:g,dta)

hth

Benedikt Niesterok schrieb:
Hello R users,
I've used the following help two compare two regression line slopes.
Wanted to test if they differ significantly:

Hi,

I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R.

I knew the method based on the following statement :
t = (b1 - b2) / sb1,b2
where b1 and b2 are the two slope coefficients and sb1,b2 the pooled standard error of the slope (b)

which can be calculated in R this way:
 > 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)

My value I get by running this test is :[1] 2.305553e-07
Does it mean the two slopes differ significantly, because this value is in
the alpha area, so that I have to reject the null- hypothesis and accept
the alternative hypothesis?
Is the  null-hypothesis: slope1=slope2?
Thanks for your help,  Benedikt
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Institut für Medizinische Biometrie und Epidemiologie
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