anova.lm() gives the sequential tests:
> set.seed(1)
> dat <- data.frame(y=rnorm(10), x1=runif(10), x2=runif(10))
> anova(lm(y ~ x1 + x2, dat))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x1 1 1.1483 1.1483 2.0943 0.1911
x2 1 0.4972 0.4972 0.9068 0.3727
Residuals 7 3.8383 0.5483 > anova(lm(y ~ x2 + x1, dat))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x2 1 0.5165 0.5165 0.9419 0.3641
x1 1 1.1291 1.1291 2.0592 0.1944
Residuals 7 3.8383 0.5483
The SS, F-stat, etc. would be invariant to order only if the terms are
orthogonal.
Andy
> From: Jarrett Byrnes
>
> I'm curious, I realize there are methods for Type II and III sums of
> squares, and yet, when I've been constructing models with lm, I've
> noticed that position of the term of the model has not mattered in
> terms of its p-value. Does lm use sequential Type I sums of squares,
> or something else?
>
> Thanks!
>
> -Jarrett
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
>
