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
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