Asking, are simple effects different from 0
Huh. Very interesting. I haven't really worked with manipulating contrast matrices before, save to do a prior contrasts. Could you explain the matrix you laid out just a bit more so that I can generalize it to my case?
Chuck Cleland wrote:
One approach would be to use glht() in the multcomp package. You
need to work out how to formulate the matrix of coefficients that give
the desired contrasts. Here is an example using the warpbreaks data
frame:
fm <- lm(breaks ~ tension*wool, data=warpbreaks)
# names(coef(fm))
# (Intercept) tensionM tensionH woolB tensionM:woolB tensionH:woolB
cm <- rbind(
"A vs. B at L" = c(0, 0, 0,-1, 0, 0),
"A vs. B at M" = c(0, 0, 0,-1,-1, 0),
"A vs. B at H" = c(0, 0, 0,-1, 0,-1),
"M vs. L at A" = c(0, 1, 0, 0, 0, 0),
"M vs. H at A" = c(0, 1,-1, 0, 0, 0),
"L vs. H at A" = c(0, 0,-1, 0, 0, 0),
"M vs. L at B" = c(0, 1, 0, 0, 1, 0),
"M vs. H at B" = c(0, 1,-1, 0, 1,-1),
"L vs. H at B" = c(0, 0,-1, 0, 0,-1))
library(multcomp)
summary(glht(fm, linfct = cm), test = adjusted(type="none"))
Simultaneous Tests for General Linear Hypotheses
Fit: lm(formula = breaks ~ tension * wool, data = warpbreaks)
Linear Hypotheses:
Estimate Std. Error t value p value
A vs. B at L == 0 16.3333 5.1573 3.167 0.002677 **
A vs. B at M == 0 -4.7778 5.1573 -0.926 0.358867
A vs. B at H == 0 5.7778 5.1573 1.120 0.268156
M vs. L at A == 0 -20.5556 5.1573 -3.986 0.000228 ***
M vs. H at A == 0 -0.5556 5.1573 -0.108 0.914665
L vs. H at A == 0 20.0000 5.1573 3.878 0.000320 ***
M vs. L at B == 0 0.5556 5.1573 0.108 0.914665
M vs. H at B == 0 10.0000 5.1573 1.939 0.058392 .
L vs. H at B == 0 9.4444 5.1573 1.831 0.073270 .
---
Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
(Adjusted p values reported -- none method)
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