Re: [R] Asking, are simple effects different from 0

From: jebyrnes <jebyrnes_at_ucdavis.edu>
Date: Wed, 05 Mar 2008 07:09:49 -0800 (PST)

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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