[R] Asking, are simple effects different from 0

[jebyrnes]

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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