[R] Asking, are simple effects different from 0

[Chuck Cleland]

On 3/5/2008 10:09 AM, jebyrnes wrote:
> 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?  

   Each column corresponds to one of the coefficients in the model, and 
each row specifies a particular contrast.  The numbers in the matrix 
indicate how the model coefficients are combined to indicate a 
particular difference in means.
   For example, the first row indicates that the third coefficient 
(woolB) is multiplied by -1.  The baseline categories are A and L for 
the wool and tension factors, so the woolB effect in fm is the simple 
effect of B vs. A in the baseline category of the tension factor. 
Multiplying this coefficient by -1 produces an A vs. B comparison in the 
baseline category of the tension factor.
   When I want to check that contrasts are as intended, I use contrast() 
in the contrast package (by Steve Weston, Jed Wing, & Max Kuhn).  That 
function allows you to specify factor levels by name to construct the 
contrast.  For example:

library(contrast)

# M vs. H at B

contrast(fm, a=list(tension = "M", wool = "B"),
              b=list(tension = "H", wool = "B"))

lm model parameter contrast

  Contrast     S.E.      Lower    Upper    t df Pr(>|t|)
        10 5.157299 -0.3694453 20.36945 1.94 48   0.0584

   It also allows you to print the design matrix for a contrast:

contrast(fm, a=list(tension = "M", wool = "B"),
              b=list(tension = "H", wool = "B"))$X

   (Intercept) tensionM tensionH woolB tensionM:woolB tensionH:woolB
1           0        1       -1     0              1             -1

> 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) 

-- 
Chuck Cleland, Ph.D.
NDRI, Inc.
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