se.contrast computes standard errors of contrasts. It does not compute the contrast itself, which is fairly simple to do directly.

c1 <- c(1,-1)[A]*c(1,-1,0)[B]
sum(c1*score)
[1] -18
se.contrast(fit, as.matrix(c1))
Contrast 1
  14.24547
(18/14.24547)^2
[1] 1.596583

so the contrast has value -18, standard error 14.24547, and the squared t-ratio is the F statistic 1.5966 you were asking for. (That's a piece of the theory of linear models.)


On Mon, 21 Feb 2005, Lorin Hochstein wrote:

Peter Dalgaard wrote:

Lorin Hochstein <[EMAIL PROTECTED]> writes:


I'd like to understand this approach as well, but I can't reproduce my
results using se.contrast. In particular, I get the same standard
error even though I tried to use different contrasts:

> c1 <- c(1,-1)[A]*c(1,-1,0)[B]
> c2 <- c(1,-1)[A]*c(1,0,-1)[B]
> c3 <- c(1,-1)[A]*c(0,1,-1)[B]
> se.contrast(fit, as.matrix(c1))
Contrast 1
 14.24547
> se.contrast(fit,as.matrix(c2))
Contrast 1
 14.24547
> se.contrast(fit,as.matrix(c3))
Contrast 1
 14.24547


They could well _be_ the same if the design is balanced...


Hmmm... One of my problems is that I don't know how to interpret the output of se.contrast.

Here's my example again.
score <- c(12, 8,10, 6, 8, 4,
     10,12, 8, 6,10,14,
      9, 7, 9, 5,11,12,
      7,13, 9, 9, 5,11,
      8, 7, 3, 8,12,10,
     13,14,19, 9,16,14)
n <- 6
A <- gl(2,3*n,labels=c("a1","a2"))
B <- rep(gl(3,n,labels=c("b1","b2","b3")),2)
contrasts(B) <- c(1,-1,0)
fit <- aov(score~A*B)
summary(fit, split=list(B=1:2), expand.split = T)
Df Sum Sq Mean Sq F value Pr(>F) A 1 18.778 18.778 2.2208 0.146606 B 2 62.000 31.000 3.6662 0.037629 *
B: C1 1 1.500 1.500 0.1774 0.676621 B: C2 1 60.500 60.500 7.1551 0.011986 *
A:B 2 81.556 40.778 4.8226 0.015274 *
A:B: C1 1 13.500 13.500 1.5966 0.216119 # <---
A:B: C2 1 68.056 68.056 8.0486 0.008085 **
Residuals 30 253.667 8.456 What I'm really looking for is that F value that's labelled A:B: C1, 1.5966 in this case. (I'm not sure what to call this term, AB interaction?)


I thought that it might be possible to use se.contrast to compute this (or at least, to get the numerator so that I could compute the F value once I had the mean square error of the residuals), but I'm not sure how to specify the contrast, and I don't know the relationship between the "standard error" output by se.contrast and the "mean square error" which is the fourth column of the output above.

-- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595

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