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 <lorin@(protected):
>>
>>
>>> 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, ripley@(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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