On Wednesday 6. March 2013 14.50.23 Ben Bolker wrote:
> Just a quick thought (sorry for removing context): what happens if
> you use sum-to-zero contrasts throughout, i.e.
> options(contrasts=c("contr.sum", "contr.poly")) ... ?
Ah, I've got it now, this pointed me in the right direction. Thanks a lot!
For future reference, the default was "contr.treatment", which implies
> contrasts(leaf$B)
+
- 0
+ 1
using contr.sum, we get
> contrasts(leaf$B)
[,1]
- 1
+ -1
which is orthogonal, and also explains the switched sign.
Kjetil
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