You can use the summary with the split argument.
See
?summary.aov
for an example.
This gives similar results to the regression coefficients approach that Ista
Zahn
suggested. I usually prefer the ANOVA table approach for factors.
On Thu, Feb 3, 2011 at 3:36 AM, Asan Ramzan wrote:
> Hello R-he
By fitting some kind of model. contr.poly doesn't fit a model or test
significance, it just sets contrasts. Here is an example
data(mtcars)
> mtcars$carb <- factor(mtcars$carb)
> contrasts(mtcars$carb) <- contr.poly(n=levels(mtcars$carb))
> contrasts(mtcars$carb)
.L .Q .C
Hello R-help
(sorry if this message gets posted twice, i think I may have accidently
postponed it)
I donât know how to interpret significance from the contr.poly() function .
From
the example below
: how can I tell if data has a significant Linear/quadratic/cubic trend?
> contr.poly(4, c(1,2,4
Hello R-help
I donât know how to interpret significance from the contr.poly() function .
From
the example below
: how can I tell if data has a significant Linear/quadratic/cubic trend?
> contr.poly(4, c(1,2,4,8))
             .L        .Q         .C
[1,] -0.5128
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