Hi,
I did a aov and used summary to obtain the p-value. I tried many ways to
extract the p-value from
the summary result but failed. Among others I tried the following:
test.summary -
summary(aov(data[,1]~time.points+Error(subject/time.points)))
test.summary
Error: subject
Df
Hi:
Try this:
test.summary[[1]][, 5]
It should return a vector of p-values, the last being NA. In your case,
since there is only one non-NA p-value, it is enough to do
test.summary[[1]][, 5][1]
You mean that wasn't obvious? :)
Explanation:
summary(aovobj) actually returns a list, but
On Sep 29, 2010, at 14:25 , Dennis Murphy wrote:
test.summary[[1]][, 5][1]
You mean that wasn't obvious? :)
Worse, it doesn't actually work...
test.summary - summary(npk.aovE)
test.summary[[1]][, 5]
Error in `[.default`(test.summary[[1]], , 5) :
incorrect number of dimensions
Hi:
On Wed, Sep 29, 2010 at 5:56 AM, peter dalgaard pda...@gmail.com wrote:
On Sep 29, 2010, at 14:25 , Dennis Murphy wrote:
test.summary[[1]][, 5][1]
You mean that wasn't obvious? :)
Worse, it doesn't actually work...
test.summary - summary(npk.aovE)
test.summary[[1]][, 5]
Hi:
Someone off-list (Josh Wiley - thank you) mentioned the Error() term in the
OP's ANOVA, which I missed in responding to the post - sorry for the
misinformation. Using the npk example with code that Josh showed me, we
have, for the following model,
npk.aov2 - aov(yield ~ N*P*K +
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