Prof Brian Ripley [EMAIL PROTECTED] writes:
drop1 is the part of R that does type II sum of squares, and it works in
your example. So does Anova in the current car:
I'm sorry, I should have included an example to clarify what I meant
(or point out my misunderstandings :-). I'll do that
On Tue, 12 Aug 2003, [iso-8859-1] Bjørn-Helge Mevik wrote:
Prof Brian Ripley [EMAIL PROTECTED] writes:
drop1 is the part of R that does type II sum of squares, and it works in
your example. So does Anova in the current car:
I'm sorry, I should have included an example to clarify what I
On Tue, 12 Aug 2003, Bjørn-Helge Mevik wrote:
Prof Brian D Ripley [EMAIL PROTECTED] writes:
On Tue, 12 Aug 2003, [iso-8859-1] Bjørn-Helge Mevik wrote:
Why should I(x^2) be regarded as subservient to x?
In polynomial regression, it is usual to first consider a linear
model, then a
Prof Brian D Ripley [EMAIL PROTECTED] writes:
On Tue, 12 Aug 2003, [iso-8859-1] Bjørn-Helge Mevik wrote:
Also, is this example (lm(y~x+I(x^2), Df)) really balanced? I think
No, and I did not use summary,aov on it!
And I didn't say you did!
This gives the SSs R(x | A, B, A:B, x^2), R(x^2
Anova != anova.
drop1 is the part of R that does type II sum of squares, and it works in
your example. So does Anova in the current car:
drop1(lm(y~x+I(x^2), Df)) # add test=F if you like
Single term deletions
Model:
y ~ x + I(x^2)
Df Sum of SqRSSAIC
none 8.3117
