[R] DF in LME

Federico Calboli f.calboli at ucl.ac.uk
Wed Jul 2 23:40:02 CEST 2003

Dear All,

I know I am quite obsessive and downright annoying (I apologize about that,
but it's the way I am), but I would like to get my understanding of the way
nlme calculates degrees of freedom straight.

For instance, on page 91 in Pinheiro and Bates (2000), on the examle of
anova(fm2Machie), how is the sum of DF *Pi* corresponding to the terms
estimated at level *i* calculated? In the example presenting
anova(fm2Machine), *P1 = 0*. I just fail to see why. Same thing for *P2 =
2* (although this seems intuitive, but intuitive could be miles off the
real reason) and *P3 = 0*.

Incidentally, should I change the grouping, putting *machine* outside and
*worker* inside, would anything change?

A second thing: is there any substantial difference between the classical
decomposition of DF for an ANOVA and the method used by lme for the
interaction between a fixed and a random effect, in case the random
variable is nominal and the fixed one continuous?

Best regards,
Federico Calboli


Federico C.F. Calboli

Department of Biology
University College London
Room 327
Darwin Building
Gower Street

Tel: (+44) 020 7679 4395 
Fax (+44) 020 7679 7096
f.calboli at ucl.ac.uk

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