[R] summary.lme() vs. anova.lme()

[Prof Brian Ripley]

On Wed, 17 Nov 2004, Dan Bebber wrote:

> I modelled changes in a variable (mconc) over time (d) for individuals
> (replicate) given one of three treatments (treatment) using:
> mconc.lme <- lme(mconc~treatment*poly(d,2), random=~poly(d,2)|replicate,
> data=my.data)
>
> summary(mconc.lme) shows that the linear coefficient of one of the
> treatments is significantly different to zero, viz.
>                            Value Std.Error  DF   t-value p-value
> 					...	     ... ...      ...
> ...
> treatmentf:poly(d, 2)1  1.3058562 0.5072409 315  2.574430  0.0105
>
> But anova(mconc.lme) gives a non-significant result for the treatment*time
> interaction, viz.
>                     numDF denDF   F-value p-value
> (Intercept)              1   315 159.17267  <.0001
> treatment                2    39   0.51364  0.6023
> poly(d, 2)               2   315  17.43810  <.0001
> treatment:poly(d, 2)     4   315   2.01592  0.0920
>
> Pinheiro & Bates (2000) only discusses anova() for single arguments briefly
> on p.90.
> I would like to know whether these results indicate that the significant
> effect found in summary(mconc.lme) is spurious (perhaps due to
> multiplicity).

Probably yes (but p values of 9% and 1% are not that different, and in 
both cases you are looking at a few p values).

But since both summary.lme and anova.lme use Wald tests, I would use a 
LRT, using anova on two fits (and I would use ML fits to get a genuine 
LRT but that is perhaps being cautious).

To  Dimitris Rizopoulos: as this is the last term in the sequential anova, 
it is the correct Wald test.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
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