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

[Dan Bebber]

For anyone following this thread in the future:
Following Prof. Ripley's advice, I compared models fitted with ML, with and
without treatment as a predictor:

> anova(mconc.lme1,mconc.lme2)
           Model df       AIC       BIC   logLik   Test L.Ratio p-value
mconc.lme1     1 10 -1366.184 -1327.240 693.0919                       
mconc.lme2     2 16 -1363.095 -1300.785 697.5475 1 vs 2 8.91124  0.1786

I can't reject the null hypothesis of no effect of treatment.

Many thanks.
Dan Bebber

Department of Plant Sciences
University of Oxford
South Parks Road
Oxford OX1 3RB
UK
Tel. 01865 275000



> -----Original Message-----
> From: Prof Brian Ripley [mailto:ripley at stats.ox.ac.uk] 
> Sent: 17 November 2004 15:32
> To: Dan Bebber
> Cc: r-help at stat.math.ethz.ch
> Subject: Re: [R] summary.lme() vs. anova.lme()
> 
> 
> 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)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>