summary.lme() vs. anova.lme()

Thu, Nov 18, 2004 1:52 AM

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:

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

summary.lme() vs. anova.lme()

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