[R-sig-ME] Wrong degrees of freedom in lmer model?

[ONKELINX, Thierry]

Dear Ryan,

Targets needs 4 degrees of freedom because you omitted the intercept
from your model. A model with intercept needs only 3 degrees of freedom
because the intercept estimates the effect of the reference level.

HTH,

Thierry

------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium 
tel. + 32 54/436 185
Thierry.Onkelinx at inbo.be 
www.inbo.be 

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey

-----Oorspronkelijk bericht-----
Van: r-sig-mixed-models-bounces at r-project.org
[mailto:r-sig-mixed-models-bounces at r-project.org] Namens Ryan Hope
Verzonden: donderdag 2 april 2009 21:52
Aan: r-sig-mixed-models at r-project.org
Onderwerp: [R-sig-ME] Wrong degrees of freedom in lmer model?

I am trying to use the lmer function to analyze the data from the
thesis experiment. The response variable is a continuous variable,
completion time. I have 2 fixed factors, number of targets (4-levels:
4,9,14,19) and displacement per second (DPS) (3-levels: 72.42, 84.99,
88.99). I also have 3 random variables, participants, block (which
group the replicates of a combination of fixed factors), and target
displacement. The R code that I am using is listed below:

library(lme4)
datafile="http://people.rit.edu/rmh3093/master1.csv"
master1=read.table(datafile,header=T)
master1$Targets=as.factor(master1$Targets)
model=lmer(log(Completion_Time)~-1+Targets+DPS+(-1+Targets|Participant_I
D)+(1|Block)+(1|Target_Displacement),data=master1)
anova(model)

Im not even sure if the lmer model I am using is 100% correct but it
seems the best to based on what I've read on
(http://www.let.uu.nl/~Hugo.Quene/personal/multilevel/jml2008/x24lmerlog
.html).
Anyway, I do not understand the degrees of freedom reported by the
anova function based on my model (shown below):

Analysis of Variance Table
        Df Sum Sq Mean Sq    F value
Targets  4  33393    8348 39411.3500
DPS      1      2       2     7.7816

Since there are 4 levels of the Target factor, shouldn't the Df for
Targets be 3 not 4?

-Ryan

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