At 01:59 PM 18/07/2005, Addison, Prue wrote:
Hi,
I am having trouble using the lme function to perform a nested ANOVA
with a random nested factor.
My design is as follows:
Location (n=6) (Random)
Site nested within each Location (n=12) (2 Sites nested within each
Location) (Random)
Dependent variable: sp (species abundance)
By using the aov function I can generate a nested ANOVA, however this
assumes that my nested factor is fixed.
summary(aov(sp~Location/Site, data=mavric))
Df Sum Sq Mean Sq F value Pr(F)
Location 4 112366 28092 1.2742 0.2962
Location:Transect 5 121690 24338 1.1039 0.3736
Residuals 40 881875 22047
Yes, this is equivalent to aov(sp ~ Location + Location:Site,...)
I have tried the following lme function to specify that Site is random:
lme1 - lme(sp~Location, random=~1|Site, data=mavric)
Here, Location is fixed, and Site is a grouping factor. There are fixed and
random components to the intercept.
lme2 - lme(sp~Location, random=~1|Location/Site, data=mavric)
Here you have Location as a fixed effect, the intercept is random and the
grouping is Site %in % Location.
anova(lme1)
numDF denDF F-value p-value
(Intercept) 140 3.418077 0.0719
Location4 5 1.152505 0.4294
How can you have 6 sites, but only 4 df for Location (should be 5?)
This gives me the correct F-value for Location from
MSLocation/MSLocation:Transect, but the p-value doesn't seem to be
correct (by my calculations in Microsoft Excel it should be 0.345)
I don't know your data, or your calculations. Microsoft Excel does not fill
me with confidence.
anova(lme2)
Warning in pf(q, df1, df2, lower.tail, log.p) :
NaNs produced
Warning: NAs introduced by coercion
numDF denDF F-value p-value
(Intercept) 140 0.459966 0.5015
Location4 0 0.155091 NaN
? I don't know what this output means
Division by zero, somewhere? :-)
anova(lme1,lme2)
Model df AIC BIClogLik Test L.Ratio p-value
lme1 1 7 603.7534 616.4000 -294.8767
lme2 2 8 605.7534 620.2067 -294.8767 1 vs 2 1.815674e-05 0.9966
? I also don't know what this output means.
you are testing the difference between the models, using a likelihood ratio
test. The difference is not significant, so the conventional wisdom is to
choose the simpler model (lme1). Note that the AIC and BIC are lower
(better) for lme1, too.
Can anyone tell me if there is a way to use the lme() function in order
to obtain the same output as the aov() function (above), but so it
correctly calculates the MS, F and p values for my main Location factor?
The following code shows agreement:
dat - data.frame(Location=factor(rep(1:6, each=2)), Site=factor(rep(1:2,
12)), sp=rnorm(12))
fit - aov(sp ~ Location + Error(Site), data=dat)
fit2 - lme(sp ~ Location, random=~1|Site, data=dat)
summary(fit)
anova(fit2)
HTH,
Simon.
Thanks,
Prue
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Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat.
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