[R-sig-ME] Mixed effects ANOVA
fernando barbero
fbarbero at bariloche.inta.gov.ar
Mon Sep 5 14:18:18 CEST 2011
Dear Carla, I think that you should run a LME model as this:
Log(Hrange)=rep (as a fixed effect)+ind (as a random effect)
Of course you should use lme4, I also think that rep might be nested under
ind,
Best regards
Fernando
-----Mensaje original-----
De: r-sig-mixed-models-bounces at r-project.org
[mailto:r-sig-mixed-models-bounces at r-project.org] En nombre de Carla Freitas
Enviado el: viernes, 26 de agosto de 2011 07:48 a.m.
Para: r-sig-mixed-models at r-project.org
Asunto: [R-sig-ME] Mixed effects ANOVA
Dear all,
I have calculated the spring home range size from a sample of tracked
animals. I would like now to test whether their home range size is
significantly different between animals of different reproductive status.
Some of the animals were tracked in two different springs and have changed
reproductive state from one year to the other. I though therefore to run a
mixed effects ANOVA, using individual identification as random effect. I
used the function aov to do this (see below) but I'm not sure if it is the
correct way to do it. My interpretation of the results is that there are
significant differences in home range size between individuals of different
reproductive status and individual identification does not significantly
affects those differences in home range size. Is this the most correct way
to test my hypothesis? Or should I instead fit a mixed effects model using
lmer?
My dataset is the following:
(rep is reproductive status, hrange is home range size and ind is individual
id)
> mydata
rep hrange ind
1 coy 138.7 2166
3 coy 426.5 2172
6 coy 102.3 2178
7 coy 49.0 2182
9 coy 404.5 9678
12 coy 17.2 9690
13 coy 378.2 9692
14 coy 392.0 9695
17 coy 53.0 2170
20 coy 67.7 2175
22 coy 48.9 2185
24 coy 45.8 9692
2 single 350.1 2170
5 single 1069.4 2175
8 single 629.5 2185
10 single 2772.5 9679
16 single 16.2 2165
18 single 506.7 2172
19 single 371.5 2174
21 single 340.5 2178
23 single 6176.3 9684
25 single 271.9 9696
4 yrlg 414.4 2174
11 yrlg 4445.5 9684
15 yrlg 323.8 9696
The code I used to fit the ANOVA:
(note that I log transformed hrange to reach normality and homogeneity of
variances)
> summary(aov(log(hrange)~rep + Error(ind)),data=mydata)
Error: ind
Df Sum Sq Mean Sq F value Pr(>F)
rep 2 12.593 6.2966 2.4796 0.1224
Residuals 13 33.011 2.5393
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
rep 2 7.3580 3.6790 5.6745 0.03429 *
Residuals 7 4.5383 0.6483
---
Thank you in advance for your help.
Carla
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