Hello,
I have some conflicting output from an aov summary and tukey contrasts
with a mixed effects model I was hoping someone could clarify. I am
comparing the abundance of a species across three willow stand types.
Since I have 2 or 3 sites within a habitat I have included site as a
random effect in the lme model. My confusion is that the F test given by
aov(model) indicates there is no difference between habitats, but the
tukey contrasts using the multcomp package shows that one pair of habits
is significantly different from each other. Why is there a discrepancy?
Have I specified my model correctly? I included the code and output
below. Thank you.
> co.lme=lme(coye~stand,data=t,random=~1|site)
> summary (co.lme)
Linear mixed-effects model fit by REML
Data: R
AIC BIC logLik
53.76606 64.56047 -21.88303
Random effects:
Formula: ~1 | site
(Intercept) Residual
StdDev: 0.3122146 0.2944667
Fixed effects: coye ~ stand
Value Std.Error DF t-value p-value
(Intercept) 0.4936837 0.2305072 60 2.1417277 0.0363
stand2 0.4853222 0.3003745 4 1.6157240 0.1815
stand3 -0.3159230 0.3251201 4 -0.9717117 0.3862
Correlation:
(Intr) stand2
stand2 -0.767
stand3 -0.709 0.544
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-2.4545673 -0.5495609 -0.3148274 0.7527378 2.5151476
Number of Observations: 67
Number of Groups: 7
> anova(co.lme)
numDF denDF F-value p-value
(Intercept) 1 60 23.552098 <.0001
stand 2 4 3.738199 0.1215
> summary(glht(co.lme,linfct=mcp(stand="Tukey")))
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Fit: lme.formula(fixed = coye ~ stand, data = R, random = ~1 | site)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
2 - 1 == 0 0.4853 0.3004 1.616 0.2385
3 - 1 == 0 -0.3159 0.3251 -0.972 0.5943
3 - 2 == 0 -0.8012 0.2994 -2.676 0.0202 *
---
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
(Adjusted p values reported -- single-step method)
Lisa Baril
Masters Candidate
Department of Ecology
Montana State University - Bozeman
406.994.2670
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