Dear Colm,
Use lm() instead of aov() and your problem is solved. The parameter of Group.L
is the linear trend along your groups. Group.Q the quandratic trend.
HTH,
Thierry
summary(lm(Mean_1 ~ Group*eventType, data=doi))
Call:
lm(formula = Mean_1 ~ Group * eventType, data = doi)
Residuals:
Min 1Q Median 3Q Max
-2.88504 -0.74862 -0.01771 0.92154 2.62031
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.1360 0.3610 14.229 3.41e-13 ***
Group.L 2.7761 0.6252 4.440 0.000172 ***
Group.Q -0.6375 0.6252 -1.020 0.318091
eventTypestops -2.0464 0.5105 -4.009 0.000515 ***
Group.L:eventTypestops -2.3123 0.8842 -2.615 0.015175 *
Group.Q:eventTypestops 1.9637 0.8842 2.221 0.036049 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05
'.' 0.1 ' ' 1
Residual standard error: 1.398 on 24 degrees of freedom
Multiple R-squared: 0.6357, Adjusted R-squared: 0.5598
F-statistic: 8.375 on 5 and 24 DF, p-value: 0.0001073
----------------------------------------------------------------------------
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
Thierry.Onkelinx at inbo.be
www.inbo.be
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experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
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-----Oorspronkelijk bericht-----
Van: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
Namens Colm Connolly
Verzonden: vrijdag 18 december 2009 16:13
Aan: r-help at stat.math.ethz.ch
Onderwerp: [R] linear contrasts for trends in an anova
Hi everybody,
I'm trying to construct contrasts for an ANOVA to determine if there is a
significant trend in the means of my groups.
In the following example, based on the type of 2x3 ANOVA I'm trying to
perform, does the linear polynomial contrast generated by contr.poly allow me to
test for a linear trend across groups?
doi=data.frame(
Group=c(
rep(1, 5), rep(2, 5), rep(3, 5), rep(1, 5), rep(2, 5), rep(3, 5)),
Mean_1=c(
rnorm(5, mean=3, sd=1), rnorm(5, mean=5.7, sd=1.2), rnorm(5, mean=7,
sd=1.3),
rnorm(5, mean=3.5, sd=1), rnorm(5, mean=3, sd=1.2), rnorm(5, mean=4.2,
sd=1.7)),
eventType=c( rep("errors", 15), rep("stops", 15)))
doi$Group=factor(doi$Group, labels=c("Control", "Short",
"Long"), ordered=T)
## construct contrasts
contrasts(doi$Group)=contr.poly(levels(doi$Group))
model1=aov(Mean_1 ~ Group*eventType, data=doi)
model1.summary=summary(model1)
print(model1.summary)
print(summary.lm(model1))
Thanks in advance,
Regards,
--
Dr Colm G. Connolly
Institute of Neuroscience
The Lloyd Building
University of Dublin
Trinity College, Dublin 2, ?ire
Fax: +353-1-671-3183
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