Sumitrajit Dhar wrote:
Hi folks,
I am trying to figure out how run a repeated measures ANOVA on the
following data set.
subject trial frequency dplvl
1 FSI052A A 1 NA
2 FSI052B B 1 NA
3 FSI053A A 1 NA
..
Both "frequency" and "trial" are factorized.
> is.factor(trial)
[1] TRUE
> is.factor(frequency)
[1] TRUE
> levels(frequency)
[1] "1" "2" "3" "4" "6"
I know that I should be using 'lme' rather than aov, but....
> summary(aov(dplvl~trial*frequency+Error(subject/
(trial*frequency)),data=tenB))
Error: subject
Df Sum Sq Mean Sq F value Pr(>F)
trial 1 7.15 7.15 0.2566 0.6203
frequency 1 40.84 40.84 1.4657 0.2461
trial:frequency 1 0.03 0.03 0.0013 0.9723
Residuals 14 390.11 27.86
Error: subject:frequency
Df Sum Sq Mean Sq F value Pr(>F)
frequency 1 183.893 183.893 11.925 0.003271 **
trial:frequency 1 16.393 16.393 1.063 0.317866
Residuals 16 246.738 15.421
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
Residuals 35 269.13 7.69
What I do not understand is why I am getting df=1 for frequency when
there are 5 levels of that factor.
Is tenB$frequency also a factor? The prototypical error would be to have
modified only a copy of it.
By the way, with that amount of NA in the response, you are not going to
undertand aov output in any case...
(and what happened with the subject:trial stratum???)
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