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
4 FSI055A A 1 NA
5 FSI055B B 1 NA
6 FSI057A A 1 NA
7 FSI057B B 1 NA
8 FSI058A A 1 NA
9 FSI058B B 1 12.93
10 FSI060A A 1 NA
11 FSI060B B 1 NA
12 FSI067A A 1 NA
13 FSI070A A 1 NA
14 FSI070B B 1 NA
15 FSI075A A 1 NA
16 FSI075B B 1 NA
17 FSI078A A 1 NA
18 FSI078B B 1 NA
19 FSI052A A 2 13.65
20 FSI052B B 2 11.71
21 FSI053A A 2 20.91
22 FSI055A A 2 9.52
23 FSI055B B 2 13.43
24 FSI057A A 2 14.57
25 FSI057B B 2 10.00
26 FSI058A A 2 14.95
27 FSI058B B 2 16.97
28 FSI060A A 2 13.06
29 FSI060B B 2 7.96
[...]
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.
Thanks for all your help.
Regards,
Sumit
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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???) -- O__ ---- Peter Dalgaard ?ster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907