[R] newbie question on contrasts and aov
Wolfgang Pauli
wolfgangpauli at web.de
Thu Jan 29 16:59:40 CET 2004
In the meantime I figured out that the Difference-contrast is not quite what I
was looking for. But I still have two questions
1) Why do I get different results for Helmert contrasts in SPSS and R. I guess
the contrast matrixes of Helmert are about the same in SPSS and R. I probably
make a mistake as i am a newbie to R. I thought that it might be, because I
have a repaeted measures design. That's why I put the Error(sub) in the
formula of aov.
2)
I tried to make my own contrast matrix, to compute comparisons between
adjacent factor levels, i.e. 1-2, 2-3 and 3-4. My matrix looks like this:
[,1] [,2] [,3]
[1,] 1 0 0
[2,] -1 1 0
[3,] 0 -1 1
[4,] 0 0 -1
But then I get the same result as with contr.helmert(4). What is wrong I
really don't get it!
Thank you,
Wolfgang Pauli
On Sunday January 11 2004 18:07, you wrote:
> Notice `SPKType III Sum of Squares'. I don't believe your contrasts are
> orthogonal, and R's are sequential sum of squares.
>
> Also, are you sure these are the same contrasts? I presume this is
> contr.sdif from MASS (in which case it is churlish not to credit it), and
> SPSS's contrasts look more like Helmert contrasts from their labelling.
>
> Since it appears all your treatments are within subjects you do seem to be
> making life difficult for yourself. Although I would have done a simple
> fixed-effects analysis, applying summary.lm to the bottom stratum would
> give you simple t-tests for each contrast, including actual estimates of
> the magnitudes.
>
> On Sun, 11 Jan 2004, Wolfgang Pauli wrote:
> > I try to move from SPSS to R/S and am trying to reproduce the results of
> > SPSS in R. I calculated a one-way anova with "spk" as experimental factor
> > and erp as depended variable.
> > The result of the Anova are the same concearning the mean square, F and p
> > values. But I also wanted to caculate the contr.sdif(4) contrast on spk.
> > The results are completely different now. I hope anybody can help me.
> >
> > Thanks, Wolfgang
> >
> > This is what I get in SPSS:
> > Tests of Within-Subjects Contrasts
> > Measure: MEASURE_1
> > Source SPKType III Sum of Squares df Mean Square F Sig.
> > SPK Level 2 vs. Level 1 3,493 1 3,493 2,026 ,178
> > Level 3 vs. Previous 20,358 1 20,358 10,168 ,007
> > Level 4 vs. Previous 18,808 1 18,808 15,368 ,002
> > Error(SPK) Level 2 vs. Level 1 22,414 13 1,724
> > Level 3 vs. Previous 26,030 13 2,002
> > Level 4 vs. Previous 15,911 13 1,224
> >
> > This is the result in R:
> > Error: sub
> > Df Sum Sq Mean Sq F value Pr(>F)
> > Residuals 13 205.79 15.83
> >
> > Error: Within
> > Df Sum Sq Mean Sq F value Pr(>F)
> > spk 3 29.425 9.808 9.4467 8.055e-05 ***
> > spk: p 1 1.747 1.747 1.6821 0.2022649
> > spk: q 1 13.572 13.572 13.0719 0.0008479 ***
> > spk: r 1 14.106 14.106 13.5861 0.0006915 ***
> > Residuals 39 40.493 1.038
> > ---
> > Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
> >
> >
> >
> > Spk.df <- data.frame(sub,spk,erp)
> > subset(Spk.df, subset=(sub!="14oddball" & sub!="18odd" & sub!="19odd" &
> > sub!="20oddball")) -> Spk.selected.df
> > contrasts(Spk.selected.df$spk) <- contr.sdif(4)
> > aov(erp ~ spk + Error(sub), data=Spk.selected.df) -> Spk.aov
> > summary(Spk.aov,data=Spk.selected.df,split=list(spk=list(p=1,q=2,r=3)))
> >
> > this is the the beginning of the dataframe, which I use:
> > sub spk erp
> > 1 10oddball spk1 2.587
> > 2 11oddball spk1 -0.335
> > 3 12oddball spk1 5.564
> > 5 15oddball spk1 0.691
> > 6 17oddball spk1 -1.846
> > 10 21oddball spk1 1.825
> > 11 22oddball spk1 0.370
> > 12 2oddball spk1 3.234
> > 13 3oddball spk1 1.462
> > 14 5oddball spk1 2.535
> > 15 6oddball spk1 9.373
> > 16 7oddball spk1 2.132
> > 17 8oddball spk1 -0.518
> > 18 9oddball spk1 2.450
> > 19 10oddball spk2 2.909
> > 20 11oddball spk2 0.708
> > 21 12oddball spk2 4.684
> > 23 15oddball spk2 3.599
> > ...
> >
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