[R] newbie question on contrasts and aov

[Wolfgang Pauli]

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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