[R] newbie question on contrasts and aov

[Prof Brian Ripley]

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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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595