Dear Nils,
This is a pretty simple design, and I wouldn't have thought that there was
much room for getting different results. More generally, but not here (since
there's only one between-subject factor), one shouldn't use
contr.treatment() with "type-III" tests, as you did. Is it possible that you
got "type-II" tests from SPSS:
------ snip ----------
> summary(Anova(betweenanova, idata=with, idesign= ~within, type = "II" ))
Type II Repeated Measures MANOVA Tests:
------------------------------------------
Term: between
Response transformation matrix:
(Intercept)
w1 1
w2 1
Sum of squares and products for the hypothesis:
(Intercept)
(Intercept) 9.6
Sum of squares and products for error:
(Intercept)
(Intercept) 18
Multivariate Tests: between
Df test stat approx F num Df den Df Pr(>F)
Pillai 1 0.347826 4.266667 1 8 0.072726 .
Wilks 1 0.652174 4.266667 1 8 0.072726 .
Hotelling-Lawley 1 0.533333 4.266667 1 8 0.072726 .
Roy 1 0.533333 4.266667 1 8 0.072726 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
------------------------------------------
Term: within
Response transformation matrix:
within1
w1 1
w2 -1
Sum of squares and products for the hypothesis:
within1
within1 0.4
Sum of squares and products for error:
within1
within1 21.33333
Multivariate Tests: within
Df test stat approx F num Df den Df Pr(>F)
Pillai 1 0.0184049 0.1500000 1 8 0.70864
Wilks 1 0.9815951 0.1500000 1 8 0.70864
Hotelling-Lawley 1 0.0187500 0.1500000 1 8 0.70864
Roy 1 0.0187500 0.1500000 1 8 0.70864
------------------------------------------
Term: between:within
Response transformation matrix:
within1
w1 1
w2 -1
Sum of squares and products for the hypothesis:
within1
within1 4.266667
Sum of squares and products for error:
within1
within1 21.33333
Multivariate Tests: between:within
Df test stat approx F num Df den Df Pr(>F)
Pillai 1 0.1666667 1.6000000 1 8 0.24150
Wilks 1 0.8333333 1.6000000 1 8 0.24150
Hotelling-Lawley 1 0.2000000 1.6000000 1 8 0.24150
Roy 1 0.2000000 1.6000000 1 8 0.24150
Univariate Type II Repeated-Measures ANOVA Assuming Sphericity
SS num Df Error SS den Df F Pr(>F)
between 4.8000 1 9.0000 8 4.2667 0.07273 .
within 0.2000 1 10.6667 8 0.1500 0.70864
between:within 2.1333 1 10.6667 8 1.6000 0.24150
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
------ snip ----------
I hope this helps,
John
------------------------------
John Fox, Professor
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web: socserv.mcmaster.ca/jfox
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Skotara
> Sent: January-23-09 12:16 PM
> To: r-help@r-project.org
> Subject: [R] Anova and unbalanced designs
>
> Dear R-list!
>
> My question is related to an Anova including within and between subject
> factors and unequal group sizes.
> Here is a minimal example of what I did:
>
> library(car)
> within1 <- c(1,2,3,4,5,6,4,5,3,2); within2 <- c(3,4,3,4,3,4,3,4,5,4)
> values <- data.frame(w1 = within1, w2 = within2)
> values <- as.matrix(values)
> between <- factor(c(rep(1,4), rep(2,6)))
> betweenanova <- lm(values ~ between)
> with <- expand.grid(within = factor(1:2))
> withinanova <- Anova(betweenanova, idata=with, idesign=
> ~as.factor(within), type = "III" )
>
> I do not know if this is the appropriate method to deal with unbalanced
> designs.
>
> I observed, that SPSS calculates everything identically except the main
> effect of the within factor, here, the SSQ and F-value are very different
> If selecting the option "show means", the means for the levels of the
> within factor in SPSS are the same as:
> mean(c(mean(values$w1[1:4]),mean(values$w1[5:10]))) and
> mean(c(mean(values$w2[1:4]),mean(values$w2[5:10]))).
> In other words, they are calculated as if both groups would have the
> same size.
>
> I wonder if this is a good solution and if so, how could I do the same
> thing in R?
> However, I think if this is treated in SPSS as if the group sizes are
> identical,
> then why not the interaction, which yields to the same result as using
> Anova()?
>
> Many thanks in advance for your time and help!
>
> ______________________________________________
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