Dear John,

thank you for your answer. You are right, I also would not have expected a divergent result.
I have double-checked it again. No, I got type-III tests.
When I use type II, I get the same results in SPSS as in 'Anova' (using also type-II tests). My guess was that the somehow weighted means SPSS shows could be responsible for this difference. Or that using 'Anova' would not be correct for unequal group n's, which was not the case I think.
Do you have any further ideas?

Thank you!
Nils

John Fox schrieb:
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!

______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Reply via email to