Re: [Rd] aov for unbalanced design (PR#7144)

From: Prof Brian Ripley <ripley_at_stats.ox.ac.uk>
Date: Fri 30 Jul 2004 - 15:07:22 EST


What do you think is the correct answer and on what authority? (These are explicitly sequential aka Type 1 anova tables.)

That the SSqs depend on the order of fitting is a feature of an unbalanced design. I believe that R is correct and your understanding is not.

On Thu, 29 Jul 2004 tlogvinenko@partners.org wrote:

> Full_Name: Tanya Logvinenko
> Version: 1.7.0

Oh, please! Don't send in bug reports from very old versions -- there have been 5 releases since then.

> OS: Windows 2000
> Submission from: (NULL) (132.183.156.125)
>
>
> For unbalanced design, I ran into problem with ANOVA (aov function). The sum of
> squares for only for the second factor and total are computed correctly, but sum
> of squares for the first factor is computed incorreclty. Changing order of
> factors in the formula changes the ANOVA table. For the balanced design, there
> is no such problem.
>
> > summary(aov(data[1,]~factor1+factor2))
> Df Sum Sq Mean Sq F value Pr(>F)
> factor1 5 1524420 304884 6.4529 0.0003229 ***
> factor2 7 1447830 206833 4.3776 0.0017808 **
> Residuals 31 1464674 47248
> ---
> Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
> > summary(aov(data[1,]~factor2+factor1))
> Df Sum Sq Mean Sq F value Pr(>F)
> factor2 7 1648225 235461 4.9836 0.0007295 ***
> factor1 5 1324025 264805 5.6046 0.0008612 ***
> Residuals 31 1464674 47248
> ---
> Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

The FAQ has a section on BUGS asking for a *reproducible* example. This is not.

-- 
Brian D. Ripley,                  ripley@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

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Received on Fri Jul 30 15:10:49 2004

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