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

[Prof Brian Ripley]

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 at 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 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