[R] Random effects ANOVA?

[Prof Brian Ripley]

I think you want print or summary rather than anova.  anova() is not very 
useful for aov() models even without error strata.  The point of the aov 
classes is that they present the results of lm() fits in ways which are 
more conventional for designed experiments, including giving conventional 
ANOVA tables without recourse to anova().

I didn't follow how the videos were chosen.  Random effects apply when the 
'treatments' were chosen from a large population (which might apply if 
each subject watched (on separate occasions) three videos chosen randomly 
from a larger pool), and if the interest is in the variability of the 
response over videos in the pool.  If subjects were observed more than 
once then I suspect you most likely want a random effect for subjects.

What I guess is more likely is that you have 60 measurements on 60 
subjects, each of whom was assigned one video and one drink.  Then there 
appear to be no random effects, and the problem caused by randomly 
selecting videos is that the experiment is likely to be unbalanced (since 
each video is not going to be selected 20 times by chance).  aov() is 
primarily intended for balanced designs, and interpretation is tricky 
without balance (see the warnings on its help page).

There are many loose ends hindering offering help by email, and the best 
advice is to consult a statistician face-to-face about an appropriate 
analysis.


On Wed, 29 Mar 2006, Chris Bergstresser wrote:

> Hi all --
>
>    So I have a very simple dataset, which consists of 60 subjects,
> who watched one of three videos, drank one of two drinks, and
> completed a task.  The response variable is the time to complete the
> task.  The ANOVA command is simple enough:  anova(aov(time ~ drink *
> video, data = df));
>
>   However, the videos were randomly selected; I need to use the
> random effects model for them.  So I tried anova(aov(time ~ drink +
> Error(video), data = df));  This gives me a "no applicable method for
> 'anova'" error.
>
>   The command aov works, but doesn't give me anything I can interpret
> effectively.  Is there a simpler command I should be using?  Am I
> doing something wrong?
>
> -- Chris

-- 
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