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anova question

2 messages · Ulf Mehlig, Peter Dalgaard

Ulf Mehlig
# 
I have a probably very naive question about the R anova/aov functions:
I've found in several text books with descriptions of anova procedures
that, in nested anova (specifying the model "X~A/B", B being a factor
specifying treatments, and A groups of samples within each treatment),
the mean square for "among treatments" is divided by the mean square
"within treatment among sample groups"; R seems to divide by the
residual mean square. So, in the following (silly) example I would
have expected a F value of 1.496/0.785=1.905732:

  > test <- data.frame(x=rnorm(12*5), a=rep(1:12,rep(5,12)), b=rep(1:5,rep(12,5)))
  > summary(aov(x ~ factor(a)/factor(b), data=test))
                      Df Sum Sq Mean Sq F value  Pr(>F)  
  factor(a)           11 16.456   1.496  1.8357 0.07647 .
  factor(a):factor(b)  4  3.142   0.785  0.9638 0.43689  
  Residuals           44 35.859   0.815                  

As I am certainly misunderstanding something, I would appreciate a
hint, and maybe a literature recommendation!

Many thanks,
Ulf

P.S.: please CC: me, I'm not on the list at the moment! Thank you!
Peter Dalgaard
# 
Ulf Mehlig <umehlig at uni-bremen.de> writes:
Um, 

If B is the treatment you want B/A, I believe?

I don't think the data frame looks like I think you think it looks:

               x  a b
1   0.9421894166  1 1
2  -0.6615090815  1 1
3   0.6659020398  1 1
4   0.6384607529  1 1
5   0.2785534174  1 1
6  -0.5093066357  2 1
7  -0.1765458342  2 1
8  -0.0003846756  2 1
9  -0.5523574783  2 1
10  1.4351543109  2 1
11  0.2338159511  3 1
12  0.2701585951  3 1
13  0.3781168982  3 2

I think you expected this:
Df Sum Sq Mean Sq
factor(b)            4  5.367   1.342
factor(b):factor(a) 55 56.619   1.029

I.e the b:a effect leaves no DF for residuals. So you remove the term
and use that SS for residual.