Nested ANOVA yields surprising results
On 30 Oct 2015, at 18:46 , Daniel Wagenaar <wagenadl at uc.edu> wrote: Dear R users: All textbook references that I consult say that in a nested ANOVA (e.g., A/B), the F statistic for factor A should be calculated as F_A = MS_A / MS_(B within A).
That would depend on which hypothesis you test in which model. If a reference tells you that you "should" do something without specifying the model, then you "should" look at a different reference. In general, having anything other than the residual MS in the denominator indicates that you think it represents an additional source of random variation. I don't think that is invariably the case in nested designs (and, by the way, notice that "nested" is used differently by different books and software). If you don't say otherwise, R assumes that there is only one source of random variation the model - a single error term if you like - and that all other terms represent systematic variations. In this mode of thinking, an A:B term represents an effect of B within A (additive and interaction effects combined), and you can test for its presence by comparing MS_A:B to MS_res. In its absence, you might choose to reduce the model and next look for an effect of A; purists would do this by comparing MS_A to the new MS_res obtained by pooling MS_A:B and MS_res, but lazy statisticians/programmers have found it more convenient to stick with the original MS_res denominator throughout (to get the pooling done, just fit the reduced model). If you want A:B to be a random term, then you need to say so, e.g. using
summary(aov(Y~A+Error(A:B-1)))
Error: A:B
Df Sum Sq Mean Sq F value Pr(>F)
A 2 0.4735 0.2367 0.403 0.7
Residuals 3 1.7635 0.5878
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
Residuals 6 4.993 0.8322
(the -1 in the Error() term prevents an error message, which as far as I can tell is spurious).
Notice that you need aov() for this; lm() doesn't do Error() terms. This _only_ works in balanced designs.
-pd
But when I run this simple example:
set.seed(1)
A <- factor(rep(1:3, each=4))
B <- factor(rep(1:2, 3, each=2))
Y <- rnorm(12)
anova(lm(Y ~ A/B))
I get this result:
Analysis of Variance Table
Response: Y
Df Sum Sq Mean Sq F value Pr(>F)
A 2 0.4735 0.23675 0.2845 0.7620
A:B 3 1.7635 0.58783 0.7064 0.5823
Residuals 6 4.9931 0.83218
Evidently, R calculates the F value for A as MS_A / MS_Residuals.
While it is straightforward enough to calculate what I think is the correct result from the table, I am surprised that R doesn't give me that answer directly. Does anybody know if R's behavior is intentional, and if so, why? Equally importantly, is there a straightforward way to make R give the answer I expect, that is:
Df Sum Sq Mean Sq F value Pr(>F)
A 2 0.4735 0.23675 0.4028 0.6999
The students in my statistics class would be much happier if they didn't have to type things like
a <- anova(...)
F <- a$`Sum Sq`[1] / a$`Sum Sq`[2]
P <- 1 - pf(F, a$Df[1], a$Df[2])
(They are not R programmers (yet).) And to be honest, I would find it easier to read those results directly from the table as well.
Thanks,
Daniel Wagenaar
--
Daniel A. Wagenaar, PhD
Assistant Professor
Department of Biological Sciences
McMicken College of Arts and Sciences
University of Cincinnati
Cincinnati, OH 45221
Phone: +1 (513) 556-9757
Email: daniel.wagenaar at uc.edu
Web: http://www.danielwagenaar.net
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