From: David Mackovjak <bejitto101_at_yahoo.com>

Date: Wed, 19 Mar 2008 16:54:47 -0700 (PDT)

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 19 Mar 2008 - 23:57:55 GMT

Date: Wed, 19 Mar 2008 16:54:47 -0700 (PDT)

I do have the values for each individual values for each cell. They are as follows:

N(0) N(20) 4.48 5.76 4.52 5.64 4.63 5.78 4.70 7.01 4.65 7.11 4.57 7.02 5.21 5.88 5.23 5.82 5.38 5.73 5.88 6.26 5.98 6.26 5.91 6.37

So how would I go about this then?

- Original Message ---- From: Rolf Turner <r.turner_at_auckland.ac.nz> To: David Mackovjak <bejitto101_at_yahoo.com> Sent: Wednesday, March 19, 2008 4:36:47 PM Subject: Re: [R] [PS] Two Way ANOVA

With the given structure of your data you CANNOT test for interaction
in the

general sense. There are no degrees of freedom left for error.

If you have access to the (three) individual values in each cell,
then you

can test for interaction.

If these individual values are lost to posterity [Expostulation: Why
the

<expletive deleted> do people ***do*** things like this? Use your
<expletive deleted>

data, not summary statistics!!!] then you can still test for a
***particular form***

of interaction using Tukey's ``1 degree of freedom for non-
additivity' test.

I don't know if it's implemented in R, but it wouldn't be hard to
roll your own.

See ``Analysis of Messy Data'' volume 2 by George A. Milliken and
Dallas E. Johnson,

van Nostrand Reinhold, 1989, page 7 ff.

On 20/03/2008, at 12:03 PM, David Mackovjak wrote:

*> Ben,
*

> I would like to test the sulfur on the clover field, nitrogen on

*> the clover field and then test for the presence of interaction.
**>
**> Sorry about the last email, seems it really screwed itself over,
**> here it is again, hopefully nicer:
**>
**> Nitrogen(0) Nitrogen(20)
**> Sulfur(0) 4.54 5.73
**> Sulfur(3) 4.64 7.05
**> Sulfur(6) 5.27 5.81
**> Sulfur(9) 5.81 6.30
**>
**> Each of those is a cell mean of 3 values.
**>
**> Would I simply do as follows?:
**>
**> yield<- c(4.54,4.64,5.27,5.81,5.73,7.05,5.81,6.30)
**> sulfur <- c(1,2,3,4,1,2,3,4)
**> nitro <- c(1,1,1,1,2,2,2,2)
*

Not quite; you need to make sulfur and nitro into ***factors***.

> summary(aov(yield~sulfur*nitro))

Better, I think, to use lm() directly rather than the aov() wrapper.

fit <- lm(yield ~ sulfur*nitro)

anova(fit)

You see that you get no F-tests. The model fits ``perfectly'' so there is no residual sum of squares (and no degrees of freedom for error).

cheers,

Rolf Turner

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