[R] [PS] Two Way ANOVA

[Daniel Malter]

You put all the responses into one vector and the appropriate independents
Sulfur and nitrogen into the other. So your table would look like this (see
below). You would test.

summary(aov(Dependent~factor(Nitrogen)*factor(Sulfur))). And your output
would look likt this (see further below) from which you can start in-depth
analyses.

Dependent	Nitrogen	Sulfur
4.48	1	1
4.52	1	1
4.63	1	1
4.7	1	2
4.65	1	2
4.57	1	2
5.21	1	3
5.23	1	3
5.38	1	3
5.88	1	4
5.98	1	4
5.91	1	4
5.76	2	1
5.64	2	1
5.78	2	1
7.01	2	2
7.11	2	2
7.02	2	2
5.88	2	3
5.82	2	3
5.73	2	3
6.26	2	4
6.26	2	4
6.37	2	4


                                Df Sum Sq Mean Sq F value    Pr(>F)    
factor(Nitrogen)                 1 7.5937  7.5937 1514.96 < 2.2e-16 ***
factor(Sulfur)                   3 3.1543  1.0514  209.76 4.717e-13 ***
factor(Nitrogen):factor(Sulfur)  3 3.8358  1.2786  255.08 1.024e-13 ***
Residuals                       16 0.0802  0.0050                      
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

 


-------------------------
cuncta stricte discussurus
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-----Ursprüngliche Nachricht-----
Von: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Im
Auftrag von David Mackovjak
Gesendet: Wednesday, March 19, 2008 7:55 PM
An: Rolf Turner; R-help at r-project.org
Betreff: Re: [R] [PS] Two Way ANOVA

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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