Btw. you can plot the effect by looking at the following graph with using either Sulfur or Nitrogen as the trace or x-axis:
interaction.plot(Sulfur,Nitrogen,Dependent) interaction.plot(Nitrogen,Sulfur,Dependent) Cheers, Daniel ------------------------- cuncta stricte discussurus ------------------------- -----Ursprüngliche Nachricht----- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von David Mackovjak Gesendet: Wednesday, March 19, 2008 7:55 PM An: Rolf Turner; R-help@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 <[EMAIL PROTECTED]> To: David Mackovjak <[EMAIL PROTECTED]> 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 ###################################################################### Attention:\ This e-mail message is privileged and confid...{{dropped:17}} ______________________________________________ R-help@r-project.org mailing list 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. ______________________________________________ R-help@r-project.org mailing list 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.