Dear Prof. Ripley and Christoph, thank you very much for your comments. You have helped me a lot.
Thanks, Tomas Goicoa >Dear Prof. Ripley > >Thank you for your email. Yes, this is of course the correct >syntax to save us the extra calculation. And I forgot the >"lower.tail = FALSE" for pf() in my example to obtain the >p-value. > >Thank you for the corrections and > >Best regards, > >Christoph Buser > > >Prof Brian Ripley writes: > > On Mon, 22 Jan 2007, Christoph Buser wrote: > > > > > Dear Tomas > > > > > > You can produce the results in Montgomery Montgomery with > > > lme. Please remark that you should indicate the nesting with the > > > levels in your nested factor. Therefore I recreated your data, > > > but used 1,...,12 for the levels of batch instead of 1,...,4. > > > > > > purity<-c(1,-2,-2,1,-1,-3,0,4, 0,-4, 1,0, 1,0,-1,0,-2,4,0,3, > > > -3,2,-2,2,2,-2,1,3,4,0,-1,2,0,2,2,1) > > > suppli<-factor(c(rep(1,12),rep(2,12),rep(3,12))) > > > batch<-factor(c(rep(1:4,3), rep(5:8,3), rep(9:12,3))) > > > material<-data.frame(purity,suppli,batch) > > > > > > As you remarked you can use aov > > > > > > summary(material.aov<-aov(purity~suppli+suppli:batch,data=material)) > > > Df Sum Sq Mean Sq F value Pr(>F) > > > suppli 2 15.056 7.528 2.8526 0.07736 . > > > suppli:batch 9 69.917 7.769 2.9439 0.01667 * > > > Residuals 24 63.333 2.639 > > > --- > > > Signif. codes: 0 $,1rx(B***$,1ry(B 0.001 $,1rx(B**$,1ry(B 0.01 > $,1rx(B*$,1ry(B 0.05 $,1rx(B.$,1ry(B 0.1 $,1rx(B $,1ry(B 1 > > > > > > Remark that the test of "suppli" is not the same as in > > > Montgomery. Here it is wrongly tested against the Residuals and > > > > I don't think so: aov() is doing the correct thing for the model > > specified. The aov() model you want is probably > > > > aov(purity ~ suppli + Error(suppli:batch), data=material) > > > > and this gives > > > > > summary(.Last.value) > > > > Error: suppli:batch > > Df Sum Sq Mean Sq F value Pr(>F) > > suppli 2 15.056 7.528 0.969 0.4158 > > Residuals 9 69.917 7.769 > > > > Error: Within > > Df Sum Sq Mean Sq F value Pr(>F) > > Residuals 24 63.333 2.639 > > > > > > > you should perform the calculate the test with: > > > > > (Fsuppi <- summary(material.aov)[[1]][1,"Mean Sq"]/ > > > summary(material.aov)[[1]][2,"Mean Sq"]) > > > pf(Fsuppi, df1 = 2, df2 = 9) > > > > You want the other tail. > > > > -- > > Brian D. Ripley, [EMAIL PROTECTED] > > Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ > > University of Oxford, Tel: +44 1865 272861 (self) > > 1 South Parks Road, +44 1865 272866 (PA) > > Oxford OX1 3TG, UK Fax: +44 1865 272595 ______________________________________________ R-help@stat.math.ethz.ch 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.