[R] Fwd: Re: aov y lme

[Tomas Goicoa]

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,                  ripley at stats.ox.ac.uk
>  > 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