[R] anova table

[George A. Dowding]

First, thanks, that worked perfectly.  I entered the data and read it
from a file using read.table.  You read my mind with the suggestion of
using Tukey's procedure. We are actually using the  book by Devore.
I am also supposed to do a contrast using a "method attributed to
Scheff\'e." to quote Devore.

Douglas Bates writes:
 > Kaspar Pflugshaupt <pflugshaupt at geobot.umnw.ethz.ch> writes:
 > 
 > > On Wednesday 04 April 2001 07:20, George A. Dowding wrote:
 > > > I am taking a stats class and need to do a single factor anova table
 > > > with four treatments for an assignment.  The instructor showed the
 > > > class how to do it with minitab, and I am unsure how to proceede in
 > > > R.
 > > >
 > > > Here is the data:
 > > >
 > > > I.   .8 .2 1.4 .9 .9 1.0 1.0 1.1 1.4 1.2 1.0
 > > > II.  1.6 1.1 .2 1.3 1.5 .7 .5 1.3 1.3 .4
 > > > III  1.8 .3 1.1 2.0 2.8 1.5 1.3 2.9 1.7 .8
 > > > IV.  2.4 2.6 2.82.9 1.7 2.6 2.7 2.1 2.1 2.3
 > > >
 > > > I have been reading the various help files and can't quite finde the
 > > > right example to get started from.
 > > 
 > > Try help(aov). Basically, you will have to  bring your data into a form like
 > > 
 > > treatment	response
 > > 1		.8
 > > 1		.2
 > > 2		1.6
 > > 2		1.1
 > > ...		...
 > > 
 > > and then use something like
 > > 
 > > model <- aov(response~treatment)
 > > summary(model)
 > > 
 > > This will give you a standard ANOVA table.
 > 
 > If the data are in the form of separate rows for each group, you could
 > read the data in as one column and use the gl function to generate the
 > treatment column.  If the data are stored as four columns in a table
 > then you could use read.table to read the data into a data frame and
 > the stack function to create one long response vector and the
 > corresponding treatment column.  (The name of the stack function is
 > taken from the similarly named command in Minitab.)
 > 
 > The simplest way of doing this is to bring in each level's data as a
 > separate vector and use stack.  That would look like
 > 
 > > grp1 <- c(.8, .2, 1.4, .9, .9, 1.0, 1.0, 1.1, 1.4, 1.2, 1.0)
 > > grp2 <- c(1.6, 1.1, .2, 1.3, 1.5, .7, .5, 1.3, 1.3, .4)
 > > grp3 <- c(1.8, .3, 1.1, 2.0, 2.8, 1.5, 1.3, 2.9, 1.7, .8)
 > > grp4 <- c(2.4, 2.6, 2.8, 2.9, 1.7, 2.6, 2.7, 2.1, 2.1, 2.3)
 > > dat <- stack(list(grp1, grp2, grp3, grp4))
 > Error in data.frame(values = unlist(unname(x)), ind = factor(rep(names(x),  : 
 > 	arguments imply differing number of rows: 41, 0
 > > dat <- stack(list(I. = grp1, II. = grp2, III. = grp3, IV. = grp4))
 > > dat[1:10, ]
 >    values ind
 > 1     0.8  I.
 > 2     0.2  I.
 > 3     1.4  I.
 > 4     0.9  I.
 > 5     0.9  I.
 > 6     1.0  I.
 > 7     1.0  I.
 > 8     1.1  I.
 > 9     1.4  I.
 > 10    1.2  I.
 > > aov(values ~ ind, data = dat)
 > Call:
 >    aov(formula = values ~ ind, data = dat)
 > 
 > Terms:
 >                      ind Residuals
 > Sum of Squares  14.05771  10.57009
 > Deg. of Freedom        3        37
 > 
 > Residual standard error: 0.5344887 
 > Estimated effects may be unbalanced
 > 
 > Notice that you can save the result of the aov function 
 > > fm1 <- aov(values ~ ind, data = dat)
 > and do things like plot it to get diagnostic plots.
 > > plot(fm1)
 > 
 > If you need to do multiple comparisons of means, use the Tukey
 > function from the Devore5 package.  That package provides the data and
 > sample analyses from an introductory text for Engineering Statistics.
 > 
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-- 
Thanks,
George

dowdingg at math.uaa.alaska.edu 
http://www.math.uaa.alaska.edu/~dowdingg

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