[R] Basic question about three factor Anova

[Bogdan Lataianu]

I understand now

On May 30, 4:04 pm, Bogdan Lataianu <bodins... at gmail.com> wrote:
>  Read the data using scan():
> #
> #          a1               a2               a3               a4
> #     -------------    -------------    -------------    -------------
> #     b1   b2   b3     b1   b2   b3     b1   b2   b3     b1   b2   b3
> #     ---  ---  ---    ---  ---  ---    ---  ---  ---    ---  ---  ---
> #
> # c1:
> #     4.1  4.6  3.7    4.9  5.2  4.7    5.0  6.1  5.5    3.9  4.4  3.7
> #     4.3  4.9  3.9    4.6  5.6  4.7    5.4  6.2  5.9    3.3  4.3  3.9
> #     4.5  4.2  4.1    5.3  5.8  5.0    5.7  6.5  5.6    3.4  4.7  4.0
> #     3.8  4.5  4.5    5.0  5.4  4.5    5.3  5.7  5.0    3.7  4.1  4.4
> #     4.3  4.8  3.9    4.6  5.5  4.7    5.4  6.1  5.9    3.3  4.2  3.9
> #
> # c2:
> #     4.8  5.6  5.0    4.9  5.9  5.0    6.0  6.0  6.1    4.1  4.9
> 4.3
> #     4.5  5.8  5.2    5.5  5.3  5.4    5.7  6.3  5.3    3.9  4.7  4.1
> #     5.0  5.4  4.6    5.5  5.5  4.7    5.5  5.7  5.5    4.3  4.9  3.8
> #     4.6  6.1  4.9    5.3  5.7  5.1    5.7  5.9  5.8    4.0  5.3  4.7
> #     5.0  5.4  4.7    5.5  5.5  4.9    5.5  5.7  5.6    4.3  4.3  3.8
> #
> # NOTE: Cut and paste the numbers without the leading # or labels
> #
>
> > Y <- scan()
> > A <- gl(4,3, 4*3*2*5, labels=c("a1","a2","a3","a4"));
> > B <- gl(3,1, 4*3*2*5, labels=c("b1","b2","b3"));
> > C <- gl(2,60, 4*3*2*5, labels=c("c1","c2"));
> > anova(lm(Y~A*B*C))   # all effects and interactions
>
> In the above example, why the number of replications for A is 3, for B
> is 1 and for C is 60?
> And why 4*3*2*5? Is the 5 because there are 5 lines in each 4*3*2
> group?
> What is the logic behind this?
>
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