I see where my confusion comes from. I counted 4 levels of Phyto, but
you have 8, being 4 in every level of Diversity. There's your
aliasing.
> table(Diversity,Phyto)
Phyto
Diversity M1 M2 M3 M4 P1 P2 P3 P4
H 0 0 0 0 6 6 6 6
L 6 6 6 6 0 0 0 0
There's no need to code them differently for every level of Diversity.
If you don't, all is fine :
> Phyto <- gsub("M","P",as.character(Phyto))
> Phyto <- as.factor(Phyto)
>
> test <- lm(C.Mean~ Mean.richness + Diversity + Zoop + Diversity/Phyto +
+ Zoop*Diversity/Phyto)
>
> Anova(test,type="III")
Anova Table (Type III tests)
Response: C.Mean
Sum Sq Df F value Pr(>F)
(Intercept) 23935609 1 10.0121 0.0034729 **
Mean.richness 49790385 1 20.8269 7.471e-05 ***
Diversity 35807205 1 14.9779 0.0005234 ***
Zoop 10794614 1 4.5153 0.0416688 *
Diversity:Phyto 118553464 6 8.2650 2.184e-05 ***
Diversity:Zoop 261789 1 0.1095 0.7429356
Diversity:Zoop:Phyto 61710162 6 4.3021 0.0028790 **
Residuals 74110938 31
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
You can check with summary(test) that the model is fitted correctly.
On Fri, Jun 4, 2010 at 12:48 AM, Anita Narwani <anitanarw...@gmail.com> wrote:
>
> You have everything right except that there are only 2 zooplankton species (C
> & D, which stand for Ceriodaphnia and Daphnia).
>
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