Hi Joris,
That seems to have worked and the contrasts look correct.
I have tried comparing the results to what SPSS produces for the same model.
The two programs produce very different results, although the model F
statistics, R squared and adjusted R squared values are identical. The
results are so different that I don't know what to trust.

For the same model you coded I got:
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)          28223311  1 11.8056  0.001701 **
Mean.richness        49790403  1 20.8269 7.471e-05 ***
Diversity            31055477  1 12.9903  0.001082 **
Zoop                  2736238  1  1.1445  0.292953
Diversity:Phyto      27943313  6  1.9481  0.104103
Diversity:Zoop         168184  1  0.0703  0.792584
Diversity:Zoop:Phyto 61710145  6  4.3021  0.002879 **
Residuals            74110911 31
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Also sightly different from your result)

and

> summary(test)

Call:
lm(formula = C.Mean ~ Mean.richness + Diversity + Zoop + Diversity/Phyto +
    +Zoop * Diversity/Phyto)

Residuals:
     Min       1Q   Median       3Q      Max
-3555.26  -479.53    49.94   423.49  4073.20

Coefficients:
                         Estimate Std. Error t value Pr(>|t|)
(Intercept)               -8562.9     2492.2  -3.436  0.00170 **
Mean.richness              4605.7     1009.2   4.564 7.47e-05 ***
DiversityL                 6576.9     1824.8   3.604  0.00108 **
ZoopD                     -1414.4     1322.1  -1.070  0.29295
DiversityH:PhytoP2        -4307.5     1824.8  -2.361  0.02472 *
DiversityL:PhytoP2         -268.4     1262.5  -0.213  0.83300
DiversityH:PhytoP3        -2233.4     1393.0  -1.603  0.11900
DiversityL:PhytoP3        -1571.4     1262.5  -1.245  0.22257
DiversityH:PhytoP4        -7914.8     2647.2  -2.990  0.00543 **
DiversityL:PhytoP4        -1612.8     1262.5  -1.277  0.21092
DiversityL:ZoopD            484.9     1828.0   0.265  0.79258
DiversityH:ZoopD:PhytoP2    683.9     1855.3   0.369  0.71493
DiversityL:ZoopD:PhytoP2   6346.4     1785.4   3.555  0.00124 **
DiversityH:ZoopD:PhytoP3   4922.8     1786.3   2.756  0.00971 **
DiversityL:ZoopD:PhytoP3   1085.4     1785.4   0.608  0.54766
DiversityH:ZoopD:PhytoP4   3261.8     1985.6   1.643  0.11055
DiversityL:ZoopD:PhytoP4    681.9     1785.4   0.382  0.70513
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1546 on 31 degrees of freedom
Multiple R-squared: 0.7858,     Adjusted R-squared: 0.6753
F-statistic: 7.109 on 16 and 31 DF,  p-value: 1.810e-06

>From SPSS I got
    Tests of Between-Subjects Effects




 Dependent Variable:C Mean




 Source Type III Sum of Squares df Mean Square F Sig.  Corrected Model
2.719E+08 16 1.700E+07 7.109 .000  Intercept 2.394E+07 1 2.394E+07 10.012
.003  Meanrichness 4.979E+07 1 4.979E+07 20.827 .000  Diversity 3.581E+07 1
3.581E+07 14.978 .001  Zoop 1.079E+07 1 1.079E+07 4.515 .042  Diversity *
Zoop 261789.172 1 261789.172 .110 .743  Phyto(Diversity) 1.186E+08 6
1.976E+07 8.265 .000  Phyto * Zoop(Diversity) 6.171E+07 6 1.029E+07 4.302
.003  Error 7.411E+07 31 2.391E+06

 Total 7.959E+08 48


 Corrected Total 3.460E+08 47




Which, gives some similar results, but a completely different F statistic
and P-value for the main effect of Zoop and the nested effect of Phyto.
Obviously SPSS is not necessarily the perfect reference, but when using the
Type I SS, the results did agree. Any thoughts on why this might be? Could
the two programs be calculating the Type III SS differently? Might it be
wise to stick to Type I SS?

Thanks very much for your time and effort. It has been very helpful.
Anita.

On Thu, Jun 3, 2010 at 4:25 PM, Joris Meys <jorism...@gmail.com> wrote:

> 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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