Re: [R] Nested ANOVA with covariate using Type III sums of squares

From: Anita Narwani <anitanarwani_at_gmail.com>
Date: Thu, 03 Jun 2010 17:19:09 -0700

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 <jorismeys_at_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 <anitanarwani_at_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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