Justin Montemarano
2012-Jun-15 19:01 UTC
[R] How do anova() and Anova(type="III") handle incomplete designs?
Hello all:
I am confused about the output from a lm() model with an incomplete
design/missing level.
I have two categorical predictors and a continuous covariate (day) that
I am using to model larval mass (l.mass):
leaf.species has three levels - map, syc, and oak
cond.time has two levels - 30 and 150.
There are no response values for Map-150, so that entire, two-way, level
is missing.
When running anova() on the model with Type I SS, the full factorial
design does not return errors; however, using package:car Anova() and
Type III SS, I receive an singularity error unless I used the argument
'singular.ok = T' (it is defaulted to F).
So, why don't I receive an error with anova() when I do with Anova(type
= "III")? How do anova() and Anova() handle incomplete designs, and
how
can interactions of variables with missing levels be interpreted?
I realize these are fairly broad questions, but any insight would be
helpful. Thanks, all.
Below is code to illustrate my question(s):
> lmMass <- lm(log(l.mass) ~ day*leaf.species + cond.time, data
growth.data) #lm() without cond.time interactions
> lmMassInt <- lm(log(l.mass) ~ day*leaf.species*cond.time, data
growth.data) #lm() with cond.time interactions
> anova(lmMass); anova(lmMassInt) #ANOVA summary of both models
with Type I SS
Analysis of Variance Table
Response: log(l.mass)
Df Sum Sq Mean Sq F value Pr(>F)
day 1 51.373 51.373 75.7451 2.073e-15
leaf.species 2 0.340 0.170 0.2506 0.7786
cond.time 1 0.161 0.161 0.2369 0.6271
day:leaf.species 2 1.296 0.648 0.9551 0.3867
Residuals 179 121.404 0.678
Analysis of Variance Table
Response: log(l.mass)
Df Sum Sq Mean Sq F value Pr(>F)
day 1 51.373 51.373 76.5651 1.693e-15
leaf.species 2 0.340 0.170 0.2533 0.77654
cond.time 1 0.161 0.161 0.2394 0.62523
day:leaf.species 2 1.296 0.648 0.9655 0.38281
day:cond.time 1 0.080 0.080 0.1198 0.72965
leaf.species:cond.time 1 1.318 1.318 1.9642 0.16282
day:leaf.species:cond.time 1 1.915 1.915 2.8539 0.09293
Residuals 176 118.091 0.671
> Anova(lmMass, type = 'III'); Anova(lmMassInt, type =
'III')
#ANOVA summary of both models with Type III SS
Anova Table (Type III tests)
Response: log(l.mass)
Sum Sq Df F value Pr(>F)
(Intercept) 39.789 1 58.6653 1.13e-12
day 3.278 1 4.8336 0.02919
leaf.species 0.934 2 0.6888 0.50352
cond.time 0.168 1 0.2472 0.61968
day:leaf.species 1.296 2 0.9551 0.38672
Residuals 121.404 179
Error in Anova.III.lm(mod, error, singular.ok = singular.ok, ...) :
there are aliased coefficients in the model
> Anova(lmMassInt, type = 'III', singular.ok = T) #Given the
error
in Anova() above, set singular.ok = T
Anova Table (Type III tests)
Response: log(l.mass)
Sum Sq Df F value Pr(>F)
(Intercept) 39.789 1 59.3004 9.402e-13
day 3.278 1 4.8860 0.02837
leaf.species 1.356 2 1.0103 0.36623
cond.time 0.124 1 0.1843 0.66822
day:leaf.species 2.783 2 2.0738 0.12877
day:cond.time 0.805 1 1.1994 0.27493
leaf.species:cond.time 0.568 1 0.8462 0.35888
day:leaf.species:cond.time 1.915 1 2.8539 0.09293
Residuals 118.091 176
>
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
<http://www.montegraphia.com/>
--
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
[[alternative HTML version deleted]]
John Fox
2012-Jun-17 01:20 UTC
[R] How do anova() and Anova(type="III") handle incomplete designs?
Dear Justin, anova() and Anova() are entirely different functions; the former is part of the standard R distribution and the second part of the car package. By default, Anova() produces an error for type-III tests conducted on rank-deficient models because the hypotheses tested aren't generally sensible.>From ?Anova:"singular.ok defaults to TRUE for type-II tests, and FALSE for type-III tests (where the tests for models with aliased coefficients will not be straightforwardly interpretable); if FALSE, a model with aliased coefficients produces an error." and "The designations "type-II" and "type-III" are borrowed from SAS, but the definitions used here do not correspond precisely to those employed by SAS. Type-II tests are calculated according to the principle of marginality, testing each term after all others, except ignoring the term's higher-order relatives; so-called type-III tests violate marginality, testing each term in the model after all of the others. This definition of Type-II tests corresponds to the tests produced by SAS for analysis-of-variance models, where all of the predictors are factors, but not more generally (i.e., when there are quantitative predictors). Be very careful in formulating the model for type-III tests, or the hypotheses tested will not make sense." I hope this helps, John ------------------------------------------------ John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Fri, 15 Jun 2012 15:01:27 -0400 Justin Montemarano <jmontema at kent.edu> wrote:> Hello all: > > I am confused about the output from a lm() model with an incomplete > design/missing level. > > I have two categorical predictors and a continuous covariate (day) that > I am using to model larval mass (l.mass): > > leaf.species has three levels - map, syc, and oak > > cond.time has two levels - 30 and 150. > > There are no response values for Map-150, so that entire, two-way, level > is missing. > > When running anova() on the model with Type I SS, the full factorial > design does not return errors; however, using package:car Anova() and > Type III SS, I receive an singularity error unless I used the argument > 'singular.ok = T' (it is defaulted to F). > > So, why don't I receive an error with anova() when I do with Anova(type > = "III")? How do anova() and Anova() handle incomplete designs, and how > can interactions of variables with missing levels be interpreted? > > I realize these are fairly broad questions, but any insight would be > helpful. Thanks, all. > > Below is code to illustrate my question(s): > > > lmMass <- lm(log(l.mass) ~ day*leaf.species + cond.time, data > growth.data) #lm() without cond.time interactions > > lmMassInt <- lm(log(l.mass) ~ day*leaf.species*cond.time, data > growth.data) #lm() with cond.time interactions > > anova(lmMass); anova(lmMassInt) #ANOVA summary of both models > with Type I SS > Analysis of Variance Table > > Response: log(l.mass) > Df Sum Sq Mean Sq F value Pr(>F) > day 1 51.373 51.373 75.7451 2.073e-15 > leaf.species 2 0.340 0.170 0.2506 0.7786 > cond.time 1 0.161 0.161 0.2369 0.6271 > day:leaf.species 2 1.296 0.648 0.9551 0.3867 > Residuals 179 121.404 0.678 > Analysis of Variance Table > > Response: log(l.mass) > Df Sum Sq Mean Sq F value Pr(>F) > day 1 51.373 51.373 76.5651 1.693e-15 > leaf.species 2 0.340 0.170 0.2533 0.77654 > cond.time 1 0.161 0.161 0.2394 0.62523 > day:leaf.species 2 1.296 0.648 0.9655 0.38281 > day:cond.time 1 0.080 0.080 0.1198 0.72965 > leaf.species:cond.time 1 1.318 1.318 1.9642 0.16282 > day:leaf.species:cond.time 1 1.915 1.915 2.8539 0.09293 > Residuals 176 118.091 0.671 > > Anova(lmMass, type = 'III'); Anova(lmMassInt, type = 'III') > #ANOVA summary of both models with Type III SS > Anova Table (Type III tests) > > Response: log(l.mass) > Sum Sq Df F value Pr(>F) > (Intercept) 39.789 1 58.6653 1.13e-12 > day 3.278 1 4.8336 0.02919 > leaf.species 0.934 2 0.6888 0.50352 > cond.time 0.168 1 0.2472 0.61968 > day:leaf.species 1.296 2 0.9551 0.38672 > Residuals 121.404 179 > Error in Anova.III.lm(mod, error, singular.ok = singular.ok, ...) : > there are aliased coefficients in the model > > Anova(lmMassInt, type = 'III', singular.ok = T) #Given the error > in Anova() above, set singular.ok = T > Anova Table (Type III tests) > > Response: log(l.mass) > Sum Sq Df F value Pr(>F) > (Intercept) 39.789 1 59.3004 9.402e-13 > day 3.278 1 4.8860 0.02837 > leaf.species 1.356 2 1.0103 0.36623 > cond.time 0.124 1 0.1843 0.66822 > day:leaf.species 2.783 2 2.0738 0.12877 > day:cond.time 0.805 1 1.1994 0.27493 > leaf.species:cond.time 0.568 1 0.8462 0.35888 > day:leaf.species:cond.time 1.915 1 2.8539 0.09293 > Residuals 118.091 176 > > > > > > - > Justin Montemarano > Graduate Student > Kent State University - Biological Sciences > > http://www.montegraphia.com > <http://www.montegraphia.com/> > -- > Justin Montemarano > Graduate Student > Kent State University - Biological Sciences > > http://www.montegraphia.com > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.