[R] Need help understanding output from aov and from anova

[Steven McKinney]

Hi Suman,

What version of R are you running?

In R 2.9.0 running your first example yields a warning

 Warning message:
 In anova.lm(lm(vtot ~ fac)) :
   ANOVA F-tests on an essentially perfect fit are unreliable

so some adept R developer has taken the time to figure
out how to warn you about such a problem.

Perhaps someone will add this to aov() at some point as well.

The only variability in this problem is that introduced
by machine precision rounding errors.

The exercise of submitting data with no variability to 
a program designed to assess variability cannot be expected 
to produce meaningful output, so there's nothing to
understand except the issue of machine precision.
Machine roundoff error is an important topic, so I'd
recommend learning about that issue, which will do most
to help understand these examples.

Best

SteveM


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> vtot=c(7.29917, 7.29917, 7.29917)  #identical values
> fac=as.factor(c(1,1,2))   #group 1 has first two elements, group 2 has
> anova(lm(vtot~fac))
Analysis of Variance Table

Response: vtot
          Df     Sum Sq    Mean Sq F value Pr(>F)
fac        1 1.6818e-30 1.6818e-30  0.3333 0.6667
Residuals  1 5.0455e-30 5.0455e-30               
Warning message:
In anova.lm(lm(vtot ~ fac)) :
  ANOVA F-tests on an essentially perfect fit are unreliable
> 
> summary(aov(vtot~fac))
            Df     Sum Sq    Mean Sq F value Pr(>F)
fac          1 1.6818e-30 1.6818e-30  0.3333 0.6667
Residuals    1 5.0455e-30 5.0455e-30               
> 
> fac=as.factor(c(1,2,2))
> anova(lm(vtot~fac))
Analysis of Variance Table

Response: vtot
          Df     Sum Sq    Mean Sq    F value    Pr(>F)    
fac        1 6.7274e-30 6.7274e-30 1.3340e+32 < 2.2e-16 ***
Residuals  1  5.043e-62  5.043e-62                         
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
Warning message:
In anova.lm(lm(vtot ~ fac)) :
  ANOVA F-tests on an essentially perfect fit are unreliable
>

 

 

Steven McKinney, Ph.D.

Statistician
Molecular Oncology and Breast Cancer Program
British Columbia Cancer Research Centre

email: smckinney at bccrc.ca
tel: 604-675-8000 x7561

BCCRC
Molecular Oncology
675 West 10th Ave, Floor 4
Vancouver B.C. 
V5Z 1L3

Canada


 

 


> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Suman Sundaresh
> Sent: Wednesday, June 03, 2009 3:55 PM
> To: r-help at r-project.org
> Subject: [R] Need help understanding output from aov and from anova
> 
> Hi all,
> 
> I noticed something strange when I ran aov and anova.
> 
> vtot=c(7.29917, 7.29917, 7.29917)  #identical values
> fac=as.factor(c(1,1,2))   #group 1 has first two elements, group 2 has
> the 3rd element
> 
> When I run:
> > anova(lm(vtot~fac))
> Analysis of Variance Table
> 
> Response: vtot
>           Df     Sum Sq    Mean Sq F value Pr(>F)
> fac        1 1.6818e-30 1.6818e-30  0.3333 0.6667
> Residuals  1 5.0455e-30 5.0455e-30
> 
> 
> I get a p-value of 0.667. This seems strange to me. I would have
> expected the p-value to be NaN.
> 
> Again, when I run:
> > summary(aov(vtot~fac))
>             Df     Sum Sq    Mean Sq F value Pr(>F)
> fac          1 1.6818e-30 1.6818e-30  0.3333 0.6667
> Residuals    1 5.0455e-30 5.0455e-30
> 
> Again same p-value.
> 
> 
> Now, if I set fac to c(1,2,2) which is essentially just switching the
> groups.
> fac=as.factor(c(1,2,2))
> 
> And run,
> > anova(lm(vtot~fac))
> Analysis of Variance Table
> 
> Response: vtot
>           Df     Sum Sq    Mean Sq    F value    Pr(>F)
> fac        1 6.7274e-30 6.7274e-30 1.3340e+32 < 2.2e-16 ***
> Residuals  1  5.043e-62  5.043e-62
> ---
> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> 
> 
> The p-value is really significant which again looks very strange.
> 
> Please could someone shed some light on what I may be missing here?
> 
> Thanks very much.
> Suman.
> 
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