I seem to be running into a machine precision issue below.  
Should I be using a different method to calculate this or should I do
some kind of validity check 
	(i.e. max(tapply(resp,data$Split,var,na.rm=TRUE),na.rm=TRUE)>0) 
in an 'if' statement before the call to anova? 
Or is there some other recommended way to get around this that I am not
thinking about?

Thanks,
Landon

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> resp
 [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
> data$Split
 [1] 03E 02E 01C 02E 01C 01C 02E 01C 02E 01C 01C 01C 02E 03E 01C 01C 01C
02E 01C
[20] 02E 01C 03E 02E 01C
Levels: 01C 02E 03E
> anova(lm(resp ~ data$Split))
Analysis of Variance Table

Response: resp
           Df     Sum Sq    Mean Sq F value  Pr(>F)  
data$Split  2 2.0907e-31 1.0454e-31  4.5937 0.02214 *
Residuals  21 4.7788e-31 2.2760e-32                  
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 
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