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$Split,var,na.rm=TRUE),na.rm=TRUE)>0)
in an 'if' statement before the call to anova?
Or is there some other recommended way to get around this that I am not
thinking about?
Thanks,
Landon
########################################################################
####> resp
[1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2> data$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
########################################################################
####
[[alternative HTML version deleted]]