Hi all,
Say I have the following data:
a<-data.frame(col1=c(rep("a",5),rep("b",7)),col2=runif(12))
a_aov<-aov(a$col2~a$col1)
summary(aov)
Note that there are 5 observations for a and 7 for b, thus is
unbalanced. What would be the correct way of doing anova for this set?
Thanks,
Sachin
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HI,
Check this link:
https://stat.ethz.ch/pipermail/r-help/2011-April/273858.html
A.K.
----- Original Message -----
From: Sachinthaka Abeywardana <sachin.abeywardana at gmail.com>
To: r-help at r-project.org
Cc:
Sent: Monday, August 13, 2012 10:09 PM
Subject: [R] anova in unbalanced data
Hi all,
Say I have the following data:
a<-data.frame(col1=c(rep("a",5),rep("b",7)),col2=runif(12))
a_aov<-aov(a$col2~a$col1)
summary(aov)
Note that there are 5 observations for a and 7 for b, thus is
unbalanced. What would be the correct way of doing anova for this set?
Thanks,
Sachin
??? [[alternative HTML version deleted]]
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> -----Original Message----- > Say I have the following data: > > a<-data.frame(col1=c(rep("a",5),rep("b",7)),col2=runif(12)) > > a_aov<-aov(a$col2~a$col1) > > summary(aov) > > > Note that there are 5 observations for a and 7 for b, thus is > unbalanced. What would be the correct way of doing anova for this set? >As this is a single factor case, the imbalance doesn't affect the interpretation. For two-way and higher models, it would affect the interpretation, and john fox's post (and a very large literature) then applies. But here, the usual variants and contrast choices will all return the same p-value, so aov works, as does anova(lm(col2~col1, data=a)) #note that the 'data' argument also works in aov S ******************************************************************* This email and any attachments are confidential. Any use...{{dropped:8}}