a naive question about aov() in R
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tony.fox2016 ▴ 30
@tonyfox2016-9796
Last seen 8.1 years ago

Hi all,

I found that in R, when applying aov() to 3 or more groups of data, even there is only one observation in some group, aov() still works. I am not sure how it calculate the within-group variance for this group without any replicates, such as the group z in the following code:

 

x<-rnorm(mean=5,10)

y<-rnorm(mean=10,10)

z<-20

f<-factor(c(rep('x',10),rep('y',10),'z'))

summary(lm(aov(c(x,y,z)~f)))

 

The output is: 

Call:

lm(formula = aov(c(x, y, z) ~ f))

 

Residuals:

     Min       1Q   Median       3Q      Max 

-1.41996 -0.51930 -0.04419  0.41832  1.75994 

 

Coefficients:

            Estimate Std. Error t value Pr(>|t|)    

(Intercept)    4.710      0.304   15.49 7.52e-12 ***

fy             5.018      0.430   11.67 7.89e-10 ***

fz            15.290      1.008   15.16 1.07e-11 ***

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

 

Residual standard error: 0.9614 on 18 degrees of freedom

Multiple R-squared:  0.9442,

Adjusted R-squared:  0.938 

F-statistic: 152.3 on 2 and 18 DF,  p-value: 5.254e-12

 

 

 

 

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In the result, group z has a p-value (in the row fz). I am curious if this p-value makes any sense.

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@james-w-macdonald-5106
Last seen 10 hours ago
United States

This is neither a Bioconductor nor an R question. Instead it's a basic stats question, and you would be better served by asking somewhere like stackoverflow. The degrees of freedom for an ANOVA model are based on the entire model, not the individual groups, so each group 'borrows information' from all other groups in the model when you test for significance. If you care to know more, here is a link to a basic ANOVA explanation.
 

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