[R] Analysis of Variance

[arun]

HI,
Try this:
 res2<-aov(time~factor(drug),data=example12_7)
 summary(res2)
#             Df Sum Sq Mean Sq F value Pr(>F)  
#factor(drug)  2  21.98  10.991   4.188 0.0345 *
#Residuals    16  41.99   2.624                 
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
A.K.





----- Original Message -----
From: David Arnold <dwarnold45 at suddenlink.net>
To: r-help at r-project.org
Cc: 
Sent: Thursday, November 29, 2012 3:39 PM
Subject: [R] Analysis of Variance

Hi, I am encountering a difficulty I don't understand. Be patient, I'm very
new to analysis of variance.

If I load this data:

example12_7=read.table("http://msemac.redwoods.edu/~darnold/math15/data/chapter12/example12_7.dat",header=TRUE)

The run the oneway.test:

oneway.test(time~drug,data=example12_7,var.equal=TRUE)

I get these results:

data:  time and drug 
F = 4.1881, num df = 2, denom df = 16, p-value = 0.03445

Now, I've done the problem by hand and this result agrees with my
calculations. Now I try aov and get these results:

res1 <- aov(time~drug,data=example12_7)
summary(res1)

            Df Sum Sq Mean Sq F value Pr(>F)
drug         1   7.96   7.964   2.417  0.138
Residuals   17  56.01   3.294

Note these do not agree with above. However, if I enter the data by hand:

Drug1=c(7.3,8.2,10.1,6.0,9.5)
Drug2=c(7.1,10.6,11.2,9.0,8.5,10.9,7.8)
Drug3=c(5.8,6.5,8.8,4.9,7.9,8.5,5.2)
boxplot(Drug1,Drug2,Drug3)

Then create a dataframe:

d=stack(list(Drug1=Drug1,Drug2=Drug2,Drug3=Drug3))

And run aov again:

res=aov(values~ind,data=d)
summary(res)

I get these results:

            Df Sum Sq Mean Sq F value Pr(>F)  
ind          2  21.98  10.991   4.188 0.0345 *
Residuals   16  41.99   2.624  

Which completely agree with my calculations. What's going on?

Thanks.

D.



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