Hi D,
R is taking drug as numeric, you ned indicate to R that drug is a factor:
> example12_7$drug <-factor(example12_7$drug)
> ej2<-aov(time~drug,data=example12_7)
> summary(ej2)
Df Sum Sq Mean Sq F value Pr(>F)
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
Regards,
Andrés
El 29/11/2012, a las 14:39, David Arnold <[email protected]> escribió:
> 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.
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Analysis-of-Variance-tp4651352.html
> Sent from the R help mailing list archive at Nabble.com.
>
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