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("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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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("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: r.789695.n4.nabble.com/Analysis-of-Variance-tp4651352.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help at r-project.org mailing list stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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 <dwarnold45@suddenlink.net> 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("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: r.789695.n4.nabble.com/Analysis-of-Variance-tp4651352.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help@r-project.org mailing list > stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.[[alternative HTML version deleted]]