# Examples using R – Analysis of Variance

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Here we have a single factor with 5 levels and 5 replicates.The data is

> cotton

$p15

[1] 7 7 15 11 9

$p20

[1] 12 17 12 18 18

$p25

[1] 14 18 18 19 23

$p30

[1] 19 25 22 19 23

$p35

[1] 7 10 11 15 11

A box plot of the data is

A histogram of data looks like

A multiple scatter plot can sometimes be used if corresponding values of the observations need comparison. The scatter plot for this data is as shown.

Analysis of Variance:

Lets use analysis of variance in the above example to find out if all means are equal or if any mean is different.

The data needs to be transformed for aov

> c(cotton_matrix[1,],cotton_matrix[2,],cotton_matrix[3,],cotton_matrix[4,],cotton_matrix[5,])->cotton_data

> cotton_data

p15 p20 p25 p30 p35 p15 p20 p25 p30 p35 p15 p20 p25 p30 p35 p15 p20 p25 p30 p35 p15 p20 p25 p30 p35

7 12 14 19 7 7 17 18 25 10 15 12 18 22 11 11 18 19 19 15 9 18 23 23 11

Analysis of variance yields

> summary(aov(cotton_data~names(cotton_data)))

from the F value we reject the null hypothesis and conclude that the means differ.

Analysis of variance uses certain assumptions and it is important to check the validity of these assumptions. The first method is to analyse the residuals for each observations. There should be no pattern in the residuals. If residuals either spread out or narrow down as time progresses then this could be an experimental error.

Here’s a plot of residuals against time (observation)

Another validation is to check the nature of the residuals themselves. One way to do is to plot of curve of residuals versus the fitted values.Here again no pattern should be present

The variance for the five sets can be compared using the Bartlett’s test

> bartlett.test(cotton_data~factor(names(cotton_data)))

Bartlett test of homogeneity of variances

data: cotton_data by factor(names(cotton_data))

Bartlett’s K-squared = 0.2801, df = 4, p-value = 0.991

The results show that the null hypothesis cannot be rejected and hence the variance of the five sets is indeed same.

We now need to do a pairwise comparison to find out which pair has a difference in mean. we use the Tukey’s test to do so.

If the assumption of normality is not met then a test known is Kruskal-Wallis test may be used

> kruskal.test(cotton_data~factor(names(cotton_data)))

Kruskal-Wallis rank sum test

data: cotton_data by factor(names(cotton_data))

Kruskal-Wallis chi-squared = 18.5513, df = 4, p-value = 0.0009626

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