Sometimes when the data that needs to be analyzed is not normally distributed. This makes it difficult to make any inferences based on the results because one of the main assumptions of parametric statistical test such as ANOVA, t-test, etc is normal distribution of the data.
Fortunately, for every parametric test there is a non-parametric test. Non-parametric test are test that make no assumptions about the normality of the data. This means that the non-normal data can still be analyzed with a certain measure of confidence in terms of the results.
This post will look at non-parametric test that are used to test the difference in means. For three or more groups we used the Kruskal-Wallis Test. The Kruskal-Wallis Test is the non-parametric version of ANOVA.
Setup
We are going to use the “ISLR” package available on R to demonstrate the use of the Kruskal-Wallis test. After downloading this package you need to load the “Auto” data. Below is the code to do all of this.
install.packages('ISLR')
library(ISLR)
data=Auto
We now need to examine the structure of the data set. This is done with the “str” function below is code followed by the results
str(Auto)
'data.frame': 392 obs. of 9 variables:
$ mpg : num 18 15 18 16 17 15 14 14 14 15 ...
$ cylinders : num 8 8 8 8 8 8 8 8 8 8 ...
$ displacement: num 307 350 318 304 302 429 454 440 455 390 ...
$ horsepower : num 130 165 150 150 140 198 220 215 225 190 ...
$ weight : num 3504 3693 3436 3433 3449 ...
$ acceleration: num 12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
$ year : num 70 70 70 70 70 70 70 70 70 70 ...
$ origin : num 1 1 1 1 1 1 1 1 1 1 ...
$ name : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
So we have 9 variables. We first need to find if any of the continuous variables are non-normal because this indicates that the Kruskal-Willis test is needed. We will look at the ‘displacement’ variable and look at the histogram to see if it is normally distributed. Below is the code followed by the histogram
hist(Auto$displacement)
This does not look normally distributed. We now need a factor variable with 3 or more groups. We are going to use the ‘origin’ variable. This variable indicates were the care was made 1 = America, 2 = Europe, and 3 = Japan. However, this variable is currently a numeric variable. We need to change it into a factor variable. Below is the code for this
Auto$origin<-as.factor(Auto$origin)
The Test
We will now use the Kruskal-Wallis test. The question we have is “is there a difference in displacement based on the origin of the vehicle?” The code for the analysis is below followed by the results.
> kruskal.test(displacement ~ origin, data = Auto) Kruskal-Wallis rank sum test data: displacement by origin Kruskal-Wallis chi-squared = 201.63, df = 2, p-value < 2.2e-16
Based on the results, we know there is a difference among the groups. However, just like ANOVA, we do not know were. We have to do a post-hoc test in order to determine where the difference in means is among the three groups.
To do this we need to install a new package and do a new analysis. We will download the “PCMR” package and run the code below
install.packages('PMCMR')
library(PMCMR)
data(Auto)
attach(Auto)
posthoc.kruskal.nemenyi.test(x=displacement, g=origin, dist='Tukey')
Here is what we did,
- Installed the PMCMR package and loaded it
- Loaded the “Auto” data and used the “attach” function to make it available
- Ran the function “posthoc.kruskal.nemenyi.test” and place the appropriate variables in their place and then indicated the type of posthoc test ‘Tukey’
Below are the results
Pairwise comparisons using Tukey and Kramer (Nemenyi) test
with Tukey-Dist approximation for independent samples
data: displacement and origin
1 2
2 3.4e-14 -
3 < 2e-16 0.51
P value adjustment method: none
Warning message:
In posthoc.kruskal.nemenyi.test.default(x = displacement, g = origin, :
Ties are present, p-values are not corrected.
The results are listed in a table. When a comparison was made between group 1 and 2 the results were significant (p < 0.0001). The same when group 1 and 3 are compared (p < 0.0001). However, there was no difference between group 2 and 3 (p = 0.51).
Do not worry about the warning message this can be corrected if necessary
Perhaps you are wondering what the actually means for each group is. Below is the code with the results
> aggregate(Auto[, 3], list(Auto$origin), mean) Group.1 x 1 1 247.5122 2 2 109.6324 3 3 102.7089
Cares made in America have an average displacement of 247.51 while cars from Europe and Japan have a displacement of 109.63 and 102.70. Below is the code for the boxplot followed by the graph
boxplot(displacement~origin, data=Auto, ylab= 'Displacement', xlab='Origin')
title('Car Displacement')
Conclusion
This post provided an example of the Kruskal-Willis test. This test is useful when the data is not normally distributed. The main problem with this test is that it is less powerful than an ANOVA test. However, this is a problem with most non-parametric test when compared to parametric test.
We are going to make a histogram of the ‘mpg’ variable in our ‘cars’ dataset. Below is the code for doing this followed by the actual histogram.
