Two sample equality of variances in R

1 F-test in R

The F test statistic can be obtained by calculating the ratio of the two variances F=VAR(A)/VAR(B). Before doing the F test, we need to check one of the major assumptions is data should be normally distributed. Normality distribution can be assessed based on the Shapiro test or visually using a QQ plot. If any violation observed from normality then it better is to use Levene’s test or Fligner-Killeen test. Levene’s test or Fligner-Killeen test is less sensitive and appropriate for when data is distributed non-normally.

str(ToothGrowth)
## 'data.frame':    60 obs. of  3 variables:
##  $ len : num  4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
##  $ supp: Factor w/ 2 levels "OJ","VC": 2 2 2 2 2 2 2 2 2 2 ...
##  $ dose: num  0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 ...

Total 60 observations and 3 variables and the variable supp contains two groups. Let’s check the variance for the same.Before that, calculate the Shapiro test for the normality assumption validation.

Shapiro Test

shapiro.test(ToothGrowth$len)
## 
##  Shapiro-Wilk normality test
## 
## data:  ToothGrowth$len
## W = 0.96743, p-value = 0.1091

The p value is greater than 0.05, we can assume the normality.

res.ftest <- var.test(len ~ supp, data = ToothGrowth)
res.ftest
## 
##  F test to compare two variances
## 
## data:  len by supp
## F = 0.6386, num df = 29, denom df = 29, p-value = 0.2331
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.3039488 1.3416857
## sample estimates:
## ratio of variances 
##          0.6385951

The p-value of 0.2331 is greater than the significance level of 0.05. We can conclude that there is no significant difference between the two variances.

Compare more than two sample variances in R

When comparing more than two samples Bartlett’s test, Levene’s test, or Fligner-Killeen’s test will be more appropriate. Coming to statistical hypotheses, Bartlett’s test, Levene’s test, or Fligner-Killeen’s test,

Ho: All populations variances are equal

H1: At least two of them different

When using Bartlett’s test one of the main assumptions data should be normally distributed. In the case of nonnormal data, the Levene test is an alternative to the Bartlett test. If data is non-normally distributed, the Fligner-Killeen test is a non-parametric test alternative.

2 Bartlett’s test in R

Let’s check the equality of variance, We are using PlantGrowth dataset contains 30 observations and 2 variables.

str(PlantGrowth)
## 'data.frame':    30 obs. of  2 variables:
##  $ weight: num  4.17 5.58 5.18 6.11 4.5 4.61 5.17 4.53 5.33 5.14 ...
##  $ group : Factor w/ 3 levels "ctrl","trt1",..: 1 1 1 1 1 1 1 1 1 1 ...

The column group contains 3 factor variables ctrl, trt1, and trt2. Before doing Bartlett’s test let’s check the normality assumption.

Shapiro Test

shapiro.test(PlantGrowth$weight)
## 
##  Shapiro-Wilk normality test
## 
## data:  PlantGrowth$weight
## W = 0.98268, p-value = 0.8915

We can assume that data is normally distributed.

res <- bartlett.test(weight ~ group, data = PlantGrowth)
res
## 
##  Bartlett test of homogeneity of variances
## 
## data:  weight by group
## Bartlett's K-squared = 2.8786, df = 2, p-value = 0.2371

The p-value is 0.2371 is greater than the significance level of 0.05. We can conclude that there is no significant difference between the tested sample variances.

3 Levene’s test in R

Levene test function is from car package, let’s load the library.

library(car)
## Loading required package: carData
leveneTest(weight ~ group, data = PlantGrowth)
## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group  2  1.1192 0.3412
##       27

The p-value is 0.3412 is greater than the significance level of 0.05. We can conclude that there is no significant difference between the tested sample variances. Levene’s test with multiple independent variables can check based on ToothGrowth dataset. ToothGrowth dataset dose column stored as numeric variable let’s convert into factor variable first.

ToothGrowth$dose <- as.factor(ToothGrowth$dose)
leveneTest(len ~ supp*dose, data = ToothGrowth)
## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group  5  1.7086 0.1484
##       54

4 Fligner-Killeen test in R

Will make use of the same data set.

fligner.test(weight ~ group, data = PlantGrowth)
## 
##  Fligner-Killeen test of homogeneity of variances
## 
## data:  weight by group
## Fligner-Killeen:med chi-squared = 2.3499, df = 2, p-value = 0.3088

The p-value is 0.3088 is greater than the significance level of 0.05. We can conclude that there is no significant difference was observed between the tested sample variances.