data <- iris
str(iris)
## 'data.frame':    150 obs. of  5 variables:
##  $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
##  $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
##  $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
##  $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
##  $ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
View(iris)
head(iris)
##   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1          5.1         3.5          1.4         0.2  setosa
## 2          4.9         3.0          1.4         0.2  setosa
## 3          4.7         3.2          1.3         0.2  setosa
## 4          4.6         3.1          1.5         0.2  setosa
## 5          5.0         3.6          1.4         0.2  setosa
## 6          5.4         3.9          1.7         0.4  setosa
tail(iris)
##     Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
## 145          6.7         3.3          5.7         2.5 virginica
## 146          6.7         3.0          5.2         2.3 virginica
## 147          6.3         2.5          5.0         1.9 virginica
## 148          6.5         3.0          5.2         2.0 virginica
## 149          6.2         3.4          5.4         2.3 virginica
## 150          5.9         3.0          5.1         1.8 virginica

#independent variable: species #dependent variable: sepal width

library(car)
## Loading required package: carData
leveneTest(Sepal.Width ~ Species, data = iris)
## Levene's Test for Homogeneity of Variance (center = median)
##        Df F value Pr(>F)
## group   2  0.5902 0.5555
##       147

#Next, we test whether the species observation for each group satisfy the homogeniety of variance assumption. The value of Pr(>F) is 0.5555, so we fail to reject the Null Hypothesis, indicating that the homogeneity of variance condition is satisfied.

result_anova <- aov(Sepal.Width ~ Species, data = iris, var.equal=TRUE)
## Warning: In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
##  extra argument 'var.equal' will be disregarded
summary(result_anova)
##              Df Sum Sq Mean Sq F value Pr(>F)    
## Species       2  11.35   5.672   49.16 <2e-16 ***
## Residuals   147  16.96   0.115                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

#We then perform an ANOVA test to examine whether the mean species of the groups are the same or different at a 5% siginificance level. The p-value is lower than 0.001. So, we reject the Null Hypothesis and accept the Alternative Hypothesis. This indicates that the mean species of the groups are significantly different.