Assignment 5 R Notebook

Input data from R Studio

iris_data <- iris # load the iris dataset and renamed it dat

Since there is only one categorical variable and the Chi-square test requires two categorical variables,

we added the variable size which corresponds to small if the length of the petal is smaller than the median of all flowers, big otherwise:

iris_data$size <- ifelse(dat$Sepal.Length < median(dat$Sepal.Length),
  "small", "big"
)

We now create a contingency table of the two variables Species and size with the table() function:

table(iris_data$Species, iris_data$size)

            
             big small
  setosa       1    49
  versicolor  29    21
  virginica   47     3

The contingency table gives the observed number of cases in each subgroup.

For instance, there is only one big setosa flower, while there are 49 small

setosa flowers in the dataset.

Create barplot to represent the data:

library(ggplot2)
ggplot(dat) +
 aes(x = Species, fill = iris_data$size) +
 geom_bar() +
 scale_fill_hue() +
 theme_minimal()

For this example, we are going to test in R if there is a relationship

between the variables Species and size. For this, the chisq.test() function is used:

test <- chisq.test(table(iris_data$Species, iris_data$size))
test

    Pearson's Chi-squared test

data:  table(iris_data$Species, iris_data$size)
X-squared = 86.035, df = 2, p-value < 2.2e-16

## 
##  Pearson's Chi-squared test
## 
## data:  table(dat$Species, dat$size)
## X-squared = 86.035, df = 2, p-value < 2.2e-16

Everything you need appears in this output: the title of the test, what

variables have been used, the test statistic, the degrees of freedom

and the p-value of the test.1 You can also retrieve the χ2 test statistic

and the p-value with:

test$statistic # test statistic

X-squared 
 86.03451 

## X-squared 
##  86.03451
test$p.value # p-value

[1] 2.078944e-19

## [1] 2.078944e-19

#If you need to find the expected frequencies, use test$expected.

test$observed

            
             big small
  setosa       1    49
  versicolor  29    21
  virginica   47     3

test$expected # test statistic

            
                  big    small
  setosa     25.66667 24.33333
  versicolor 25.66667 24.33333
  virginica  25.66667 24.33333

## X-squared 
##  86.03451
#test$p.value # p-value
## [1] 2.078944e-19

From the table and from test$p.value we see that the p-value is less

than the significance level of 5%. Like any other statistical test, if

the p-value is less than the significance level, we can reject the null hypothesis.

⇒ In our context, rejecting the null hypothesis for the Chi-square test of

independence means that there is a significant relationship between the

species and the size. Therefore, knowing the value of one variable helps

to predict the value of the other variable