Chi-squared test in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a basic Chi-squared test in R

What is the output of this R code performing a Chi-squared test on a 2x2 table?

R Programming

data <- matrix(c(10, 20, 20, 40), nrow = 2)
result <- chisq.test(data)
result$p.value

A 1.0000000

B 0.9876543

C 0.0012345

D 0.0455003

💡 Hint

Check the p-value returned by chisq.test for the given matrix.

The chisq.test function calculates the Chi-squared test and returns a p-value. For this data, the observed frequencies do not exactly match the expected frequencies under independence, so the p-value is approximately 0.0455, indicating borderline evidence against independence.

🧠 Conceptual

intermediate

Understanding Chi-squared test assumptions

Which of the following is NOT an assumption of the Chi-squared test in R?

A Observations must be independent

B Data must be normally distributed

C Data should be categorical

D Expected frequencies should be at least 5 in each cell

💡 Hint

Think about the type of data Chi-squared test is designed for.

The Chi-squared test does not require data to be normally distributed. It is used for categorical data and assumes independence and sufficient expected frequencies.

🔧 Debug

advanced

Identify the error in Chi-squared test code

What error will this R code produce?

R Programming

data <- matrix(c(5, 3, 2), nrow = 3)
chisq.test(data)

A Error: 'data' must have at least 2 columns and 2 rows

B Error: 'data' must be a square matrix

C Error: 'x' must be a matrix or a table

D No error, test runs successfully

💡 Hint

Check the shape of the matrix and requirements for chisq.test.

chisq.test requires a matrix with at least 2 rows and 2 columns. The given matrix has 3 rows but only 1 column, causing an error.

🚀 Application

advanced

Calculate Chi-squared statistic manually

Given the observed counts matrix in R: obs <- matrix(c(30, 10, 20, 40), nrow=2) Which code correctly calculates the Chi-squared test statistic manually (without using chisq.test)?

A

expected <- outer(rowSums(obs), colSums(obs)) / sum(obs)
chisq <- sum((obs - expected)^2 / expected)
chisq

B

expected <- rowSums(obs) * colSums(obs) / sum(obs)
chisq <- sum((obs - expected)^2 / expected)
chisq

C

expected <- outer(rowSums(obs), colSums(obs)) * sum(obs)
chisq <- sum((obs - expected)^2 / expected)
chisq

D

expected <- outer(rowSums(obs), colSums(obs)) / sum(obs)
chisq <- sum((expected - obs)^2 / obs)
chisq

💡 Hint

Expected counts are calculated by multiplying row and column sums divided by total sum.

The expected counts are calculated as the outer product of row sums and column sums divided by total sum. The Chi-squared statistic sums the squared differences divided by expected counts.

❓ Predict Output

expert

Output of Chi-squared test with simulated data

What is the output of this R code snippet?

R Programming

set.seed(123)
data <- matrix(sample(1:100, 9), nrow=3)
result <- chisq.test(data)
round(result$statistic, 2)

A 4.67

B 7.89

C 5.12

D 2.34

💡 Hint

Run the code with the given seed to get the Chi-squared statistic.

Using set.seed(123) ensures reproducible random data. The Chi-squared statistic calculated by chisq.test on this matrix is approximately 5.12.