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.

Practice

1. What does the Chi-squared test in R primarily check for?

easy

A. Whether two categorical variables are related

B. Whether two numerical variables have the same mean

C. Whether a variable follows a normal distribution

D. Whether the data contains missing values

Solution

1. Step 1: Understand the purpose of the Chi-squared test

   The Chi-squared test is used to check if there is an association between two categorical variables.

2. Step 2: Compare with other options

   Options A and B relate to numerical data tests, and D is about data quality, not relationships.

3. Final Answer:

   Whether two categorical variables are related -> Option A

4. Quick Check:

   Chi-squared test checks categorical variable relationship = C [OK]

Hint: Chi-squared tests relationships between categories, not numbers [OK]

Common Mistakes:

• Confusing Chi-squared with t-test or ANOVA
• Thinking it tests numerical data means
• Assuming it checks data normality

2. Which of the following is the correct way to perform a Chi-squared test on a table named tbl in R?

easy

A. chisq(tbl)

B. chisq.test(tbl)

C. chi.test(tbl)

D. test.chisq(tbl)

Solution

1. Step 1: Recall the correct function name in R

   The function to perform a Chi-squared test in R is chisq.test().

2. Step 2: Check the syntax

   The correct syntax is calling chisq.test() with the table as the argument.

3. Final Answer:

   chisq.test(tbl) -> Option B

4. Quick Check:

   Correct function is chisq.test() = B [OK]

Hint: Use chisq.test() function for Chi-squared test in R [OK]

Common Mistakes:

• Using incorrect function names like chisq() or chi.test()
• Placing the function name after the table
• Forgetting parentheses after function name

3. What will be the output of the following R code?

tbl <- matrix(c(10, 20, 30, 40), nrow=2)
result <- chisq.test(tbl)
round(result$p.value, 3)

medium

A. 0.504

B. 0.025

C. 0.000

D. Error: object 'result' not found

Solution

1. Step 1: Understand the input table

   The matrix has counts: 10, 20 in first row and 30, 40 in second row.

2. Step 2: Perform Chi-squared test and check p-value

   Running chisq.test on this table gives a p-value around 0.504, meaning no strong association.

3. Final Answer:

   0.504 -> Option A

4. Quick Check:

   Chi-squared p-value rounded = 0.504 [OK]

Hint: Check p-value from chisq.test() to judge significance [OK]

Common Mistakes:

• Confusing p-value with test statistic
• Expecting zero p-value for small counts
• Mistaking error message for output

4. Identify the error in this R code snippet for Chi-squared test:

data <- matrix(c(5, 15, 20, 10), ncol=2)
chisq.test(data)

medium

A. Matrix should have rows and columns swapped

B. chisq.test() requires a data frame, not a matrix

C. No error, code runs correctly

D. Matrix must be named 'tbl' to work

Solution

1. Step 1: Check matrix creation

   The matrix is created with 4 values and 2 columns, which is valid.

2. Step 2: Check chisq.test() input type

   chisq.test() accepts a matrix of counts, so this input is valid and runs without error.

3. Final Answer:

   No error, code runs correctly -> Option C

4. Quick Check:

   chisq.test() accepts matrix input = A [OK]

Hint: chisq.test() works with matrices or tables of counts [OK]

Common Mistakes:

• Thinking chisq.test() only accepts data frames
• Assuming matrix dimensions must be square
• Believing variable name affects function

5. You have survey data of 3 groups choosing among 3 products. How do you apply a Chi-squared test in R to check if product preference depends on group?

group <- c('A', 'A', 'B', 'B', 'C', 'C')
product <- c('X', 'Y', 'X', 'Z', 'Y', 'Z')
# What next?

hard

A. Convert group and product to numeric and run chisq.test()

B. Run chisq.test(group, product) directly without table

C. Use t.test() on group and product vectors

D. Create a table with table(group, product) then run chisq.test() on it

Solution

1. Step 1: Create a contingency table

   Use table(group, product) to get counts of each group-product combination.

2. Step 2: Apply Chi-squared test on the table

   Run chisq.test() on the table to test independence between group and product preference.

3. Final Answer:

   Create a table with table(group, product) then run chisq.test() on it -> Option D

4. Quick Check:

   Table then chisq.test() = A [OK]

Hint: Always make a table before chisq.test() for categorical data [OK]

Common Mistakes:

• Trying to run chisq.test() on raw vectors
• Using t.test() instead of chisq.test() for categories
• Converting categories to numbers incorrectly