Chi-squared test in R Programming - Step-by-Step Execution

Concept Flow - Chi-squared test

Start with observed data

↓

Calculate expected data

↓

Compute (Observed - Expected)^2 / Expected

↓

Sum all values to get Chi-squared statistic

↓

Compare statistic to Chi-squared distribution

↓

Decide if difference is significant

↓

End

The test compares observed counts to expected counts, calculates a statistic, then checks if the difference is significant.

Execution Sample

R Programming

observed <- c(50, 30, 20)
expected <- c(40, 40, 20)
chisq_stat <- sum((observed - expected)^2 / expected)
p_value <- 1 - pchisq(chisq_stat, df=2)
p_value

This code calculates the chi-squared statistic and p-value for observed vs expected counts.

Execution Table

Step	Calculation	Value	Explanation
1	Observed counts	50, 30, 20	Given data from experiment
2	Expected counts	40, 40, 20	Hypothesized distribution
3	(O - E)	10, -10, 0	Difference between observed and expected
4	(O - E)^2	100, 100, 0	Square differences
5	(O - E)^2 / E	2.5, 2.5, 0	Divide squared differences by expected
6	Sum statistic	5.0	Sum of all values from step 5
7	Degrees of freedom	2	Number of categories minus 1
8	p-value	0.0821	Probability of observing this or more extreme
9	Decision	Not significant	p-value > 0.05, fail to reject null hypothesis

💡 Test ends after calculating p-value and making decision based on significance level

Variable Tracker

Variable	Start	After Step 3	After Step 5	Final
observed	c(50,30,20)	c(50,30,20)	c(50,30,20)	c(50,30,20)
expected	c(40,40,20)	c(40,40,20)	c(40,40,20)	c(40,40,20)
difference	NA	c(10,-10,0)	c(10,-10,0)	c(10,-10,0)
squared_diff	NA	c(100,100,0)	c(2.5,2.5,0)	c(2.5,2.5,0)
chisq_stat	NA	NA	NA	5.0
p_value	NA	NA	NA	0.0821

Key Moments - 3 Insights

Why do we square the difference (O - E) before dividing by E?

Squaring makes all differences positive and emphasizes larger differences, as shown in step 4 of the execution_table.

What does the p-value tell us in this test?

The p-value (step 8) tells us how likely it is to see the observed data if the null hypothesis is true. A high p-value means no strong evidence against the null.

Why do we subtract 1 from the number of categories to get degrees of freedom?

Degrees of freedom (step 7) reflect the number of independent comparisons. Since total counts are fixed, one category is dependent on others.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the value of (O - E)^2 / E for the second category?

A 100

B 2.5

C 10

D -10

💡 Hint

Check the row labeled '5' under the 'Value' column in execution_table.

At which step does the test calculate the total chi-squared statistic?

A Step 6

B Step 4

C Step 8

D Step 2

💡 Hint

Look for the step where the sum of all values from step 5 is computed.

If the observed counts were exactly equal to expected counts, what would be the chi-squared statistic?

A Equal to number of categories

B 1

C 0

D Cannot be determined

💡 Hint

Refer to the formula in execution_sample and consider when (O - E) is zero.

Concept Snapshot

Chi-squared test compares observed and expected counts.
Calculate sum of (O - E)^2 / E.
Degrees of freedom = categories - 1.
Use p-value to decide significance.
If p < 0.05, difference is significant.

Full Transcript

The Chi-squared test starts with observed and expected counts. We find the difference between them, square it, and divide by expected counts. Summing these values gives the chi-squared statistic. We then compare this statistic to a chi-squared distribution with degrees of freedom equal to the number of categories minus one. The p-value tells us if the observed differences are likely due to chance. If the p-value is small (usually less than 0.05), we say the difference is significant and reject the null hypothesis. Otherwise, we do not reject it.

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