🔗 Module 3: Chi-Square Test of Independence

Testing Relationships Between Two Categorical Variables

📚 Learning Objectives

By the end of this module, you will be able to:

Understand what independence means statistically
Create and interpret contingency tables
Calculate expected frequencies from row and column totals
Run chi-square tests of independence in R
Interpret standardized residuals to identify which cells drive significance
Calculate and interpret effect sizes (Cramér's V)

🔗 What Does "Independence" Mean?

Two variables are independent when knowing the value of one variable tells you nothing about the other.

✅ Independent Variables

Example: Coin flip outcome and dice roll

Knowing you flipped heads tells you NOTHING about what number the die will show.

In research: "Treatment response is independent of sex" means males and females respond equally.

❌ Associated/Dependent Variables

Example: Species and habitat preference

Knowing the species DOES tell you about likely habitat (e.g., fish prefer water!).

In research: "Diagnosis is associated with treatment type" means different diagnoses get different treatments.

Chi-square test of independence asks:
"Are these two categorical variables related, or are they independent?"

📊 Understanding Contingency Tables

A contingency table (also called a cross-tabulation or crosstab) displays frequencies for two categorical variables simultaneously.

Example: Treatment Response by Sex

Research Question: Is treatment response independent of patient sex?

	Treatment Response		Row Total
Sex	Improved	No Improvement
Male	45	15	60
Female	25	35	60
Column Total	70	50	120

Reading this table:

45 males improved, 15 males did not improve
25 females improved, 35 females did not improve
Total sample size (n) = 120
Row totals: Number in each sex category
Column totals: Number in each response category

Visual pattern: Males seem more likely to improve (75%) than females (42%). Is this difference statistically significant?

🧮 Calculating Expected Frequencies

Under the null hypothesis (independence), we calculate what frequencies we'd EXPECT if the variables were truly unrelated.

Expected Frequency Formula

Expected = (Row Total × Column Total) / Grand Total

For each cell in the contingency table

📐 Calculate Expected Frequencies

Using the treatment response example:

	Improved	No Improvement
Male (n=60)	Observed: 45 Expected: (60 × 70) / 120 = 35	Observed: 15 Expected: (60 × 50) / 120 = 25
Female (n=60)	Observed: 25 Expected: (60 × 70) / 120 = 35	Observed: 35 Expected: (60 × 50) / 120 = 25

Key insight: If sex and response were independent, we'd expect equal success rates for males and females. The expected values reflect the OVERALL success rate (70/120 = 58%) applied to each group.

Step-by-step for Male/Improved cell:

Row total for Males = 60
Column total for Improved = 70
Grand total = 120
Expected = (60 × 70) / 120 = 4200 / 120 = 35

Interpretation: If there's no relationship between sex and improvement, we'd expect 35 of the 60 males to improve (the same proportion as in the overall sample).

💻 Running Chi-Square Test of Independence in R

1Create the Contingency Table

# Method 1: Enter data as a matrix
data <- matrix(c(45, 15, 25, 35), 
               nrow = 2, byrow = TRUE)
rownames(data) <- c("Male", "Female")
colnames(data) <- c("Improved", "No Improvement")

# View the table
data

# Method 2: If you have raw data
# data <- table(sex, response)

2Run the Chi-Square Test

# Simple test
result <- chisq.test(data)
result

# To see expected frequencies
result$expected

# To see standardized residuals (more on this soon!)
result$stdres

3Interpret the Output

Pearson's Chi-squared test with Yates' continuity correction

data:  data
X-squared = 15.63, df = 1, p-value = 7.743e-05

What this tells you:

X-squared = 15.63: Chi-square statistic
df = 1: (rows - 1) × (columns - 1) = (2-1) × (2-1) = 1
p-value = 0.00007743: Extremely strong evidence against independence
Yates' correction: Applied automatically for 2×2 tables (makes test more conservative)

Conclusion: We reject H₀. Treatment response is NOT independent of sex (p < .001). Males show significantly higher improvement rates than females.

🔢 Degrees of Freedom for Test of Independence

df = (r - 1) × (c - 1)

r = number of rows
c = number of columns

Examples:

Table Size	Calculation	df
2 × 2	(2-1) × (2-1)	1
2 × 3	(2-1) × (3-1)	2
3 × 3	(3-1) × (3-1)	4
4 × 3	(4-1) × (3-1)	6

⚠️ Why (r-1) × (c-1)?

Once you know frequencies in (r-1) rows and (c-1) columns, plus all the marginal totals, the remaining cells are determined. This creates constraints on the data, reducing degrees of freedom.

🎯 Practice Problem 1: Species and Habitat

Research Question: Is habitat preference independent of species?

Data: Three species observed in two habitat types

	Habitat		Total
Species	Forest	Grassland
Species A	35	15	50
Species B	20	30	50
Species C	25	25	50
Total	80	70	150

# Create the table
habitat_data <- matrix(c(35, 15, 20, 30, 25, 25),
                       nrow = 3, byrow = TRUE)
rownames(habitat_data) <- c("Species A", "Species B", "Species C")
colnames(habitat_data) <- c("Forest", "Grassland")

# Run test
result <- chisq.test(habitat_data)
result

# Check expected frequencies
result$expected

Pearson's Chi-squared test

data:  habitat_data
X-squared = 8.259, df = 2, p-value = 0.01611

Expected Frequencies:

	Forest	Grassland
Species A	26.67	23.33
Species B	26.67	23.33
Species C	26.67	23.33

Interpretation: χ²(2) = 8.26, p = .016. We reject H₀. Habitat preference is NOT independent of species. Looking at the data: Species A prefers forest (70%), Species B prefers grassland (60%), and Species C shows no preference (50/50).

🔍 Standardized Residuals: Finding the Pattern

When Chi-square is significant, the next question is: "WHICH cells are driving the effect?"

Standardized residuals tell you how far each cell deviates from expectation, in standard deviation units.

Standardized Residual = (Observed - Expected) / √Expected

Interpretation guidelines:

|residual| > 2: Cell contributes notably to chi-square
|residual| > 3: Cell contributes strongly
Positive residual: More frequent than expected
Negative residual: Less frequent than expected

# Get standardized residuals
result$stdres

# Visualize them
library(corrplot)
corrplot(result$stdres, is.corr = FALSE, 
         method = "color", addCoef.col = "black")

Example: Treatment Response Residuals

             Improved  No Improvement
Male            2.69        -2.69
Female         -2.69         2.69

Interpretation:

Male/Improved: residual = +2.69 → Males improved MORE than expected
Male/No Improvement: residual = -2.69 → Males had FEWER non-improvements than expected
Female cells: Opposite pattern - fewer improvements, more non-improvements than expected

All |residuals| > 2, confirming that ALL cells contribute to the significant chi-square!

📏 Effect Size: Cramér's V

Chi-square tells you IF variables are related, but not HOW STRONG the relationship is. That's where effect size comes in!

Cramér's V measures the strength of association, ranging from 0 (no association) to 1 (perfect association).

V = √[χ² / (n × (k - 1))]

k = smaller of (number of rows, number of columns)
n = total sample size

Interpretation guidelines (for df=1):

Cramér's V	Interpretation
0.00 - 0.10	Negligible association
0.10 - 0.30	Weak association
0.30 - 0.50	Moderate association
> 0.50	Strong association

# Calculate Cramér's V in R
library(rcompanion)
cramerV(data)

# Or calculate manually:
chi_sq <- result$statistic
n <- sum(data)
k <- min(nrow(data), ncol(data))
V <- sqrt(chi_sq / (n * (k - 1)))
V