The chi-square test of independence checks whether two categorical variables are associated.
Why it matters
It is the standard test for questions like "is smoking related to lung disease?" when both are categories.
Key ideas
- Data arranged in a contingency table
- H0: the variables are independent
- Expected count = row total × column total ÷ N
- χ² = Σ (O − E)² ÷ E, df = (r − 1)(c − 1)
- Needs mostly expected counts ≥ 5
- Report Cramér's V
Real-life research example
Smoking (yes/no) vs respiratory symptoms (yes/no) in 200 adults.
When to use
- Two categorical variables, independent observations
When NOT to use
- Don't use with small expected counts (use Fisher for 2×2)
- Don't use on paired data
Mini example
Smokers: 30 with symptoms, 20 without; Non-smokers: 25 with, 125 without
Quick takeaway
Compare what you observe with what independence would predict.
Try it yourself
Enter the 2×2 counts above. Are smoking and symptoms associated? Check expected counts and Cramér's V.
Model answer
The association is strong (small p). Check that all expected counts are ≥ 5 and read V for the size of the association.