Chi-square test for a 2×2 table
Observed counts are compared with those expected under independence; divide by the expected values.
Worked example
Are subject and passing related? Test independence with the χ² test (α = 5%, 1 degree of freedom, critical value 3.841).
| passed | failed | |
|---|---|---|
| Subject A | 37 | 17 |
| Subject B | 16 | 30 |
Worked solution
- 1.Expected count: eᵢⱼ = row total · column total / grand total
- 2.e = 28.62; 25.38; 24.38; 21.62
- 3.χ² = Σ (o − e)² / e = 11.35
- 4.11.35 > 3.841 → reject H₀ (independence)
Typical mistake: Dividing by the observed instead of the expected counts.
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task
Are subject and passing related? Test independence with the χ² test (α = 5%, 1 degree of freedom, critical value 3.841).
| passed | failed | |
|---|---|---|
| Subject A | 37 | 18 |
| Subject B | 48 | 18 |
χ² (2 decimal places)
Decision
Comma or point; fractions like 1/6 work too.
Formulas for this topic
- Expected: eᵢⱼ = row total · column total / grand total
- χ² = Σ (o − e)² / e
- Degrees of freedom: (rows − 1) · (columns − 1); 1 for 2×2
- χ² > critical value → reject H₀ (independence)
Typical mistake: Dividing by the observed instead of the expected counts.
Frequently asked questions
What is the typical mistake in "Chi-square test for a 2×2 table"?
Dividing by the observed instead of the expected counts.
When is the χ² test valid?
Rule of thumb: all expected counts should be at least 5.