Test of Independence Calculators

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The chi-square test of independence assesses whether two categorical variables are statistically associated or independent. It is applied to a contingency table (cross-tabulation) of observed frequencies and compares them to expected frequencies that would occur if the variables were independent. The test statistic is χ² = Σ[(O−E)²/E], where O is observed and E is expected. A significant result (p < 0.05) means the variables are associated; non-significant means there is insufficient evidence to reject independence. It is one of the most widely used statistical tests in the social sciences, medical research, and biology.

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Chi-Square Test of Independence Formula

χ² = Σ [(O − E)² / E]

O = observed frequency; E = expected frequency = (row total × column total) / grand total.

Degrees of freedom: df = (rows − 1) × (columns − 1)

Step-by-Step Example

Does smoking status relate to lung cancer? Observed (2×2 table):

CancerNo Cancer
Smoker4060
Non-smoker1090

Row totals: 100, 100. Column totals: 50, 150. Grand total: 200.

Expected E(smoker, cancer) = (100×50)/200 = 25. E(smoker, no cancer) = 75. E(non-smoker, cancer) = 25. E(non-smoker, no cancer) = 75.

χ² = (40−25)²/25 + (60−75)²/75 + (10−25)²/25 + (90−75)²/75 = 9 + 3 + 9 + 3 = 24. df = 1. p < 0.001. Conclusion: significant association between smoking and cancer.

Assumptions

  • All expected counts ≥ 5 (use Fisher's exact test for smaller expected counts)
  • Observations are independent
  • Data are categorical counts, not continuous

Glossary

Chi-Square Test of Independence
χ² = Σ[(O−E)²/E]; tests whether two categorical variables in a contingency table are associated; df = (rows−1)(columns−1); significant result = variables are not independent.
Expected Frequency (E)
The cell frequency predicted if the two variables were independent: E = (row total × column total) / grand total; must be ≥ 5 in all cells for valid chi-square approximation.
Cramér's V
Effect size for chi-square test of independence: V = √(χ²/n×min(r−1,c−1)); ranges 0–1; small ~0.1, medium ~0.3, large ~0.5; measures strength of association independent of sample size.

Frequently Asked Questions

The chi-square test of independence tests whether two categorical variables are associated. It compares observed frequencies in a contingency table to expected frequencies calculated under the null hypothesis of independence (E = row total × column total / grand total). χ² = Σ[(O−E)²/E]; df = (rows−1)(columns−1). Large χ² → small p-value → reject independence. A significant result means the variables are associated; it does not tell you the direction or magnitude of the association (use odds ratio or Cramér's V for that).

Expected frequency for each cell = (row total × column total) / grand total. Example: in a 3×2 table, if row 1 total = 50, column 1 total = 40, grand total = 100: E(row1, col1) = (50 × 40)/100 = 20. All expected frequencies must be ≥ 5 for the chi-square approximation to be valid. If any E < 5, merge categories, use Fisher's exact test for 2×2 tables, or use a simulation-based test for larger tables.

Goodness of fit: tests whether one categorical variable matches a specified theoretical distribution (e.g., 3:1 Mendelian ratio). One variable; expected values from the theoretical model; df = categories − 1. Test of independence: tests whether two categorical variables in a contingency table are associated. Two variables; expected values from the marginal totals; df = (rows−1)(columns−1). Both use χ² = Σ[(O−E)²/E] but differ in how expected values are calculated and in the degrees of freedom formula.

A significant chi-square (p < 0.05) means there is evidence that the two variables are associated — their distribution is not independent. However, chi-square alone does not tell you: (1) the direction of the association; (2) the strength of the association; (3) which specific cells drive the significance. Follow up with: Cramér's V (effect size: V = √(χ²/n×min(r−1,c−1)) — ranges 0 to 1); standardized residuals for each cell (which cells deviate most from expected); and odds ratios for 2×2 tables (direction and magnitude of association).