## Chi-Square Test of Independence: Application and Rationale **Key Point:** The chi-square test is the appropriate statistical test for assessing the association between two categorical variables in independent samples. ### When to Use Chi-Square Test - **Both variables are categorical** (nominal or ordinal) - **Independent samples** (no pairing or matching) - **Contingency table format** (rows × columns) - Tests the **null hypothesis of independence** (no association) ### Test Structure for This Question | | Lung Cancer Present | Lung Cancer Absent | Total | |---|---|---|---| | **Smoker** | a | b | a+b | | **Non-smoker** | c | d | c+d | | **Total** | a+c | b+d | 500 | ### Chi-Square Test Statistic $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ where O = observed frequency, E = expected frequency under independence. **High-Yield:** Chi-square tests **association only**, not causation. A significant chi-square indicates the variables are **not independent**, but does not prove smoking causes cancer. ### Assumptions 1. **Categorical data** in both variables 2. **Independent observations** (no pairing) 3. **Expected frequency ≥ 5** in at least 80% of cells (if violated, use Fisher's exact test for 2×2 tables) 4. **Adequate sample size** (generally n > 20) **Mnemonic:** **CCAT** — **C**ategorical data, **C**ontingency table, **A**ssociation test, **T**wo independent variables → **Chi-square** **Clinical Pearl:** In a 2×2 contingency table with small expected frequencies (< 5), Fisher's exact test is preferred over chi-square.
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