A public health researcher in Delhi conducted a study to compare the effectiveness of two different health education interventions on reducing tobacco use among 200 factory workers. Group A (n=100) received intensive counseling, while Group B (n=100) received standard written materials. After 6 months, the mean reduction in cigarettes per day was 8.5 (SD=2.1) in Group A and 5.2 (SD=2.8) in Group B. The researcher wants to determine if this difference is statistically significant. Which test of significance is most appropriate, and what is the underlying assumption required?

Explanation

## Test Selection for Continuous Outcome Data ### Study Design Analysis This study compares **two independent groups** (Group A vs Group B) with a **continuous outcome** (mean reduction in cigarettes per day measured in absolute numbers). The data are presented as means with standard deviations, indicating interval/ratio scale measurement. ### Why Unpaired t-test is Correct **Key Point:** The unpaired (independent samples) t-test is used when: 1. Comparing means between **two independent groups** 2. Outcome variable is **continuous** (interval/ratio scale) 3. Sample sizes are reasonably large (n=100 each) 4. Data are **approximately normally distributed** 5. **Variances are approximately equal** between groups (Levene's test can verify) ### Assumptions of Unpaired t-test | Assumption | Explanation | How to Check | |---|---|---| | **Independence** | Observations in one group are not related to observations in the other | Study design (random allocation) | | **Normality** | Data in each group approximately follow normal distribution | Shapiro-Wilk test; Q-Q plot; n>30 often sufficient | | **Equal variances** | Variance in Group A ≈ Variance in Group B | Levene's test; SD ratio <2 is acceptable | | **Continuous outcome** | Data measured on interval or ratio scale | Check measurement units | **High-Yield:** With n=100 in each group, the Central Limit Theorem ensures the sampling distribution of means is approximately normal even if individual data are slightly non-normal. ### Formula Context $$t = \frac{\bar{X}_1 - \bar{X}_2}{SE_{diff}} = \frac{8.5 - 5.2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$ where $SE_{diff}$ is the standard error of the difference between means. ### Clinical Pearl If Levene's test shows unequal variances (p<0.05), use **Welch's t-test** (a modification that does not assume equal variances) rather than abandoning the t-test entirely.