1. **State the problem:** We want to test if the treatment (Drug or Placebo) is independent of the side effect of nausea at significance level $\alpha=0.10$ using the given contingency table. 2. **Null hypothesis $H_0$:** The treatment and nausea are independent. 3. **Alternative hypothesis $H_a$:** The treatment and nausea are not independent. 4. **Level of significance:** $\alpha=0.10$. 5. **Degrees of freedom:** For a $2 \times 2$ table, $df = (rows-1)(columns-1) = (2-1)(2-1) = 1$. 6. **Calculate expected counts:** - Total patients = 565 - Row totals: Nausea = 49, No nausea = 516 - Column totals: Drug = 290, Placebo = 275 Expected count formula: $$E = \frac{(row\ total)(column\ total)}{grand\ total}$$ | Result | Drug (E) | Placebo (E) | |-----------|-------------------|-------------------| | Nausea | $\frac{49 \times 290}{565} = 25.13$ | $\frac{49 \times 275}{565} = 23.87$ | | No nausea | $\frac{516 \times 290}{565} = 264.87$ | $\frac{516 \times 275}{565} = 251.13$ | 7. **Calculate test statistic $\chi^2$:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Where $O$ is observed count, $E$ is expected count. Calculate each term: - Drug-Nausea: $\frac{(36 - 25.13)^2}{25.13} = \frac{(10.87)^2}{25.13} = 4.70$ - Placebo-Nausea: $\frac{(13 - 23.87)^2}{23.87} = \frac{(-10.87)^2}{23.87} = 4.95$ - Drug-No nausea: $\frac{(254 - 264.87)^2}{264.87} = \frac{(-10.87)^2}{264.87} = 0.45$ - Placebo-No nausea: $\frac{(262 - 251.13)^2}{251.13} = \frac{(10.87)^2}{251.13} = 0.47$ Sum: $$\chi^2 = 4.70 + 4.95 + 0.45 + 0.47 = 10.57$$ 8. **Critical value:** For $df=1$ and $\alpha=0.10$, critical value from chi-square table is approximately $2.71$. 9. **Decision:** Since $\chi^2 = 10.57 > 2.71$, we reject the null hypothesis. **Conclusion:** There is sufficient evidence at the 0.10 significance level to conclude that the treatment and nausea are not independent. | Result | Drug (O) | Placebo (O) | Drug (E) | Placebo (E) | |-----------|----------|-------------|----------|-------------| | Nausea | 36 | 13 | 25.13 | 23.87 | | No nausea | 254 | 262 | 264.87 | 251.13 | Test statistic $\chi^2 = 10.57$, critical value $= 2.71$, reject $H_0$.