1. **State the problem:** We want to test if the population mean cholesterol level of people taking the medication (group 1) is lower than that of people taking the placebo (group 2). 2. **Given data:** - Group 1 (medication): $n_1=31$, mean $\bar{x}_1=229$, population standard deviation $\sigma_1=28$ - Group 2 (placebo): $n_2=39$, mean $\bar{x}_2=239$, population standard deviation $\sigma_2=35$ - Significance level $\alpha=0.08$ 3. **Hypotheses:** - Null hypothesis $H_0: \mu_1 = \mu_2$ - Alternative hypothesis $H_a: \mu_1 < \mu_2$ (medication lowers cholesterol) 4. **Test type:** Since population standard deviations are known, use a Z-test for difference of means. 5. **Formula for Z-test:** $$Z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$$ 6. **Calculate standard error:** $$SE = \sqrt{\frac{28^2}{31} + \frac{35^2}{39}} = \sqrt{\frac{784}{31} + \frac{1225}{39}} = \sqrt{25.29 + 31.41} = \sqrt{56.7} \approx 7.53$$ 7. **Calculate Z statistic:** $$Z = \frac{229 - 239}{7.53} = \frac{-10}{7.53} \approx -1.33$$ 8. **Find critical value:** For a left-tailed test at $\alpha=0.08$, critical Z value is approximately $-1.41$ (from Z-tables). 9. **Decision:** Since $Z = -1.33$ is greater than $-1.41$, we fail to reject $H_0$. 10. **Conclusion:** There is not enough evidence at the 0.08 significance level to conclude that the medication lowers cholesterol compared to placebo.