1. **Problem Statement:** Test if the difference in tranquilizer dosage (1.0 mg vs 1.5 mg) affects the time to fall asleep using a t-test for independent samples at significance level $\alpha=0.01$. 2. **Formula and Rules:** Use the independent samples t-test formula: $$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$ where $\bar{x}_1, \bar{x}_2$ are sample means, $s_1^2, s_2^2$ are sample variances, and $n_1, n_2$ are sample sizes. 3. **Calculate sample statistics for 1.0 mg dose:** Data: 9.7, 7.9, 9.8, 13.0, 11.2, 9.8, 10.2, 12.3, 12.1, 9.5, 13.2 $n_1=11$ Mean: $$\bar{x}_1 = \frac{9.7 + 7.9 + 9.8 + 13.0 + 11.2 + 9.8 + 10.2 + 12.3 + 12.1 + 9.5 + 13.2}{11} = \frac{118.7}{11} = 10.79$$ Variance: Calculate each squared deviation, sum, then divide by $n_1-1=10$: $$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{10} = 2.68$$ 4. **Calculate sample statistics for 1.5 mg dose:** Data: 13.5, 9.8, 13.0, 9.8, 12.0, 10.5, 12.5, 11.5, 7.4 $n_2=9$ Mean: $$\bar{x}_2 = \frac{13.5 + 9.8 + 13.0 + 9.8 + 12.0 + 10.5 + 12.5 + 11.5 + 7.4}{9} = \frac{100}{9} = 11.11$$ Variance: $$s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{8} = 3.10$$ 5. **Calculate t-statistic:** $$t = \frac{10.79 - 11.11}{\sqrt{\frac{2.68}{11} + \frac{3.10}{9}}} = \frac{-0.32}{\sqrt{0.243 + 0.344}} = \frac{-0.32}{\sqrt{0.587}} = \frac{-0.32}{0.766} = -0.42$$ 6. **Degrees of freedom (approximate):** $$df = \min(n_1 - 1, n_2 - 1) = \min(10, 8) = 8$$ 7. **Critical t-value for two-tailed test at $\alpha=0.01$ and $df=8$:** $$t_{critical} = \pm 3.355$$ 8. **Decision:** Since $|t|=0.42 < 3.355$, we fail to reject the null hypothesis. **Conclusion:** There is no significant difference in sleep time between the two dosages at the 0.01 significance level.