1. **Problem Statement:** We are testing if the difference in tranquilizer dosage (1.0 mg vs 1.5 mg) affects the time it takes rats to fall asleep using an independent samples t-test at significance level $\alpha=0.01$. 2. **Formula and Rules:** The independent samples t-test formula is: $$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$ 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] - $n_1=11$ - $\bar{x}_1 = \frac{\sum x_i}{n_1} = \frac{9.7+7.9+...+13.2}{11} = 10.82$ - Calculate variance $s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1} = 2.56$ - For 1.5 mg dose: data = [13.5, 9.8, 13.0, 9.8, 12.0, 10.5, 12.5, 11.5, 7.4] - $n_2=9$ - $\bar{x}_2 = \frac{\sum x_i}{n_2} = \frac{13.5+9.8+...+7.4}{9} = 11.11$ - Calculate variance $s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1} = 3.10$ 4. **Calculate t-statistic:** $$t = \frac{10.82 - 11.11}{\sqrt{\frac{2.56}{11} + \frac{3.10}{9}}} = \frac{-0.29}{\sqrt{0.2327 + 0.3444}} = \frac{-0.29}{\sqrt{0.5771}} = \frac{-0.29}{0.76} = -0.38$$ 5. **Degrees of freedom (approximate):** $$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} = \frac{0.5771^2}{\frac{0.2327^2}{10} + \frac{0.3444^2}{8}} \approx 16$$ 6. **Decision:** For $\alpha=0.01$ and two-tailed test, critical t-value $\approx \pm 2.921$. Since $|t|=0.38 < 2.921$, we **fail to reject the null hypothesis**. **Conclusion:** There is no significant evidence at the 0.01 level that the dosage affects the time to fall asleep.