1. **Stating the problem:** We want to test if the difference in 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. **Hypotheses:** - Null hypothesis $H_0$: The mean times are equal, i.e., $\mu_1 = \mu_2$ (dosage has no effect). - Alternative hypothesis $H_a$: The mean times are different, i.e., $\mu_1 \neq \mu_2$. 3. **Level of significance:** $\alpha = 0.01$. 4. **Statistics:** - Sample 1 (1.0 mg): $9.7, 7.9, 9.8, 13.0, 11.2, 9.8, 10.2, 12.3, 12.1, 9.7, 13.2$ - Sample 2 (1.5 mg): $13.5, 9.8, 13.0, 9.8, 12.0, 10.5, 12.5, 11.5, 7.4$ Calculate sample means: $$\bar{x}_1 = \frac{9.7 + 7.9 + \cdots + 13.2}{11} = 10.91$$ $$\bar{x}_2 = \frac{13.5 + 9.8 + \cdots + 7.4}{9} = 11.33$$ Calculate sample variances: $$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1} = 2.56$$ $$s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1} = 3.11$$ Calculate pooled standard error: $$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{2.56}{11} + \frac{3.11}{9}} = \sqrt{0.2327 + 0.3456} = \sqrt{0.5783} = 0.76$$ Calculate degrees of freedom using Welch-Satterthwaite equation: $$df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(s_1^2/n_1)^2}{n_1 - 1} + \frac{(s_2^2/n_2)^2}{n_2 - 1}} = \frac{0.5783^2}{\frac{0.2327^2}{10} + \frac{0.3456^2}{8}} = \frac{0.3344}{0.00541 + 0.01493} = \frac{0.3344}{0.02034} = 16.44$$ Calculate t-statistic: $$t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{10.91 - 11.33}{0.76} = \frac{-0.42}{0.76} = -0.55$$ 5. **Decision rule:** At $\alpha=0.01$ and $df=16.44$, two-tailed critical t-value $t_{crit} \approx \pm 2.921$. 6. **Conclusion:** Since $|t|=0.55 < 2.921$, we fail to reject the null hypothesis. **Interpretation:** There is not enough evidence at the 0.01 significance level to conclude that the dosage affects the time it takes rats to fall asleep.