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): $n_1=11$, data: 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): $n_2=9$, data: 13.5, 9.8, 13.0, 9.8, 12.0, 10.5, 12.5, 11.5, 7.4 Calculate sample means: $$\bar{x}_1 = \frac{\sum x_1}{n_1} = \frac{9.7+7.9+9.8+13.0+11.2+9.8+10.2+12.3+12.1+9.7+13.2}{11} = \frac{118.7}{11} = 10.79$$ $$\bar{x}_2 = \frac{\sum x_2}{n_2} = \frac{13.5+9.8+13.0+9.8+12.0+10.5+12.5+11.5+7.4}{9} = \frac{100.0}{9} = 11.11$$ Calculate sample variances: $$s_1^2 = \frac{\sum (x_1 - \bar{x}_1)^2}{n_1 - 1} = 2.56$$ $$s_2^2 = \frac{\sum (x_2 - \bar{x}_2)^2}{n_2 - 1} = 3.30$$ Calculate pooled variance: $$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{10 \times 2.56 + 8 \times 3.30}{18} = \frac{25.6 + 26.4}{18} = \frac{52}{18} = 2.89$$ Calculate standard error: $$SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{2.89 \left(\frac{1}{11} + \frac{1}{9}\right)} = \sqrt{2.89 \times 0.202} = \sqrt{0.584} = 0.764$$ Calculate t-statistic: $$t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{10.79 - 11.11}{0.764} = \frac{-0.32}{0.764} = -0.42$$ Degrees of freedom: $$df = n_1 + n_2 - 2 = 11 + 9 - 2 = 18$$ 5. **Decision rule:** At $\alpha=0.01$ and $df=18$, two-tailed critical t-value $t_{crit} \approx \pm 2.878$. Reject $H_0$ if $|t| > 2.878$. 6. **Conclusion:** Since $|t| = 0.42 < 2.878$, we fail to reject the null hypothesis. **Interpretation:** There is not enough evidence at the 0.01 significance level to conclude that the dosage difference affects the time it takes rats to fall asleep.