1. **State the problem:** We want to test if the mean appreciation rate of houses in Sun Beach ($\mu_1$) is higher than that in North Arden ($\mu_2$) at a significance level $\alpha=0.01$. This is a one-tailed test with hypotheses: $$H_0: \mu_1 \leq \mu_2$$ $$H_a: \mu_1 > \mu_2$$ 2. **Given data:** Sun Beach appreciation rates: $12.9, 10.9, 11, 8.6, 12.1, 8.8, 8.6, 9.5, 12.5$ North Arden appreciation rates: $10.5, 9.3, 12, 8.4, 7.9, 9.9, 10, 4.6, 10.6$ Sample sizes: $n_1 = n_2 = 9$ 3. **Calculate sample means:** $$\bar{x}_1 = \frac{12.9 + 10.9 + 11 + 8.6 + 12.1 + 8.8 + 8.6 + 9.5 + 12.5}{9} = \frac{94.9}{9} = 10.544$$ $$\bar{x}_2 = \frac{10.5 + 9.3 + 12 + 8.4 + 7.9 + 9.9 + 10 + 4.6 + 10.6}{9} = \frac{82.2}{9} = 9.133$$ 4. **Calculate sample variances:** For Sun Beach: $$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1} = \frac{(12.9-10.544)^2 + \cdots + (12.5-10.544)^2}{8} = 2.799$$ For North Arden: $$s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1} = \frac{(10.5-9.133)^2 + \cdots + (10.6-9.133)^2}{8} = 4.066$$ 5. **Calculate pooled variance:** $$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{8 \times 2.799 + 8 \times 4.066}{16} = \frac{22.392 + 32.528}{16} = 3.806$$ 6. **Calculate standard error:** $$SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{3.806 \times \left(\frac{1}{9} + \frac{1}{9}\right)} = \sqrt{3.806 \times \frac{2}{9}} = \sqrt{0.8457} = 0.9197$$ 7. **Calculate t-statistic:** $$t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{10.544 - 9.133}{0.9197} = \frac{1.411}{0.9197} = 1.534$$ 8. **Degrees of freedom:** $$df = n_1 + n_2 - 2 = 9 + 9 - 2 = 16$$ 9. **Find p-value:** Using a t-distribution table or calculator for $t=1.534$ with $df=16$ in a one-tailed test: $$p \approx 0.074$$ 10. **Conclusion:** Since $p = 0.074 > \alpha = 0.01$, we fail to reject the null hypothesis. There is not enough evidence at the 0.01 significance level to conclude that the mean appreciation rate in Sun Beach is higher than in North Arden.