1. **State the problem:** We want to find a 90% confidence interval for the mean difference in wing lengths between birds from northern and southern regions, given sample statistics. 2. **Given data:** - Sample size $n = 30$ - Mean difference $\bar{d} = -2.0$ - Standard deviation of differences $s_d = 1.5$ 3. **Formula for confidence interval for mean difference:** $$\bar{d} \pm t_{\alpha/2, n-1} \times \frac{s_d}{\sqrt{n}}$$ where $t_{\alpha/2, n-1}$ is the critical t-value for a 90% confidence level and $n-1=29$ degrees of freedom. 4. **Find the critical t-value:** For 90% confidence, $\alpha = 0.10$, so $\alpha/2 = 0.05$. Using a t-table or calculator, $$t_{0.05, 29} \approx 1.699$$ 5. **Calculate the margin of error:** $$ME = 1.699 \times \frac{1.5}{\sqrt{30}} = 1.699 \times 0.2739 \approx 0.4655$$ 6. **Calculate the confidence interval:** $$\text{Lower bound} = -2.0 - 0.4655 = -2.4655$$ $$\text{Upper bound} = -2.0 + 0.4655 = -1.5345$$ 7. **Interpretation:** The 90% confidence interval for the mean difference is $$(-2.4655, -1.5345)$$. 8. **Answer to part (b):** Since the entire interval is below 0, there is enough evidence to support the belief that northern birds have shorter wing lengths than southern birds. **Final answers:** (a) $(-2.4655, -1.5345)$ (b) Yes, since the interval is completely below 0.