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 data. 2. **Given data:** - Sample size $n = 31$ - Mean difference $\bar{d} = -0.6$ - Standard deviation of differences $s_d = 1.1$ 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 two-tailed test with $n-1$ degrees of freedom. 4. **Find the critical t-value:** For 90% confidence, $\alpha = 0.10$, so $\alpha/2 = 0.05$. Degrees of freedom $df = 31 - 1 = 30$. From t-tables or calculator, $t_{0.05,30} \approx 1.697$ 5. **Calculate the margin of error:** $$ME = 1.697 \times \frac{1.1}{\sqrt{31}} = 1.697 \times \frac{1.1}{5.5678} \approx 1.697 \times 0.1976 = 0.3355$$ 6. **Calculate the confidence interval:** $$\text{Lower bound} = -0.6 - 0.3355 = -0.9355$$ $$\text{Upper bound} = -0.6 + 0.3355 = -0.2645$$ 7. **Interpretation:** The 90% confidence interval for the mean difference is $(-0.9355, -0.2645)$. 8. **Conclusion about the belief:** Since the entire interval is below 0, it suggests that northern birds have significantly shorter wings than southern birds at the 90% confidence level. **Answer to (a):** $(-0.9355, -0.2645)$ **Answer to (b):** Yes, since the interval is completely below 0.