1. **State the problem:** We have a sample of 400 adults, with 180 believing their student loan debt is unmanageable. We want to find the degree of confidence for a confidence interval given as $0.45 \pm 0.04$. 2. **Identify the sample proportion:** The sample proportion $\hat{p}$ is calculated as: $$\hat{p} = \frac{180}{400} = 0.45$$ 3. **Recall the confidence interval formula for a proportion:** $$\hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $z$ is the z-score corresponding to the confidence level, and $n=400$ is the sample size. 4. **Given the margin of error (ME):** $$ME = 0.04 = z \times \sqrt{\frac{0.45 \times 0.55}{400}}$$ 5. **Calculate the standard error (SE):** $$SE = \sqrt{\frac{0.45 \times 0.55}{400}} = \sqrt{\frac{0.2475}{400}} = \sqrt{0.00061875} \approx 0.02488$$ 6. **Solve for $z$:** $$z = \frac{ME}{SE} = \frac{0.04}{0.02488} \approx 1.607$$ 7. **Find the confidence level corresponding to $z=1.607$:** The confidence level is $2 \times P(Z \leq 1.607) - 1$. Using standard normal distribution tables or a calculator: $$P(Z \leq 1.607) \approx 0.946$$ So, $$\text{Confidence level} = 2 \times 0.946 - 1 = 0.892 = 89.2\%$$ **Final answer:** The degree of confidence is approximately **89.2%**.