1. **State the problem:** We want to find a 90% confidence interval for the percentage of all college seniors who would have chosen to attend a different college based on a sample where 34 out of 100 seniors said yes. 2. **Formula used:** The confidence interval for a population proportion $p$ is given by $$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p}$ is the sample proportion, $z^*$ is the critical value for the confidence level, and $n$ is the sample size. 3. **Identify values:** - Sample proportion $\hat{p} = \frac{34}{100} = 0.34$ - Sample size $n = 100$ - For 90% confidence, $z^* = 1.645$ (from z-tables) 4. **Calculate standard error:** $$SE = \sqrt{\frac{0.34(1-0.34)}{100}} = \sqrt{\frac{0.34 \times 0.66}{100}} = \sqrt{0.002244} \approx 0.0474$$ 5. **Calculate margin of error:** $$ME = 1.645 \times 0.0474 \approx 0.0780$$ 6. **Calculate confidence interval:** $$0.34 \pm 0.0780$$ which gives $$0.34 - 0.0780 = 0.262$$ $$0.34 + 0.0780 = 0.418$$ 7. **Convert to percentages:** $$26.2\% \text{ to } 41.8\%$$ **Final answer:** The 90% confidence interval is 26.2% to 41.8%, which corresponds to option C.