1. **State the problem:** We want to find how many more students need to be surveyed to achieve a margin of error (ME) of 4% when the proportion of students who traveled abroad is 57%. 2. **Formula for margin of error in proportion:** $$ ME = z \times \sqrt{\frac{p(1-p)}{n}} $$ where $p$ is the sample proportion, $n$ is the sample size, and $z$ is the z-score for the confidence level. 3. **Given:** - $p = 0.57$ - $ME = 0.04$ - Initial sample size $n_0 = 265265$ 4. **Find the z-score for 95% confidence (common choice):** $z = 1.96$ 5. **Calculate the required sample size $n$ to achieve $ME=0.04$:** $$ 0.04 = 1.96 \times \sqrt{\frac{0.57 \times (1-0.57)}{n}} $$ 6. **Isolate $n$:** $$ \frac{0.04}{1.96} = \sqrt{\frac{0.57 \times 0.43}{n}} $$ $$ \left(\frac{0.04}{1.96}\right)^2 = \frac{0.57 \times 0.43}{n} $$ $$ n = \frac{0.57 \times 0.43}{\left(\frac{0.04}{1.96}\right)^2} $$ 7. **Calculate numerator:** $$ 0.57 \times 0.43 = 0.2451 $$ 8. **Calculate denominator:** $$ \left(\frac{0.04}{1.96}\right)^2 = \left(0.020408\right)^2 = 0.0004165 $$ 9. **Calculate $n$:** $$ n = \frac{0.2451}{0.0004165} \approx 588.5 $$ 10. **Determine how many more students to survey:** $$ \text{Additional students} = \lceil 588.5 \rceil - 265265 = 589 - 265265 = -264676 $$ Since the required sample size to achieve a 4% margin of error is much smaller than the current sample size, no additional students are needed. **Final answer:** You do not need to survey more students because your current sample size already achieves a margin of error less than 4%.