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. **Identify values:** - $p = 0.57$ - $ME = 0.04$ - For a 95% confidence level, $z = 1.96$ - Current sample size $n_0 = 265265$ 4. **Rearrange formula to solve for $n$:** $$ n = \frac{z^2 \times p(1-p)}{ME^2} $$ 5. **Calculate $n$:** $$ n = \frac{1.96^2 \times 0.57 \times (1-0.57)}{0.04^2} $$ 6. **Calculate numerator:** $$ 1.96^2 = 3.8416 $$ $$ 0.57 \times 0.43 = 0.2451 $$ $$ 3.8416 \times 0.2451 = 0.9413 $$ 7. **Calculate denominator:** $$ 0.04^2 = 0.0016 $$ 8. **Calculate $n$:** $$ n = \frac{0.9413}{0.0016} = 588.3125 $$ 9. **Determine how many more students to survey:** Current surveyed: 265265 Needed total: 589 (rounded up from 588.3125) Since the current sample size is much larger than needed, no additional students are required to achieve the margin of error of 4%.