1. **Problem statement:** Construct a 95% confidence interval for the percentage of university students who own a laptop based on a sample of 400 students, where 340 own a laptop. 2. **Formula and explanation:** The confidence interval for a population proportion $p$ is given by: $$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level. 3. **Calculate sample proportion:** $$\hat{p} = \frac{340}{400} = 0.85$$ 4. **Find critical value:** For 95% confidence, $z_{\alpha/2} = 1.96$. 5. **Calculate standard error:** $$SE = \sqrt{\frac{0.85 \times (1 - 0.85)}{400}} = \sqrt{\frac{0.85 \times 0.15}{400}} = \sqrt{\frac{0.1275}{400}} = \sqrt{0.00031875} \approx 0.01785$$ 6. **Calculate margin of error:** $$ME = 1.96 \times 0.01785 \approx 0.035$$ 7. **Construct confidence interval:** $$0.85 \pm 0.035 = (0.815, 0.885)$$ 8. **Interpretation:** We are 95% confident that the true percentage of university students who own a laptop is between 81.5% and 88.5%.