1. **State the problem:** We want to find a 99% confidence interval (CI) for the population proportion $p$ based on the sample proportion $\hat{p}$. 2. **Given data:** Sample proportion $\hat{p} = \frac{157}{n}$ and $n(1 - \hat{p}) = 157$. This implies $n\hat{p} = 157$ and $n(1-\hat{p}) = 157$, so $n = 157 + 157 = 314$ and $\hat{p} = \frac{157}{314} = 0.5$. 3. **Formula for one-sample z confidence interval for proportion:** $$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $z^*$ is the critical z-value for the confidence level. 4. **Find $z^*$ for 99% confidence:** The critical value $z^*$ for 99% confidence is approximately 2.576. 5. **Calculate the standard error (SE):** $$SE = \sqrt{\frac{0.5 \times (1 - 0.5)}{314}} = \sqrt{\frac{0.25}{314}} = \sqrt{0.000796} \approx 0.0282$$ 6. **Calculate margin of error (ME):** $$ME = 2.576 \times 0.0282 \approx 0.0726$$ 7. **Calculate confidence interval:** $$0.5 \pm 0.0726 = (0.5 - 0.0726, 0.5 + 0.0726) = (0.4274, 0.5726)$$ 8. **Interpretation:** We are 99% confident that the true proportion of all U.S. residents who think life ever existed on Mars is between approximately 0.427 and 0.573. 9. **Check conditions:** Since $n\hat{p} = 157 \geq 10$ and $n(1-\hat{p}) = 157 \geq 10$, the normal approximation is appropriate.