1. **State the problem:** We want to find the lower limit of a 95% confidence interval for the proportion $p$ of Jamaican nationals who select jogging as a recreational activity. 2. **Given data:** Sample size $n=100$, number selecting jogging $x=30$, sample proportion $\hat{p} = \frac{x}{n} = \frac{30}{100} = 0.3$. 3. **Formula for confidence interval for a proportion:** $$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $z_{\alpha/2}$ is the critical value from the standard normal distribution for a 95% confidence level. 4. **Find $z_{\alpha/2}$:** For 95% confidence, $\alpha=0.05$, so $z_{0.025} = 1.96$. 5. **Calculate the standard error (SE):** $$SE = \sqrt{\frac{0.3 \times (1-0.3)}{100}} = \sqrt{\frac{0.3 \times 0.7}{100}} = \sqrt{0.0021} \approx 0.0458$$ 6. **Calculate the margin of error (ME):** $$ME = 1.96 \times 0.0458 \approx 0.0897$$ 7. **Calculate the lower limit of the confidence interval:** $$0.3 - 0.0897 = 0.2103$$ **Final answer:** The lower limit of the 95% confidence interval for $p$ is approximately $0.210$.