1. **State the problem:** We need to calculate the lower limit of a 90% confidence interval for the mean systolic blood pressure after the community health program. 2. **Identify the data:** - Sample mean $\bar{x} = 132$ mmHg - Population mean from last study (initial mean) = 145 mmHg (not directly used for CI calculation) - Sample standard deviation $s = 9$ mmHg - Sample size $n = 28$ - Confidence level = 90% 3. **Find the t-critical value:** Degrees of freedom $df = n - 1 = 27$. For a 90% confidence level, the significance level $\alpha = 0.10$, so each tail has $0.05$. Using a t-distribution table or calculator, $t_{0.05,27} \approx 1.703$. 4. **Calculate the standard error (SE):** $$SE = \frac{s}{\sqrt{n}} = \frac{9}{\sqrt{28}} \approx \frac{9}{5.2915} \approx 1.701$$ 5. **Calculate the margin of error (ME):** $$ME = t_{\alpha/2, df} \times SE = 1.703 \times 1.701 \approx 2.895$$ 6. **Calculate the lower limit of the confidence interval:** $$\text{Lower limit} = \bar{x} - ME = 132 - 2.895 = 129.105$$ **Final answer:** The lower limit of the 90% confidence interval for the mean systolic blood pressure is approximately $129.11$ mmHg.