1. **State the problem:** We want to construct a 99% confidence interval for the population proportion of mothers who give birth to low birth weight babies, based on a sample where 12.8% of 86 mothers had low birth weight babies. 2. **Identify the sample proportion and sample size:** - Sample proportion $\hat{p} = 0.128$ - Sample size $n = 86$ 3. **Formula for confidence interval for a population proportion:** $$\hat{p} \pm z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level. 4. **Find the critical value for 99% confidence:** For 99% confidence, $\alpha = 0.01$, so $\alpha/2 = 0.005$. The critical value is $$z_{0.005} = 2.576$$ 5. **Calculate the standard error:** $$SE = \sqrt{\frac{0.128 \times (1 - 0.128)}{86}} = \sqrt{\frac{0.128 \times 0.872}{86}} = \sqrt{\frac{0.111616}{86}} = \sqrt{0.001298} \approx 0.0360$$ 6. **Calculate the margin of error:** $$ME = 2.576 \times 0.0360 \approx 0.0927$$ 7. **Construct the confidence interval:** $$0.128 \pm 0.0927 = (0.128 - 0.0927, 0.128 + 0.0927) = (0.0353, 0.2207)$$ 8. **Interpretation:** We are 99% confident that the true population proportion of mothers with low birth weight babies is between 3.53% and 22.07%.