1. **Problem Statement:** A doctor wants to estimate the mean HDL cholesterol of 20- to 29-year-old females within 2 points margin of error. We need to find the required sample size for 99% confidence and then for 90% confidence, given the standard deviation $s=13.8$. 2. **Formula for sample size $n$ when estimating a mean:** $$n = \left(\frac{z^* \cdot s}{E}\right)^2$$ where: - $z^*$ is the critical z-value for the confidence level, - $s$ is the standard deviation, - $E$ is the margin of error. 3. **Find $z^*$ values:** - For 99% confidence, $z^* \approx 2.576$ (from z-tables). - For 90% confidence, $z^* \approx 1.645$. 4. **Calculate sample size for 99% confidence:** $$n = \left(\frac{2.576 \times 13.8}{2}\right)^2 = \left(\frac{35.5728}{2}\right)^2 = (17.7864)^2 = 316.32$$ Round up to nearest whole number: $n = 317$ subjects. 5. **Calculate sample size for 90% confidence:** $$n = \left(\frac{1.645 \times 13.8}{2}\right)^2 = \left(\frac{22.701}{2}\right)^2 = (11.3505)^2 = 128.82$$ Round up: $n = 129$ subjects. 6. **Interpretation:** Decreasing the confidence level from 99% to 90% reduces the required sample size from 317 to 129. This is because a lower confidence level requires a smaller critical value $z^*$, thus reducing the sample size needed to achieve the same margin of error. **Final answers:** - At 99% confidence, $n=317$ subjects. - At 90% confidence, $n=129$ subjects. This shows the trade-off between confidence level and sample size in estimating a population mean.