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Section 9.1 -
Question 14, 9.2.39
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A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 2 points with $99 \%$ confidence assuming $s=19.3$ based on earlier studies? Suppose the doctor would be content with $95 \%$ confidence. How does the decrease in confidence affect the sample size required?
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A $99 \%$ confidence level requires $\square$ subjects. (Round up to the nearest subject.)
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Step 1 ：To solve this problem, we need to use the formula for sample size in estimating a population mean: \(n = (Z_{\alpha/2} * \sigma / E)^2\)

Step 2 ：Where: \(n\) is the sample size, \(Z_{\alpha/2}\) is the z-score corresponding to the desired confidence level, \(\sigma\) is the standard deviation of the population, and \(E\) is the desired margin of error.

Step 3 ：For a 99% confidence level, the z-score (\(Z_{\alpha/2}\)) is approximately 2.576. The standard deviation (\(\sigma\)) is given as 19.3, and the desired margin of error (\(E\)) is 2.

Step 4 ：Substituting these values into the formula, we get: \(n = (2.576 * 19.3 / 2)^2\)

Step 5 ：Calculating the above expression, we get: \(n = (49.8688)^2\)

Step 6 ：Further simplifying, we get: \(n = 2486.88\)

Step 7 ：Since we can't have a fraction of a subject, we round up to the nearest whole number. So, the doctor needs a sample size of approximately 2487 subjects for a 99% confidence level.

Step 8 ：So, the final answer is \(\boxed{2487}\)