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 $s=15.7$ based on earlier studies? Suppose the doctor would be content with $90 \%$ confidence. How does the decrease in confidence affect the sample size required?
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A 99\% confidence level requires 409 subjects. (Round up to the nearest subject.)
A $90 \%$ confidence level requires subjects. (Round up to the nearest subject.)

Step 1 ：The doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. The doctor wants to estimate the mean within 2 points. The standard deviation based on earlier studies is 15.7.

Step 2 ：The sample size needed to estimate a population mean with a certain level of confidence can be calculated using the formula: \(n = (Z*σ/E)^2\), where: n is the sample size, Z is the z-score (which depends on the desired level of confidence), σ is the standard deviation of the population, and E is the desired margin of error.

Step 3 ：In this case, we know that σ = 15.7, E = 2, and we want to find n for two different confidence levels: 99% and 90%.

Step 4 ：The z-scores for these confidence levels can be found in a standard z-table or calculated using a statistical function. For a 99% confidence level, the z-score is approximately 2.576. For a 90% confidence level, the z-score is approximately 1.645.

Step 5 ：We can plug these values into the formula to find the required sample sizes. For a 99% confidence level, the sample size is \(n = (2.576*15.7/2)^2\), which rounds up to 409. For a 90% confidence level, the sample size is \(n = (1.645*15.7/2)^2\), which rounds up to 167.

Step 6 ：Final Answer: The doctor needs \(\boxed{409}\) subjects for a 99% confidence level and \(\boxed{167}\) subjects for a 90% confidence level.