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 $\mathrm{s}=18.2$ based on earlier studies? Suppose the doctor would be content with $90 \%$ confidence. How does the decrease in confidence affect the sample size required? Click the icon to view a partial table of critical values. A $99 \%$ confidence level requires $\square$ subjects. (Round up to the nearest subject) Partial Critical Value Table \begin{tabular}{ccc} \begin{tabular}{c} Level of Confidence, \\ $(1-\boldsymbol{\alpha}) \cdot \mathbf{1 0 0} \%$ \end{tabular} & Area in Each Tail, $\frac{\boldsymbol{\alpha}}{\mathbf{2}}$ & Critical Value, $\mathbf{z}_{\boldsymbol{\alpha} / \mathbf{2}}$ \\ \hline $90 \%$ & 0.05 & 1.645 \\ \hline $95 \%$ & 0.025 & 1.96 \\ \hline $99 \%$ & 0.005 & 2.575 \\ \hline \end{tabular}
Solution
Step 1 ：The problem is asking for the sample size needed to estimate the mean HDL cholesterol within 2 points with 99% confidence. This is a problem of sample size determination for estimating a population mean. The formula for this is: \(n = (Z*σ/E)^2\), where: n is the sample size, Z is the z-score for the desired confidence level, σ is the standard deviation of the population, and E is the desired margin of error.
Step 2 ：From the problem, we know that σ = 18.2, E = 2, and from the table, we know that the z-score for a 99% confidence level is 2.575. We can substitute these values into the formula to find the sample size.
Step 3 ：Substituting the values into the formula, we get \(n = (2.575*18.2/2)^2\).
Step 4 ：Calculating the above expression, we find that the sample size needed is 550.
Step 5 ：Final Answer: The doctor needs to sample \(\boxed{550}\) subjects to estimate the mean HDL cholesterol within 2 points with 99% confidence.
