Problem

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}

Answer

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Answer

Final Answer: The doctor needs to sample \(\boxed{550}\) subjects to estimate the mean HDL cholesterol within 2 points with 99% confidence.

Steps

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.

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