6. Suppose that we are interested in estimating the mean age u, of all people in the civil labor force in the United States. How large a sample should we use to estimate u to within 0.33 year at a 99% confidence level? Use o = 16.25. 5750/6125/2 E 0.33 n=16078.04788 I6078 2=2.575 モ-0.33 9990 C.I. J= 16.25
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![### Estimating the Mean Age in the Civil Labor Force
**Problem Statement:**
Suppose that we are interested in estimating the mean age \(\mu\) of all people in the civil labor force in the United States. How large a sample should we use to estimate \(\mu\) to within \(0.33\) year at a \(99\%\) confidence level? Use \(\sigma = 16.25\).
**Solution Approach:**
To determine the required sample size (\(n\)), we use the following formula for the margin of error (\(E\)) in estimating a population mean:
\[ E = Z_{c} \frac{\sigma}{\sqrt{n}} \]
Where:
- \(E\) = Margin of error = \(0.33\)
- \(Z_{c}\) = Z-score for the confidence level
- \(\sigma\) = Population standard deviation = \(16.25\)
- \(n\) = Required sample size
### Detailed Steps:
1. **Identify the Z-score for a 99% Confidence Level:**
The Z-score (Z_{c}) corresponding to a 99% confidence level is approximately \(2.575\).
2. **Rearrange the Equation to Solve for \(n\):**
\[ n = \left( \frac{Z_{c} \sigma}{E} \right)^2 \]
3. **Substitute the Known Values into the Equation:**
\[ n = \left( \frac{2.575 \times 16.25}{0.33} \right)^2 = \left( \frac{41.59375}{0.33} \right)^2 = (126.04198)^2 \]
\[ \therefore n = 16078.04788 \]
4. **Round to the Nearest Whole Number:**
Since the sample size must be a whole number,
\[ n \approx 16078 \]
Thus, a sample size of \(16078\) people is needed to estimate the mean age \(\mu\) to within \(0.33\) year at a \(99\%\) confidence level.
### Explanation of Graphs or Diagrams:
This transcription does not include any graphical or diagrammatic content. The steps involve calculations and algebraic manipulation to derive the required sample size.
### Summary:
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