To determine the required sample size \( n \) to estimate a population proportion with a specified confidence level and margin of error, we can use the formula for the sample size for proportions when no prior estimate is available:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
where:
- \( Z \) is the critical value from the Z-distribution corresponding to the desired confidence level,
- \( p \) is the estimated population proportion (when no prior estimate is available, we can use \( p = 0.5 \) for maximum variability),
- \( E \) is the margin of error.
Step 1: Determine the values
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Critical Value \( Z \) for 99.9% Confidence Level:
- The confidence level is 99.9%, which means the significance level \( \alpha = 1 - 0.999 = 0.001 \).
- We need \( Z \) such that \( P(Z < z) = 0.999 \).
- Looking up in the Z-table or using a calculator, we find that \( Z \approx 3.291 \) (this might slightly vary based on the source, but it should be around this value).
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Estimated Population Proportion \( p \):
- Since we have no prior estimate, we use \( p = 0.5 \) (which maximizes the overall sample size).
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Margin of Error \( E \):
- Given in the problem, \( E = 0.01 \) (which is 1%).
Step 2: Substitute into the formula
We plug in the values into the formula:
\[ n = \frac{(3.291)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.01)^2} \]
Step 3: Calculate
Now we perform the calculations:
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Calculate \( Z^2 \): \[ Z^2 = (3.291)^2 \approx 10.814581 \]
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Calculate \( p(1 - p) \): \[ p(1 - p) = 0.5 \times 0.5 = 0.25 \]
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Calculate \( E^2 \): \[ E^2 = (0.01)^2 = 0.0001 \]
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Substitute into the formula: \[ n = \frac{10.814581 \cdot 0.25}{0.0001} = \frac{2.70364525}{0.0001} = 27036.4525 \]
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Round up (sample size must be a whole number): \[ n \approx 27037 \]
Final Answer
Thus, the required sample size is:
\[ \boxed{27037} \]
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