To determine the necessary sample size (n) for estimating a population proportion with a given margin of error (E) and confidence level, you can use the formula for the sample size for proportions:
\[ n = \left( \frac{Z^2 \times p(1 - p)}{E^2} \right) \]
Where:
- \( n \) = required sample size
- \( Z \) = Z-value corresponding to the desired confidence level
- \( p \) = estimated population proportion
- \( E \) = margin of error
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Identify the values:
- For a confidence level of 99%, the Z-value is approximately 2.576 (you can find this value in Z-tables or standard normal distribution tables).
- The sample proportion \( p = 0.35 \).
- The margin of error \( E = 0.05 \).
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Calculate \( p(1 - p) \): \[ p(1 - p) = 0.35 \times (1 - 0.35) = 0.35 \times 0.65 = 0.2275 \]
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Substitute the values into the sample size formula: \[ n = \left( \frac{(2.576)^2 \times 0.2275}{(0.05)^2} \right) \]
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Calculate \( Z^2 \): \[ (2.576)^2 \approx 6.635776 \]
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Calculate \( E^2 \): \[ (0.05)^2 = 0.0025 \]
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Combine everything: \[ n = \left( \frac{6.635776 \times 0.2275}{0.0025} \right) \]
\[ n = \left( \frac{1.51041436}{0.0025} \right) \]
\[ n \approx 604.165744 \]
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Round up to ensure sufficient sample size: Since you can't have a fraction of a sample, round up to the next whole number: \[ n \approx 605 \]
Thus, a sample size of 605 is needed to achieve a 5% margin of error with 99% confidence level when the sample proportion is 0.35.