To determine the required sample size for an 80% confidence interval with a specified margin of error, we can use the formula for the sample size \( n \) needed to achieve a margin of error \( E \):
\[ n = \left( \frac{Z \cdot \sqrt{p(1-p)}}{E} \right)^2 \]
Where:
- \( Z \) is the z-score corresponding to the desired confidence level,
- \( p \) is the estimated proportion,
- \( E \) is the margin of error.
Step 1: Identify the parameters
- The estimated proportion \( p = 0.41 \).
- The margin of error \( E = 0.04 \).
Step 2: Find the z-score for an 80% confidence level
For an 80% confidence level, the z-score can be found using standard z-tables or a calculator. The critical value \( Z \) for 80% confidence is approximately:
\[ Z \approx 1.28 \]
Step 3: Plug the values into the sample size formula
Now we substitute the values we have into the formula:
\[ n = \left( \frac{1.28 \cdot \sqrt{0.41(1 - 0.41)}}{0.04} \right)^2 \]
Calculate \( p(1-p) \):
\[ p(1 - p) = 0.41 \cdot (1 - 0.41) = 0.41 \cdot 0.59 = 0.2419 \]
Now compute \( \sqrt{0.2419} \):
\[ \sqrt{0.2419} \approx 0.4918 \]
Next, calculate the numerator:
\[ 1.28 \cdot 0.4918 \approx 0.6291 \]
Now divide it by the margin of error \( E = 0.04 \):
\[ \frac{0.6291}{0.04} \approx 15.7275 \]
Finally, square this result to find \( n \):
\[ n = (15.7275)^2 \approx 247.24 \]
Step 4: Round up to the nearest whole number
Since sample sizes must be whole numbers, we round up to the next whole number:
\[ n \approx 248 \]
Conclusion
The required sample size needed to obtain an 80% confidence interval with a margin of error of \( 0.04 \) is 248.
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