To calculate the required sample size for a given margin of error and confidence level, we can use the formula for the sample size \( n \) when estimating a proportion:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
where:
- \( Z \) = Z-score corresponding to the desired confidence level
- \( p \) = sample proportion
- \( E \) = margin of error
Step 1: Determine the Z-score
For a 99% confidence level, the Z-score can be found using a Z-table or calculator. The Z-score for a 99% confidence level is approximately:
\[ Z \approx 2.576 \]
Step 2: Plug in the values
Given:
- \( p = 0.35 \)
- \( E = 0.05 \)
We can now substitute these values into the formula:
\[ n = \left( \frac{(2.576)^2 \cdot 0.35 \cdot (1 - 0.35)}{(0.05)^2} \right) \]
Calculating \( 1 - p \):
\[ 1 - p = 1 - 0.35 = 0.65 \]
Calculating \( n \):
\[ n = \left(\frac{(2.576)^2 \cdot 0.35 \cdot 0.65}{(0.05)^2}\right) \]
Now calculating each part:
- \( (2.576)^2 \approx 6.635776 \)
- \( 0.35 \cdot 0.65 = 0.2275 \)
- \( (0.05)^2 = 0.0025 \)
Substituting these values back into the formula:
\[ n = \left( \frac{6.635776 \cdot 0.2275}{0.0025} \right) \]
Calculating the numerator:
\[ 6.635776 \cdot 0.2275 \approx 1.507299 \]
Now divide by the denominator:
\[ n = \frac{1.507299}{0.0025} \approx 602.9196 \]
Step 3: Round up
Since sample size must be a whole number, we round up to the next whole number:
\[ n \approx 603 \]
Conclusion
The required sample size needed to achieve a 5% margin of error and 99% confidence level is 603.