To determine the required sample size for estimating a population proportion \( p \) with a specified level of confidence and margin of error, you can use the formula for the sample size \( n \):
\[ n = \left( \frac{Z^2 \cdot p^* \cdot (1 - p^*)}{E^2} \right) \]
Where:
- \( Z \) is the z-score corresponding to the desired confidence level.
- \( p^* \) is the estimated population proportion.
- \( E \) is the desired margin of error.
Step 1: Find the Z-score for a 98% confidence level
For a 98% confidence level, the critical value \( Z \) can be found using a Z-table or calculator. The Z-score for 98% confidence (two-tailed) is approximately:
\[ Z \approx 2.326 \]
Step 2: Substitute the known values into the formula
Given:
- \( p^* = 0.59 \)
- \( E = 0.05 \)
Step 3: Calculate \( n \)
Substituting the values into the formula:
\[ n = \frac{(2.326)^2 \cdot 0.59 \cdot (1 - 0.59)}{(0.05)^2} \]
Calculating each part:
-
Calculate \( Z^2 \): \[ (2.326)^2 \approx 5.419 \]
-
Calculate \( p^(1 - p^) \): \[ 0.59 \cdot (1 - 0.59) = 0.59 \cdot 0.41 \approx 0.2419 \]
-
Calculate \( E^2 \): \[ (0.05)^2 = 0.0025 \]
Putting these into the sample size formula:
\[ n = \frac{5.419 \cdot 0.2419}{0.0025} \]
Now calculate the numerator:
\[ 5.419 \cdot 0.2419 \approx 1.3106 \]
Now calculate \( n \):
\[ n = \frac{1.3106}{0.0025} \approx 524.24 \]
Since you cannot have a fraction of a sample, round up to the next whole number:
\[ n \approx 525 \]
Final Answer:
\[ n = 525 \]
Thus, the required sample size is 525.