To determine the sample size needed to estimate a population proportion with a specified confidence level and margin of error, you can use the formula for sample size in proportion estimation:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
Where:
- \( n \) = required sample size
- \( Z \) = Z-value corresponding to the desired confidence level
- \( p \) = estimated population proportion (since we have no reasonable estimate, we use \( p = 0.5 \) for maximum variability)
- \( E \) = margin of error
Given:
- Confidence level = 90%
- Margin of error (E) = 0.02 (2%)
Step 1: Find the Z-value
For a 90% confidence level, the Z-value can be found in a Z-table or using statistical software. The critical value for a two-tailed test at 90% confidence is:
\[ Z \approx 1.645 \]
Step 2: Plug values into the formula
We will use \( p = 0.5 \) to ensure the sample size is sufficient (since it provides the largest sample size estimate).
Now substitute the values into the formula:
\[ n = \left( \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.02)^2} \right) \]
Calculating each component step-by-step:
-
Calculate \( Z^2 \): \[ Z^2 = (1.645)^2 \approx 2.706 \]
-
Calculate \( p \cdot (1 - p) \): \[ p \cdot (1 - p) = 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25 \]
-
Calculate \( E^2 \): \[ E^2 = (0.02)^2 = 0.0004 \]
Now substitute these values into the formula: \[ n = \frac{2.706 \cdot 0.25}{0.0004} \]
Calculating that: \[ n = \frac{0.6765}{0.0004} = 1691.25 \]
Step 3: Round up to the next whole number
Since sample size must be a whole number, we round up:
\[ n = 1692 \]
Conclusion
The required sample size to estimate the population proportion with 90% confidence and a margin of error of 2% is:
\[ \boxed{1692} \]
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