To determine the needed sample size to obtain a 90% confidence interval estimate with the same level of margin of error (±4 percentage points), we can use the formula for the sample size for estimating proportions:
\[ n = \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \]
Where:
- \( n \) is the sample size,
- \( Z \) is the z-score corresponding to the desired confidence level,
- \( p \) is the estimated proportion (from the poll),
- \( E \) is the margin of error.
Step 1: Identify the values:
- For a 90% confidence level, the z-score \( Z \) is approximately \( 1.645 \) (this can be found from standard normal distribution tables).
- The estimated proportion \( p \) is 0.50 (50%).
- The margin of error \( E \) is 0.04 (4 percentage points).
Step 2: Plug the values into the formula:
\[ n = \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.04)^2} \]
Calculating the components step by step:
\[ Z^2 = (1.645)^2 \approx 2.706 \]
\[ p \cdot (1 - p) = 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25 \]
Now substituting those values back into our formula:
\[ n = \frac{2.706 \cdot 0.25}{0.0016} \]
Calculating the numerator:
\[ 2.706 \cdot 0.25 = 0.6765 \]
Now calculate:
\[ n = \frac{0.6765}{0.0016} \approx 422.8125 \]
Since the sample size must be a whole number, we round up to ensure the sample is sufficiently large:
\[ n \approx 423 \]
Conclusion: The needed sample size to obtain a 90% confidence interval estimate of the percentage of the targeted young adults who will purchase the new game with a margin of error of ±4 percentage points is 423.
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