To determine the minimum sample size needed for using normal approximations in estimating a population proportion, we use the formula related to the sample size \( n \) for a proportion:
\[ n = \frac{(Z^2 \cdot p \cdot (1-p))}{E^2} \]
Where:
- \( Z \) is the Z-value corresponding to the desired confidence level,
- \( p \) is the estimated proportion (probability of success),
- \( E \) is the margin of error.
However, to use the normal approximation reliably, certain conditions should be met, which typically include:
- \( np \geq 10 \)
- \( n(1 - p) \geq 10 \)
Given \( p = 0.25 \), we can calculate:
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For \( np \geq 10 \): \[ n \cdot 0.25 \geq 10 \implies n \geq \frac{10}{0.25} = 40 \]
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For \( n(1 - p) \geq 10 \): \[ n \cdot (1 - 0.25) \geq 10 \implies n \cdot 0.75 \geq 10 \implies n \geq \frac{10}{0.75} \approx 13.33 \]
The more stringent condition here is \( n \geq 40 \).
Thus, the minimum sample size needed to use normal approximations for an estimate about this population, given a probability of success of 25 percent, is 40.
So, the answer is 40.