To calculate the minimum sample size needed for a researcher to estimate the population proportion with a certain confidence level and margin of error, we can use the following formula:
\[ n = \left( \frac{Z_{\alpha/2} \cdot \sqrt{p \cdot (1-p)}}{E} \right)^2 \]
Where:
- \( n \) is the sample size,
- \( Z_{\alpha/2} \) is the z-score corresponding to the desired confidence level,
- \( p \) is the estimated proportion of the population,
- \( E \) is the margin of error.
(a) No preliminary estimate is available. In such a case, we use \( p = 0.5 \) because this value of \( p \) maximizes the product \( p \cdot (1-p) \), and hence, the sample size. This is because we want to ensure that our sample size is large enough irrespective of the actual population proportion, which is unknown at this point.
Given:
- Confidence level = 99%
- Margin of error = 3% or 0.03
Firstly, we need to find the z-score \( Z_{\alpha/2} \) that corresponds to a 99% confidence level. Since the normal distribution is symmetric, we are looking for the z-score such that the area to the left of it under the standard normal distribution curve is 0.995 (because 99% confidence splits 1% of the probability into two tails, thus \( 0.5 + 0.99/2 = 0.995 \)).
Looking up 0.995 in the standard normal distribution table or using a calculator, we find that the z-score is approximately 2.576.
Now, we can calculate the sample size:
\[ n = \left( \frac{2.576 \cdot \sqrt{0.5 \cdot (1-0.5)}}{0.03} \right)^2 \]
\[ n = \left( \frac{2.576 \cdot \sqrt{0.25}}{0.03} \right)^2 \]
\[ n = \left( \frac{2.576 \cdot 0.5}{0.03} \right)^2 \]
\[ n = \left( \frac{1.288}{0.03} \right)^2 \]
\[ n = \left( 42.933 \right)^2 \]
\[ n = 1844.322 \]
Since we can't survey a fraction of a person, we'll need to round up to the nearest whole number. So, the minimum sample size needed assuming that no prior information is available is:
\[ n = 1845 \]
