Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p
-Margin of error: $0.002$; confidence level: $93 \% ; \hat { \mathrm { p } }$ and $\hat{\mathrm { q }}$ unknown
A) 410
B) 204,757
C) 409
D) 204,756

Question

Quizplus · Accepted Answer

The formula for the sample size needed to estimate a population proportion with a given margin of error (E) and confidence level is $n = \frac{{\hat{p} \hat{q} Z^2}}{{E^2}}$, where $\hat{p}$ is the sample proportion, $\hat{q} = 1 - \hat{p}$, and $Z$ is the Z-score corresponding to the confidence level. Without $\hat{p}$ and $\hat{q}$, we use $\hat{p} = 0.5$ and $\hat{q} = 0.5$ for the maximum possible sample size (worst-case scenario). The Z-score for a 93% confidence level is approximately 1.81 (from Z-tables or calculators). Substituting these values and the given margin of error $E = 0.002$ into the formula, we get $n = \frac{0.5 \times 0.5 \times 1.81^2}{0.002^2}$, which calculates to approximately 204,756.25. Therefore, the closest option is 204,757.

Use the Given Degree of Confidence and Sample Data to Construct