Use the given data to find the minimum sample size required to estimate the population proportion.
-$\text { Margin of error: 0.009; confidence level: } 99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }$
A) 20,465
B) 21,442
C) 9642
D) 19,566

Question

Quizplus · Accepted Answer

The minimum sample size required to estimate the population proportion is given by the formula:$$n = \frac{z^2 \hat{p} \hat{q}}{e^2}$$where:* z is the z-score corresponding to the desired confidence level* $\hat{p}$ is the estimated population proportion* $\hat{q}$ is the estimated population proportion* e is the margin of errorIn this case, we are given the following information:* Margin of error: 0.009* Confidence level: 99%* $\hat{p}$ and $\hat{q}$ are unknownThe z-score corresponding to a 99% confidence level is 2.576. Substituting this value into the formula, we get:$$n = \frac{2.576^2 \hat{p} \hat{q}}{0.009^2}$$Since $\hat{p}$ and $\hat{q}$ are unknown, we can use the conservative estimate of 0.5 for both values. Substituting these values into the formula, we get:$$n = \frac{2.576^2 (0.5)(0.5)}{0.009^2} = 20,465$$Therefore, the minimum sample size required to estimate the population proportion with a margin of error of 0.009 and a confidence level of 99% is 20,465.