Provide an appropriate response.
-A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation. How large a sample is needed in order to be $99 \%$ confident that the sample proportion will not differ from the true proportion by more than $4 \%$ ?
A) 1037
B) 17
C) 2073
D) 849

Question

Quizplus · Accepted Answer

To estimate the sample size needed for a proportion with a specific margin of error (E) at a certain confidence level, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (for 99% confidence, $Z \approx 2.576$), $p$ is the estimated proportion (if unknown, 0.5 is used as it maximizes the sample size), and $E$ is the margin of error (0.04 in this case). Plugging in the values: $n = \frac{(2.576)^2 \cdot 0.5 \cdot (1-0.5)}{(0.04)^2} \approx 1037$.

Provide an Appropriate Response 99%99 \%99% Confident That the Sample Proportion Will Not Differ from the Wishes

Provide an Appropriate Response $99 \%$ Confident That the Sample Proportion Will Not Differ from the Wishes