The unemployment rate of persons with a disability is typically higher than for those with no disability.Recent statistics report that the national rate of unemployment among this group is 14.5%.The local government of a small city wants to determine the rate in its region to within ±5%, with 95% confidence.How many individuals with disabilities should they sample?
A)135
B)191
C)354
D)678
E)981

Question

Quizplus · Accepted Answer

To determine the sample size needed for a proportion with a specified margin of error and confidence level, the formula is $n = \frac{Z^2 \times p(1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the margin of error. For a 95% confidence level, $Z = 1.96$. Given $p = 0.145$ (14.5%) and $E = 0.05$ (5%), substituting these values gives $n = \frac{(1.96)^2 \times 0.145(1-0.145)}{(0.05)^2}$, which calculates to approximately 191. Therefore, they should sample 191 individuals.

The Unemployment Rate of Persons with a Disability Is Typically