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 $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z = 1.96$), $p$ is the estimated proportion (0.145 in this case), and $E$ is the margin of error (0.05). Plugging in the values: $n = \frac{(1.96)^2 \cdot 0.145 \cdot (1-0.145)}{(0.05)^2}$ results in approximately 191. This calculation assumes a simple random sample and uses the normal approximation for the distribution of the sample proportion.

The Unemployment Rate of Persons with a Disability Is Typically