A human resources manager at a large company wants to estimate the proportion of employees that would be interested in reimbursement for college courses.If she wishes to be 95% confident that her estimate is within 5% of the true proportion, how many employees would need to be sampled?
A)271
B)385
C)543
D)646
E)1234

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 $n = \frac{Z^2 \times p(1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z = 1.96$), $p$ is the estimated proportion (if unknown, use 0.5 for maximum sample size), and $E$ is the margin of error (in decimal form, so 5% = 0.05). Plugging in the values: $n = \frac{1.96^2 \times 0.5(1-0.5)}{0.05^2} = 384.16$. Since you can't sample a fraction of a person, round up to the nearest whole number, which is 385.

A Human Resources Manager at a Large Company Wants to Estimate