Question

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?

Answer 1

n ≥ 385

Explanation:

Solution:-

- The %error allowed for the estimate of true proportion "p", E = 0.05

- The Confidence Level ( 1 - α ) = 0.95

- The number of employees sampled = n

- Use the look table for Z-critical value corresponding to the significance level "α" = 0.05:

Z-critical = Z-α/2 = Z-0.025

Z-0.025 = 1.96

- Assume the true population to be p = 0.5

- We will use the error formula for estimation of true population proportion:

`E = Z-critical *\sqrt{(p*(1-p))/(n) } \\\\n = ((Z-critical)/(E))^2*p*(1-p)\\\\n = ((1.96)/(0.05))^2*0.5*(1-0.5)\\\\n = 385`

Answer: The sample size for employees must be n≥385 to keep the estimation within the margin of 5% error