Question

Sameer polled a random selection of 120 of the 250 workers at his office to determine whether they were satisfied with their office supply ordering process. The survey showed that 90 of the workers polled were satisfied. With a desired confidence level of 99%, which has a z*-score of 2.58, what is the margin of error of Sameer’s survey?

Answer 1

The answer that I found is 10% or B. I'll comment again to let you know if it is correct.

Answer 2

The margin of error can be calculated with the formula:

ME = z · √(p(1-p)/n)

where:
p = sample proportion
n = sample size
z = z-score

In your case:
p = 90 / 120 = 0.75

ME = 2.58 · √(0.75·0.25/120)
= 0.10
= 10%

The margin of error will be 10%.