Question

Given n = 2500 and p ^ = 0.86, find the margin of error E that corresponds to a 99% confidence level.

Answer 1

Sample size = n = 2500
Sample Proportion = p = 0.86
Confidence Level = 99%
Z-value for the confidence level = z = 2.576 (This Value comes from the z-table)

The formula for Margin of Error is:

$E=z* \sqrt{ \frac{p(1-p)}{n} }$

Using the given values, we get:

$E=2.576* \sqrt{ \frac{0.86(0.14)}{2500} } \\ \\ E=0.0179$

Thus, the margin of error would be 0.0179 or 1.79%