Question

A certain survey conducted had a population proportion of at most 22% for people who agreed. What is the confidence level if 20% of the people sampled agreed, and the standard error of the proportion is ±1%?

Answer 1

The solution would be like this for this specific problem:

$p \pm Z_c \sqrt{\frac{p(1-p)}{n}} \\ \, \\ .20 + Z_c (.01) = .22 \\ \, \\ Z_c (.01) = .22 - .20 \\ \, \\ Z_c (.01) = .02 \\ \, \\ Z_c = \frac{.02}{.01} \\ \, \\ Z_c = 2 \\ \, \\ c = .95$