Question

There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 92% confidence interval for the true proportion of Americans who support the gun control in 2018 with a 0.24 margin of error, how many randomly selected Americans must be surveyed?

Answer 1

When no prior estimate for proportion is specified then p = 0.50

Sample size = Z² $\alpha$ /2 * p( 1 - p) / E²

= 1.7507² * 0.50 * 0.50 / 0.24²

= 13.3

n = 14 (Rounded up to nearest integer)