Question

There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support the gun control in 2018 with a 0.2 margin of error, how many randomly selected Americans must be surveyed?

(Round up your answer to nearest whole number)

Answer 1

Sample size formula is

n = 0.25*(z/E)^2

where z = 1.96 [From z table for 95% confidence level]
E = margin of error = 0.2

this gives us

n = 0.25*(1.96/0.2)^2

= 0.25*96.04

= 24.01

=25 (rounded up to nearest integer)

Required sample size is 25