Question

There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 90% 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?

(Round up your answer to nearest whole number)

Answer 1

Solution :

Given that,

= 1 - = 0.5

margin of error = E = 0.24

At the confidence level the z is,

= 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.05

Z_/2 = Z0.05 = 1.645

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= ( 1.645 / 0.24)² * 0.5* 0.5

= 11.74

sample size = n = 12