Question

The United States has over 310 million residents. Suppose that you want to estimate the proportion of Americans who ate breakfast this morning to within a margin-of-error of 3 percentage points with 95% confidence. About how many people would you need to randomly sample? (Assume all selected will respond to the survey.)

Answer 1

Solution :

Given that,

= 0.5 ( when estimate is not given then use 0.5)

1 - = 1 - 0.5 = 0.5

margin of error = E = 3% = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.96 /0.03)² * 0.5 * 0.5

= 1067.11 = 1067

Sample size = 1067