Question

​It's believed that as many as 20​%of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group

a) How many of this younger age group must we survey in order to estimate the proportion of​ non-grads to within 88​% with​ 90% confidence?

​b) Suppose we want to cut the margin of error to 55​%. What is the necessary sample​ size?

Answer 1

Solution,

Given that,

= 0.20

1 - = 1 - 0.80

margin of error = E = 0.08

At 90% confidence level

= 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.025

Z_/2 = Z_0.025 = 1.645

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

= (1.645 / 0.08)² * 0.20 * 0.80

= 67.65

sample size = n = 68

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

= (1.645 * 0.05)² * 0.20 * 0.80

= 173.19

sample size = n = 174