Question

Assume we want to estimate with a confidence interval what percent of Americans are planning on going on a vacation this year. If we desire a 95% confidence interval with a margin of error no larger than +/- 5%, how many people must we survey

Answer 1

Solution :

Given that,

= 0.5

1 - = 0.5

margin of error = E = 0.05

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 / 0.05)² * 0.5*0.5

= 384.16

sample size = 384

384 people must we survey.