Question

1-A travel agent wants to estimate the proportion of vacationers who plan to travel outside the United States in the next 12 months. A random sample of 150 vacationers revealed that 45 had plans for foreign travel in that time frame. a)Suppose (at the 95% confidence level) you need to have a margin of error no more than 4 percentage points, but you have no estimate of ^p. How many vacationers would you have to sample?

Answer 1

Solution,

Given that,

=  1 - = 0.5

margin of error = E = 0.04

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

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

= (1.96 / 0.04 )2 * 0.5 * 0.5

= 600.25

sample size = n = 601