Answer to Question #275409 in Statistics and Probability for abdikani

Question

A U.S. Travel Data Center survey conducted for Better Homes and Gardens of 1500 adults found that 39% said that they would take more vacations this year than last year. Find the 95% confidence interval for the true proportion of adults who said that they will travel more this year.

 

Answer 1

Sample size (n) =1500

Sample proportion (p) = 0.39

z value at 95% confidence interval =1.96

Following formula is used to find the 95% confidence interval for the true proportion of adults who said that they will travel more this year:

$p\pm z\sqrt{\frac{p(1-p)}{n}}$

$0.39\pm 1.96\sqrt{\frac{0.39(1-0.39)}{1500}}$

$0.39\pm 1.96\sqrt{0.000159}$

$0.39\pm 0.02468$

$0.365<p<0.415$