Question

1.) An online site presented this​ question, "Would the recent norovirus outbreak deter you from taking a​ cruise?" Among the 34,705 people who​ responded, 69​% answered​ "yes." Use the sample data to construct a 90​% confidence interval estimate for the proportion of the population of all people who would respond​ "yes" to that question. Does the confidence interval provide a good estimate of the population​ proportion?

______<p<______

Answer 1

sample size          n=		34705
sample proportion p̂ =x/n=		0.6900
std error se= √(p*(1-p)/n) =		0.0025
for 90 % CI value of z=		1.645
margin of error E=z*std error   =		0.0041
lower bound=p̂ -E                       =		0.6859
Upper bound=p̂ +E                     =		0.6941

from above 90% confidence interval for population proportion =(0.6856 <p< 0.6941)