An online site presnted this question Would the resent norov

An online site presnted this question. Would the resent norovirus outbreak deter you from taking the cruise. Among the 34,646 people who responded 63% answered \"yes\". Use sample data to constrct a 90% confidence interval estimate for the proportion of the population all people who would respond \'yes\' to that question. Does the confidence interval provide a good estimate of the population proportion. ____< p < ____. show my work.

Solution

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Sample Size(n)=346446
Sample proportion = x/n =0.63
Confidence Interval = [ 0.63 ±Z a/2 ( Sqrt ( 0.63*0.37) /346446)]
= [ 0.63 - 1.64* Sqrt(0) , 0.63 + 1.64* Sqrt(0) ]
= [ 0.6287,0.6313]