An online site presented this question Would the recent noro

An online site presented this question, \'Would the recent norovirus outbreak deter you from taking a cruise?\' Among the 34,131

people who responded,

67%

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?

Solution

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
Mean(x)=22867.77
Sample Size(n)=34131
Sample proportion = x/n =0.67
Confidence Interval = [ 0.67 ±Z a/2 ( Sqrt ( 0.67*0.33) /34131)]
= [ 0.67 - 1.64* Sqrt(0) , 0.67 + 1.64* Sqrt(0) ]
= [ 0.666,0.674]