Question

An online presented this question "Would the recent norovirus outbreak deter you from takind a cruise?". Among the 34,271 people who responded, 62% answered "Yes". Use the sample data to construct a 95% confidence interval estimate for the proportion of the population of all the people who would respond yes to that question. Does the confidence interval provide a good estimate of the population proportion?

Answer 1

95% C.I. = 0.62 + or - 1.96 * sqrt[0.62(1 - 0.62)/34,271] = 0.62 + or - sqrt(6.875 x 10^-6) = 0.62 + or - 0.002622 = 0.62 - 0.002622 to 0.62 + 0.002622 = 0.6174 to 0.6226 = 61.74% to 62.26%.