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?

Question

Mathematics

Makhi Hill

13 June, 22:25

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?

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Genesis Walls · 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%.