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faust18 [17]
3 years ago
11

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?
Mathematics
1 answer:
umka2103 [35]3 years ago
3 0
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%.
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