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 34,219 people who responded 65% 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 given question does the confidence interval provide good estimate of the population proportion?

____<p<____ (round to three decimal places as needed).

Does the confidence interval provide a good estimate of the population proportion?

A) No the responses are not independent

B) No, the sample is a voluntary sample and might not be representative of the population

C)Yes all the assumptions for a confidence interval are satisfied

D) Yes the sample is large enough to provide a good estimate of the population

Solution

Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.65 -1.645*sqrt(0.65*(1-0.65)/34219) =0.646

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.65 +1.645*sqrt(0.65*(1-0.65)/34219) =0.654

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Yes, the confidence interval provide a good estimate of the population proportion because the sample size is large.