Question

According to a health organization in a certain​ country, 21.8​% of the population smoked in 2008....

According to a health organization in a certain​ country, 21.8​% of the population smoked in 2008. In​ 2014, a random sample of 650 citizens of this country was​ selected, 129 of whom smoked. Complete parts a through c below. a. Construct a 90​% confidence interval to estimate the actual proportion of people who smoked in this country in 2014. A 90​% confidence interval to estimate the actual proportion has a lower limit of nothing and an upper limit of nothing. Please help!!

Homework Answers

Answer #1

Given :-

N=650

x=129

P^ = (x/N) = (129/650)

Here using ti-83 calculator.

90​% confidence interval to estimate the actual proportion has a lower limit of 0.1725 and an upper limit of 0.22442

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