Question

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!!

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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