SOLUTION: In a random sample of 250 students at a university, 202 stated that they were nonsmokers. Based on this sample, compute a 91% confidence interval for the proportion of all students
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Question 1159646: In a random sample of 250 students at a university, 202 stated that they were nonsmokers. Based on this sample, compute a 91% confidence interval for the proportion of all students at the university who are nonsmokers.
Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.
What is the lower limit of the 90% confidence interval?
You can put this solution on YOUR website! Will use a 90% interval, not a 91% one
half-interval for a proportion is z(0.95)*sqrt(p*(1-p)/n)
p hat is used here which is 202/250=0.808
half-interval is 1.645*sqrt(0.808*0.192/250)=1.645*0.0249=0.0410
90%CI is (0.767, 0.849) or (0.77, 0.85)
