SOLUTION: 131 students at a college were asked whether they had completed their required English 101 course, and 101 students said yes.
Construct the 80% confidence interval for the proport
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-> SOLUTION: 131 students at a college were asked whether they had completed their required English 101 course, and 101 students said yes.
Construct the 80% confidence interval for the proport
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Question 1182770: 131 students at a college were asked whether they had completed their required English 101 course, and 101 students said yes.
Construct the 80% confidence interval for the proportion of students at the college who have completed their required English 101 course.
Answer as a tri-linear inequality using decimals (not percents) accurate to 3 decimal places.
_________ Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 101/131=0.7710
80% CI half-interval is z(0.90)*sqrt (p*(1-p)/n)
=1.282*sqrt(0.771*0.229/131)
=0.0471
(0.724 < mu < 0.818) is the 80% CI for the true proportion of students.
