SOLUTION: The upper limit of the 90% confidence interval for the population proportion p, given that n = 100; and = 0.20

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Question 978068: The upper limit of the 90% confidence interval for the population proportion p, given that n = 100; and = 0.20
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
CI=1.645 +/- SE
SE= sqrt {(0.2)(0.8)/100}
=0.4/10, because sqrt (0.16)=0.4
=0.04
CI=1.645*0.04=0.0658
upper limit is 0.2658