# Estimation, Confidence Interval for proportions

A random sample of 90 prisoners in a large state prison revealed that 20% were college graduates. Find the 99% confidence interval in the population sampled for the percentage of prisoners who are college graduates.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

p=proportion of college graduates= 20.00%

q=1-p= 80.00%

n=sample size= 90

s p=standard error of proportion= square root of (pq/n)= 4.22% = square root of ( 20.% * 80.% / 90)

Confidence level= 99%

Significance level= a (alpha) = 1% =100% -99%

No of tails= 2

This is a 2 tailed test because we are calculating upper and lower confidence level

Since sample size= 90 > = 30 use normal distribution

Z at the 0.01 level of significance 2 tailed test = 2.5758

Upper confidence limit= p+z*s p= 30.86% =20.%+2.5758*4.22%

Lower confidence limit= p-z*s p= 9.14% =20.%-2.5758*4.22%

The proportion is between 9.14% and 30.86% for 99% Confidence level

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