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# Estimation, Confidence Interval for proportions

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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.

© BrainMass Inc. brainmass.com October 2, 2022, 7:35 pm ad1c9bdddf
https://brainmass.com/statistics/confidence-interval/estimation-confidence-interval-proportions-19016

## 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

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 2, 2022, 7:35 pm ad1c9bdddf>
https://brainmass.com/statistics/confidence-interval/estimation-confidence-interval-proportions-19016