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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

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    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

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