Explore BrainMass

Explore BrainMass

    Confidence Interval for Proportion

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

    As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants, 14 failed the test. Develop a 99 percent interval for the proportion of applicants that fail the test.
    Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test?
    In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year.
    Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test?

    © BrainMass Inc. brainmass.com June 3, 2020, 10:55 pm ad1c9bdddf
    https://brainmass.com/statistics/confidence-interval/confidence-interval-for-proportion-256252

    Solution Preview

    The answers are in the attached file.

    As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants, 14 failed the test. Develop a 99 percent interval for the proportion of applicants that fail the test.

    99% Confidence limits for proportions
    Data
    Sample proportion=p= 6.36% =14/220
    Therefore, q=1-p= 93.64% = 100 % - 6.36%
    n=sample size= 220
    Confidence level= 99%
    Therefore, Significance level=alpha (a) = 1% =100% -99%

    No of tails= 2 (This is 2 tailed because we are calculating the confidence interval)
    Since sample size= 220 > 30 use normal distribution
    Z at the 0.01 level of significance 2 tailed test = 2.5758
    sp=standard error of proportion=square root of (pq/n)= 1.6453% =square root of ( 6.36% * 93.64% / 220)
    z*sx= 4.2380% =2.5758 x 1.6453%
    Confidence interval = sample proportion ±z*sp= 6.3600% ± 4.2380%
    Upper confidence limit= p+t*sp= 10.5980% =6.36%+4.238%
    Lower confidence ...

    Solution Summary

    Develops a 99 percent interval for the proportion of applicants that fail a drug test.

    $2.19

    ADVERTISEMENT