# Confidence Interval for Proportion

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?

https://brainmass.com/statistics/confidence-interval/confidence-interval-for-proportion-256252

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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.