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?

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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.

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.

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B. wider for a sample size of 100 than for a sample size of 50
C. narrower for 90% confidence than for 95% confidence
D. narrower when the sample proportion is 0.50 than when the sample proportion is

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