Question 267411
As a condition of employment, Fashion Industries applicants must pass a drug test. 
Of the last 220 applicants 14 failed the test. 
So sample proportion = 14/220 = 0.64
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Develop a 99 percent confidence interval for the proportion
of applicants that fail the test. 
invNorm(0.995) = 2.5758
E = standard error = 2.5758*sqrt[0.64*0.36/220) = 0.0834
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99% CI: 0.64-0.0834 < p < 0.64+0.0834
99% CI: 0.5566 < p < 0.7234
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Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? 
Relating the question to the confidence interval the answer would
be no.
Cheers,
Stan H.