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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% confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10% of the applicants are now failing the test?

User DogEatDog
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Answer:

Of the last 220 applicants, 14 failed the test.

So, we have the sample proportion =
(14)/(220) = 0.064

At 99% confidence interval, z score is 2.576

Standard error or E =
0.064+2.576* ((√(0.064*(1-0.064)) )/(√(220) ) ) and
0.064-2.576* ((√(0.064*(1-0.064)) )/(√(220) ) )

=
0.064+0.04250 and
0.064-0.04250

= 0.0215, 0.1065

Rounding off we get 0.02,0.11

In percentage the applicants who failed lies between interval 2% and 11%.

So, yes it would be reasonable to conclude that more than 10% of the applicants are now failing the test.

User TheWildSushii
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