Question 320805: 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 confidence 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?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! As a condition of employment, Fashion Industries applicants must pass a drug test.
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Of the last 220 applicants 14 failed the test.
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sample proportion = 14/220 = 0.6364
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Develop a 99 percent confidence interval for the proportion
of applicants that fail the test.
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I ran a 1-Prop ZInt and got the following:
(0.0212, 0.1060)
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Would it be reasonable to conclude that more than
10 percent of the applicants are now failing the test?
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No, > 10% is not in the 99% confidence interval
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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.
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p-hat = 14/400 = 0.0350
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Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test?
Ho: p >= 0.05
Ha: p < 0.05 (claim)
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I ran a 1-Prop ZTest and got the following results:
test statistic: z = -1.3765
p-value = 0.0843
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Since the p-value is greater than 5%, fail to reject Ho.
Test results do not support the claim at the 5% significance level.
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Cheers,
stan H.
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