Download As a condition of employment, Fashion Industries

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?
(a) Here it is given that,
The sample proportion, p = 14/220 = 0.063636 and q = 1 - p = 0.936364
Now, the standard error
SE = Sqrt(pq/n)
= Sqrt(0.063636 * 0.936364/220)
= Sqrt(0.000271)
= 0.016457
For 99% confidence, z = 2.576
Thus a 99 percent confidence interval for the proportion of applicants that fail the test is
given by
(p – 2.576 * SE, p + 2.576 * SE)
= (0.063636 - 2.576 * 0.016457, 0.063636 + 2.576 * 0.016457)
= (0.063636 – 0.042394, 0.063636 + 0.042394)
= (0.021242, 0.106031)
Thus the 99 percent confidence interval for the proportion of applicants that fail the test is
from 2.12% to 10.60%.
Since 10% falls within the 99% confidence interval but near the upper limit of 10.6%, it
may not be reasonable to conclude that more than 10% are now failing the test.
(b) Here it is given that,
The sample proportion, p = 14/400 = 0.035 and q = 1 - p = 0.965
Now, the standard error
SE = Sqrt(pq/n)
= Sqrt(0.035 * 0.965/400)
= 0.009189
For 99% confidence, z = 2.576
Thus a 99 percent confidence interval for the proportion of applicants that fail the test is
given by
(p – 2.576 * SE, p + 2.576 * SE)
= (0.035 - 2.576 * 0.009189, 0.035 + 2.576 * 0.009189)
= (0.035 – 0.02367, 0.035 + 0.02367)
= (0.01133, 0.05867)
Thus the 99 percent confidence interval for the proportion of applicants that fail the test is
from 1.13% to 5.87%.
Since 5% falls within the 99% confidence interval, we may conclude that less than 5
percent of the employees are not able to pass the random drug test.
A recent survey of 50 executives who were laid off from their previous position revealed
it took a mean of 26 weeks for them to find another position. The standard deviation of
the sample was 6.2 weeks. Construct a 95 percent confidence interval for the population
mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.
Let μ be the population mean.
Here it is given that,
Sample mean, x-bar = 26
Sample standard deviation, s = 6.2
Sample size, n = 50.
Since we are using large samples (n > 30), the 95% confidence interval for the population
mean is
(x-bar – 1.96*s/√n, x-bar + 1.96*s/√n)
= (26 – 1.96*6.2/√50, 26 + 1.96*6.2/√50)
= (26 – 1.7186, 26 – 1.7186)
= (24.2814, 27.7186)
Thus the 95% confidence interval for the population mean is from 24.28 weeks to 27.72
weeks.
Since 28 weeks falls outside the 95% confidence interval, it is not reasonable to assume
that the population mean is 28 weeks.