Confidence Interval for Proportion
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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.
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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Solution Summary
Develops a 99 percent interval for the proportion of applicants that fail a 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.
99% Confidence limits for proportions
Data
Sample proportion=p= 6.36% =14/220
Therefore, q=1-p= 93.64% = 100 % - 6.36%
n=sample size= 220
Confidence level= 99%
Therefore, Significance level=alpha (a) = 1% =100% -99%
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 ...
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