SOLUTION: HighTech Inc. randomly tests its employees about company policies. Last year in the 390 random tests conducted, 12 employees failed the test.
Develop a 95% confidence interval f
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Develop a 95% confidence interval f
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Question 1206695: HighTech Inc. randomly tests its employees about company policies. Last year in the 390 random tests conducted, 12 employees failed the test.
Develop a 95% confidence interval for the proportion of applicants that fail the test. (Round to three decimal places.)
standard error of a proportion = sqrt(p*q/n) = sqrt(12/390 * 378/390 / 390) = .0087445932.
95% two tail confidence interval required a critical z-score of plus or minus 1.959963986.
on the low side of the confidence interval, the z-score formula of z = (x - m) / s becomes -1.959963986 = (x - 12/390) / .0087445932.
solve for x to get x = .013630143.
on the high side of the confidence interval, the z-score formula of z = (x - m) / s becomes 1.959963986 = (x - 12/390) / .0087445932.
solve for x to get x = .0479083186.
the calculator truncates inputs to 6 or 7 decimal digits, so i didn't get the full value in there.
the rounding / truncating is usually good enough to get an accurate answer.
from the calculator, you can see that the confidence interval is the area between those two x-scores = .95.
n = 390 = sample size
x = 12 employees failed
phat = sample proportion of those who failed
phat = x/n
phat = 12/390
phat = 0.03076923 approximately
Around 3.077% of the sample of employees failed the test.
This is the center of the confidence interval.
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.03076923*(1-0.03076923)/390)
E = 0.01713940 approximately
This helps determine how wide or spread out the confidence interval is.
L = lower bound of confidence interval
L = phat - E
L = 0.03076923 - 0.01713940
L = 0.01362983
L = 0.014
U = upper bound of confidence interval
U = phat + E
U = 0.03076923 + 0.01713940
U = 0.04790863
U = 0.048