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

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.)

Found 2 solutions by Theo, math_tutor2020:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
test size = 390
12 out of 390 failed the test
failure proportion = 12/390

p = 12/390
q = 1 - p = 378/390

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.

that should be your 95% confidence interval.
i used the calculator at https://davidmlane.com/hyperstat/z_table.html

here's what it looks like using that calculator.

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.

Answer by math_tutor2020(3820) (Show Source): You can put this solution on YOUR website!

p = population proportion
phat = sample proportion
The job of phat is to estimate p.

At 95% confidence, the z critical value is roughly z = 1.96
This is something to memorize or have on a reference sheet.

You can use a table such as this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to determine the z critical values. Look at the bottom row and at the value just above the 95% confidence level.

What this means is that P(-1.96 < z < 1.96) = 0.95 approximately.

Another way to determine this z critical value is to use a stats calculator such as a TI84.

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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

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Answer:
The confidence interval in the format (L, U) would be approximately (0.014, 0.048)

This is equivalent to writing 0.014 < p < 0.048 which provides more context of which parameter we're trying to estimate.

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