SOLUTION: ighTech Incorporated randomly tests its employees about company policies. Last year in the 510 random tests conducted, 16 employees failed the test. a. What is the point estimat

Question 1202823: ighTech Incorporated randomly tests its employees about company policies. Last year in the 510 random tests conducted, 16 employees failed the test.
a. What is the point estimate of the population proportion that failed the test?
b. What is the margin of error for a 95% confidence interval estimate?
c. Compute the 95% confidence interval for the population proportion.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
best i can make out of this based on what's given is:

p = mean failure rate = 16 / 510 = .03137.
q = (1 - p) - (1 - 16 / 510) = (1 - .03137) = .96863.

point estimate of the population proportion that failed the test is 16 / 510 = .03137 rounded to 5 decimal places.

standard error for a proportion is equal to sqrt (p * q / n) which is equal to sqrt(.03137 * .96863 / 510) = .00772.

margin of error for 95% confidence interval of the mean failure rate would be based on the critical z-score of plus or minus 1.96.

formula becomes 1.96 = (x - .03137) / .00772.
solve for (x - .03137) to get:
(x - .03137) = 1.96 * .00772 = .01513.
that's your margin of error.

solve for x in that same formula to get:
x = 1.96 * .00772 + .03157 = .0465012.
that would be the high side of the interval.

the low side of the interval would be equal to -1.96 * .00772 + .031357 = .01623388.

answers to your questions would be, as far as i can tell.

a. What is the point estimate of the population proportion that failed the test?
.03137

b. What is the margin of error for a 95% confidence interval estimate?
.01513

c. Compute the 95% confidence interval for the population proportion.
.0162388 to .0465012.

this is the best that i can do, based on what i know.
the mean failure rate was 16/510.
the 95% confidence interval was based on that as the mean.

let me know is you have any questions.
theo

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