Question 1202823
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