SOLUTION: A random sample of 225 adults was given an IQ test. It was found that 117 of them scored higher than 100. Based on this, compute a 99% confidence interval for the proportion of all

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Question 1199185: A random sample of 225 adults was given an IQ test. It was found that 117 of them scored higher than 100. Based on this, compute a 99% confidence interval for the proportion of all adults whose IQ score is greater than 100. Then find the lower limit and upper limit of the 99% confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
sample size is 225.
117 of them scored higher than 100.
your sample mean proportion is 117/225 = .52
let p = .52
q = 1 - p = .48
let n = 225
standard error = sqrt(p * q / n) = sqrt(.52 * .48 / 225) = .0333
99% confidence interval z-score = plus or minus 2.5758
z-score formula = (x - m) / s
x is the raw score.
m is the mean
s is the standard error.

on the low side of the confidence interval, your formula becomes:
-2.5758 = (x - .52) / .0333
solve for x to get x = -2.5758 * .0333 + .52 = .4342

on the high side of the confidence interval, your formula becomes:
2.5758 = (x - .52) / .0333
solve for x to get x = 2.5758 * .0333 + .52 = .6058

your 99% confidence interval is from .4342 to .6058
round to 2 decimal places to get .43 to .61