Question 1206894: Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 341 with 74% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 341. .023
p = .74
q = 1 - .74 = .26
sample proportion mean is .74
sample standard error = sqrt(p * q / n) = sqrt(.74*.26/341) = .02375 rounded to 5 decimal places.
critical z-score at 90% two tail confidence interval is equal to plus or minus 1.64485 rounded to 5 decimal places.
z-score formula of z = (x-m)/s is used.
in this problem:
z = z-score z-score
x = critical raw score
m = mean proportion
s = standard error
on the high side, z-score formula becomes 1.64485 = (x - .74) / .02375.
solve for x to get x = 1.64485 * .02375 + .74 = .779 rounded to 3 decimal places.
on the low side, z-score formula becomes -1.64485 = (x - .74) / .02375.
solve for x to get x = -1.64485 * .02375 + .75 = .701 rounded to 3 decimal places.
your 90% interval is from .701 to .779.
that's your solution.
i believe that's going to be (.701,.779) in interval form.
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