SOLUTION: Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 203 with 39% successes. Enter your answer as an open-inter

Question 1200442: Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 203 with 39% 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!
p = .39
q = 1 - .39 = .61
sample size is 203.
standard error = sqrt(.39 * .61 / 203) = .0342333344.
95% confidence interval is equal to plus or minus z = 1.959963986.
z-score formula is z = (x - m) / s
z is the z-score
x is the sample mean proportion.
m is the population mean proportion.
s is the standard error.
solve for x to get x = z * s + m which becomes:
x = -1.959963986 * .0342333344 + .39 = .3229038975 at the lower limit.
solve for x = 1.959963986 * .0342333344 + .39 = .4570961025 at the upper limit.
round to 3 decial places to get:
(.323,.457) as your answer.
here's what it looks like, using the calculator at https://davidmlane.com/hyperstat/z_table.html

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