Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 224 with 17% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

Answer 1

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.17 - 2.5758 * sqrt [ 0.17 ( 1 - 0.17) / 224 ] < p < 0.17 + 2.5758 * sqrt [ 0.17 ( 1 - 0.17) / 224 ]

0.105 < p < 0.235​​​​​​​