Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 288 with 143 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

_____< p < ______

Answer 1

Sample proportion = 143 / 288 = 0.497

99% confidence interval for p is

- Z/2 * Sqrt ( ( 1 - ) / n) < p < + Z/2 * Sqrt ( ( 1 - ) / n)

0.497 - 2.576 * sqrt( 0.497 * ( 1 - 0.497) / 288) < p < 0.497 + 2.576 * sqrt( 0.497 * ( 1 - 0.497) / 288)

0.421 < p < 0.573