Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 238 with 159 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 95% C.I. =

Answer 1

Sample proportion = 0.668

95% confidence interval for p is

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

0.668 - 1.96 * sqrt ( 0.668 * ( 1 - 0.668) / 238 ) < p < 0.668 + 1.96 * sqrt ( 0.668 * ( 1 - 0.668) / 238 )

0.608 < p < 0.728

95% CI is ( 0.608 , 0.728 )