Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 262 with 137 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

C.I. =

Answer #1

Sample proportion = 137 / 262 = 0.523

99% confidence interval for p is

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

0.523 - 2.5758 * sqrt [ 0.523 * ( 1 - 0.523) / 262] < p < 0.523 + 2.5758 * sqrt [ 0.523 * ( 1 - 0.523) / 262]

0.444 < p < 0.602

CI **( 0.444 , 0.602 )**

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