Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 273 with 41% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

____ < p < ____  

** Enter a decimal or integral number.

Answer 1

n = 273

p = 0.41

80 % CI , z score for 0.8 is 1.282

80% confidence interval for p is

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

0.41 - 1.282 * sqrt( 0.41 * 0.59 / 273) < p < 0.41 + 1.282 * sqrt( 0.41 * 0.59 / 273)

0.372 < p < 0.448