Question

During a national debate on changes to health care, a cable news service performs an opinion poll of 460 small-business owners. It shows that 41% of small-business owners do not approve of the changes. Develop a 95% confidence interval for the proportion opposing health care changes.

Answer 1

95% confidence interval for p is

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

0.41 - 1.96 * sqrt [ 0.41 * ( 1 - 0.41) / 460] < p < 0.41 + 1.96 * sqrt [ 0.41 * ( 1 - 0.41) / 460]

0.365 < p < 0.455

95% CI Is ( 0.365 , 0.455 )