Question

ind the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed.

A 99% confidence interval for a proportion p if the sample has n=300 with ,p-hat=.73 and the standard error isSE = .03 . Round your answers to three decimal places. The confidence interval is

Answer 1

Given,

n =300

p^ = 0.73

SE = 0.03

99% confidence interval is

CI = p^ z*(p^*(1-p^)/n)

z value for 99% confidence interval is 2.58

(p^*(1-p^)/n) = SE = 0.03

CI = 0.73 2.58*0.03

= 0.73 0.0774

CI = 0.73 - 0.0774 and CI = 0.73 + 0.0774

= 0.653 and CI = 0.807

Therefore 99% confidence interval is

(0.653, 0.807).