Question

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 144 with 62% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

_ < p < _

Answer 1

Given:

Sample size = n = 144

Sample proportion = = 0.62

Significance level = = 1-0.98 = 0.02

At 98% confidence level, the critical value of Z is

Z/2 = Z0.02/2 = 2.33

98% confidence interval:

CI = Z/2 * √[ (1-)/n]

= 0.62 2.33 * √[0.62(1-0.62)/144]

= 0.62 0.0942

= (0.526, 0.714)

Therefore 98% confidence interval is 0.526 < p < 0.714