Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 118 with 82 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

answer ____ < p < answer _____

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 82 / 118 = 0.695

1 - = 1 - 0.695 = 0.305

Z/2 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.695 * 0.305) / 118)

= 0.083

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.695 - 0.083 < p < 0.695 + 0.083

0.612 < p < 0.778

The 95% confidence interval for the population proportion p is : 0.612 , 0.778