Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 322 with 219 successes.

Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
Confidence interval =

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 219 / 322 = 0.680

1 - = 1 - 0.680 = 0.32

Z/2 = 2.576

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

= 2.576 * (((0.680 * 0.32) / 322)

Margin of error = E = 0.067

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.680 - 0.067 < p < 0.680 + 0.067

0.613 < p < 0.747

Confidence interval = 0.613 , 0.747