Question

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 321 with 230 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. C.I. =

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 230 / 321 = 0.717

1 - = 1 - 0.717 = 0.283

Z/2 = 2.576

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

= 2.576 * (((0.717 * 0.283) / 321)

Margin of error = E = 0.065

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.717 - 0.065 < p < 0.717 + 0.065

0.652 < p < 0.782

99% C. I. = (0.652 , 0.782)