Question

Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 391 with 304 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 304 / 391 = 0.777

1 - = 1 - 0.777 = 0.223

Z_/2 = 3.291

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

= 3.291 * (((0.777 * 0.223) / 391)

Margin of error = E = 0.069

A 99.9% confidence interval for population proportion p is ,

- E < p < + E

0.777 - 0.069 < p < 0.777 + 0.069

0.708 < p < 0.846

The 99.9% confidence interval for the population proportion p is : 0.708 , 0.846