Question

Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 85 with 49 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 = 49 / 85 = 0.576

1 - = 1 - 0.576 = 0.424

Z_/2 = 2.807

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

= 2.807 * (((0.576 * 0.424) / 85)

Margin of error = E = 0.150

A 99.5% confidence interval for population proportion p is ,

- E < p < + E

0.576 - 0.150 < p < 0.576 + 0.150

0.426 < p < 0.726

The 99.5% confidence interval for the population proportion p is : 0.426 , 0.726