Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 304 with 100 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 = 100 / 304 = 0.329

1 - = 1 - 0.329 = 0.671

Z/2 = 1.28

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

= 1.28 * (((0.329 * 0.671) / 304)

Margin of error = E = 0.034

A 80% confidence interval for population proportion p is ,

- E < p < + E

0.329 - 0.034 < p < 0.329 + 0.034

0.295 < p < 0.363

The 80% confidence interval for the population proportion p is : 0.295 , 0.363