Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 227 with 112 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

____< p < ____

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 112 / 227 = 0.493

1 - $\hat p$ = 1n - 0.493 = 0.507

Z$\alpha$/2 = 1.282

Margin of error = E = Z$\alpha$ / 2 * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.282 * ($\sqrt$((0.493 * 0.507) / 227)

Margin of error = E = 0.043

A 80% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.493 - 0.043 < p < 0.493 + 0.043

0.450 < p < 0.536