Question

97 students at a college were asked whether they had completed their required English 101 course, and 54 students said "yes". Construct the 80% confidence interval for the proportion of students at the college who have completed their required English 101 course. 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 = $\hat p$ = x / n = 54 / 97 = 0.558

1 - $\hat p$ = 0.442

Z$\alpha$/2 = 1.282

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

= 1.282 * $\sqrt$ ((0.558 * 0.442) / 97)

= 0.065

A 80 % confidence interval for population proportion p is ,

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

0.558 - 0.065 < p < 0.558 + 0.065

0.493 < p < 0.623

(0.493, 0.623)