Question

A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 530 college students showed that ​21% of them​ had, or intended​ to, cheat on examinations. Find the​ 95% confidence interval for the population proportion. Round to four decimal places.

Answer 1

(0.1753 ; 0.2447)

Explanation:

Using the relation :

Confidence interval :

p ± margin of error

p = 0.21

1 - p = 0.79

n = 530

Zcritical at 95% confidence level = 1.96

Margin of Error : Zcritical * sqrt[(p(1-p))/n]

1.96 * sqrt[(0.21(0.79))/530]

1.96 * 0.0176923 = 0.0346769

Margin of Error = 0.0346769

Lower boundary = 0.21 - 0.0347 = 0.1753

Upper boundary = 0.21 + 0.0347 = 0.2447

Confidence interval :

(0.1753 ; 0.2447)