Question

A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 560 college students showed that 15% of them had, or intended to, cheat on examinations. Find the 95% confidence interval.

Answer 1

0.1198 < p < 0.1802

Explanation:

Percentage who had, or intended to cheat (p) = 15% = 0.15.

1 - p = 1 - 0.15 = 0.85

Confidence interval = 95%, z = 1.96 = 2

number of observation= 560

p ± z * √(p * (1 -p)/n)

Lower limit:

0.15 - 2 * √0.15 * 0.85/560

0.15 - 0.0301780 = 0.119822

= 0.1198

Upper limit:

0.15 + 2 * √0.15 * 0.85/560

0.15 + 0.0301780 = 0.180178

= 0.1802

0.1198 < p < 0.1802