A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 490 college students showed that 33% of them had, or intended to, cheat on examinations. Find the margin of error for the 95% confidence interval.

Answer 1

Margin of error = (1.96)[√(pq/n)]
...where p = .33, q = 1 - p, and n = 490

Note: 1.96 represents 95% confidence interval.

Plug the values into the formula and calculate.

I let you take it from here.

Answer 2

4.16%