Homework Help: MATH##$$&&^^**

Posted by Bridget on Friday, April 20, 2012 at 5:44pm.

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.

MATH##$$&&^^** - MathGuru, Friday, April 20, 2012 at 6:59pm
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.
MATH##$$&&^^** - john, Saturday, July 21, 2012 at 8:06pm
4.16%

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