Question 628652
Problem 7)

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.

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We have  p = 0.33  and q = 1- 0.33 = 0.67 , n = 490, the critical value is 1.96 (95% confidence interval)

standard deviation = sqrt(0.33 * 0.67 * 490) =  10.41

E = (zs)/sqrt(n)= (1.96 * 10.41)/sqrt(490) = 0.92

Hope it helps:)

John10