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A researcher wishes to estimate the proportion of college students who cheat on exams. A poll...

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 margin of error for the​ 95% confidence interval of the population proportion. Round to two decimal places.

Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

Point estimate = sample proportion = = 0.15

Z/2 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.15 * 0.85) / 560)

= 0.03

Margin of error = 0.03

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