Find the margin of error for the survey results described.
-In a poll of 1093 college students, 19% said that they had cheated at least once on an exam. Give your answer as a percentage to one decimal place.
A)0.1%
B)3.0%
C)1.5%
D)6.0%

Question

Quizplus · Accepted Answer

The margin of error for a proportion in a survey can be calculated using the formula $ME = Z \times \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (typically 1.96 for 95% confidence), $p$ is the proportion (in this case, 0.19), and $n$ is the sample size (1093). Plugging in the values, $ME = 1.96 \times \sqrt{\frac{0.19 \times (1-0.19)}{1093}}$, which calculates to approximately 0.028 or 2.8%, which rounds to 3.0% when given as a percentage to one decimal place.

Find the Margin of Error for the Survey Results Described