Find the margin of error for the survey results described.
-In a survey of 602 employees of a large company, 25% said that they had high job satisfaction. Give your answer as a decimal to three decimal places.
A)0.041
B)0.002
C)0.082
D)0.020

Question

Quizplus · Accepted Answer

The margin of error for a proportion is 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 sample proportion, and $n$ is the sample size. Without specifying a confidence level, we assume the standard 95% confidence level, so $z = 1.96$, $p = 0.25$, and $n = 602$. Plugging these values into the formula gives $ME = 1.96 \times \sqrt{\frac{0.25(1-0.25)}{602}} \approx 0.041$.

Find the Margin of Error for the Survey Results Described