Use the confidence level and sample data to find the margin of error E
-College students' annual earnings: $99 \%$ confidence; $n = 71 , \bar { x } = \$ 3660 , \sigma = \$ 879$
A) $\$ 1118$
B) $\$ 243$
C) $\$ 269$
D) $\$ 8$

Question

Quizplus · Accepted Answer

The margin of error (E) for a given confidence level and sample data can be calculated using the formula $E = Z \cdot \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 99% confidence level, the Z-score is approximately 2.576. Plugging in the given values: $E = 2.576 \cdot \frac{879}{\sqrt{71}} \approx 269$.

Use the Confidence Level and Sample Data to Find the Margin