Question

A college surveyed 2,000 students and asked what their GPA was the previous semester. The mean GPA was 3.2. The standard deviation for the population was 0.3. Which of the following correctly calculates the margin of error for a 98% confidence level, rounded to three decimal places?(1 point)
Responses

0.016
0.016

0.029
0.029

0.013
0.013

0.019

Answer 1

To calculate the margin of error for a confidence interval, you can use the formula:

\[
\text{Margin of Error} = Z \times \left( \frac{\sigma}{\sqrt{n}} \right)
\]

where:
- \( Z \) is the Z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.

For a 98% confidence level, the Z-score is approximately 2.33 (you can find this using a Z-table or calculator).

Given the information provided:

- \( \sigma = 0.3 \)
- \( n = 2000 \)

Now, we calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.3}{\sqrt{2000}} \approx \frac{0.3}{44.721} \approx 0.00671
\]

Next, we calculate the margin of error:

\[
\text{Margin of Error} = 2.33 \times SE = 2.33 \times 0.00671 \approx 0.01566
\]

Rounding this to three decimal places gives approximately **0.016**.

Thus, the correct response is:

**0.016**