Question

Which of the following correctly calculates the margin of error of population mean estimate if the sample size is 1,503, the confidence level is 98%, and the standard deviation is 0.5?(1 point)
Responses

0.025
0.025

0.028
0.028

0.04
0.04

0.03

Answer 1

To calculate the margin of error for the population mean estimate, we can use the following formula:

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

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

1. **Determine the z-score for a 98% confidence level:**
The z-score for a 98% confidence level can generally be found from z-tables or standard normal distribution charts. For 98% confidence, the z-score is approximately 2.33.

2. **Use the given values:**
- \( n = 1503 \)
- \( \sigma = 0.5 \)

3. **Calculate the margin of error:**

\[
\text{Margin of Error} = 2.33 \left( \frac{0.5}{\sqrt{1503}} \right)
\]

Calculating the sample size's square root:

\[
\sqrt{1503} \approx 38.73
\]

Now substituting that into the equation:

\[
\text{Margin of Error} = 2.33 \left( \frac{0.5}{38.73} \right) \approx 2.33 \left( 0.01292 \right) \approx 0.0301
\]

Rounding this results in approximately 0.03.

**Final Answer: 0.03**