Question

A sample of size n=57 is drawn from a population whose standard deviation is o=9. Find the margin of error for a 99% confidence interval for u. Round the answer to at least three decimal places.

Answer 1

To find the margin of error for a confidence interval for the population mean (\( \mu \)), we can use the formula:

\[
E = z^* \cdot \frac{\sigma}{\sqrt{n}}
\]

Where:
- \( E \) is the margin of error,
- \( z^* \) is the z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.

1. **Identify the parameters:**
- Sample size (\( n \)) = 57
- Population standard deviation (\( \sigma \)) = 9
- Confidence level = 99%

2. **Find the z-score for a 99% confidence level:**
The z-score for a 99% confidence level is approximately 2.576 (you can find this value in a z-table or using a calculator that provides z-scores).

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

\[
E = 2.576 \cdot \frac{9}{\sqrt{57}}
\]

First, calculate \( \sqrt{57} \):

\[
\sqrt{57} \approx 7.550
\]

Now calculate \( \frac{9}{\sqrt{57}} \):

\[
\frac{9}{\sqrt{57}} \approx \frac{9}{7.550} \approx 1.192
\]

Now multiply by the z-score:

\[
E = 2.576 \cdot 1.192 \approx 3.070752
\]

4. **Round to at least three decimal places:**

\[
E \approx 3.071
\]

Thus, the margin of error for a 99% confidence interval for \( \mu \) is approximately **3.071**.