A study was conducted to find the mean weight of the residents in a certain town. The study examined a random sample of 46 residents and found the mean weight to be 163 pounds with a standard deviation of 23 pounds.

At the 95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write ±).

To determine the margin of error for the mean weight of residents in the town, let's follow these steps:

1. Identify the key values:
- Sample size ([tex]\(n\)[/tex]) = 46
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 163 pounds
- Standard deviation ([tex]\(\sigma\)[/tex]) = 23 pounds
- Confidence level = 95%

2. Determine the z-score for the 95% confidence level:
- For a 95% confidence level, the z-score is approximately 1.96. This is because, in a normal distribution, 95% of the data falls within approximately 1.96 standard deviations from the mean.

3. Calculate the standard error (SE):
- The standard error of the mean is calculated using the formula:

[tex]\[
\text{Standard Error (SE)} = \frac{\text{Standard Deviation}}{\sqrt{\text{Sample Size}}}
\][/tex]

- Substitute the values:

[tex]\[
\text{SE} = \frac{23}{\sqrt{46}} \approx 3.391
\][/tex]

4. Calculate the margin of error (ME):
- The margin of error can be found using the formula:

[tex]\[
\text{Margin of Error (ME)} = z \times \text{SE}
\][/tex]

- Substitute the values:

[tex]\[
\text{ME} = 1.96 \times 3.391 \approx 6.6
\][/tex]

5. Conclusion:
- The margin of error for the mean weight at the 95% confidence level is approximately 6.6 pounds.

This means that you can be 95% confident that the true mean weight of all residents in the town falls within 6.6 pounds of 163 pounds.