To solve the problem of finding a 90% confidence interval for the true population mean income for households in Charleston County, we'll use the following formula for the confidence interval of the mean:

$\text{Confidence Interval} = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right)$

Where:

• $\bar{x}$ is the sample mean.
• $Z$ is the Z-score corresponding to the desired confidence level.
• $\sigma$ is the standard deviation of the sample.
• $n$ is the sample size.

Given values:

• Sample mean ($\bar{x}$) = 24,477
• Standard deviation ($\sigma$) = 5,389
• Sample size ($n$) = 81
• Confidence level = 90%, which corresponds to a Z-score of approximately 1.645.

Step 1: Calculate the Standard Error (SE)

$\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{5,389}{\sqrt{81}} = \frac{5,389}{9} \approx 598.78$

Step 2: Calculate the Margin of Error (ME)

$\text{ME} = Z \times \text{SE} = 1.645 \times 598.78 \approx 984.13$

Step 3: Calculate the Confidence Interval

$\text{Lower Bound} = \bar{x} - \text{ME} = 24,477 - 984.13 \approx 23,493$ $\text{Upper Bound} = \bar{x} + \text{ME} = 24,477 + 984.13 \approx 25,461$

Conclusion

The 90% confidence interval for the true population mean income for households in Charleston County is approximately: $23,493 < \mu < 25,461$

Would you like more details or have any questions about this solution?

Here are some related questions to consider:

1. What would the confidence interval be if the confidence level was increased to 95%?
2. How does the sample size affect the width of the confidence interval?
3. What assumptions do we make about the population distribution when constructing this confidence interval?
4. How do we determine the appropriate Z-score for different confidence levels?
5. What happens to the confidence interval if the sample mean changes while the other values remain the same?

Tip: Remember that increasing the sample size decreases the standard error, leading to a narrower confidence interval for the same confidence level.