To calculate the 90% confidence interval for the mean additional tax owed, we use the formula for the confidence interval of the mean:

$CI = \bar{x} \pm z_{\alpha/2} \cdot \left( \frac{\sigma}{\sqrt{n}} \right)$

Where:

• $\bar{x} = 3413$ (the sample mean)
• $\sigma = 2535$ (the standard deviation)
• $n = 81$ (the sample size)
• $z_{\alpha/2}$ is the critical value from the standard normal distribution for a 90% confidence level, which is approximately 1.645.

Now, calculating the margin of error (MOE):

$MOE = 1.645 \cdot \left( \frac{2535}{\sqrt{81}} \right) = 1.645 \cdot \left( \frac{2535}{9} \right) = 1.645 \cdot 281.67 \approx 463.55$

Thus, the confidence interval is:

$CI = 3413 \pm 463.55$

This gives:

$\text{Lower limit} = 3413 - 463.55 = 2949.45 \approx 2950$ $\text{Upper limit} = 3413 + 463.55 = 3876.55 \approx 3876$

Therefore, the 90% confidence interval is approximately between $2950 and $3876.

The correct choice in the multiple-choice options is:

• C: "One can be 90% confident that the mean additional tax owed is between $2950 and $3876."

Would you like further clarification or additional details?

Related questions:

1. What is the significance of the 90% confidence level in this context?
2. How would the confidence interval change with a larger sample size?
3. How does the standard deviation affect the width of the confidence interval?
4. What is the interpretation of a 95% confidence interval in comparison to a 90% interval?
5. Can we interpret a confidence interval as having a 90% probability that the true mean lies within the interval?

Tip: Confidence intervals provide a range of plausible values for the population mean, not a definitive range where the mean is guaranteed to fall.