To compute the 99% confidence interval for the true mean resale value of the sedan, we use the formula for the confidence interval when the population standard deviation is unknown, and the sample size is small ($n < 30$). This involves the t-distribution.

The formula for the confidence interval is:

$\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}$

Where:

• $\bar{x} = 12,800$ (sample mean resale value),
• $s = 600$ (sample standard deviation),
• $n = 17$ (sample size),
• $t^*$ is the critical value from the t-distribution for $n-1 = 16$ degrees of freedom and a 99% confidence level.

Step 1: Find $t^*$

Using a t-table or calculator, for 16 degrees of freedom and a 99% confidence level (two-tailed test, so $\alpha = 0.01$), $t^* \approx 2.921$.

Step 2: Compute the standard error

The standard error of the mean is: $SE = \frac{s}{\sqrt{n}} = \frac{600}{\sqrt{17}} \approx \frac{600}{4.123} \approx 145.53$

Step 3: Compute the margin of error

The margin of error is: $ME = t^* \cdot SE = 2.921 \cdot 145.53 \approx 424.92$

Step 4: Calculate the confidence interval

The confidence interval is: $\bar{x} \pm ME = 12,800 \pm 424.92$

Thus, the 99% confidence interval is approximately: $(12,375.08, 13,224.92)$

Final Answer:

The 99% confidence interval for the true mean resale value of the sedan is (12,375.08, 13,224.92).

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Would you like further clarification or help with this process?

Here are some related questions to consider:

1. What is the difference between a t-distribution and a z-distribution?
2. How does increasing the confidence level affect the width of the confidence interval?
3. Why is the sample standard deviation used here instead of the population standard deviation?
4. How would the interval change if the sample size were larger?
5. What role does the critical value ($t^*$) play in determining the interval?

Tip: When dealing with small sample sizes, always check whether the population is approximately normal, as the t-distribution assumes this.