We are given the following information to construct a 99% confidence interval for the average number of kilometers driven by an automobile annually:

• Sample mean $(\bar{x}) = 23,500 \text{ km/year}$
• Population standard deviation $(\sigma) = 3900 \text{ km/year}$
• Sample size $(n) = 100$
• Confidence level = 99%

Step 1: Determine the z-score for a 99% confidence level

For a 99% confidence interval, the z-score corresponding to the tail area of 0.005 (which is $\frac{1 - 0.99}{2} = 0.005$) is approximately $z = 2.576$.

Step 2: Compute the standard error (SE)

The standard error of the mean is calculated as: $SE = \frac{\sigma}{\sqrt{n}} = \frac{3900}{\sqrt{100}} = \frac{3900}{10} = 390 \text{ km/year}$

Step 3: Calculate the margin of error (MOE)

The margin of error is given by: $MOE = z \times SE = 2.576 \times 390 = 1004.64 \text{ km/year}$

Step 4: Calculate the confidence interval

The 99% confidence interval for the population mean is: $\text{Confidence interval} = \bar{x} \pm MOE = 23,500 \pm 1004.64$ This gives:

• Lower boundary: $23,500 - 1004.64 = 22,495.36$
• Upper boundary: $23,500 + 1004.64 = 24,504.64$

Thus, the 99% confidence interval is approximately:

$(22,495.36 \text{ km/year}, 24,504.64 \text{ km/year})$

Answers:

• Lower Confidence Interval boundary: 22,495.36 km/year
• Upper Confidence Interval boundary: 24,504.64 km/year

Would you like further details on how to interpret these results, or any follow-up questions?

5 Related Questions:

1. How would the confidence interval change if we increased the sample size?
2. What is the effect of changing the confidence level to 95% on the interval?
3. How would the confidence interval be affected if the population standard deviation were larger?
4. Why do we use the z-score instead of the t-score in this calculation?
5. What are the assumptions required for constructing this confidence interval?

Tip:

A higher confidence level results in a wider confidence interval, meaning greater certainty but less precision in the estimate.