To solve this problem, we need to calculate a 99.9% confidence interval for the population mean, based on the given sample data. Here are the steps involved:

Step 1: Calculate the Sample Mean ($\bar{x}$) and Sample Standard Deviation (s)

The sample data provided are: $94.3, 84.5, 73.5, 78.9, 86.1, 82.9, 91, 81.1$

Sample Mean

$\bar{x} = \frac{\sum x}{n}$

Sample Standard Deviation

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$

Step 2: Determine the Critical Value (t*) for 99.9% Confidence Level

Since the sample size is 8, we have $n - 1 = 7$ degrees of freedom. For a 99.9% confidence interval, we will use the t-distribution critical value for $7$ degrees of freedom and a 0.999 confidence level.

Step 3: Calculate the Margin of Error

$\text{Margin of Error} = t^* \cdot \frac{s}{\sqrt{n}}$

Step 4: Find the Confidence Interval

The confidence interval is: $\left(\bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error}\right)$

Let's calculate these values.The calculations are as follows:

1. Sample Mean ($\bar{x}$): 84.04
2. Sample Standard Deviation (s): 6.61
3. Critical t-value for 99.9% confidence (with 7 degrees of freedom): 5.408
4. Margin of Error: 12.63

99.9% Confidence Interval:

The 99.9% confidence interval for the population mean is: $(71.40, 96.67)$

This interval, written in open-interval form, is: $(71.40, 96.67)$

Would you like more details on any step, or have further questions?

Here are some related questions that could help deepen your understanding:

1. How does the choice of confidence level (e.g., 95% vs 99.9%) affect the confidence interval?
2. Why do we use the t-distribution instead of the z-distribution here?
3. How would the confidence interval change with a larger sample size?
4. What are degrees of freedom, and why are they relevant in finding the t-value?
5. How would you interpret this confidence interval in practical terms?

Tip: Remember that the margin of error increases as confidence level increases because higher confidence requires a wider range.