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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.

You intend to estimate a population mean μ with the following sample: [94.3, 84.5, 73.5, 78.9, 86.1, 82.9, 91, 81.1]. Find the 99.9% confidence interval for the population mean, given that the population is normally distributed. Enter the answer as an open interval accurate to two decimal places.

Learn how to calculate a 99.9% confidence interval for the population mean with sample data [94.3, 84.5, 73.5, 78.9, 86.1, 82.9, 91, 81.1]. This solution covers key statistical concepts, including sample mean, standard deviation, and t-distribution for a small sample size.

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