To calculate a 99% confidence interval for the population mean $\mu$ based on a sample, we follow these steps:

Step 1: Calculate the sample mean ($\bar{x}$)

The sample data is: $82.7, 95.2, 93.2, 56.9, 81.6, 84.7, 85.1, 94.5, 109.5, 72.3$

$\bar{x} = \frac{82.7 + 95.2 + 93.2 + 56.9 + 81.6 + 84.7 + 85.1 + 94.5 + 109.5 + 72.3}{10}$

Step 2: Calculate the sample standard deviation ($s$)

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

Step 3: Find the t-score for a 99% confidence level with $n - 1 = 9$ degrees of freedom

Using the t-distribution table for a 99% confidence level and 9 degrees of freedom, we find the t-score.

Step 4: Calculate the margin of error (ME)

$\text{ME} = t \times \frac{s}{\sqrt{n}}$

Step 5: Determine the confidence interval

$\text{Confidence Interval} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right)$

I'll compute these values.The 99% confidence interval for the population mean $\mu$ based on the given sample is approximately:

$(70.93, 100.21)$

This interval suggests that we are 99% confident that the population mean lies within this range.

Would you like further details on this process or have any questions?

Here are five related questions to explore further:

1. How would the confidence interval change with a 95% confidence level?
2. What would the confidence interval be if we had a larger sample size?
3. How does the standard deviation of the sample affect the confidence interval width?
4. What assumptions are made when using a t-distribution for confidence intervals?
5. How would the interval differ if we assumed the population standard deviation was known?

Tip: Remember, increasing the confidence level increases the interval width, as higher confidence requires capturing more possible values for the population mean.