The problem asks to construct a 99% confidence interval for two dependent data sets (A and B), assuming the paired data comes from a normally distributed population. Let's work through this step by step:

Data:

• Set A: $5.8, 6.8, 8.7, 5.7, 5.8$
• Set B: $8.2, 7.1, 7.0, 6.9, 8.3$

Step 1: Compute the differences

The differences between the paired observations (A - B) are: $d = (5.8 - 8.2), (6.8 - 7.1), (8.7 - 7.0), (5.7 - 6.9), (5.8 - 8.3)$ $d = -2.4, -0.3, 1.7, -1.2, -2.5$

Step 2: Calculate the mean of differences

$\bar{d} = \frac{-2.4 + (-0.3) + 1.7 + (-1.2) + (-2.5)}{5} = \frac{-4.7}{5} = -0.94$

Step 3: Compute the standard deviation of the differences

$s_d = \sqrt{\frac{(d_1 - \bar{d})^2 + (d_2 - \bar{d})^2 + \cdots + (d_n - \bar{d})^2}{n-1}}$ Substituting the values: $s_d = \sqrt{\frac{(-2.4 - (-0.94))^2 + (-0.3 - (-0.94))^2 + (1.7 - (-0.94))^2 + (-1.2 - (-0.94))^2 + (-2.5 - (-0.94))^2}{5-1}}$ $s_d = \sqrt{\frac{( -1.46)^2 + 0.64^2 + 2.64^2 + (-0.26)^2 + (-1.56)^2}{4}} = \sqrt{\frac{2.1316 + 0.4096 + 6.9696 + 0.0676 + 2.4336}{4}}$ $s_d = \sqrt{\frac{12.011}{4}} = \sqrt{3.00275} \approx 1.732$

Step 4: Find the critical t-value

For a 99% confidence interval with $n - 1 = 4$ degrees of freedom, the critical t-value $t_{\alpha/2}$ is approximately 4.604 from the t-distribution table.

Step 5: Calculate the margin of error

$\text{Margin of Error} = t_{\alpha/2} \times \frac{s_d}{\sqrt{n}} = 4.604 \times \frac{1.732}{\sqrt{5}} = 4.604 \times 0.774 \approx 3.561$

Step 6: Confidence interval

The 99% confidence interval is: $\bar{d} \pm \text{Margin of Error} = -0.94 \pm 3.561$ This gives the interval: $(-0.94 - 3.561, -0.94 + 3.561) = (-4.501, 2.621)$

Conclusion:

The correct answer choice is: $\boxed{B. (-4.508, 2.628)}$

Would you like further details or explanations on any step? Here are five related questions to explore:

1. How do you calculate the standard deviation of paired differences?
2. What is the significance of the t-distribution in constructing confidence intervals?
3. How would the confidence interval change if we used a 95% confidence level instead of 99%?
4. What are the assumptions behind using a paired t-test?
5. How would you interpret the result of this confidence interval in a practical context?

Tip: The margin of error increases with a higher confidence level due to the wider range of values that need to be covered.