To calculate the margin of error (ME) for the 95% confidence interval for the difference between the means of two dependent random samples, we proceed with the following steps:

Step 1: Compute the differences ($d_i$) between the paired values

Let $d_i = \text{Population 1 value} - \text{Population 2 value}$.

[ d = [30-44, 15-28, 28-36, 44-35, 26-18, 47-35, 32-22] = [-14, -13, -8, 9, 8, 12, 10] ]

Step 2: Calculate the mean of the differences ($\bar{d}$)

$\bar{d} = \frac{\sum d_i}{n}$

$\bar{d} = \frac{-14 - 13 - 8 + 9 + 8 + 12 + 10}{7} = \frac{4}{7} \approx 0.571429$

Step 3: Calculate the standard deviation of the differences ($s_d$)

The formula for $s_d$ is:

$s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}$

First, calculate each $(d_i - \bar{d})^2$: $(-14 - 0.571429)^2 = 212.469388, \quad (-13 - 0.571429)^2 = 179.755102, \quad (-8 - 0.571429)^2 = 73.469388$ $(9 - 0.571429)^2 = 70.612245, \quad (8 - 0.571429)^2 = 55.755102, \quad (12 - 0.571429)^2 = 129.183673, \quad (10 - 0.571429)^2 = 89.183673$

Sum these values: $\sum (d_i - \bar{d})^2 = 212.469388 + 179.755102 + 73.469388 + 70.612245 + 55.755102 + 129.183673 + 89.183673 = 810.428571$

Now calculate $s_d$: $s_d = \sqrt{\frac{810.428571}{7-1}} = \sqrt{\frac{810.428571}{6}} = \sqrt{135.071429} \approx 11.621158$

Step 4: Calculate the standard error of the mean difference ($SE_d$)

The formula for $SE_d$ is: $SE_d = \frac{s_d}{\sqrt{n}}$

$SE_d = \frac{11.621158}{\sqrt{7}} \approx \frac{11.621158}{2.645751} \approx 4.389150$

Step 5: Find the critical value for a 95% confidence level ($t^*$)

For $n-1 = 7-1 = 6$ degrees of freedom and a 95% confidence level, the critical $t$-value is approximately $t^* = 2.447$ (from a $t$-distribution table).

Step 6: Calculate the margin of error (ME)

The formula for $ME$ is: $ME = t^* \cdot SE_d$

$ME = 2.447 \cdot 4.389150 \approx 10.739613$

Final Answer:

The margin of error is: $\boxed{10.739613}$

Let me know if you'd like further clarification or assistance with this problem. Here are five related questions to expand on this topic:

1. How is the $t$-value chosen for constructing the confidence interval?
2. Why is the paired sample standard deviation used instead of individual standard deviations?
3. What changes in the calculation if the samples were independent instead of paired?
4. How does increasing the sample size affect the margin of error?
5. Can the confidence interval be interpreted as a probability statement?

Tip: Always verify whether the assumptions (normality and paired data) are satisfied before using $t$-distribution methods for confidence intervals.