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Let's analyze the problem step-by-step based on the provided information.

### Given Data
- **Dataset 1**: 
  - Sample mean (\(\bar{x}_1\)) = 69.2
  - Standard deviation (\(s_1\)) = 6.24
  - Sample size (\(n_1\)) = 18

- **Dataset 2**:
  - Sample mean (\(\bar{x}_2\)) = 80.3
  - Standard deviation (\(s_2\)) = 7.13
  - Sample size (\(n_2\)) = 18

- **Difference between datasets**:
  - Mean difference (\(\bar{d}\)) = -11.1
  - Standard deviation of differences (\(s_d\)) = 9.04

We need to find a 96% confidence interval for the mean difference \(\mu_d = \mu_1 - \mu_2\).

### Step-by-Step Solution

1. **Determine whether to use \( t \) or \( z \)**:
   - Since we are dealing with a sample (not population) and the sample size is relatively small (\(n = 18\)), we should use the **\( t \)-distribution**.

2. **Calculate the \( t \)-value**:
   - For a 96% confidence level with \(n - 1 = 17\) degrees of freedom, we need the critical \( t \)-value. For 96%, the two-tailed confidence level corresponds to 2% in each tail, so we are looking for the 0.98 quantile of the \( t \)-distribution with 17 degrees of freedom.
   - Using a \( t \)-table or calculator, the \( t \)-value approximately is:
     \[
     t_{0.98, 17} \approx 2.11
     \]

3. **Calculate the Margin of Error (ME)**:
   - The margin of error is given by:
     \[
     \text{ME} = t \times \frac{s_d}{\sqrt{n}}
     \]
   - Substituting the values:
     \[
     \text{ME} = 2.11 \times \frac{9.04}{\sqrt{18}} \approx 2.11 \times 2.13 = 4.4883 \approx 4.488
     \]

4. **Construct the 96% Confidence Interval**:
   - The confidence interval is:
     \[
     \bar{d} \pm \text{ME} = -11.1 \pm 4.488
     \]
   - Calculating the bounds:
     - Lower bound: \(-11.1 - 4.488 = -15.588\)
     - Upper bound: \(-11.1 + 4.488 = -6.612\)
   - Therefore, the 96% confidence interval is:
     \[
     (-15.588, -6.612)
     \]

### Summary of Answers
1. **Should you use \( t \) or \( z \)?**
   - Answer: **\( t \)**

2. **State the value of \( t \) to 3 decimals**:
   - Answer: **2.110**

3. **State the value of the margin of error (to 3 decimals)**:
   - Answer: **4.488**

4. **Find the 96% confidence interval**:
   - Answer: **(-15.588, -6.612)**

Would you like further details on any of these calculations, or do you have any questions?

### Follow-Up Questions
1. How would the confidence interval change with a larger sample size?
2. What effect would a higher confidence level, such as 99%, have on the interval?
3. How does the standard deviation of the differences impact the margin of error?
4. What assumptions must be met to use the \( t \)-distribution in this case?
5. How would you calculate the interval if the population variances were known?

### Tip
Remember that the choice between \( t \) and \( z \) depends on both sample size and whether the population standard deviation is known.

Suppose we have the following sample statistics. Dataset 1: x̄ = 69.2, s = 6.24, n = 18; Dataset 2: x̄ = 80.3, s = 7.13, n = 18; Dataset 1 - Dataset 2: d̄ = -11.1, s_d = 9.04. Find a 96% confidence interval for μ_d = μ_1 - μ_2. To do this, answer the following questions: 1) Should you use z or t? 2) State the value of z (to 2 decimals) or t (to 3 decimals). 3) State the value of the margin of error (to 3 decimals). 4) Find the 96% confidence interval.

Calculate a 96% confidence interval for the mean difference between two datasets, given sample means, standard deviations, and sample sizes. Includes step-by-step calculation of t-value, margin of error, and confidence interval bounds.

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