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Let's go through each part of the question step-by-step:

### 1. Should you use \( z \) or \( t \)?
Since the sample size \( n = 10 \) is relatively small, and we typically assume a \( t \)-distribution for smaller sample sizes (especially under 30), we should use \( t \).

**Answer:** \( t \)

### 2. State the value of \( t \) (to 3 decimals).
To find the \( t \)-value for a 91% confidence level with \( n - 1 = 9 \) degrees of freedom, we look up the critical value in the \( t \)-distribution table or calculate it using statistical software.

For a 91% confidence level:
- Confidence level of 91% corresponds to a significance level \( \alpha = 1 - 0.91 = 0.09 \).
- This is a two-tailed test, so \( \alpha/2 = 0.045 \) for each tail.

Using a \( t \)-table or software for 9 degrees of freedom and a cumulative probability of 0.955, the \( t \)-value is approximately:

**Answer:** \( t \approx 1.833 \) (to 3 decimals).

### 3. State the value of the margin of error (to 3 decimals).
The margin of error (ME) for the difference \( \mu_d = \mu_1 - \mu_2 \) is calculated as:

\[
\text{ME} = t \times \frac{s_d}{\sqrt{n}}
\]

Given:
- \( t = 1.833 \)
- \( s_d = 8.46 \)
- \( n = 10 \)

Substitute these values:

\[
\text{ME} = 1.833 \times \frac{8.46}{\sqrt{10}} \approx 4.905
\]

**Answer:** Margin of error \( \approx 4.905 \)

### 4. Find the 91% confidence interval.
The confidence interval for \( \mu_d \) is calculated as:

\[
\mu_d \pm \text{ME} = -10.8 \pm 4.905
\]

So:

\[
-10.8 - 4.905 < \mu_d < -10.8 + 4.905
\]

This gives:

\[
-15.705 < \mu_d < -5.895
\]

**Answer:** \( -15.705 < \mu_d < -5.895 \)

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Let me know if you have any questions or if you'd like further details on any of these steps.

### Related Questions:
1. How is the \( t \)-distribution different from the \( z \)-distribution?
2. What conditions must be met to use a \( t \)-distribution?
3. Why do we divide \( \alpha \) by 2 for two-tailed tests?
4. What is the margin of error, and why is it important in confidence intervals?
5. How would the confidence interval change if the sample size was increased?

### Tip:
For smaller sample sizes, especially under 30, the \( t \)-distribution provides a more accurate estimate than the \( z \)-distribution due to its adjustment for sample variability.

Find a 91% confidence interval for the difference in means (μ₁ - μ₂) using sample data for two datasets with sample sizes of n=10.

Learn to calculate a 91% confidence interval for the difference in means (μ₁ - μ₂) between two datasets using t-distribution. This problem involves small sample sizes, margin of error, and confidence intervals, making it suitable for college-level statistics.

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