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: $-15.705 < \mu_d < -5.895$

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.