To construct the 99% confidence interval for the mean, we will use the following formula for the confidence interval bounds:

$\text{CI} = \bar{X}_D \pm t \times s_{\bar{X}_D}$

where:

• $\bar{X}_D = 1.25$ is the sample mean,
• $s_{\bar{X}_D} = 0.45$ is the standard error,
• $t$ is the t-score corresponding to the desired confidence level and degrees of freedom (df = 10).

For a 99% confidence level with $df = 10$, the t-score is approximately 3.169.

Step 1: Calculate the Margin of Error

$\text{Margin of Error} = t \times s_{\bar{X}_D} = 3.169 \times 0.45 \approx 1.426$

Step 2: Determine the Lower Bound of the Confidence Interval

The lower bound of the confidence interval is: $\bar{X}_D - \text{Margin of Error} = 1.25 - 1.426 \approx -0.176$

Answer:

The lower bound of the 99% confidence interval is approximately -0.176.

Would you like further details or have any questions?

---

Here are some related questions to deepen understanding:

1. How would the confidence interval change with a 95% confidence level?
2. What would the upper bound of this confidence interval be?
3. How does the t-score affect the width of the confidence interval?
4. How is the confidence interval affected by a larger standard error?
5. What is the interpretation of this confidence interval in practical terms?

Tip: The t-score varies with both the confidence level and degrees of freedom—lower confidence levels or higher sample sizes reduce the t-score.