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Let's work through this problem. The goal is to construct an 80% confidence interval for the difference in mean emissions between trucks running at 40°F and 80°F.

### Step 1: Calculate the differences
For each truck, calculate the difference \( d \) between the emissions at 40°F and 80°F:
\[
d_i = \text{emission at 40°F} - \text{emission at 80°F}
\]

Here are the differences for each truck:
\[
\begin{aligned}
d_1 &= 838.2 - 814.3 = 23.9 \\
d_2 &= 754.3 - 764.4 = -10.1 \\
d_3 &= 854.4 - 843.5 = 10.9 \\
d_4 &= 902.5 - 796.6 = 105.9 \\
d_5 &= 786.6 - 763.7 = 22.9 \\
d_6 &= 861.7 - 821.8 = 39.9 \\
d_7 &= 883.8 - 783.9 = 99.9 \\
d_8 &= 739.9 - 695.0 = 44.9 \\
d_9 &= 748.0 - 771.1 = -23.1 \\
d_{10} &= 848.1 - 795.2 = 52.9 \\
\end{aligned}
\]

### Step 2: Find the sample mean and standard deviation of the differences
Next, calculate the sample mean \( \bar{d} \) and the standard deviation \( s_d \) of these differences:

1. **Mean of the differences** \( \bar{d} \):
\[
\bar{d} = \frac{23.9 + (-10.1) + 10.9 + 105.9 + 22.9 + 39.9 + 99.9 + 44.9 + (-23.1) + 52.9}{10}
\]
\[
\bar{d} = \frac{367.2}{10} = 36.72
\]

2. **Standard deviation of the differences** \( s_d \):
First, calculate the squared differences from the mean:
\[
\begin{aligned}
(23.9 - 36.72)^2 &= 162.53 \\
(-10.1 - 36.72)^2 &= 2203.35 \\
(10.9 - 36.72)^2 &= 664.33 \\
(105.9 - 36.72)^2 &= 4806.63 \\
(22.9 - 36.72)^2 &= 190.49 \\
(39.9 - 36.72)^2 &= 10.12 \\
(99.9 - 36.72)^2 &= 4012.17 \\
(44.9 - 36.72)^2 &= 67.15 \\
(-23.1 - 36.72)^2 &= 3587.49 \\
(52.9 - 36.72)^2 &= 262.43 \\
\end{aligned}
\]

Now sum the squared differences:
\[
162.53 + 2203.35 + 664.33 + 4806.63 + 190.49 + 10.12 + 4012.17 + 67.15 + 3587.49 + 262.43 = 15966.68
\]

Then divide by \( n - 1 \) (degrees of freedom):
\[
s_d = \sqrt{\frac{15966.68}{9}} = \sqrt{1774.08} \approx 42.11
\]

### Step 3: Construct the confidence interval
The confidence interval is given by:
\[
\bar{d} \pm t_{\alpha/2} \times \frac{s_d}{\sqrt{n}}
\]
Where:
- \( \bar{d} = 36.72 \)
- \( s_d = 42.11 \)
- \( n = 10 \)
- \( t_{\alpha/2} \) is the critical value for a t-distribution with 9 degrees of freedom at 80% confidence. From t-tables, \( t_{0.10} \approx 1.383 \).

The margin of error is:
\[
1.383 \times \frac{42.11}{\sqrt{10}} \approx 1.383 \times 13.32 \approx 18.42
\]

So, the 80% confidence interval for the difference in emissions is:
\[
36.72 \pm 18.42
\]
This gives the interval:
\[
(18.30, 55.14)
\]

Thus, the 80% confidence interval for the difference in mean emissions is \( (18.30, 55.14) \).

Let me know if you'd like further details, or have any questions!

---

### Relative questions:
1. How do you interpret the confidence interval result in the context of emissions?
2. What assumptions are made in constructing this confidence interval?
3. How would the confidence interval change if the sample size were larger?
4. How do you calculate the t-critical value for different confidence levels?
5. Can you explain the difference between a confidence interval and a prediction interval?

**Tip:** The wider the confidence interval, the less certain you are about the estimate of the population mean difference.

Construct an 80% confidence interval for the difference in mean emissions between trucks running at 40°F and trucks running at 80°F. Let d represent the emissions at 40°F minus the emissions at 80°F.

Learn how to construct an 80% confidence interval for the difference in mean NOx emissions between trucks running at 40°F and 80°F. This problem involves paired data analysis, t-distribution, and calculation of standard deviation and mean. Suitable for undergraduate statistics or advanced high school math students.

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