Let's solve the problem step by step to construct the 99.8% confidence interval for the difference in mean emissions between trucks running at 40°F and 80°F.

Step 1: Define the Problem

The data represents emissions ($NO_x$) for 10 trucks at two temperatures, 40°F and 80°F. The difference $d$ is calculated as: $d = \text{Emission at } 40^\circ\text{F} - \text{Emission at } 80^\circ\text{F}$

Step 2: Compute the Differences ($d$)

We will calculate $d$ for each truck and then compute the mean ($\bar{d}$) and standard deviation ($s_d$) of these differences.

Step 3: Compute the Confidence Interval

Using the formula for a confidence interval of the mean difference: $\bar{d} \pm t^* \frac{s_d}{\sqrt{n}}$ Where:

• $t^*$ is the critical value from the $t$-distribution for 99.8% confidence level with $n-1$ degrees of freedom.
• $n$ is the number of trucks (10).
• $\bar{d}$ is the mean of the differences.
• $s_d$ is the standard deviation of the differences.

Let me calculate this for you.import numpy as np from scipy.stats import t

Data for emissions at 40°F and 80°F

emissions_40F = np.array([836.0, 756.1, 855.2, 903.3, 784.4, 863.5, 881.6, 740.7, 749.8, 847.9]) emissions_80F = np.array([817.1, 766.2, 841.3, 797.4, 764.5, 820.6, 785.7, 693.8, 773.9, 794.0])

Calculate differences

differences = emissions_40F - emissions_80F

Mean and standard deviation of differences

mean_diff = np.mean(differences) std_diff = np.std(differences, ddof=1) # Sample standard deviation n = len(differences)

99.8% confidence level

confidence_level = 0.998 degrees_freedom = n - 1 t_critical = t.ppf((1 + confidence_level) / 2, degrees_freedom)

Margin of error

margin_of_error = t_critical * (std_diff / np.sqrt(n))

Confidence interval

lower_bound = mean_diff - margin_of_error upper_bound = mean_diff + margin_of_error

(mean_diff, std_diff, lower_bound, upper_bound)(36.4, 41.81506905411014, -20.417030257104102, 93.2170302571041)(36.4, 41.81506905411014, -20.417030257104102, 93.2170302571041)successThe mean difference in emissions between trucks running at 40°F and 80°F is $36.4$, with a standard deviation of $41.82$. The 99.8% confidence interval for the difference in mean emissions is:

$-20.42 \leq \mu_d \leq 93.22$

Would you like a step-by-step breakdown of the calculations or have any questions?

Here are 5 related questions you might find interesting:

1. How does the critical $t$-value change with different confidence levels?
2. What are the implications of a wide confidence interval in this context?
3. How would the confidence interval change if the sample size were larger?
4. Why is the sample standard deviation used instead of the population standard deviation?
5. How does this analysis help evaluate the effect of temperature on emissions?

Tip: Always check if the data meets assumptions for $t$-tests, such as normality of differences or sufficient sample size.