To construct the confidence interval for the difference in means ($\mu_1 - \mu_2$) for the given data, we need to use the following steps:

Steps:

1. Identify the data:
   • Regular Coke ($x_1$) and Diet Coke ($x_2$) weights.
2. Calculate sample statistics:
   • Sample means ($\bar{x}_1$ and $\bar{x}_2$).
   • Sample standard deviations ($s_1$ and $s_2$).
   • Sample sizes ($n_1$ and $n_2$).
3. Determine the degrees of freedom using Welch's formula: $df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}$
4. Calculate the standard error (SE): $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
5. Find the critical t-value ($t^*$) for the given confidence level.
6. Construct the confidence interval: $(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot SE$

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

Data for Regular Coke and Diet Coke

regular_coke = np.array([0.81915, 0.81498, 0.81628, 0.82111, 0.81809, 0.82473, 0.80622, 0.81283, 0.81722, 0.81102, 0.82513, 0.82644]) diet_coke = np.array([0.77731, 0.77582, 0.78960, 0.78684, 0.78441, 0.78606, 0.78062, 0.78303, 0.78522, 0.78785, 0.78812, 0.78264, 0.79227, 0.78522, 0.78719, 0.78127])

Sample statistics

mean1 = np.mean(regular_coke) mean2 = np.mean(diet_coke) std1 = np.std(regular_coke, ddof=1) std2 = np.std(diet_coke, ddof=1) n1 = len(regular_coke) n2 = len(diet_coke)

Degrees of freedom (Welch-Satterthwaite equation)

numerator = (std12 / n1 + std22 / n2)2 denominator = ((std12 / n1)2 / (n1 - 1)) + ((std22 / n2)**2 / (n2 - 1)) df = numerator / denominator

Standard error

SE = np.sqrt((std12 / n1) + (std22 / n2))

Confidence interval (95% confidence level)

confidence_level = 0.95 t_critical = t.ppf(1 - (1 - confidence_level) / 2, df) margin_of_error = t_critical * SE

Confidence interval

lower_limit = (mean1 - mean2) - margin_of_error upper_limit = (mean1 - mean2) + margin_of_error

lower_limit, upper_limit(0.028859793020363705, 0.03748979031296926)(0.028859793020363705, 0.03748979031296926)successThe 95% confidence interval for the difference in means ($\mu_1 - \mu_2$) is:

$0.02886 < \mu_1 - \mu_2 < 0.03749$

(Rounded to five decimal places as needed.)

Let me know if you need a breakdown of any step or additional help with the hypothesis test. Here are some related questions you might consider:

1. How to interpret this confidence interval in the context of the hypothesis test?
2. What are the assumptions required for this calculation to be valid?
3. How to calculate the exact p-value for the hypothesis test?
4. How does the choice of confidence level affect the interval width?
5. Can this method be applied if the variances are assumed to be equal?

Tip: Always ensure your data meets the assumptions (normality, independence) before performing parametric tests.