To solve this problem and construct a 99.8% confidence interval for the mean reduction in cholesterol levels, let's follow these steps:

Step 1: Calculate the differences

For each individual, calculate the difference $d = \text{Before} - \text{After}$:

1. $238 - 218 = 20$
2. $240 - 214 = 26$
3. $230 - 180 = 50$
4. $245 - 205 = 40$
5. $241 - 168 = 73$
6. $281 - 201 = 80$
7. $241 - 170 = 71$
8. $285 - 182 = 103$

Differences: $[20, 26, 50, 40, 73, 80, 71, 103]$

Step 2: Find the sample mean ($\bar{d}$) and standard deviation ($s_d$):

1. Mean $\bar{d}$: $\bar{d} = \frac{\sum d}{n} = \frac{20 + 26 + 50 + 40 + 73 + 80 + 71 + 103}{8}$

2. Standard deviation $s_d$: Use the formula: $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}$

Step 3: Calculate the standard error ($SE$):

$SE = \frac{s_d}{\sqrt{n}}$

Step 4: Find the critical value $t^*$ for 99.8% confidence level with $n-1$ degrees of freedom ($df = 8-1 = 7$).

Step 5: Construct the confidence interval:

$\bar{d} \pm t^* \cdot SE$

Let me calculate these values for you. from scipy.stats import t

Given data

differences = [20, 26, 50, 40, 73, 80, 71, 103] n = len(differences)

Calculate mean and standard deviation of differences

mean_d = np.mean(differences) std_d = np.std(differences, ddof=1)

Standard error

SE = std_d / np.sqrt(n)

Degrees of freedom

df = n - 1

Critical t-value for 99.8% confidence level

alpha = 1 - 0.998 t_critical = t.ppf(1 - alpha / 2, df)

Confidence interval

margin_of_error = t_critical * SE lower_bound = mean_d - margin_of_error upper_bound = mean_d + margin_of_error

mean_d, std_d, SE, t_critical, (lower### Results:

1. Mean difference ($\bar{d}$): $57.9$ (rounded to one decimal place)
2. Standard deviation ($s_d$): $28.7$
3. Standard error ($SE$): $10.1$
4. Critical t-value ($t^*$): $4.79$
5. 99.8% Confidence Interval: $(9.3, 106.4)$

Part (b):

The confidence interval does include values greater than 90, meaning it does not contradict the claim that the mean reduction in cholesterol levels is more than 90 milligrams per deciliter.

Would you like further clarification or calculations? Here are 5 related questions to expand:

1. How do we interpret confidence intervals in hypothesis testing?
2. What would change in the interval if we used a 95% confidence level instead of 99.8%?
3. How does sample size influence the width of the confidence interval?
4. What does the critical $t^*$-value represent in this context?
5. How would the results change if there were an additional individual included?

Tip: Always check whether the interval's range supports the hypothesis or contradicts it!