To solve the problem of obtaining a 90% confidence interval for the difference between the mean number of acute postoperative days ($\mu_1 - \mu_2$) in the dynamic and static systems, we'll use the provided information for a two-sample t-interval:

Given data:

• $\bar{x}_1 = 7.25$, $s_1 = 1.71$, $n_1 = 7$ (dynamic system)
• $\bar{x}_2 = 10.67$, $s_2 = 4.18$, $n_2 = 7$ (static system)

Formula for the confidence interval:

$(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

Where:

• $(\bar{x}_1 - \bar{x}_2)$ is the difference in sample means.
• $t^*$ is the critical value from the t-distribution for a 90% confidence level with degrees of freedom calculated using the Welch-Satterthwaite equation.
• $s_1^2 / n_1 + s_2^2 / n_2$ is the standard error squared.

Step-by-step:

1. Difference in sample means: $\bar{x}_1 - \bar{x}_2 = 7.25 - 10.67 = -3.42$

2. Standard error (SE): $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{1.71^2}{7} + \frac{4.18^2}{7}}$

3. Degrees of freedom: Use the Welch-Satterthwaite 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. Critical t-value ($t^*$): Find the $t^*$ value for 90% confidence and the calculated $df$.

Let me compute these values.import math from scipy.stats import t

Given data

x1, s1, n1 = 7.25, 1.71, 7 # Dynamic system x2, s2, n2 = 10.67, 4.18, 7 # Static system

Step 1: Difference in means

mean_difference = x1 - x2

Step 2: Standard error (SE)

se = math.sqrt((s12 / n1) + (s22 / n2))

Step 3: Degrees of freedom (df) using Welch-Satterthwaite equation

numerator = (s12 / n1 + s22 / n2) ** 2 denominator = ((s12 / n1) ** 2 / (n1 - 1)) + ((s22 / n2) ** 2 / (n2 - 1)) df = numerator / denominator

Step 4: Critical t-value (t*)

confidence_level = 0.90 alpha = 1 - confidence_level t_critical = t.ppf(1 - alpha / 2, df)

Step 5: Confidence interval

margin_of_error = t_critical * se ci_lower = mean_difference - margin_of_error ci_upper = mean_difference + margin_of_error

ci_lower, ci_upper(-6.596618950598896, -0.24338104940110394)(-6.596618950598896, -0.24338104940110394)successThe 90% confidence interval for the difference in mean postoperative days between the dynamic and static systems is:

$(-6.597, -0.243)$

This means the mean difference in postoperative days is likely between -6.597 and -0.243.

Would you like a detailed explanation of any step, or do you have additional questions? Here are some related questions to consider:

1. How does the Welch-Satterthwaite formula work for degrees of freedom?
2. Why is the t-distribution used instead of the normal distribution?
3. What would change if the confidence level was increased to 95%?
4. How do standard deviations impact the width of the confidence interval?
5. What assumptions are necessary for this test to be valid?

Tip: Always ensure sample sizes and variances are adequate when using non-pooled t-tests, as the test is sensitive to deviations in variance.