To construct a 90% confidence interval for the difference in means between two independent groups, we will use the formula for the confidence interval in a two-sample $t$-test:

$CI = (\bar{x}_1 - \bar{x}_2) \pm t^* \cdot SE$

Where:

• $\bar{x}_1$ and $\bar{x}_2$ are the sample means for the two groups.
• $t^*$ is the critical $t$-value for the specified confidence level and degrees of freedom.
• $SE$ is the standard error of the difference in means.

The formula for the standard error of the difference is:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

Step-by-Step Solution:

1. Given Data:

   • $\bar{x}_1 = 7.42$, $s_1 = 1.88$, $n_1 = 12$
   • $\bar{x}_2 = 11.00$, $s_2 = 3.58$, $n_2 = 6$
   • Confidence level = 90%, so $t^*$ will correspond to $\alpha/2 = 0.05$ for the degrees of freedom.

2. Calculate the standard error ($SE$): $SE = \sqrt{\frac{1.88^2}{12} + \frac{3.58^2}{6}}$

3. Determine degrees of freedom using the Welch–Satterthwaite equation: $df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}$

4. Find the critical $t^*$-value for the calculated degrees of freedom.

5. Compute the confidence interval using: $CI = (\bar{x}_1 - \bar{x}_2) \pm t^* \cdot SE$

Let me calculate this step-by-step.import math from scipy.stats import t

Given data

x1 = 7.42 s1 = 1.88 n1 = 12 x2 = 11.00 s2 = 3.58 n2 = 6 confidence_level = 0.90

Calculate standard error (SE)

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

Calculate degrees of freedom using Welch–Satterthwaite equation

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

Critical t-value for the confidence level

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

Confidence interval

mean_diff = x1 - x2 margin_of_error = t_star * SE CI_lower = mean_diff - margin_of_error CI_upper = mean_diff + margin_of_error

CI_lower, CI_upper, SE, df(-6.574485076898247, -0.5855149231017536, 1.5590381650235507, 6.4184522879071775)(-6.574485076898247, -0.5855149231017536, 1.5590381650235507, 6.4184522879071775)successThe 90% confidence interval for the difference in means ($\mu_1 - \mu_2$) is:

$(-6.574, -0.586)$

Additional Details:

• Standard Error (SE): $1.559$
• Degrees of Freedom (df): $6.418$

If you have further questions or need clarification, let me know!

5 Related Questions:

1. How does changing the confidence level (e.g., to 95%) affect the confidence interval?
2. What assumptions are required for using a nonpooled $t$-procedure?
3. Why is the Welch–Satterthwaite equation used for degrees of freedom in this case?
4. How would increasing the sample sizes affect the confidence interval?
5. Can this method be used if the variances between groups are equal?

Tip:

When constructing confidence intervals, always verify the underlying assumptions (e.g., normality and independence) for more accurate results.