Let's walk through the steps for constructing a 95% confidence interval for the difference between two population means, $\mu_1 - \mu_2$, given the data.

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Formula for Confidence Interval for Two Means (Unpooled)

The formula for the confidence interval is:
$(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ where:

• $\bar{x}_1 = 2.41$, $s_1 = 0.99$, $n_1 = 25$
• $\bar{x}_2 = 2.56$, $s_2 = 0.73$, $n_2 = 27$
• $t^*$ is the critical value from the t-distribution at 95% confidence with appropriate degrees of freedom (df).

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Step 1: Calculate the Difference in Sample Means

$\bar{x}_1 - \bar{x}_2 = 2.41 - 2.56 = -0.15$

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Step 2: Calculate the Standard Error (SE) of the Difference

The standard error for the difference is: $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

$s_1^2 = (0.99)^2 = 0.9801, \quad s_2^2 = (0.73)^2 = 0.5329$

$SE = \sqrt{\frac{0.9801}{25} + \frac{0.5329}{27}}$

$SE = \sqrt{0.039204 + 0.019739} = \sqrt{0.058943} \approx 0.2428$

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Step 3: Determine the Critical Value $t^*$

To find the critical value $t^*$, we need the degrees of freedom (df). Since we are not pooling variances, we use the Welch-Satterthwaite equation for $df$:

$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}$

$df = \frac{(0.039204 + 0.019739)^2}{\frac{(0.039204)^2}{24} + \frac{(0.019739)^2}{26}}$

$df \approx \frac{(0.058943)^2}{\frac{0.001537}{24} + \frac{0.000389}{26}} = \frac{0.003473}{0.000064 + 0.000015} = \frac{0.003473}{0.000079} \approx 43.97$

Since $df \approx 44$, the critical value $t^*$ for a 95% confidence interval is approximately 2.015.

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Step 4: Construct the Confidence Interval

$CI = -0.15 \pm 2.015 \cdot 0.2428$

$2.015 \cdot 0.2428 \approx 0.4893$

$CI = -0.15 \pm 0.4893$

$\text{Lower bound} = -0.15 - 0.4893 = -0.6393 \approx -0.64$ $\text{Upper bound} = -0.15 + 0.4893 = 0.3393 \approx 0.34$

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Step 5: Final Answer

The 95% confidence interval for $\mu_1 - \mu_2$ is: $(-0.64, 0.34)$

This means we are 95% confident that the true difference in mean weight loss between the two groups is between -0.64 and 0.34 units.

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Let me know if you need further clarification!

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Here are 5 follow-up questions for practice:

1. What happens to the confidence interval if we increase the sample size?
2. How would the interval change if the two groups had similar standard deviations?
3. How do we interpret the confidence interval if it includes zero?
4. Why is the t-distribution used instead of the z-distribution in this case?
5. How would pooling the variances affect the result?

Tip: If you struggle with degrees of freedom calculations, some tools and calculators can automate it!