To find the 95% confidence interval for the mean difference between two independent samples, we’ll use the following approach.

Given Data

• Sample 1: $n_1 = 18$, $\text{SS}_1 = 102$, $M_1 = 12$
• Sample 2: $n_2 = 5$, $\text{SS}_2 = 171$, $M_2 = 8$

Steps

1. Calculate the pooled variance ($s_p^2$): The pooled variance is calculated as: $s_p^2 = \frac{\text{SS}_1 + \text{SS}_2}{n_1 + n_2 - 2}$

2. Compute the standard error for the difference in means ($\text{SE}_{M_1 - M_2}$): $\text{SE}_{M_1 - M_2} = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$

3. Calculate the t-value for a 95% confidence interval: The degrees of freedom are $n_1 + n_2 - 2$, which is $18 + 5 - 2 = 21$.

4. Determine the confidence interval for the mean difference: The mean difference is $M_1 - M_2 = 12 - 8 = 4$. The confidence interval is: $(M_1 - M_2) \pm t \cdot \text{SE}_{M_1 - M_2}$

Let's calculate each step.

Step 1: Calculate $s_p^2$

$s_p^2 = \frac{102 + 171}{18 + 5 - 2} = \frac{273}{21} = 13$

Step 2: Compute $\text{SE}_{M_1 - M_2}$

$\text{SE}_{M_1 - M_2} = \sqrt{13 \left(\frac{1}{18} + \frac{1}{5}\right)} = \sqrt{13 \left(0.0556 + 0.2\right)} = \sqrt{13 \times 0.2556} = \sqrt{3.3228} \approx 1.822$

Step 3: Find the t-value

For a 95% confidence interval with 21 degrees of freedom, the critical t-value is approximately 2.080.

Step 4: Calculate the confidence interval

$\text{CI} = (M_1 - M_2) \pm t \cdot \text{SE}_{M_1 - M_2}$ $= 4 \pm 2.080 \cdot 1.822$ $= 4 \pm 3.79$

Final Answer

• Lower limit: $4 - 3.79 = 0.21$
• Upper limit: $4 + 3.79 = 7.79$

Thus, the 95% confidence interval for the mean difference is approximately: $(0.21, 7.79)$

Would you like more details on any of these steps, or do you have other questions?

Here are some related questions to explore:

1. How is the pooled variance calculated when sample sizes differ?
2. Why is the t-distribution used instead of the z-distribution here?
3. How does increasing sample sizes affect the confidence interval width?
4. What are degrees of freedom, and why are they calculated as $n_1 + n_2 - 2$?
5. How would the confidence interval change with a different confidence level, like 99%?

Tip: In statistics, wider confidence intervals typically indicate more uncertainty about the population parameter, while narrower intervals suggest more precision.