Solution:

1. Given Information:

   • First Population:
     • $n_1 = 6$, $\bar{x}_1 = 51.9$, $s_1 = 6.6$
   • Second Population:
     • $n_2 = 18$, $\bar{x}_2 = 48.3$, $s_2 = 7.6$
   • Significance level: $\alpha = 0.10$
   • Hypotheses:
     • $H_0: \mu_1 = \mu_2$
     • $H_a: \mu_1 \neq \mu_2$ (two-tailed test)

2. Calculate the Test Statistic: The test statistic formula for the difference of means is:

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

   • Variance estimates: $\frac{s_1^2}{n_1} = \frac{6.6^2}{6} = \frac{43.56}{6} = 7.26$ $\frac{s_2^2}{n_2} = \frac{7.6^2}{18} = \frac{57.76}{18} = 3.21$

   • Standard error: $SE = \sqrt{7.26 + 3.21} = \sqrt{10.47} = 3.236$

   • Test statistic: $t = \frac{51.9 - 48.3}{3.236} = \frac{3.6}{3.236} = 1.113$

3. Degrees of Freedom: Using the conservative estimate for degrees of freedom (smallest $n - 1$): $df = n_1 - 1 = 6 - 1 = 5$

4. Find the Critical t-value: For a two-tailed test at $\alpha = 0.10$ with $df = 5$: $t_{\text{critical}} = \pm 2.015$

5. Confidence Interval: The formula for the confidence interval is: $(\bar{x}_1 - \bar{x}_2) \pm t_{\text{critical}} \cdot SE$

   • Compute the margin of error: $ME = t_{\text{critical}} \cdot SE = 2.015 \cdot 3.236 = 6.516$

   • Confidence interval: $CI = (3.6 - 6.516, 3.6 + 6.516) = (-2.916, 10.116)$

6. Decision:

   • The confidence interval contains 0, so we fail to reject the null hypothesis.

7. Conclusion: There is not sufficient evidence to support the claim that the first population mean is not equal to the second population mean.

Final Answers:

1. Confidence Interval: $(-2.916, 10.116)$
2. Test Statistic: $t = 1.113$
3. Confidence Interval Contains Zero: Yes
4. Decision: Fail to reject the null hypothesis
5. Conclusion: There is not sufficient evidence to warrant rejection of the claim that the first population mean is not equal to the second population mean.

Would you like me to clarify any part of the calculation? Here are related questions:

1. How does the conservative degree of freedom approach impact results?
2. Why is the confidence interval key to the decision process?
3. How do different significance levels affect conclusions?
4. Can this test be generalized to unequal sample sizes beyond this case?
5. What are alternative methods to calculate degrees of freedom for t-tests?

Tip: Always confirm the degrees of freedom method used when reporting test statistics, as it can influence critical values and final conclusions.