To solve this problem, let's go through the steps for conducting a two-sample t-test assuming equal variances. We’ll find the test statistic for Part a and then construct the confidence interval for Part b.

Given Data

• Sample 1:

  • $n_1 = 18$
  • $\bar{x}_1 = 5.3$
  • $s_1 = 3.7$

• Sample 2:

  • $n_2 = 12$
  • $\bar{x}_2 = 7.9$
  • $s_2 = 4.5$

• Hypotheses:

  • $H_0 : (\mu_1 - \mu_2) = 0$
  • $H_a : (\mu_1 - \mu_2) \neq 0$

• Significance level $\alpha = 0.10$

Part (a): Finding the Test Statistic

Since we assume equal variances, we will pool the standard deviations.

1. Calculate the pooled standard deviation $s_p$: $s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}$

2. Calculate the test statistic $t$: The formula for the test statistic $t$ is: $t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

Step-by-Step Calculation

1. Pooled Standard Deviation: $s_p = \sqrt{\frac{(18 - 1) \cdot (3.7)^2 + (12 - 1) \cdot (4.5)^2}{18 + 12 - 2}}$ Simplifying: $s_p = \sqrt{\frac{17 \cdot 13.69 + 11 \cdot 20.25}{28}} = \sqrt{\frac{232.73 + 222.75}{28}} = \sqrt{\frac{455.48}{28}} \approx 4.04$

2. Test Statistic $t$: $t = \frac{5.3 - 7.9}{4.04 \sqrt{\frac{1}{18} + \frac{1}{12}}}$ Simplifying further: $t = \frac{-2.6}{4.04 \cdot \sqrt{0.0556 + 0.0833}} = \frac{-2.6}{4.04 \cdot \sqrt{0.1389}} = \frac{-2.6}{4.04 \cdot 0.3732} = \frac{-2.6}{1.508} \approx -1.72$

Part (b): 90% Confidence Interval for $\mu_1 - \mu_2$

The formula for the confidence interval is: $(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \cdot s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

For a 90% confidence interval and 28 degrees of freedom (since $n_1 + n_2 - 2 = 28$), the critical $t$-value $t_{0.05} \approx 1.701$.

1. Margin of Error: $\text{Margin of Error} = 1.701 \cdot 4.04 \cdot 0.3732 \approx 2.56$

2. Confidence Interval: $(5.3 - 7.9) \pm 2.56$ $-2.6 \pm 2.56$ So, the confidence interval is approximately: $(-5.16, -0.04)$

Summary

• Test Statistic: $t \approx -1.72$
• 90% Confidence Interval: $(-5.16, -0.04)$

Would you like additional details or have any further questions?

Related Questions to Explore

1. What are the assumptions behind using a two-sample t-test with equal variances?
2. How would the analysis change if we assumed unequal variances?
3. What does a 90% confidence interval mean in practical terms?
4. How do we interpret the result if the confidence interval includes zero?
5. How can sample size affect the test statistic and confidence interval?

Tip

In hypothesis testing, always confirm the assumptions (e.g., normality, equal variances) to ensure valid results.