To compute the test statistic for comparing the means of two populations, we use a two-sample $t$-test formula:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

where:

• $\bar{x}_1, \bar{x}_2$: sample means
• $s_p^2$: pooled variance
• $n_1, n_2$: sample sizes

Step 1: Compute the pooled variance ($s_p^2$):

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

Given:

• $s_1 = 3.4$, $s_2 = 5.9$
• $n_1 = 15$, $n_2 = 18$

First, calculate $s_1^2$ and $s_2^2$: $s_1^2 = (3.4)^2 = 11.56, \quad s_2^2 = (5.9)^2 = 34.81$

Now substitute into the pooled variance formula: $s_p^2 = \frac{(15 - 1)(11.56) + (18 - 1)(34.81)}{15 + 18 - 2}$ $s_p^2 = \frac{14(11.56) + 17(34.81)}{31}$ $s_p^2 = \frac{161.84 + 591.77}{31} = \frac{753.61}{31} \approx 24.310$

Step 2: Compute the standard error (SE):

$SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$ $SE = \sqrt{24.310 \left(\frac{1}{15} + \frac{1}{18}\right)}$ $SE = \sqrt{24.310 \left(0.0667 + 0.0556\right)} = \sqrt{24.310 \cdot 0.1223}$ $SE = \sqrt{2.9717} \approx 1.724$

Step 3: Compute the test statistic ($t$):

$t = \frac{\bar{x}_1 - \bar{x}_2}{SE}$

Given: $\bar{x}_1 = 81.2, \quad \bar{x}_2 = 83.8$

$t = \frac{81.2 - 83.8}{1.724} = \frac{-2.6}{1.724} \approx -1.508$

Final Answer:

The test statistic is approximately: $t = -1.508$

Let me know if you would like further clarification or details.

Follow-up Questions:

1. What is the critical value for the $t$-test at $\alpha = 0.10$?
2. What is the interpretation of the test statistic in this context?
3. How does assuming equal variances affect the calculation?
4. What would happen if the population variances were unequal?
5. How do sample sizes impact the pooled variance?

Tip: Always check assumptions (normality, equal variances) before conducting a $t$-test to ensure validity.