To compute the value of the test statistic, we will use the two-sample t-test formula. Since the population variances are assumed to be equal, we use the pooled standard deviation.

Step 1: Write down the given values

1. $\bar{X}_1 = 83.3$, $s_1 = 6.1$, $n_1 = 14$
2. $\bar{X}_2 = 87.3$, $s_2 = 5.3$, $n_2 = 12$
3. $\alpha = 0.05$
4. Assume $\sigma_1^2 = \sigma_2^2$ (equal population variances)

Step 2: Calculate the pooled standard deviation ($S_p$)

The formula for the pooled standard deviation is:

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

$S_p = \sqrt{\frac{(14 - 1)(6.1^2) + (12 - 1)(5.3^2)}{14 + 12 - 2}}$

$S_p = \sqrt{\frac{13(37.21) + 11(28.09)}{24}}$

$S_p = \sqrt{\frac{483.73 + 309.01}{24}} = \sqrt{\frac{792.74}{24}} \approx \sqrt{33.03} \approx 5.747$

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

The formula for the test statistic is:

$t = \frac{\bar{X}_1 - \bar{X}_2}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

$t = \frac{83.3 - 87.3}{5.747 \sqrt{\frac{1}{14} + \frac{1}{12}}}$

First, calculate the denominator:

$\sqrt{\frac{1}{14} + \frac{1}{12}} = \sqrt{0.07143 + 0.08333} = \sqrt{0.15476} \approx 0.393$

$t = \frac{-4}{5.747 \times 0.393} = \frac{-4}{2.260} \approx -1.770$

Final Answer:

The value of the test statistic is approximately:

$t = -1.770$

Would you like to proceed with further steps to interpret this result? Here are 5 relative questions to consider:

1. What is the critical t-value for a one-tailed test with a significance level of 0.05?
2. Should we reject the null hypothesis based on this test statistic?
3. How does assuming equal variances impact the result?
4. What are the implications of a negative test statistic in this context?
5. How would the result change if the variances were unequal?

Tip: Always verify the assumptions of normality and equal variances when performing a two-sample t-test.