This problem requires performing a two-sample t-test for the hypothesis:

Hypotheses

1. Null Hypothesis ($H_0$): $\mu_1 \leq \mu_2$ (The mean time without the tool is less than or equal to the mean time with the tool.)
2. Alternative Hypothesis ($H_a$): $\mu_1 > \mu_2$ (The mean time without the tool is greater than the mean time with the tool.)

Formula for Test Statistic

$t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$ where:

• $\bar{x}_1 = 4.5$, $\bar{x}_2 = 4.2$ are the sample means,
• $n_1 = 6$, $n_2 = 8$ are the sample sizes,
• $s_1 = 0.3$, $s_2 = 0.2$ are the sample standard deviations,
• $s_p$ is the pooled standard deviation.

Step 1: Compute the pooled standard deviation ($s_p$)

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

Substitute the values: $s_p = \sqrt{\frac{(6 - 1)(0.3^2) + (8 - 1)(0.2^2)}{6 + 8 - 2}}$ $s_p = \sqrt{\frac{5(0.09) + 7(0.04)}{12}} = \sqrt{\frac{0.45 + 0.28}{12}} = \sqrt{\frac{0.73}{12}} = \sqrt{0.060833} \approx 0.246$

Step 2: Compute the t-statistic

$t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$ Substitute the values: $t = \frac{4.5 - 4.2}{0.246 \sqrt{\frac{1}{6} + \frac{1}{8}}}$ $t = \frac{0.3}{0.246 \sqrt{0.1667 + 0.125}} = \frac{0.3}{0.246 \sqrt{0.2917}} = \frac{0.3}{0.246 \times 0.5402} = \frac{0.3}{0.1329} \approx 2.258$

Final Answer:

The test statistic is approximately $t = 2.258$.

Would you like help with interpreting this result or calculating the critical value?