To compute the test statistic for comparing the means of two independent populations with equal variances, we use the formula for the two-sample t-test:

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

Where:

• $\bar{X}_1 = 4.9$ (mean of Population 1)
• $\bar{X}_2 = 4.7$ (mean of Population 2)
• $n_1 = 9$ (sample size of Population 1)
• $n_2 = 7$ (sample size of Population 2)
• $s_1 = 0.5$ (standard deviation of Population 1)
• $s_2 = 0.3$ (standard deviation of Population 2)

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

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

Substitute the values:

$s_p = \sqrt{\frac{(9 - 1)(0.5^2) + (7 - 1)(0.3^2)}{9 + 7 - 2}}$

$s_p = \sqrt{\frac{8(0.25) + 6(0.09)}{14}} = \sqrt{\frac{2 + 0.54}{14}} = \sqrt{\frac{2.54}{14}} \approx \sqrt{0.1814} \approx 0.426$

Step 2: Calculate the test statistic:

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

Substitute the values:

$t = \frac{4.9 - 4.7}{0.426 \sqrt{\frac{1}{9} + \frac{1}{7}}}$

$t = \frac{0.2}{0.426 \sqrt{0.1111 + 0.1429}} = \frac{0.2}{0.426 \sqrt{0.254}}$

$t = \frac{0.2}{0.426 \times 0.504} = \frac{0.2}{0.2147} \approx 0.932$

Final Answer:

The value of the test statistic is approximately 0.932.

Would you like to proceed with interpreting the result or calculating the critical value? Here are some related questions:

1. What is the critical value for the two-tailed test at a 0.01 significance level?
2. Can we conclude that the tool has significantly reduced the painting time?
3. What assumptions must be satisfied for the two-sample t-test to be valid?
4. How can the equal variance assumption affect the test results?
5. How does sample size impact the test statistic?

Tip: Always check the assumptions of normality and equal variances before conducting hypothesis tests.