2. A credit and insurance organization has developed a new, high-tech method of training new sales staff. The company obtained a sample of 20 employees trained in the original way and found average daily sales of L16,512, with a sample standard deviation of L783.12. They also took a sample of 18 employees trained with the new method and found an average daily sales of L16,944, with a sample standard deviation of L596.16. For α = 0.01, can the company conclude that average daily sales increase with the new plan?

Given data: $$n_1 = 20, \quad \bar{x}_1 = 16512, \quad s_1 = 783.12$$ $$n_2 = 18, \quad \bar{x}_2 = 16944, \quad s_2 = 596.16 $$ $$\alpha = 0.01$$ **Hypotheses:** $$H_0 : \mu_1 = \mu_2$$ $$H_1 : \mu_1 < \mu_2$$ **Test Statistic:** The test statistic for the two-sample t-test is calculated as: $$\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]$$ **Degrees of Freedom:** The degrees of freedom for the t-test is calculated as: $$\[ \text{df} = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} \]$$ **Critical Value:** Using a significance level of $$\(\alpha = 0.01\)$$ and the degrees of freedom, find the critical value from the t-distribution. **Conclusion:** Compare the calculated test statistic with the critical value. If the test statistic is less than the critical value, we reject the null hypothesis and conclude that there is evidence that average daily sales increase with the new plan.