1. **State the problem:** We want to test if the gardener's average daily profit is higher when bouquets are sold at 10 dollars compared to 15 dollars. 2. **Given data:** - For $15 price: sample size $n_1=10$, mean $\bar{x}_1=171$, standard deviation $s_1=26$ - For $10 price: sample size $n_2=5$, mean $\bar{x}_2=198$, standard deviation $s_2=29$ 3. **Hypotheses:** - Null hypothesis $H_0$: $\mu_2 \leq \mu_1$ (average profit at $10 is not higher) - Alternative hypothesis $H_a$: $\mu_2 > \mu_1$ (average profit at $10 is higher) 4. **Test type:** Two-sample t-test for difference of means with unequal variances (Welch's t-test). 5. **Test statistic formula:** $$ t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$ 6. **Calculate the test statistic:** $$ s_1^2 = 26^2 = 676, \quad s_2^2 = 29^2 = 841 $$ $$ SE = \sqrt{\frac{676}{10} + \frac{841}{5}} = \sqrt{67.6 + 168.2} = \sqrt{235.8} \approx 15.36 $$ $$ t = \frac{198 - 171}{15.36} = \frac{27}{15.36} \approx 1.757 $$ 7. **Degrees of freedom (Welch-Satterthwaite equation):** $$ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} = \frac{(67.6 + 168.2)^2}{\frac{67.6^2}{9} + \frac{168.2^2}{4}} = \frac{235.8^2}{\frac{4569.76}{9} + \frac{28288.84}{4}} = \frac{55600.64}{507.75 + 7072.21} = \frac{55600.64}{7579.96} \approx 7.34 $$ 8. **Critical value:** For a one-tailed test at $\alpha=0.10$ and $df \approx 7$, the critical t-value $t_{0.10,7} \approx 1.415$ 9. **Decision:** Since calculated $t=1.757 > 1.415$, we reject the null hypothesis. 10. **Conclusion:** There is sufficient evidence at the 10% significance level to conclude the gardener’s average daily profit is higher when bouquets are sold at 10 dollars each. **Final answer:** A. Data provide sufficient evidence that average daily profit is higher if the bouquets are sold at 10 each.