1. **State the problem:** We want to test if the new fertilizer increases the mean grape yield by more than 5 pounds compared to the traditional method. 2. **Set hypotheses:** - Null hypothesis: $H_0: \mu_1 - \mu_2 = 5$ - Alternative hypothesis: $H_a: \mu_1 - \mu_2 \neq 5$ 3. **Given data:** - Fertilized group: $n_1=44$, $\bar{x}_1=58.4$, $s_1=3.7$ - Unfertilized group: $n_2=47$, $\bar{x}_2=52.1$, $s_2=3.4$ 4. **Calculate the test statistic:** The formula for the two-sample t-test statistic when testing $\mu_1 - \mu_2 = d_0$ is: $$ t = \frac{(\bar{x}_1 - \bar{x}_2) - d_0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$ where $d_0=5$. 5. **Compute numerator:** $$ (58.4 - 52.1) - 5 = 6.3 - 5 = 1.3 $$ 6. **Compute denominator:** $$ \sqrt{\frac{3.7^2}{44} + \frac{3.4^2}{47}} = \sqrt{\frac{13.69}{44} + \frac{11.56}{47}} = \sqrt{0.3111 + 0.246} = \sqrt{0.5571} \approx 0.7464 $$ 7. **Calculate t:** $$ t = \frac{1.3}{0.7464} \approx 1.742 $$ 8. **Degrees of freedom (approximate using Welch's formula):** $$ df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} = \frac{(0.3111 + 0.246)^2}{\frac{0.3111^2}{43} + \frac{0.246^2}{46}} = \frac{0.5571^2}{\frac{0.0968}{43} + \frac{0.0605}{46}} = \frac{0.3104}{0.00225 + 0.00131} = \frac{0.3104}{0.00356} \approx 87.2 $$ 9. **Find p-value:** For $t=1.742$ with $df \approx 87$, two-tailed test, p-value is about 0.085 (using t-distribution tables or software). 10. **Conclusion:** Since $p=0.085 > 0.05$, we fail to reject the null hypothesis at the 5% significance level. There is not enough evidence to conclude that the fertilizer increases the mean yield by more than 5 pounds. **Final answer:** The data do not provide sufficient evidence to support the agronomist's claim that the fertilizer increases grape yield by more than 5 pounds per vine.