Problem Statement
An agricultural scientist tests whether a new fertilizer increases crop yield (kg/plot). Two independent random samples yield the following summary statistics:
• Standard fertilizer ($n_1 = 12$): $\bar{x}_1 = 45.2$, $s_1 = 3.8$
• New fertilizer ($n_2 = 10$): $\bar{x}_2 = 49.1$, $s_2 = 4.2$
Assuming underlying populations are normally distributed with equal variances, perform a two-sample pooled $t$-test at the $5\%$ significance level to test whether the new fertilizer produces a significantly higher yield. State hypotheses, $t$-statistic, $p$-value, and conclusion.
Verified Solution & Marking Scheme
State null and alternative hypotheses
$$H_0: \mu_1 = \mu_2 \quad (\text{or } \mu_2 - \mu_1 = 0)$$
$$H_1: \mu_2 > \mu_1 \quad (\text{one-tailed test})$$
Pooled degrees of freedom and standard deviation
$$\text{df} = n_1 + n_2 - 2 = 12 + 10 - 2 = 20$$
$$s_p = \sqrt{\frac{(11)(3.8)^2 + (9)(4.2)^2}{20}} = \sqrt{\frac{158.84 + 158.76}{20}} = \sqrt{15.88} \approx 3.985$$
Calculate t-statistic and p-value on GDC
$$t = \frac{49.1 - 45.2}{3.985 \sqrt{\frac{1}{12} + \frac{1}{10}}} = \frac{3.9}{3.985(0.4282)} = \frac{3.9}{1.706} \approx 2.285$$
From GDC 2-SampTTest (Pooled = Yes, $\mu_1 < \mu_2$):
$$t = -2.285 \implies p\text{-value} = 0.0166$$
Compare with significance level and conclude
Since $p\text{-value} = 0.0166 < 0.05$, we reject $H_0$.
There is sufficient evidence at the $5\%$ significance level to conclude that the new fertilizer produces a significantly higher mean crop yield.