1. **State the problem:** We want to test if the new organic fertilizer increases the mean height of flowering plants compared to the historical mean height of untreated plants, which is 18.5 cm. 2. **Formulate hypotheses:** - Parameter of interest: population mean height $\mu$ of plants treated with the fertilizer. - Null hypothesis $H_0$: $\mu = 18.5$ cm (fertilizer has no effect). - Alternative hypothesis $H_a$: $\mu > 18.5$ cm (fertilizer increases height). 3. **Identify the suitable test:** - Sample size $n=15$ is small (less than 30). - Population standard deviation unknown; sample standard deviation $s=2.8$ cm used. - Data are continuous and assumed approximately normal. - Therefore, a one-sample one-tailed t-test is appropriate. 4. **Carry out the test at 5% significance level:** - Sample mean $\bar{x} = 20.1$ cm. - Historical mean $\mu_0 = 18.5$ cm. - Sample standard deviation $s=2.8$ cm. - Sample size $n=15$. Calculate the test statistic: $$ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{20.1 - 18.5}{2.8/\sqrt{15}} = \frac{1.6}{2.8/3.873} = \frac{1.6}{0.722} \approx 2.216 $$ Degrees of freedom: $df = n-1 = 14$. Critical t-value for one-tailed test at $\alpha=0.05$ and $df=14$ is approximately $1.761$. Since $t=2.216 > 1.761$, we reject the null hypothesis. 5. **Statistical decision:** There is sufficient evidence at the 5% significance level to conclude that the fertilizer increases the mean height of plants. 6. **Interpretation for farmers:** The test shows that plants treated with the new fertilizer tend to grow taller than untreated plants. This suggests the fertilizer is effective in improving plant height. 7. **Assumption and its importance:** - Assumption: The plant heights are approximately normally distributed. - If this assumption is violated (e.g., data are heavily skewed), the t-test results may be invalid, leading to incorrect conclusions and possibly recommending a fertilizer that does not truly improve plant height.