To test whether the mean phosphorus amount is greater than 0.55 kg/hectare/year, we can perform a one-sample t-test. The null hypothesis $H_0$ is that the mean phosphorus amount is less than or equal to 0.55 kg/hectare/year, and the alternative hypothesis $H_a$ is that the mean phosphorus amount is greater than 0.55 kg/hectare/year.

Steps:

1. Null hypothesis ($H_0$): $\mu \leq 0.55 \, \text{kg/hectare/year}$
2. Alternative hypothesis ($H_a$): $\mu > 0.55 \, \text{kg/hectare/year}$
3. We'll calculate the mean of the given data, the sample standard deviation, and then the t-statistic.
4. We compare the t-statistic to a t-distribution to determine the p-value.

The phosphorus data is:

• 0.3886, 0.3176, 0.3651, 0.3175, 0.6457, 0.5521, 0.8097, 0.7781, 0.7025, 0.8449, 0.6191, 1.0916, 0.4813, 0.5143, 0.4794, 0.5262, 0.5495, 0.5312

Let’s calculate the necessary statistics and perform the t-test.The calculated t-statistic is approximately $0.716$, and the corresponding p-value is approximately $0.242$.

Interpretation:

• Since the p-value ($0.242$) is greater than the common significance level (e.g., $\alpha = 0.05$), we fail to reject the null hypothesis.
• This means there is not enough evidence to conclude that the mean phosphorus amount is greater than 0.55 kg/hectare/year.

Would you like to explore any of the following details?

1. How to compute a confidence interval for the mean phosphorus amount.
2. Understanding the assumptions of the t-test in more detail.
3. Exploring a different significance level for the hypothesis test.
4. Examining the distribution of the data.
5. Reviewing the calculation steps for mean and standard deviation.

Tip: Ensure that the sample data is normally distributed or the sample size is large enough when performing a t-test for better reliability.