using the values below, test whether the mean phosphorus amount is greater than 0.55 kilograms/hectare/year. calculate and report an appropriate test statistic and p-value upper delaware's totalp is 0.3886
east branch delaware's totalp is 0.3176
lackawaxen's totalp is 0.3651
middle delaware-mongaup-brodhead's totalp is 0.3175
middle delaware-musconetcong's totalp is 0.6457
Leigh's totalp is 0.5521
Crosswicks-Neshaminy's totalp is 0.8097
Lower Delaware's totalp is 0.7781
Schuylkill's totalp is 0.7025
Brandywine-Christina's totalp is 0.8449
Cohansey-Maurice's totalp is 0.6191
Broadkill-Smryna's totalp is 1.0916
Upper Susquehanna's totalp is 0.4813
Chenango's totalp is 0.5143
Owego-Wappasening's totalp is 0.4794
Tioga's totalp is 0.5262
Chemung's totalp is 0.5495
Upper Susquehanna-Tunkhannock's totalp is 0.5312

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.

Using the values below, test whether the mean phosphorus amount is greater than 0.55 kilograms/hectare/year. Calculate and report an appropriate test statistic and p-value. Upper Delaware's totalp is 0.3886, East Branch Delaware's totalp is 0.3176, Lackawaxen's totalp is 0.3651, Middle Delaware-Mongaup-Brodhead's totalp is 0.3175, Middle Delaware-Musconetcong's totalp is 0.6457, Leigh's totalp is 0.5521, Crosswicks-Neshaminy's totalp is 0.8097, Lower Delaware's totalp is 0.7781, Schuylkill's totalp is 0.7025, Brandywine-Christina's totalp is 0.8449, Cohansey-Maurice's totalp is 0.6191, Broadkill-Smyrna's totalp is 1.0916, Upper Susquehanna's totalp is 0.4813, Chenango's totalp is 0.5143, Owego-Wappasening's totalp is 0.4794, Tioga's totalp is 0.5262, Chemung's totalp is 0.5495, Upper Susquehanna-Tunkhannock's totalp is 0.5312.

This problem involves performing a one-sample t-test to determine if the mean phosphorus level across multiple regions exceeds 0.55 kg/hectare/year. Includes a detailed solution with the calculation of the test statistic and p-value. Suitable for college-level statistics students.

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