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MATH 201 Statistics for Environmental Professionals ● Worksheet 4
Name Section 1.
Suppose that you suspect nutrient overload in stream water, especially phosphorus. It is generally considered that the upper limit of phosphorus amount in natural water is 0.05 mg/L. You want to test whether the stream has a larger amount of phosphorus than this upper limit. You randomly take a water sample of size 16 from the stream. The mean phosphorus amount in the sample is 0.053 mg/L, and the standard deviation is 0.008 mg/L. Perform the hypothesis test using the significance level of 0.05 by completing the following steps. Refer to the examples in textbook 9.5.
Question 1
. Write down the null and alternative hypothesis using words. (10 pts.)
In this scenario, it seeks to determine whether the stream has a larger amount of phosphorus than this upper limit, and thus, we consider the following hypothesis:
Null Hypothesis, H
0
: The mean phosphorus in the stream is the same as the natural water
Alternative Hypothesis, H
a
: The mean phosphorus in the stream is higher than natural water
Question 2
. Write down the null and alternative hypothesis using symbols. (10 pts.)
Null Hypothesis, H
0
: µ = 0.05 mg/L
Alternative Hypothesis, H
a
: µ > 0.05 mg/L
Question 3
. What is the value of α in this hypothesis test? (10 pts.)
In this scenario, I will use an alpha value of 5%.
Question 4.
Determine an appropriate distribution (standard normal distribution or t-distribution)
to use. State the reason to support your choice.
(10 pts.)
Since the sample size is less than 30, we shall conduct a t-test distribution in this scenario. Also, Since the alternative hypothesis uses a greater sign, we shall conduct a one-sample t-test to the right.
Question 5.
Sketch the distribution (by hand or digitally) selected above and shade the area where your null hypothesis is rejected. (10 pts.)
From the information provided, the sample size is 16, and thus the degrees of freedom will be 15(n-1), with a t critical value being 1. 753. The figure presented below shows where our null hypothesis will be rejected.
Question 6.
Calculate the value of the test statistic. Show your work.
(10 pts.)
Table 1: One-Sample t Test
From the findings presented above, the test statistic is 1.5.
Question 7. Determine the p-value using an appropriate table or other technology (e.g. calculator). (10 pts.)
The results in Table 1 above show that the p-value is 0.077.
Question 8.
Compare the p-value and the value of α. Then determine whether you should reject or not reject the null hypothesis. (10 pts.)
The p-value was 0.08, greater than 0.05; we do not reject the null hypothesis at 5%.
Question 9.
State the conclusion of the hypothesis testing in 10-20 words. (10 pts.)
From the results presented, since the p-value was greater than 0.05, we do not reject the Ho, and the conclusion is that the mean phosphorus in the stream is the same as the natural water at 5%.
Section 2.
The accompanied data is part of a dataset of measured phosphorus amount (kilograms/hectare/year) by watershed. We want to test whether the mean phosphorus amount is greater than 0.55 kilograms/hectare/year. Using the provided Excel spreadsheet, calculate and report an appropriate test statistic and p-value. (10 pts.)
In this scenario, it seeks to determine whether the mean phosphorus amount is greater than 0.55 kgs per hectare per year, and thus, we consider the following hypothesis:
Null Hypothesis, H
0
: µ = 0.55 Alternative Hypothesis, H
a
: µ > 0.55
From the above hypothesis, since the alternative hypothesis uses a greater sign, we shall conduct a one-sample t-test to the right, and the results are as follows.
Table 1: One-Sample t Test
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Test statistic: -0.4457
P-value: 0.3304
Since the p-value > 0.05, we fail to reject the null hypothesis and conclude that the mean phosphorus amount is not greater than 0.55 kgs per hectare per year.
Source: U.S. Environmental Protection Agency (2016) Total phosphorus in surface water. https://catalog.data.gov/dataset/total-phosphorus-in-surface-water (accessed 7/14/2020)