MATH 201 Statistics for Environmental Professionals ● Worksheet 4
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Jan 9, 2024
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MATH 201 Statistics for Environmental Professionals ● Worksheet 4
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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 the textbook 9.5.
Question 1. Write down the null and alternative hypothesis using words. (10 pts.)
Null hypothesis: Upper limit of phosphorous amount in natural water is 0.05mg/L Alternative
hypothesis: Amount of phosphorous in natural water is larger than 0.05mg/L
Question 2. Write down the null and alternative hypothesis using symbols. (10 pts.)
H0: µ ≤ 0.05 mg/L
Ha: µ > 0.05 mg/L
Question 3. What is the value of α in this hypothesis test? (10 pts.)
For this hypothesis test, 0.05 is the significance level (α).
Question 4. Determine an appropriate distribution (standard normal distribution or
t-distribution) to use. State the reason to support your choice. (10 pts.)
The t-distribution will be used because the sample size is small (n = 16) and the population
standard deviation is unknown. When the sample size is small and the population standard
deviation is unknown, the t-distribution is utilized.
Question 5. Sketch the distribution (by hand or digitally) selected above and shade the
area where your null hypothesis is rejected. (10 pts.)
Question 6. Calculate the value of the test statistic. Show your work. (10 pts.)
Question 7. Determine the p-value using an appropriate table or other technology (e.g.
calculator). (10 pts.)
Based on question 6, t value/score = 1.5, and degrees of freedom = n-1 = 16-1 = 15.
P-value = 0.077183
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.)
Since p-value (0.077183) is > than the level of significance, we fail to reject the null
hypothesis at 0.05.
Question 9. State the conclusion of the hypothesis testing in 10-20 words.(10 pts.)
The p-value is higher than the level of significance. We fail to reject the null hypothesis and
conclude that the stream contains no more phosphorous than the maximum limit of 0.05mg/L.
Section 2. The accompanied data is a 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.)
Test Statistic= 0.45
p-value= 0.330
We know that if the p-value is less than 0.05, the null hypothesis is rejected. Because the p-value
is more than 0.05, we cannot reject the null hypothesis and conclude that the mean phosphorus
level is less than 0.55 kg 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)
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