Question

In: Math

n Ecological Engineering, the potential for floating aquatic plants to treat dairy manure wastewater was investigated....

n Ecological Engineering, the potential for floating aquatic plants to treat dairy manure wastewater was investigated. For one part of the study, 16 treated wastewater samples were randomly divided into two groups- a control algal was cultured in half the samples and the water hyacinth was cultured in the other half. The rate of increase in the amount of total phosphorus was measured in each water sample. The control algal had a sample mean of 0.036 with a standard deviation of 0.008 while the water hyacinth had a sample mean of 0.026 with a standard deviation of 0.006. Conduct a test to determine if there is a difference in mean rates of increase of total phosphorus for the two aquatic plants. Use alpha = 0.05.

1) What type of test should be conducted?

Independent t-test

Dependent t-test

One-tailed test

One-sample t-test

2. State the null hypothesis in equation format.

3. State the alternative hypothesis in equation format

4. What is the calculated t-value (to 3 significant digits)?

5. What is the critical t-value (to 3 significant digits)? Use alpha = 0.05.

6. Is the null hypothesis accepted or rejected? Use alpha = 0.05.

7. Is there sufficient evidence to conclude that there is a difference between the mean rate of increase of total phosphorus of the control algal and the water hyacinth? Use alpha = 0.05. Explain in one sentence.

Solutions

Expert Solution

1) Independent t-test.

2) H0:

3) H1:

4) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                                 = (0.036 - 0.026)/sqrt((0.008)^2/8 + (0.006)^2/8)

                                 = 2.828

5) DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

          = ((0.008)^2/8 + (0.006)^2/8)^2/(((0.008)^2/8)^2/7 + ((0.006)^2/8)^2/7)

          = 13

At alpha = 0.05, the critical values are t0.025, 13 = +/- 2.160

6) Since the test statistic value is greater than the upper critical value(2.828 > 2.160), we should reject the null hypothesis.

7) So at alpha = 0.05, there is sufficient evidence to conclude that there is a difference between the mean rate of increase of total phosphorus of the control algal and the water hyacinth.


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