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Chapter 15: Problem 12

Dissolved oxygen content is a measure of the ability of a river, lake, or stream to support aquatic life, with high levels being better. A pollution- control inspector who suspected that a river community was releasing semitreated sewage into a river, randomly selected five specimens of river water at a location above the town and another five below. These are the dissolved oxygen readings (in parts per million): $$ \begin{array}{l|lllll} \text { Above Town } & 4.8 & 5.2 & 5.0 & 4.9 & 5.1 \\ \hline \text { Below Town } & 5.0 & 4.7 & 4.9 & 4.8 & 4.9 \end{array} $$ a. Use a one-tailed Wilcoxon rank sum test with \(\alpha=.05\) to confirm or refute the theory. b. Use a Student's \(t\) -test (with \(\alpha=.05\) ) to analyze the data. Compare the conclusion reached in part a.

Short Answer

Expert verified
Answer: Based on the results from both the Wilcoxon rank sum test and the Student's t-test, we cannot confirm that there is a significant difference in dissolved oxygen content above and below the town.

Step by step solution

01

Organize the data

First, we need to organize the data into two arrays for easy manipulation: Above Town: [4.8, 5.2, 5.0, 4.9, 5.1] Below Town: [5.0, 4.7, 4.9, 4.8, 4.9]
02

a. Wilcoxon Rank Sum Test

03

Calculate ranks

Combine both samples into a single array, and calculate the ranks for each observation. Combined array: [4.8, 5.2, 5.0, 4.9, 5.1, 5.0, 4.7, 4.9, 4.8, 4.9] Ranks: [2, 10, 7.5, 5.5, 9, 7.5, 1, 5.5, 2, 5.5]
04

Calculate Rank Sum for each group

Calculate the Rank Sum for the Above Town (W1) and Below Town(W2) groups. W1 (Above Town): 2 + 10 + 7.5 + 5.5 + 9 = 34 W2 (Below Town): 7.5 + 1 + 5.5 + 2 + 5.5 = 21.5
05

Determine critical value and test statistic

We are conducting a one-tailed test with 伪 = 0.05 and n1 = n2 = 5. From the Wilcoxon Rank Sum table, the critical value for W is 15. Since W2(21.5) > W_critical(15), we fail to reject the null hypothesis, and we cannot confirm that there is a significant difference in dissolved oxygen content above and below the town.
06

b. Student's t-test

07

Calculate means and variances

Calculate the means and variances for Above Town (M1) and Below Town (M2) groups. M1 (Above Town): (4.8 + 5.2 + 5.0 + 4.9 + 5.1) / 5 = 4.96 M2 (Below Town): (5.0 + 4.7 + 4.9 + 4.8 + 4.9) / 5 = 4.86 Var1 (Above Town): 0.028 Var2 (Below Town): 0.012
08

Perform t-test

Calculate the test statistic t, and the degrees of freedom (df). t = (M1 - M2) / sqrt((Var1 / n1) + (Var2 / n2)) = (4.96 - 4.86) / sqrt((0.028 / 5) + (0.012 / 5)) = 1.041 df = (Var1 / n1 + Var2 / n2)^2 / (((Var1 / n1)^2) / (n1 - 1) + ((Var2 / n2)^2) / (n2 - 1)) = 7.52
09

Determine critical value and compare with t statistic

Using a one-tailed t-test with 伪 = 0.05 and df = 7.52, we find that the critical value for the t-test is approximately 1.90. Since our test statistic (1.041) is less than the critical value (1.90), we fail to reject the null hypothesis in this case as well. We cannot confirm that there is a significant difference in dissolved oxygen content above and below the town.
10

Conclusion

Based on both the Wilcoxon rank sum test and the Student's t-test, the data does not provide enough evidence to confirm that there is a significant difference in dissolved oxygen content above and below the town. The conclusions reached in both tests are in agreement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wilcoxon Rank Sum Test
The Wilcoxon Rank Sum Test is a non-parametric statistical hypothesis test used when comparing two independent samples to determine whether there are significant differences between the two groups. It is particularly useful when the data cannot be assumed to be normally distributed, which is why it's an excellent alternative to the Student's t-test in such circumstances.

The procedure involves ranking all the observations from both groups together, calculating the sum of ranks for each group, and then determining whether these ranks are significantly different from each other. The test statistic is the smaller of the two rank sums, and conclusions about statistical significance are drawn by comparing this statistic to a critical value obtained from a reference table.

For the dissolved oxygen content example, after ranking the combined samples and summing the ranks for each group, the test did not show a significant difference between the two locations, failing to refute the null hypothesis.
Student's t-test

Understanding the Student's t-test

The Student's t-test is a parametric test used to compare the means of two independent groups and determine if they are statistically different from each other. It's suitable for normally distributed data with known or assumed variances.

In the dissolved oxygen content example, the t-test involves calculating the means and variances of both samples, then determining the test statistic 't' which shows if the difference between the two means is significant relative to the spread of the groups. Although it's a robust test, a critical assumption is the normality of the data, which wasn't a noted concern in the exercise provided.

Using the obtained t statistic and comparing it to a critical value based on the determined degrees of freedom, the dissolved oxygen content data analysis does not indicate a significant difference, agreeing with the outcome of the Wilcoxon test.
One-tailed test
A one-tailed test in statistics is a hypothesis test where the alternative hypothesis specifies a direction of the statistical difference. This means that the test looks for evidence of an effect in one specific direction only, either higher or lower, but not both.

In the context of the dissolved oxygen content problem, a one-tailed test is appropriate when the inspector has a specific suspicion that the dissolved oxygen content below the town is lower than above the town due to pollution. Using a one-tailed test increases the statistical power to detect an effect in the specified direction but would not be sensitive to a difference in the opposite direction. However, in both the Wilcoxon Rank Sum and Student's t-test applied to the samples, the one-tailed test yielded no significant evidence to support the alternative hypothesis that the dissolved oxygen is lower below the town compared to above.

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Most popular questions from this chapter

The data were collected using a randomized block design. For each data set, use the Friedman \(F\) -test to test for differences in location among the treatment distributions using \(\alpha=.05 .\) Bound the \(p\) -value for the test using Table 5 of Appendix \(I\) and state your conclusions. $$ \begin{array}{crrrr} \hline \quad &&& {\text { Treatment }} \\ \text { Block } & 1 & 2 & 3 & 4 \\ \hline 1 & 89 & 81 & 84 & 85 \\ 2 & 93 & 86 & 86 & 88 \\ 3 & 91 & 85 & 87 & 86 \\ 4 & 85 & 79 & 80 & 82 \\ 5 & 90 & 84 & 85 & 85 \\ 6 & 86 & 78 & 83 & 84 \\ 7 & 87 & 80 & 83 & 82 \\ 8 & 93 & 86 & 88 & 90 \\ \hline \end{array} $$

In an investigation of the visual scanning behavior of deaf children, measurements of eye movement were taken on nine deaf and nine hearing children. The table gives the eye movement rates and their ranks (in parentheses). Does it appear that the distributions of eye-movement rates for deaf children and hearing children differ? $$ \begin{array}{llc} \hline & \text { Deaf Children } & \text { Hearing Children } \\ \hline & 2.75(15) & .89(1) \\ & 2.14(11) & 1.43(7) \\ & 3.23(18) & 1.06(4) \\ & 2.07(10) & 1.01(3) \\ & 2.49(14) & .94(2) \\ & 2.18(12) & 1.79(8) \\ & 3.16(17) & 1.12(5.5) \\ & 2.93(16) & 2.01(9) \\ & 2.20(13) & 1.12(5.5) \\ \hline \text { Rank Sum } & 126 & 45 & \\ \hline \end{array} $$

The data given result from experiments run in completely randomized designs. Use the Kruskal-Wallis H statistic to determine whether there are significant differences between at least two of the treatment groups at the \(5 \%\) level of significance. You can use a computer program if one is available. Summarize your results. $$ \begin{array}{lcc} \hline & \text { Treatment } & \\ \hline 1 & 2 & 3 \\ \hline 26 & 27 & 25 \\ 29 & 31 & 24 \\ 23 & 30 & 27 \\ 24 & 28 & 22 \\ 28 & 29 & 24 \\ 26 & 32 & 20 \\ & 30 & 21 \\ & 33 & \\ \hline \end{array} $$

Use the information given in Exercises \(4-7\) to calculate Spearman's rank correlation coefficient, where \(x_{i}\) and \(y_{i}\) are the ranks of the ith pair of observations and \(d_{i}=x_{i}-y_{i} .\) Assume that there are no ties in the ranks. \(d_{i}=\\{-6,-3,-3,-4,2,5,5,4\\}\)

Use the Wilcoxon rank sum test to determine whether population 1 lies to the left of population 2 by (1) stating the null and alternative hypotheses to be tested, (2) calculating the values of \(T_{1}\) and \(T_{l}^{*},(3)\) finding the rejection region for \(\alpha=.05,\) and (4) stating your conclusions. $$ \begin{array}{l|ccccc} \text { Sample } 1 & 6 & 7 & 3 & 1 & \\ \hline \text { Sample } 2 & 4 & 4 & 9 & 2 & 7 \end{array} $$
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