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7.6 A Nonparametric Test for Comparing Two Populations: Paired Difference Experiment (Optional)

Nonparametric techniques may also be employed to compare two probability distributions when a paired difference design is used. For example, consumer preferences for two competing products are often compared by having each of a sample of consumers rate both products. Thus, the ratings have been paired on each consumer. Following is an example of this type of experiment.

For some paper products, softness is an important consideration in determining consumer acceptance. One method of determining softness is to have judges give a sample of the products a softness rating. Suppose each of 10 judges is given a sample of two products that a company wants to compare. Each judge rates the softness of each product on a scale from 1 to 20, with higher ratings implying a softer product. The results of the experiment are shown in Table 7.7.

Since this is a paired difference experiment, we analyze the differences between the measurements. (See Section 7.3.) However, a nonparametric approach developed by Wilcoxon requires that we calculate the ranks of the absolute values of the differences between the measurements (i.e., the ranks of the differences after removing any minus signs). Note that tied absolute differences (e.g., the two differences of 4) are assigned the average of the ranks they would receive if they were unequal, but successive, measurements (e.g., 4.5, the average of the ranks 4 and 5). After the absolute differences are ranked, the sum of the ranks of the positive differences of the original measurements, $T_{+},$ $T_{+},$ and the sum of the ranks of the negative differences of the original measurements, $T_{-},$ $T_{-},$ are computed. (The ranks of the negative differences are highlighted in Table 7.7.)

We are now prepared to test the nonparametric hypotheses:

$H_{0} :$ $H_{0} :$ The probability distributions of the ratings for products A and B are identical.
$H_{a} :$ $H_{a} :$ The probability distributions of the ratings differ (in location) for the two products. (Note that this is a two-sided alternative and that it implies a two-tailed test.)
Test statistic: $T = Smaller$ $T = Smaller$ of the positive and negative rank sums $T_{+}$ $T_{+}$ and $T_{-}$ $T_{-}$

Alternate View

Product Difference

Judge A B $(A - B)$ $(A - B)$ Absolute Value of Difference Rank of $(A - B)$ $(A - B)$ Absolute Value

1 12 8 4 4 4.5

2 16 10 6 6 7

3 8 9 $- 1$ $- 1$ 1 1

4 10 8 2 2 2

5 19 12 7 7 8

6 14 17 $- 3$ $- 3$ 3 3

7 12 4 8 8 9

8 10 6 4 4 4.5

9 12 17 $- 5$ $- 5$ 5 6

10 16 4 12 12 10

$T_{\pm} = Sum of positive ranks = 45$ $T_{\pm} = Sum of positive ranks = 45$

$T_{-} = Sum of negative ranks = 10$ $T_{-} = Sum of negative ranks = 10$

Data Set: SOFTPAP

	Product	Difference
1	12	8	4	4	4.5
2	16	10	6	6	7
3	8	9	$- 1$ $- 1$	1	1
4	10	8	2	2	2
5	19	12	7	7	8
6	14	17	$- 3$ $- 3$	3	3
7	12	4	8	8	9
8	10	6	4	4	4.5
9	12	17	$- 5$ $- 5$	5	6
10	16	4	12	12	10
				$T_{\pm} = Sum of positive ranks = 45$ $T_{\pm} = Sum of positive ranks = 45$
				$T_{-} = Sum of negative ranks = 10$ $T_{-} = Sum of negative ranks = 10$

The smaller the value of T, the greater is the evidence indicating that the two probability distributions differ in location. The rejection region for T can be determined by consulting Table VI in AppendixB (part of which is shown in Table 7.8). This table gives a value $T_{0}$ $T_{0}$ for both one-tailed and two-tailed tests for each value of n, the number of matched pairs. For a two-tailed test with $α = .05,$ $α = .05,$ we will reject $H_{0}$ $H_{0}$ if $T \leq T_{0} .$ $T \leq T_{0} .$ You can see in Table 7.8 that the value of $T_{0}$ $T_{0}$ that locates the boundary of the rejection region for the judges’ ratings for $α = .05$ $α = .05$ and $n = 10$ $n = 10$ pairs of observations is 8. Thus, the rejection region for the test (see Figure 7.18) is

\begin{array}{l} R e j e c t i o n r e g i o n : & T \leq 8 & for α = .05 \end{array}

$\begin{array}{l} R e j e c t i o n r e g i o n : & T \leq 8 & for α = .05 \end{array}$

Rejection region for paired difference experiment

Since the smaller rank sum for the paper data, $T_{-} = 10,$ $T_{-} = 10,$ does not fall within the rejection region, the experiment has not provided sufficient evidence indicating that the two paper products differ with respect to their softness ratings at the $α = .05$ $α = .05$ level.

Note that if a significance level of $α = .10$ $α = .10$ had been used, the rejection region would have been $T \leq 11$ $T \leq 11$ and we would have rejected $H_{0} .$ $H_{0} .$ In other words, the samples do provide evidence that the probability distributions of the softness ratings differ at the $α = .10$ $α = .10$ significance level.

The Wilcoxon signed rank test is summarized in the next box. Note that the difference measurements are assumed to have a continuous probability distribution so that the absolute differences will have unique ranks. Although tied (absolute) differences can be assigned ranks by averaging, in order to ensure the validity of the test, the number of ties should be small relative to the number of observations.

Nonparametric Wilcoxon Signed Rank Test: Matched Pairs Experiment

Let D₁ and D₂ represent the probability distributions for Populations 1 and 2, respectively.

Rank sums: Calculate the measurement differences within each of the n matched pairs. Then rank the absolute values of the n differences from smallest (rank 1) to largest (rank n).

$T_{+} = the sum$ $T_{+} = the sum$ of the ranks of the positive differences
$T_{-} = the sum$ $T_{-} = the sum$ of the ranks of the negative differences

[Note: Differences equal to 0 are eliminated and not ranked.]

Ties: Assign tied absolute differences the average of the ranks they would receive if they were unequal but occurred in successive order. For example, if the third-ranked and fourth-ranked absolute differences are tied, assign each a rank of $(3 + 4) / 2 = 3.5.$ $(3 + 4) / 2 = 3.5.$

	One-Tailed Tests		Two-Tailed Test
	$H_{0} : D_{1} = D_{2}$ $H_{0} : D_{1} = D_{2}$	$H_{0} : D_{1} = D_{2}$ $H_{0} : D_{1} = D_{2}$	$H_{0} : D_{1} = D_{2}$ $H_{0} : D_{1} = D_{2}$
	`H`_a: `D`₁ shifted to the left of `D`₂	`H`_a: `D`₁ shifted to the right of `D`₂	`H`_a: `D`₁ shifted to left or right of `D`₂
Test statistic:	$T_{+}$ $T_{+}$	$T_{-}$ $T_{-}$	`T`, the smaller of $T_{+}$ $T_{+}$ or $T_{-}$ $T_{-}$
Rejection region:	$T_{+} \leq T_{0}$ $T_{+} \leq T_{0}$	$T_{-} \leq T_{0}$ $T_{-} \leq T_{0}$	$T \leq T_{0}$ $T \leq T_{0}$

Decision: Reject H₀ if test statistic falls into the rejection region where T₀ is obtained from Table VI of Appendix B.

Conditions Required for a Valid Signed Rank Test

The sample of differences is randomly selected from the population of differences.
The probability distribution from which the sample of paired differences is drawn is continuous.

Example 7.9 Applying the Signed Rank Test—Comparing Two Crime Prevention Plans

Problem

Suppose the police commissioner in a small community must choose between two plans for patrolling the town’s streets. Plan A, the less expensive plan, uses voluntary citizen groups to patrol certain high-risk neighborhoods. In contrast, plan B would utilize police patrols. As an aid in reaching a decision, both plans are examined by 10 trained criminologists, each of whom is asked to rate the plans on a scale from 1 to 10. (High ratings imply a more effective crime prevention plan.) The city will adopt plan B (and hire extra police) only if the data provide sufficient evidence that criminologists tend to rate plan B more effective than plan A. The results of the survey are shown in Table 7.9. Do the data provide evidence at the $α = .05$ $α = .05$ level that the distribution of ratings for plan B lies above that for plan A? Use the Wilcoxon signed rank test to answer the question.

Alternate View

Plan Difference

Crime Prevention Expert A B $(A - B)$ $(A - B)$ Rank of Absolute Difference

1 7 9 $- 2$ $- 2$ 4.5

2 4 5 $- 1$ $- 1$ 2

3 8 8 0 (Eliminated)

4 9 8 1 2

5 3 6 $- 3$ $- 3$ 6

6 6 10 $- 4$ $- 4$ 7.5

7 8 9 $- 1$ $- 1$ 2

8 10 8 2 4.5

9 9 4 5 9

10 5 9 $- 4$ $- 4$ 7.5

$Positive rank sum = T_{+} = 15.5$ $Positive rank sum = T_{+} = 15.5$

Data Set: CRIME

	Plan	Difference
1	7	9	$- 2$ $- 2$	4.5
2	4	5	$- 1$ $- 1$	2
3	8	8	0	(Eliminated)
4	9	8	1	2
5	3	6	$- 3$ $- 3$	6
6	6	10	$- 4$ $- 4$	7.5
7	8	9	$- 1$ $- 1$	2
8	10	8	2	4.5
9	9	4	5	9
10	5	9	$- 4$ $- 4$	7.5
	$Positive rank sum = T_{+} = 15.5$ $Positive rank sum = T_{+} = 15.5$

Solution

The null and alternative hypotheses are as follows:
- $H_{0} :$ $H_{0} :$ The two probability distributions of effectiveness ratings are identical
- $H_{a} :$ $H_{a} :$ The effectiveness ratings of the more expensive plan (B) tend to exceed those of plan A
Observe that the alternative hypothesis is one sided (i.e., we only wish to detect a shift in the distribution of the B ratings to the right of the distribution of A ratings); therefore, it implies a one-tailed test of the null hypothesis. (See Figure 7.19.) If the alternative hypothesis is true, the B ratings will tend to be larger than the paired A ratings, more negative differences in pairs will occur, $T_{-}$ $T_{-}$ will be large, and $T_{+}$ $T_{+}$ will be small. Because Table VI is constructed to give lower-tail values of $T_{0},$ $T_{0},$ we will use $T_{+}$ $T_{+}$ as the test statistic and reject $H_{0}$ $H_{0}$ for $T_{+} \leq T_{0} .$ $T_{+} \leq T_{0} .$

Figure 7.19

The alternative hypothesis for Example 7.9

Because a paired difference design was used (both plans were evaluated by the same criminologists), we first calculate the difference between the rating for each expert. The differences in ratings for the pairs $(A - B)$ $(A - B)$ are shown in Table 7.9. Note that one of the differences equals 0. Consequently, we eliminate this pair from the ranking and reduce the number of pairs to $n = 9.$ $n = 9.$ Looking in Table VI, we have $T_{0} = 8$ $T_{0} = 8$ for a one-tailed test with $α = .05$ $α = .05$ and $n = 9.$ $n = 9.$ Therefore, the test statistic and rejection region for the test are

$\begin{array}{l} \begin{matrix} T e s t s t a t i s t i c : T_{+}, the positive rank sum \end{matrix} \\ \begin{matrix} R e j e c t i o n r e g i o n : T_{+} \leq 8 \end{matrix} \end{array}$ $\begin{array}{l} \begin{matrix} T e s t s t a t i s t i c : T_{+}, the positive rank sum \end{matrix} \\ \begin{matrix} R e j e c t i o n r e g i o n : T_{+} \leq 8 \end{matrix} \end{array}$

Summing the ranks of the positive differences (highlighted) in Table 7.9, we find that $T_{+} = 15.5.$ $T_{+} = 15.5.$ Since this value exceeds the critical value, $T_{0} = 8,$ $T_{0} = 8,$ we conclude that the sample provides insufficient evidence at the $α = .05$ $α = .05$ level to support the alternative hypothesis. The commissioner cannot conclude that the plan utilizing police patrols tends to be rated higher than the plan using citizen volunteers. That is, on the basis of this study, extra police will not be hired.

Look Back

An SPSS printout of the analysis, shown in Figure 7.20, confirms the preceding conclusion. Both the test statistic and two-tailed p-value are highlighted on the printout. Since the one-tailed p-value, $.404 / 2 = .202,$ $.404 / 2 = .202,$ exceeds $α = .05,$ $α = .05,$ we fail to reject $H_{0} .$ $H_{0} .$

Now Work Exercise 7.89

As is the case for the rank sum test for independent samples, the sampling distribution of the signed rank statistic can be approximated by a normal distribution when the number n of paired observations is large (say, $n \geq 25$ $n \geq 25$ ). The large-sample z-test is summarized in the following box:

Large-Sample Wilcoxon Signed Rank Test: Matched Pairs Experiment $(n \geq 25)$ $(n \geq 25)$

Let D₁ and D₂ represent the probability distributions for Populations 1 and 2, respectively.

Rank sums: Calculate the measurement differences within each of the n matched pairs. Then rank the absolute values of the n differences from smallest (rank 1) to largest (rank n).

T_{+} = the sum of theranks of the positive differences

$T_{+} = the sum of theranks of the positive differences$

[Note: Differences equal to 0 are eliminated and not ranked.]

	One-Tailed Tests		Two-Tailed Test
	$H_{0} : D_{1} = D_{2}$ $H_{0} : D_{1} = D_{2}$	$H_{0} : D_{1} = D_{2}$ $H_{0} : D_{1} = D_{2}$	$H_{0} : D_{1} = D_{2}$ $H_{0} : D_{1} = D_{2}$
	`H`_a: `D`₁ shifted to the left of `D`₂	`H`_a: `D`₁ shifted to the right of `D`₂	`H`_a: `D`₁ shifted to left or right of `D`₂
Test statistic:	$z_{c} = \frac{T_{+} - [n (n + 1) / 4]}{\sqrt{[n (n + 1) (2 n + 1)] / 24}}$ $z_{c} = \frac{T_{+} - [n (n + 1) / 4]}{\sqrt{[n (n + 1) (2 n + 1)] / 24}}$
Rejection region:	$z_{c} < - z_{α}$ $z_{c} < - z_{α}$	$z_{c} > z_{α}$ $z_{c} > z_{α}$	$\| z_{c} \| > z_{α / 2}$ $\| z_{c} \| > z_{α / 2}$
p-value:	$P (z < z_{c})$ $P (z < z_{c})$	$P (z > z_{c})$ $P (z > z_{c})$	$2 P (z > \| z_{c} \|)$ $2 P (z > \| z_{c} \|)$

Decision: Reject H₀ if $α > p -value$ $α > p -value$ or test statistic falls into the rejection region where $α = P (Type I error) = P (Reject H_{0} | H_{0} true)$ $α = P (Type I error) = P (Reject H_{0} | H_{0} true)$ and tabulated z values are found in Table II of Appendix B.

Exercises 7.86–7.102

Understanding the Principles

7.86 Explain the difference between the one- and two-tailed versions of the Wilcoxon signed rank test for the paired difference experiment.
7.87 In order to conduct the Wilcoxon signed rank test, why do we need to assume that the probability distribution of differences is continuous?

Learning the Mechanics

7.88 Specify the test statistic and the rejection region for the Wilcoxon signed rank test for the paired difference design in each of the following situations:
1. $H_{0} :$ $H_{0} :$ Two probability distributions, A and B, are identical
  
  $H_{a} :$ $H_{a} :$ The probability distribution for population A is shifted to the right or left of the probability distribution for population B $n = 20, α = .10$ $n = 20, α = .10$
2. $H_{0} :$ $H_{0} :$ Two probability distributions, A and B, are identical
  
  $H_{a} :$ $H_{a} :$ The probability distribution for population A is shifted to the right of the probability distribution for population B $n = 39, α = .05$ $n = 39, α = .05$
3. $H_{0} :$ $H_{0} :$ Two probability distributions, A and B, are identical
  
  $H_{a} :$ $H_{a} :$ The probability distribution for population A is shifted to the left of the probability distribution for population B $n = 7, α = .005$ $n = 7, α = .005$
L07089 7.89 Suppose you want to test a hypothesis that two treatments, A and B, are equivalent against the alternative hypothesis that the responses for A tend to be larger than those for B. You plan to use a paired difference experiment and to analyze the resulting data with the Wilcoxon signed rank test.
1. Specify the null and alternative hypotheses you would test.
2. Suppose the paired difference experiment yielded the data in the accompanying table. Conduct the test of part a. Test using $α = .025 .$ $α = .025 .$
Pair A B

1 54 45

2 60 45

3 98 87

4 43 31

5 82 71

6 77 75

7 74 63

8 29 30

9 63 59

10 80 82
7.90 Suppose you wish to test a hypothesis that two treatments, A and B, are equivalent against the alternative that the responses for A tend to be larger than those for B.
1. If the number of pairs equals 25, give the rejection region for the large-sample Wilcoxon signed rank test for $α = .05 .$ $α = .05 .$
2. Suppose that $T_{+} = 273.$ $T_{+} = 273.$ State your test conclusions.
3. Find the p-value for the test and interpret it.

Pair	A	B
1	54	45
2	60	45
3	98	87
4	43	31
5	82	71
6	77	75
7	74	63
8	29	30
9	63	59
10	80	82

Applying the Concepts—Basic

7.91 Treating psoriasis with the “Doctorfish of Kangal.” Refer to the Evidence-Based Research in Complementary and Alternative Medicine (Dec. 2006) study of treating psoriasis with ichthyotherapy, presented in Exercise 2.146 (p. 92). (Recall that the therapy is also known as the “Doctorfish of Kangal,” since it uses fish from the hot pools of Kangal, Turkey, to feed on skin scales.) In the study, 67 patients diagnosed with psoriasis underwent three weeks of ichthyotherapy. The Psoriasis Area Severity Index (PASI) of each patient was measured both before and after treatment. (The lower the PASI score, the better is the skin condition.) Before- and after-treatment PASI scores were compared with the use of the Wilcoxon signed rank test.
1. Explain why the PASI scores should be analyzed with a test for paired differences.
2. Refer to the box plots shown in Exercise 2.146 . Give a reason that the researchers opted to use a nonparametric test to compare the PASI scores.
3. The p-value for the Wilcoxon signed rank test was reported as $p < .0001 .$ $p < .0001 .$ Interpret this result, and comment on the effectiveness of ichthyotherapy in treating psoriasis.
SHALLOW 7.92 Settlement of shallow foundations. Refer to the Environmental & Engineering Geoscience (Nov. 2012) study of the settlement of structures built on shallow foundations, Exercise 7.44 (p. 396). Recall that actual settlement data (measured in millimeters) for a sample of 13 structures built on a shallow foundation were collected, and these values were compared to settlement predictions made using a formula that accounts for dimension, rigidity, and embedment depth of the foundation. The data are reproduced in the table. Because it is unlikely that the difference between actual and predicted is normally distributed, the researchers want to make the comparison using a nonparametric procedure.

Structure Actual Predicted

1 11 11

2 11 11

3 10 12

4 8 6

5 11 9

6 9 10

7 9 9

8 39 51

9 23 24

10 269 252

11 4 3

12 82 68

13 250 264

Source: Ozur, M. “Comparing methods for predicting immediate settlement of shallow foundations on cohesive soils based on hypothetical and real cases.” Environmental & Engineering Geoscience, Vol. 18, No. 4, Nov. 2012 (from Table 4).
1. Which nonparametric test should be applied to the data? Why?
2. Specify the null and alternative hypotheses for the test.
3. Compute the difference between actual and predicted settlement values for each of the 13 sampled structures.
4. Rank the absolute values of the differences, part c.
5. Use the ranks, part d, to find the appropriate test statistic.
6. Find the rejection region for the test using $α = .01$ $α = .01$ .
7. Make the appropriate conclusion in the words of the problem.

Structure	Actual	Predicted
1	11	11
2	11	11
3	10	12
4	8	6
5	11	9
6	9	10
7	9	9
8	39	51
9	23	24
10	269	252
11	4	3
12	82	68
13	250	264

MUSEUM 7.93 Healing potential of handling museum objects. Refer to the Museum & Society (Nov. 2009) study of the healing potential of handling museum objects, Exercise 7.41 (p. 396). Recall that the health status of each of 32 hospital patients was recorded both before and after handling a museum object (such as an archaeological artifact or brass etching). The simulated data (measured on a 100-point scale) are reproduced in the next table below. The Wilcoxon signed rank test was applied to the data, with the results shown in the SPSS printout in the next column.

Session	Before	After
1	52	59
2	42	54
3	46	55
4	42	51
5	43	42
6	30	43
7	63	79
8	56	59
9	46	53
10	55	57
11	43	49
12	73	83
13	63	72
14	40	49
15	50	49
16	50	64
17	65	65
18	52	63
19	39	50
20	59	69
21	49	61
22	59	66
23	57	61
24	56	58
25	47	55
26	61	62
27	65	61
28	36	53
29	50	61
30	40	52
31	65	70
32	59	72

Use the information in the printout to find the large-sample Wilcoxon signed rank test statistic.
Does handling a museum object have a positive impact on a sick patient’s well-being? Test, using $α = .01$ $α = .01$ .

REDLIT 7.94 Impact of red light cameras on car crashes. Refer to the June 2007 Virginia Department of Transportation (VDOT) study of a newly adopted photo-red-light enforcement program, Exercise 7.51 (p. 398). Recall that the VDOT provided crash data both before and after installation of red light cameras at several intersections. The data (measured as the number of crashes caused by red light running per intersection per year) for 13 intersections in Fairfax County, Virginia, are reproduced in the table on p. 419. The VDOT wants to determine if the photo-red enforcement program is effective in reducing red-light-running crash incidents at intersections. Use the nonparametric Wilcoxon signed rank test (and the accompanying MINITAB printout) to analyze the data for the VDOT.

Data for Exercise 7.94
Intersection	Before Camera	After Camera
1	3.60	1.36
2	0.27	0
3	0.29	0
4	4.55	1.79
5	2.60	2.04
6	2.29	3.14
7	2.40	2.72
8	0.73	0.24
9	3.15	1.57
10	3.21	0.43
11	0.88	0.28
12	1.35	1.09
13	7.35	4.92

Based on Virginia Transportation Research Council, “Research Report: The Impact of Red Light Cameras (Photo-Red Enforcement) on Crashes in Virginia,” June 2007.

READERS 7.95 Reading comprehension strategies of elementary school children. An investigation of the reading comprehension strategies employed by good and average elementary school readers was the topic of research published in The Reading Matrix (Apr. 2004). Both good and average readers were recruited on the basis of their scores on a midterm language test. Each group was evaluated on how often its members employed each of eight different reading strategies. The accompanying table gives the proportion of times the reading group used each strategy (called the Factor Specificity Index, or FSI score). The researchers conducted a Wilcoxon signed rank test to compare the FSI score distributions of good and average readers.

	FSI Scores
Strategy	Good Readers	Average Readers
Word meaning	.38	.32
Words in context	.29	.25
Literal comprehension	.42	.25
Draw inference from single string	.60	.26
Draw inference from multiple string	.45	.31
Interpretation of metaphor	.32	.14
Find salient or main idea	.21	.03
Form judgment	.73	.80

Based on Ahmed, S., and Asraf, R. M. “Making sense of text: Strategies used by good and average readers.” The Reading Matrix, Vol. 4, No. 1, Apr. 2004 (Table 2).

State $H_{0}$ $H_{0}$ and $H_{a}$ $H_{a}$ for the desired test of hypothesis.
For each strategy, compute the difference between the FSI scores of good and average readers.
Rank the absolute values of the differences.
Calculate the value of the signed rank test statistic.
Find the rejection region for the test, using $α = .05 .$ $α = .05 .$
Make the appropriate inference in the words of the problem.

SOLAR 7.96 Solar energy generation along highways. Refer to the International Journal of Energy and Environmental Engineering (Dec. 2013) study of solar panels constructed above national highways to generate energy, Exercise 7.45 (p. 397). Recall that solar panels were constructed above sections of both east-west and north-south highways in India. The amount of energy (kilowatt-hours) supplied to the country’s grid by the solar panels above the two types of highways was determined each month. The data for several randomly selected months are reproduced in the table. Use a nonparametric test to determine if the distribution of monthly solar energy levels for north-south oriented highways is shifted above the corresponding distribution for east-west oriented highways. Test, using $α = .05 .$ $α = .05 .$

Month East-West North-South

February 8658 8921

April 7930 8317

July 5120 5274

September 6862 7148

October 8608 8936

Source: Sharma, P., and Harinarayana, T. “Solar energy generation potential along national highways.” International Journal of Energy and Environmental Engineering, Vol. 49, No. 1, Dec. 2013 (Table 3).

Month	East-West	North-South
February	8658	8921
April	7930	8317
July	5120	5274
September	6862	7148
October	8608	8936

Applying the Concepts—Intermediate

7.97 TOLER Ethical sensitivity of teachers towards racial intolerance. Refer to the Journal of Moral Education (Mar. 2010) study of the effectiveness of a program to encourage teachers to embrace racial tolerance, Exercise 7.50 (p. 398). Recall that the level of racial tolerance was measured for each teacher before (pretest) and after (posttest) the teachers participated in an all-day workshop on cultural competence. The original sample included 238 high school teachers. The table below lists the pretest and posttest scores for a smaller sample of 10 high school teachers. The researchers conducted a paired-difference test to gauge the effectiveness of the program. Use the smaller sample to conduct the appropriate nonparametric test at $α = .01$ $α = .01$ . What do you conclude?

Teacher Pretest Posttest

1 53 74

2 73 80

3 70 94

4 72 78

5 77 78

6 81 84

7 73 71

8 87 88

9 61 63

10 76 83
7.98 Sea turtles and beach nourishment. According to the National Oceanic and Atmospheric Administration’s Office of Protected Species, sea turtle nesting rates have declined in all parts of the southeastern United States over the past 10 years. Environmentalists theorize that beach nourishment may improve the nesting rates of these turtles. (Beach nourishment involves replacing the sand on the beach in order to extend the high-water line seaward.) A study was undertaken to investigate the effect of beach nourishment on sea turtle nesting rates in Florida (Aubry Hershorin, unpublished doctoral dissertation, University of Florida, 2010). For one part of the study, eight beach zones were sampled in Jacksonville, Florida. Each beach zone was nourished by the Florida Fish and Wildlife Conservation Commission between 2000 and 2008. Nesting densities (measured as nests per linear meter) were recorded both before and after nourishing at each of the eight beach zones. The data are listed in the following table. Conduct a Wilcoxon signed rank test to compare the sea turtle nesting densities before and after beach nourishing. Use $α = .05$ $α = .05$ .

Beach Zone Before Nourishing After Nourishing

401 0 0.003595

402 0.001456 0.007278

403 0 0.003297

404 0.002868 0.003824

405 0 0.002198

406 0 0.000898

407 0.000626 0

408 0 0

Teacher	Pretest	Posttest
1	53	74
2	73	80
3	70	94
4	72	78
5	77	78
6	81	84
7	73	71
8	87	88
9	61	63
10	76	83

Beach Zone	Before Nourishing	After Nourishing
401	0	0.003595
402	0.001456	0.007278
403	0	0.003297
404	0.002868	0.003824
405	0	0.002198
406	0	0.000898
407	0.000626	0
408	0	0

SCHEALTH 7.99 Food availability at middle schools. Most schools offer a la carte food items in the cafeteria for students at lunch. To encourage students to eat healthfully, the U.S. Department of Agriculture (USDA) requires schools to offer nutritional food items. Two methods for identifying and quantifying food items in the a la carte line—a detailed inventory approach and a checklist approach—were compared in the Journal of School Health (Dec. 2009). Data were collected for a sample of 36 middle schools. For each school, the accompanying table gives the percentage of a la carte food items deemed healthy as determined by both methods. The researchers used a nonparametric analysis to determine if the distribution of healthy food item percentages using the inventory method is shifted above or below the distribution of healthy food item percentages using the checklist method. If no significant difference is detected, the checklist method will be recommended because it is simpler and requires fewer resources. A MINITAB analysis of the difference (inventory value minus checklist value) is displayed below. Use this printout to conduct the appropriate analysis at $α = .05$ $α = .05$ . Which method do you recommend?

School	Percentage Inventory	Healthy Checklist
A	100.0	100.0
B	95.5	66.7
C	90.6	62.5
D	77.8	71.4
E	66.7	66.7
F	64.5	50.0
G	62.5	50.0
H	55.1	63.6
I	54.3	58.3
J	54.3	55.6
K	53.8	58.3
L	53.7	58.3
M	52.9	66.7
N	52.0	54.5
O	51.5	50.0
P	50.0	50.0
Q	50.0	100.0
R	50.0	66.7
S	50.0	62.5
T	46.3	60.0
U	44.2	70.0
V	43.8	50.0
W	43.5	60.0
X	42.2	40.0
Y	41.3	54.5
Z	40.7	55.6
AA	39.0	55.6
BB	38.5	42.9
CC	35.8	44.4
DD	32.4	50.0
EE	29.2	50.0
FF	28.9	45.5
GG	27.8	50.0
HH	25.0	12.5
II	25.0	100.0
JJ	7.7	66.7
KK	6.3	66.7

Source: Hearst, M. O., et al. “Inventory versus checklist approach to assess middle school a la carte food availability.” Journal of School Health, Vol. 79, No. 12, Dec. 2009 (Table 3).

7.100 POWVEP Neurological impairment of POWs. Eleven prisoners of war during the war in Croatia were evaluated for neurological impairment after their release from a Serbian detention camp (Collegium Antropologicum, June 1997). All 11 experienced blows to the head and neck and/or loss of consciousness during imprisonment. Neurological impairment was assessed by measuring the amplitude of the visual evoked potential (VEP) in both eyes at two points in time: 157 days and 379 days after their release. (The higher the VEP value, the greater the neurological impairment.) The data on the 11 POWs are shown in the accompanying table. Determine whether the VEP measurements of POWs 379 days after their release tend to be greater than the VEP measurements of POWs 157 days after their release. Test, using $α = .05 .$ $α = .05 .$

POW	157 Days after Release	379 Days after Release
1	2.46	3.73
2	4.11	5.46
3	3.93	7.04
4	4.51	4.73
5	4.96	4.71
6	4.42	6.19
7	1.02	1.42
8	4.30	8.70
9	7.56	7.37
10	7.07	8.46
11	8.00	7.16

Based on Vrca, A., et al. “The use of visual evoked potentials to follow-up prisoners of war after release from detention camps.” Collegium Antropologicum, Vol. 21, No. 1, June 1997, p. 232. (Data simulated from information provided in Table 3.)

7.101 TENDON Treatment for tendon pain. In a British Journal of Sports Medicine (Feb. 1, 2004) study of chronic Achilles tendon pain, each in a sample of 25 patients with chronic Achilles tendinosis was treated with heavy-load eccentric calf muscle training. Tendon thickness (in millimeters) was measured both before and following the treatment of each patient. The experimental data are provided in the table. Use a nonparametric test to determine whether the treatment for tendonitis tends to reduce the thickness of tendons. Test, using $α = .10 .$ $α = .10 .$

Patient	Before Thickness (millimeters)	After Thickness (millimeters)
1	11.0	11.5
2	4.0	6.4
3	6.3	6.1
4	12.0	10.0
5	18.2	14.7
6	9.2	7.3
7	7.5	6.1
8	7.1	6.4
9	7.2	5.7
10	6.7	6.5
11	14.2	13.2
12	7.3	7.5
13	9.7	7.4
14	9.5	7.2
15	5.6	6.3
16	8.7	6.0
17	6.7	7.3
18	10.2	7.0
19	6.6	5.3
20	11.2	9.0
21	8.6	6.6
22	6.1	6.3
23	10.3	7.2
24	7.0	7.2
25	12.0	8.0

Based on Ohberg, L., et al. “Eccentric training in patients with chronic Achilles tendinosis: Normalized tendon structure and decreased thickness at follow up.” British Journal of Sports Medicine, Vol. 38, No. 1, Feb. 1, 2004 (Table 2).

Applying the Concepts—Advanced

BOWLER 7.102 Bowler’s hot hand. Is the probability of a bowler rolling a strike higher after he has thrown four consecutive strikes? An investigation into the phenomenon of a “hot hand” in bowling was published in The American Statistician (Feb. 2004). Frame-by-frame results were collected on 43 professional bowlers from a recent Professional Bowlers Association (PBA) season. For each bowler, the researchers calculated the proportion of strikes rolled after bowling four consecutive strikes and the proportion after bowling four consecutive nonstrikes. The data (proportion of strikes after four strikes and after four nonstrikes, respectively) on 4 of the 43 bowlers are as follows: Bowler 1: .683 and .432; Bowler 2: .684 and .400; Bowler 3: .632 and .421; Bowler 4: .610 and .529.
1. Do the data on the sample of four bowlers provide support for the “hot hand” theory in bowling? Explain. No
2. When the data on all 43 bowlers are used, the p-value for the hypothesis test is approximately 0. Interpret this result.

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Table of Contents for
7.6 A Nonparametric Test for Comparing Two Populations: Paired Difference Experiment (Optional)

7.6 A Nonparametric Test for Comparing Two Populations: Paired Difference Experiment (Optional)

Table 7.7 Softness Ratings of Paper

Table 7.8 Reproduction of Part of Table VI of Appendix B: Critical Values for the Wilcoxon Paired Difference Signed Rank Test

Figure 7.18

Nonparametric Wilcoxon Signed Rank Test: Matched Pairs Experiment

Conditions Required for a Valid Signed Rank Test

Example 7.9 Applying the Signed Rank Test—Comparing Two Crime Prevention Plans

Problem

Table 7.9 Effectiveness Ratings by 10 Qualified Crime Prevention Experts

Solution

Figure 7.19

Look Back

Figure 7.20

Large-Sample Wilcoxon Signed Rank Test: Matched Pairs Experiment $(n \geq 25)$ $(n \geq 25)$

Exercises 7.86–7.102

Understanding the Principles

Learning the Mechanics

Applying the Concepts—Basic

Data for Exercise 7.94

Applying the Concepts—Intermediate

Applying the Concepts—Advanced

Table of Contents for 7.6 A Nonparametric Test for Comparing Two Populations: Paired Difference Experiment (Optional)

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Table of Contents for
7.6 A Nonparametric Test for Comparing Two Populations: Paired Difference Experiment (Optional)