1. © 2011 Pearson Education, Inc

2. Statistics for Business and Economics
Chapter 14
Nonparametric Statistics
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3. © 2011 Pearson Education, Inc
Content
14.1 Introduction: Distribution-Free Tests
14.2 Single Population Inferences
14.3 Comparing Two Populations: Independent Samples
14.4 Comparing Two Populations: Paired Difference Experiment
14.5 Comparing Three or More Populations: Completely Randomized Design
As a result of this class, you will be able to...
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4. © 2011 Pearson Education, Inc
Content
14.6 Comparing Three or More Populations: Randomized Block Design
14.7 Rank Correlation
As a result of this class, you will be able to...
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5. © 2011 Pearson Education, Inc
Learning Objectives
Develop the need for inferential techniques that require fewer, or less stringent, assumptions than the methods of earlier chapters
Introduce nonparametric tests that are based on ranks (i.e., on an ordering of the sample measurements according to their relative magnitudes)
As a result of this class, you will be able to ...
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6. Introduction: Distribution-Free Tests
14.1
Introduction: Distribution-Free Tests
:1, 1, 3
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7. Parametric Test Procedures
Involve population parameters
Example: population mean
Require interval scale or ratio scale
Whole numbers or fractions
Example: height in inches (72, 60.5, 54.7)
Have stringent assumptions
Example: normal distribution
Examples: z-test, t-test, F-test, 2-test
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8. Nonparametric Test Procedures
Do not involve population parameters
Example: probability distributions, independence
Data measured on any scale
Ratio or interval
Ordinal
Example: good-better-best
Nominal
Example: male-female
Example: Wilcoxon rank sum test
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9. Nonnormal Distributions - t-Statistic is Invalid
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10. Distribution-Free Tests
Distribution-free tests are statistical tests that do not rely on any underlying assumptions about the probability distribution of the sampled population.
The branch of inferential statistics devoted to distribution-free tests is called nonparametrics.
Nonparametric statistics (or tests) based on the ranks of measurements are called rank statistics (or rank tests).
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11. Advantages of Nonparametric Tests
Used with all scales
Easier to compute
Developed originally before wide computer use
Make fewer assumptions
Need not involve population parameters
Results may be as exact as parametric procedures
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12. Disadvantages of Nonparametric Tests
May waste information
If data permit using parametric procedures
Example: converting data from ratio to ordinal scale
Difficult to compute by hand for large samples
Tables not widely available
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13. Frequently Used Nonparametric Tests
Sign Test
Wilcoxon Rank Sum Test
Wilcoxon Signed Rank Test
Kruskal Wallis H-Test
Spearman’s Rank Correlation Coefficient
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14. Single Population Inferences
14.2
Single Population Inferences
:1, 1, 3
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Sign Test
Tests one population median,  (eta)
Corresponds to t-test for one mean
Assumes population is continuous
Small sample test statistic: Number of sample values above (or below) median
Alternative hypothesis determines
Can use normal approximation if n  30
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16. Sign Test Uses p-Value to Make Decision
Binomial: n = 8 p = 0.5
Distribution has different shapes.
Example 1:
If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 0 defective item is about 0.6 (60%).
If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 1 defective items is about .35 (35%).
Example 2:
If inspecting 5 items & the Probability of a defect is 0.5 (50%), the Probability of finding 1 defective items is about .18 (18%).
Note:
Could use formula or tables at end of text to get Probabilities.
P-value is the probability of getting an observation at least as extreme as we got. If 7 of 8 observations ‘favor’ Ha, then p-value = p(x  7) = = If  = .05, then reject H0 since p-value  .
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17. Sign Test for a Population Median 
One-Tailed Test
H0:  = 0
Ha:  > 0 [or Ha:  < 0 ]
Test statistic:
S = Number of sample measurements greater than0 [or S = number of measurements less than 0]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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18. Sign Test for a Population Median 
Observed significance level:
p-value = P(x ≥ S)
where x has a binomial distribution with parameters n and p = .5
(Use Table II, Appendix B)
Rejection region: Reject H0 if p-value ≤ .05
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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19. Sign Test for a Population Median 
Two-Tailed Test
H0:  = 0
Ha:  ≠ 0
Test statistic:
S = Larger of S1 and S2, where S1 is the number of sample measurements less than0 and S2 is the number of measurements greater than 0
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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20. Sign Test for a Population Median 
Observed significance level:
p-value = 2P(x ≥ S)
where x has a binomial distribution with parameters n and p = .5
(Use Table II, Appendix B)
Rejection region: Reject H0 if p-value ≤ .05
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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21. Conditions Required for Valid Application of the Sign Test
The sample is selected randomly from a continuous probability distribution.
[Note: No assumptions need to be made about the shape of the probability distribution.]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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22. Large Sample Sign Test for a Population Median 
One-Tailed Test
H0:  = 0
Ha:  > 0 [or Ha:  < 0 ]
Test statistic:
S = Number of sample measurements greater than0 [or S = number of measurements less than 0] The “– .5” is the “correction for continuity.”
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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23. Large Sample Sign Test for a Population Median 
The null hypothesized mean value is np = .5n, and the standard deviation is
Rejection region: z > z
where tabulated z-values can be found in Table IV of Appendix B.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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24. Large Sample Sign Test for a Population Median 
Two-Tailed Test
H0:  = 0
Ha:  ≠ 0
Test statistic:
S = Larger of S1 and S2, where S1 is the number of sample measurements less than0 and S2 is the number of measurements greater than 0
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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25. Large Sample Sign Test for a Population Median 
The null hypothesized mean value is np = .5n, and the standard deviation is
Rejection region: z > z
where tabulated z-values can be found in Table IV of Appendix B.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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26. © 2011 Pearson Education, Inc
Sign Test Example
You’re an analyst for Chef-Boy-R-Dee. You’ve asked 7 people to rate a new ravioli on a 5-point Likert scale (1 = terrible to 5 = excellent). The ratings are: At the .05 level of significance, is there evidence that the median rating is less than 3?
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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27. © 2011 Pearson Education, Inc
Sign Test Solution
H0:
Ha:
 =
Test Statistic:
 = 3
 < 3
p-value:
Decision:
Conclusion:
P(x  2) = 1 – P(x  1) = .937 (Binomial Table, n = 7, p = 0.50)
.05
Note: More than 5 have been sold (6.4), but not enough to be significant.
S = 2 (Ratings 1 & 2 are less than  = 3: 2, 5, 3, 4, 1, 4, 5)
Do not reject at  = .05
There is no evidence median is less than 3
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28. Comparing Two Populations: Independent Samples
14.3
Comparing Two Populations: Independent Samples
:1, 1, 3
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29. © 2011 Pearson Education, Inc
Wilcoxon Rank Sum Test
Tests two independent population probability distributions
Corresponds to t-test for two independent means
Assumptions
Independent, random samples
Populations are continuous
Can use normal approximation if ni  10
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30. Wilcoxon Rank Sum Test: Independent Samples
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
One-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the right of D2
[or D1 is shifted to the left of D2]
Test statistic:
T1, if n1 < n2; T2, if n2 < n1
(Either rank sum can be used if n1 = n2.)
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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31. Wilcoxon Rank Sum Test: Independent Samples
Rejection region:
T1: T1 ≥ TU [or T1 ≤ TL]
T2: T2 ≤ TL [or T2 ≥ TU]
where TL and TU are obtained from Table XIV of Appendix B
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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32. Wilcoxon Rank Sum Test: Independent Samples
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
Two-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the left or to the right of D2
Test statistic:
T1, if n1 < n2; T2, if n2 < n1
(Either rank sum can be used if n1 = n2.) We will denote this rank sum as T.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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33. Wilcoxon Rank Sum Test: Independent Samples
Rejection region:
T ≤ TL or T ≥ TU
where TL and TU are obtained from Table XIV of Appendix B
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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34. Conditions Required for Valid Wilcoxon Rank Sum Test
1. The two samples are random and independent.
2. The two probability distributions from which the samples are drawn are continuous.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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35. Wilcoxon Rank Sum Test Procedure
Assign ranks, Ri, to the n1 + n2 sample observations
If unequal sample sizes, let n1 refer to smaller- sized sample
Smallest value = 1
Average ties
Sum the ranks, Ti, for each sample
Test statistic is Ti (smallest sample)
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36. Wilcoxon Rank Sum Test Example
You’re a production planner. You want to see if the operating rates for two factories is the same. For factory 1, the rates (% of capacity) are 85, 82, 94, and 97. For factory 2, the rates are 71, 82, 77, 92, and Do the factory rates have the same probability distributions at the .10 level of significance?
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37. Wilcoxon Rank Sum Test Solution
H0:
Ha:
 =
n1 = n2 =
Critical Value(s):
Identical Distrib.
Shifted Left or Right
.10 4 5
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38. Wilcoxon Rank Sum Table (Portion)
 = .05 one-tailed;  = .10 two-tailed n1 3 4 5 .. TL TU TL TU TL TU .. 3 6 15 7 17 7 20 .. n2 4 7 17 12 24 13 27 .. 5 7 20 13 27 19 36 .. : : : : : : : :
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39. Wilcoxon Rank Sum Test Solution
H0: Identical Distrib.
Ha: Shifted Left or Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Reject H0
Do Not Reject H0
Reject H0 13 27
 Ranks
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40. Wilcoxon Rank Sum Test Computation Table
Factory 1
Factory 2
Rate
Rank
Rate
Rank 85 5 71 1 82
3.5 3 82 4 94 8 77 2 97 9 92 7
... 88 6
Rank Sum
19.5
25.5
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41. Wilcoxon Rank Sum Test Solution
H0: Identical Distrib.
Ha: Shifted Left or Right
 = .10
n1 = 4 n2 = 5
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
T1 = = 25.5 (Smallest sample)
Do not reject at  = .10
Reject H0
Do Not Reject H0
Reject H0
There is no evidence distrib. are not identical 13 27
 Ranks
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42. Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10)
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
One-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the right of D2
[or D1 is shifted to the left of D2]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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43. Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10)
Test statistic:
Rejection region:
z > z (or z < –z)
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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44. Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10)
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
Two-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the right or to the left of D2
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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45. Wilcoxon Rank Sum Test for Large Samples (n1 ≥ 10 and n2 ≥ 10)
Test statistic:
Rejection region:
| z | > z/2
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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46. Comparing Two Populations: Paired Differences Experiment
14.4
Comparing Two Populations:
Paired Differences Experiment
:1, 1, 3
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47. Wilcoxon Signed Rank Test
Tests probability distributions of two related populations
Corresponds to t-test for dependent (paired) means
Assumptions
Random samples
Both populations are continuous
Can use normal approximation if n  25
n’ is the number of non-zero difference scores.
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48. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
One-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the right of D2
[or D1 is shifted to the left of D2]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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49. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Calculate the difference within each of the n matched pairs of observations. Then rank the absolute value of the n differences from the smallest (rank 1) to the highest (rank n) and calculate the rank sum T– of the negative differences and the rank sum T+ of the positive differences. [Note: Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly.]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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50. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Test statistic:
T–, the rank sum of the negative differences [or T+, the rank sum of the positive differences] Rejection region:
T– ≤ T0 [or T+ ≤ T0]
where T0 is given in Table XV of Appendix B
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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51. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
Two-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the left or to the right of D2
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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52. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Calculate the difference within each of the n matched pairs of observations. Then rank the absolute value of the n differences from the smallest (rank 1) to the highest (rank n) and calculate the rank sum T– of the negative differences and the rank sum T+ of the positive differences. [Note: Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly.]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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53. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Test statistic:
T–, the rank sum of the negative differences [or T+, the rank sum of the positive differences] Rejection region:
T ≤ T0
where T0 is given in Table XV of Appendix B
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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54. Conditions Required for a Valid Signed Rank Test
1. The sample of differences is randomly selected from the population of differences.
2. The probability distribution from which the sample of paired differences is drawn is continuous.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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55. Signed Rank Test Procedure
Obtain difference scores, Di = X1i – X2i
Take absolute value of differences, |Di|
Delete differences with 0 value
Assign ranks, Ri, where smallest = 1
Assign ranks same signs as Di
Sum ‘+’ ranks (T+) and ‘–’ ranks (T–)
Test statistic is T– or T+ (one-tail test)
Test statistic is T = smaller of T– or T+ (two-tail test)
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56. Signed Rank Test Computation Table
X1i
X2i
Di = X1i - X2i
|Di| Ri
Sign
Sign Ri
X11
X21
D1 = X11 - X21
|D1| R1
± R1
X12
X22
D2 = X12 - X22
|D2| R2
± R2
X13
X23
D3 = X13 - X23
|D3| R3
± R3 : : : : : : :
X1n
X2n
Dn = X1n - X2n
|Dn| Rn Rn
Total T+
& T–
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57. Signed Rank Test Example
You work in the finance department. Is the new financial package faster (α = .05)? You collect the following data entry times:
User Current New
Donna
Santosha
Sam
Tamika
Brian
Jorge
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58. Signed Rank Test Solution
H0:
Ha:
 =
n' =
Critical Value(s):
Identical Distrib.
Current Shifted Right
.05
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59. Signed Rank Test Computation Table
X1i
X2i Di
|Di | Ri
Sign
Sign Ri
9.98
9.88
+0.10
+0.02
+0.07
+0.19
0.00
0.10
0.02
0.07
0.19
0.00 4 +
... +4 +1
+2.5 +5
Discard
9.88
9.86 1
9.90
9.83
2 2.5
9.99
9.80 5
9.94
9.87
3 2.5
9.84
9.84
...
Total
T+= 15, T–= 0
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60. Signed Rank Test Solution
H0:
Ha:
 =
n' = 5 (not 6; 1 elim.)
Critical Value(s):
Identical Distrib.
Current Shifted Right
.05
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61. Wilcoxon Signed Rank Table (Portion)
One-Tailed
Two-Tailed
n = 5
n = 6
n = 7 .. a
= .05 a
= .10 1 2 4 .. a
= .025 a
= .05 1 2 .. a
= .01 a
= .02 .. a
= .005 a
= .01 ..
n = 11
n = 12
n = 13 : : : :
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62. Signed Rank Test Solution
H0:
Ha:
 =
n' = 5 (not 6; 1 elim.)
Critical Value(s):
Identical Distrib.
Current Shifted Right
Test Statistic:
Decision:
Conclusion:
T– = 0
.05
Reject at  = .05
Do Not Reject H0
Reject H0
There is evidence new package is faster 1 T0
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63. Wilcoxon Signed Rank Test for Large Samples (n ≥ 25)
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
One-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the right of D2
[or D1 is shifted to the left of D2]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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64. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Test statistic:
Rejection region:
z > z [or z < –z ]
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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65. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Assumptions:
The sample size n is greater than or equal to 25. Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly. Tied absolute differences receive ranks equal to the average of the ranks they would have received had they not been tied.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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66. Wilcoxon Signed Rank Test for Large Samples (n ≥ 25)
Let D1 and D2 represent the probability distributions for populations 1 and 2, respectively.
Two-Tailed Test
H0: D1 and D2 are identical
Ha: D1 is shifted to the left or to the right of D2
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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67. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Test statistic:
Rejection region:
| z | > z
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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68. Wilcoxon Signed Rank Test for a Paired Difference Experiment
Assumptions:
The sample size n is greater than or equal to 25. Differences equal to 0 are eliminated, and the number n of differences is reduced accordingly. Tied absolute differences receive ranks equal to the average of the ranks they would have received had they not been tied.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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69. Comparing Three or More Populations: Completely Randomized Design
14.5
Comparing Three or More Populations:
Completely Randomized Design
:1, 1, 3
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70. Kruskal-Wallis H-Test
Tests the equality of more than two (p) population probability distributions
Corresponds to ANOVA for more than two means
Used to analyze completely randomized experimental designs
Uses 2 distribution with p – 1 df
if sample size nj ≥ 5
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71. Kruskal-Wallis H-Test for Comparing k Probability Distributions
H0: The k probability distributions are identical
Ha: At least two of the k probability distributions differ in location.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
Test statistic:
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72. Kruskal-Wallis H-Test for Comparing k Probability Distributions
where
nj = Number of measurements in sample j
Rj = Rank sum for sample j, where the rank of each measurement is computed according to its relative magnitude in the totality of data for the k samples
Rj = Rj/nj = Mean rank sum for jth sample
R = Mean of all ranks = (n + 1)/2
n = Total sample size = n1 + n nk
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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73. Kruskal-Wallis H-Test for Comparing k Probability Distributions
Rejection region:
H > with (k – 1) degrees of freedom
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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74. Conditions Required for the Validity of the Kruskal-Wallis H-Test
1. The k samples are random and independent.
2. There are five or more measurements in each sample.
3. The k probability distributions from which the samples are drawn are continuous
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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75. Kruskal-Wallis H-Test Procedure
Assign ranks, Ri , to the n combined observations
Smallest value = 1; largest value = n
Average ties
Sum ranks for each group
Compute test statistic
Squared total of each group
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111

76. Kruskal-Wallis H-Test Example
As production manager, you want to see if three filling machines have different filling times. You assign 15 similarly trained and experienced workers, 5 per machine, to the machines. At the .05 level of significance, is there a difference in the distribution of filling times?
Mach1 Mach2 Mach
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112

77. Kruskal-Wallis H-Test Solution
Ha:
 =
df =
Critical Value(s):
Identical Distrib.
At Least 2 Differ
.05
p – 1 = 3 – 1 = 2 c 2
5.991
 = .05
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78. Kruskal-Wallis H-Test Solution
Raw Data
Mach1 Mach2 Mach
Ranks
Mach1 Mach2 Mach3
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79. Kruskal-Wallis H-Test Solution
Raw Data
Mach1 Mach2 Mach
Ranks
Mach1 Mach2 Mach
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80. Kruskal-Wallis H-Test Solution
Raw Data
Mach1 Mach2 Mach
Ranks
Mach1 Mach2 Mach
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81. Kruskal-Wallis H-Test Solution
Raw Data
Mach1 Mach2 Mach
Ranks
Mach1 Mach2 Mach
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82. Kruskal-Wallis H-Test Solution
Raw Data
Mach1 Mach2 Mach
Ranks
Mach1 Mach2 Mach
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83. Kruskal-Wallis H-Test Solution
Raw Data
Mach1 Mach2 Mach
Ranks
Mach1 Mach2 Mach
Total
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84. Kruskal-Wallis H-Test Solution
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115

85. Kruskal-Wallis H-Test Solution
Ha:
 =
df =
Critical Value(s):
Identical Distrib.
At Least 2 Differ
Test Statistic:
Decision:
Conclusion:
H = 11.58
.05
p – 1 = 3 – 1 = 2 c 2
Reject at  = .05
5.991
 = .05
There is evidence population distrib. are different
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86. Comparing Three or More Populations: Randomized Block Design
14.6
Comparing Three or More Populations:
Randomized Block Design
:1, 1, 3
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87. Friedman Fr-statistic
provides another method for testing to detect a shift in location of a set of k populations that have the same spread (or, scale)
is based on the rank sums of the treatments, measures the extent to which the k samples differ with respect to their relative ranks within the blocks
n’ is the number of non-zero difference scores.
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88. Friedman Fr-Test for a Randomized Block Design
H0: The probability distributions for the k treatments are identical
Ha: At least two of the probability distributions differ in location
Test statistic:
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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89. Friedman Fr-Test for a Randomized Block Design
where
b = Number of blocks
k = Number of treatments
Rj = Rank sum of the jth treatment, where the rank of each measurement is computed relative to its position within its own block
Test statistic:
Fr > with (k – 1) degrees of freedom
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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90. Friedman Fr-Test for a Randomized Block Design
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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91. Friedman Fr -Test Example
Consider the data in the table. A pharmaceutical firm wants to compare the reaction times of subjects under the influence of three different drugs that it produces. Apply the Friedman Fr-test to the data. What conclusion can you draw? Test using  = .05.
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92. Friedman Fr -Test Example
H0: The population distributions of reaction times are identical for the three drugs
Ha: At least two of the three drugs have reaction time distributions that differ in location
From the table: k = 3 treatments, b = 6 blocks, R1 = 7, R2 = 12, R3 = 17; so
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93. Friedman Fr -Test Example
For  = .05,  2.05 = , therefore
Rejection region: H >
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94. Friedman Fr -Test Example
Conclusion: Because H = 8.33 exceeds the critical value of 5.99, we reject the null hypothesis and conclude that at least two of the three drugs have distributions of reaction times that differ in location. That is, at least one of the drugs tends to yield reaction times that are faster than the others.
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95. © 2011 Pearson Education, Inc
14.7
Rank Correlation
:1, 1, 3
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96. Spearman’s Rank Correlation Coefficient
Measures correlation between ranks
Corresponds to Pearson product moment correlation coefficient
Values range from –1 to +1
Formula (shortcut)
di = ui – vi (difference in ranks of ith observation
for samples 1 and 2)
n = number of pairs of observations
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97. Spearman’s Rank Correlation Procedure
Assign ranks, Ri , to the observations of each variable separately
Calculate differences, di , between each pair of ranks
Square differences, di2, between ranks
Sum squared differences for each variable
Use shortcut approximation formula
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98. Spearman’s Rank Correlation Example
You’re a research assistant for the FBI. You’re investigating the relationship between a person’s attempts at deception and percent changes in their pupil size. You ask subjects a series of questions, some of which they must answer dishonestly. At the .05 level of significance, what is the correlation coefficient?
Subj. Deception Pupil
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112

99. Spearman’s Rank Correlation Table2
Subj.
Decep.
R1i
Pupil
R2i di di 1 87 10 2 63 6 3 95 11 4 50 7 5 43 4 4 1 -1 1 2 3 2 5 5 2 3 1 1
Σdi2=
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100. Spearman’s Rank Correlation Solution
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101. Spearman’s Nonparametric Test for Rank Correlation
One-Tailed Test
H0:  = 0
Ha:  > 0 (or Ha:  < 0)
Test statistic:
Rejection region: rs > rs, or rs <–rs, when s < 0 where rs, is the value from Table XVI corresponding to the upper-tail area  and n pairs of observations
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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102. Spearman’s Nonparametric Test for Rank Correlation
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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103. Spearman’s Nonparametric Test for Rank Correlation
Two-Tailed Test
H0:  = 0
Ha:  ≠ 0
Test statistic:
Rejection region: |rs | > rs,
where rs, is the value from Table XVI corresponding to the upper-tail area  and n pairs of observations
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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104. Spearman’s Nonparametric Test for Rank Correlation
Ties: Assign tied measurements 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 measurements are tied, assign each a rank of (3 + 4)/2 = 3.5. The number should be small relative to the total number of observations.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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105. Conditions Required for a Valid Spearman’s Test
1. The sample of experimental units on which the two variables are measured is randomly selected.
2. The probability distributions of the two variables are continuous.
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.
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106. © 2011 Pearson Education, Inc
Key Ideas
Distribution-Free Tests
Do not rely on assumptions about the probability distribution of the sampled population
As a result of this class, you will be able to...
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107. © 2011 Pearson Education, Inc
Key Ideas
Nonparametrics
Distribution-free tests that are based on rank statistics
One-sample nonparametric test for the population median – sign test
Nonparametric test for two independent samples – Wilcoxon rank sum test
Nonparametric test for matched pairs – Wilcoxon signed rank test
As a result of this class, you will be able to...
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108. © 2011 Pearson Education, Inc
Key Ideas
Nonparametrics
Nonparametric test for a completely randomized design – Kruskal-Wallis test
Nonparametric test for a randomized block design – Friedman test
Nonparametric test for rank correlation – Spearman’s test
As a result of this class, you will be able to...
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