1. Chapter 11
Nonparametric Tests

2. § 11.1
The Sign Test

3. Sign Test for a Population Median
A nonparametric test is a hypothesis test that does not require any specific conditions concerning the shape of the population or the value of any population parameters.
The sign test is a nonparametric test that can be used to test a population median against a hypothesized value k.
The sign test for a population median can be left tailed, right tailed, or two tailed.
Left-tailed test:
H0: median  k and Ha: median < k
Right-tailed test:
H0: median  k and Ha: median > k
Two-tailed test:
H0: median = k and Ha: median  k

4. Sign Test for a Population Median
To use the sign test, each entry is compared with the hypothesized median. If the entry is below the median, a  sign is assigned; if above the median, a + sign is assigned.
Test Statistic for the Sign Test
When n  25, the test statistic x for the sign test is the smaller number of + or  signs.
When n > 25, the test statistic for the sign test is
where x is the smaller number of + or  signs and n is the sample size, i.e., the total number of + or  signs.

5. Sign Test for a Population Median
Performing a Sign Test for a Population Median
In Words In Symbols
State the claim. Identify the null and alternative hypotheses.
Specify the level of significance.
Determine the sample size n by assigning + signs and – signs to the sample data .
Determine the critical value.
State H0 and Ha.
Identify .
n = total number of + and – signs
If n  25, use Table 8. If n > 25, use Table 4.
Continued.

6. Sign Test for a Population Median
Performing a Sign Test for a Population Median
In Words In Symbols
Calculate the test statistic.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If n  25, use x.
If n > 25, use
If x or z is in the rejection region, reject H0. Otherwise, fail to reject H0.

7. Sign Test for a Population Median
Example:
A college statistics professor claims that the median test score for his students’ last test is 58. The scores for 18 randomly selected tests are listed below. At  = 0.01, can you reject the professor’s claim? 58 62 55 53 52 59 60 56 57 61 63
H0: median = 58 (Claim)
Ha: median  58
Determine the values that are above and below the median of 58.
There are 6 + signs and 10  signs. + 
Continued.

8. Sign Test for a Population Median
Example continued:
Since there are 6 + signs and 10  signs, n = = 16.
Using Table 8 with  = 0.01 (two tailed) and n = 16, the critical value is 2.
Because n  25, the test statistic x is the smaller number of + signs or  signs, so x = 6.
6 is greater than the critical value, so we fail to reject H0.
There is not enough evidence at the 1% level to reject the professor’s claim that the median test score is 58.

9. The Paired-Sample Sign Test
The paired-sample sign test is used to test the difference between two population medians when the populations are not normally distributed.
For the paired-sample sign test to be used, the following must be true.
A sample must be randomly selected from each population.
The samples must be dependent (paired).
The difference between corresponding data entries is found and the sign of the difference is recorded.

10. The Paired-Sample Sign Test
Performing a Paired-Sample Sign Test
In Words In Symbols
Identify the claim. State the null and alternative hypotheses.
Specify the level of significance.
Determine the sample size n by finding the difference for each data pair. Assign a + sign for a positive difference, a – sign for a negative difference, and a 0 for no difference.
State H0 and Ha.
Identify .
n = total number of + and – signs
Continued.

11. The Paired-Sample Sign Test
Performing a Paired-Sample Sign Test
In Words In Symbols
Determine the critical value.
Find the test statistic.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
Use Table 8 in Appendix B.
x = lesser number of + and – signs
If the test statistic is less than or equal to the critical value, reject H0. Otherwise, fail to reject H0.

12. The Paired-Sample Sign Test
Example:
Students at a certain school are required to take the SAT twice. The table shows both verbal SAT scores for 12 students. At  = 0.05, can you conclude that the scores improved the second time they took the SAT?
Student 1 2 3 4 5 6
Score on first SAT
308
456
352
433
306
471
Score on second SAT
300
524
409
419
304
483
Sign  +
There are 8 + signs and 4  signs.
Student 7 8 9 10 11 12
Score on first SAT
538
207
205
351
360
251
Score on second SAT
708
253
399
350
480
303
Sign + 
Continued.

13. The Paired-Sample Sign Test
Example continued:
H0: The SAT scores have not improved.
Ha: The SAT scores have improved. (Claim)
Since there are 8 + signs and 4  signs, n = = 12.
Using Table 8 with  = 0.05 (one tailed) and n = 12, the critical value is 2.
The test statistic x is the smaller number of + signs or  signs, so x = 4.
4 is greater than the critical value, so we fail to reject H0.
There is not enough evidence at the 5% level to support the claim that verbal SAT scores improved.

14. § 11.2
The Wilcoxon Tests

15. The Wilcoxon Signed-Rank Test
The Wilcoxon signed-rank test is a nonparametric test that can be used to determine whether two dependent samples were selected from populations having the same distribution.
Performing a Wilcoxon Signed-Rank Test
In Words In Symbols
Identify the claim. State the null and alternative hypotheses.
Specify the level of significance.
Determine the sample size n.
State H0 and Ha.
Identify .
Continued.

16. The Wilcoxon Signed-Rank Test
Performing a Wilcoxon Signed-Rank Test
In Words In Symbols
Determine the critical value.
Calculate the test statistic ws.
Complete a table with the following headers:
Find the sum of the positive ranks and the sum of the negative ranks.
Select the smaller of absolute values of the sums.
Use Table 9 in Appendix B.
Headers: Sample 1, Sample 2, Difference, Absolute value, Rank, and Signed rank.
Continued.

17. The Wilcoxon Signed-Rank Test
Performing a Wilcoxon Signed-Rank Test
In Words In Symbols
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If ws is less than or equal to the critical value, reject H0. Otherwise, fail to reject H0.

18. The Wilcoxon Signed-Rank Test
Example:
A medical researcher want to determine whether a new drug affects the number of headache hours experienced by headache sufferers. To do so, he selects seven patients and asks each to give the number of headache hours (per day) each experiences before and after taking the drug. The results are shown in the table. At  = 0.05, can the researcher conclude that the new drug affects the number of hours?
Patient 1 2 3 4 5 6 7
Headache Hours (before)
0.8
2.4
2.8
2.6
2.7
0.9
1.2
Headache Hours (after)
1.6
1.3
1.4
1.5
1.7
H0: The drug does not affect the number of headache hours.
Ha: The drug does affect the number of headache hours. (Claim)
This is a two-tailed signed-rank test with  = 0.05 and n = 7. From Table 9, the critical value is 2.
Continued.

19. The Wilcoxon Signed-Rank Test
Example continued:
Hours (before)
Hours (after)
Difference
Absolute value
Rank
Signed rank
0.8
1.6
0.8 3 3
2.4
1.3
1.1 4
2.8
1.2 6
2.6
1.4
2.7
1.5
0.9
0.7
0.7 2 2
1.7
0.5
0.5 1 1
The average of rank 5, 6, and 7 is used for these.
The sum of the negative ranks is 6.
The sum of the positive ranks is 22.
Continued.

20. The Wilcoxon Signed-Rank Test
Example continued:
The test statistic is the smaller of the absolute value of the two sums.
|6| = 6
|22| = 22
ws = 6 which is greater than the critical value of 2.
Fail to reject H0.
There is not enough evidence at the 5% level to support the claim that the drug affects the number of headache hours.

21. The Wilcoxon Rank Sum Test
The Wilcoxon rank sum test is a nonparametric test that can be used to determine whether two independent samples were selected from populations having the same distribution.
A requirement for the Wilcoxon rank sum test is that the sample size of both samples must be at least 10.
n1 represents the size of the smaller sample and n2 represents the size of the larger sample.
When calculating the sum of the ranks R, use the ranks for the smaller of the two samples.

22. The Wilcoxon Rank Sum Test
Test Statistic for the Wilcoxon Rank Sum Test
Given two independent samples, the test statistic z for the Wilcoxon rank sum test is
where
R = sum of the ranks for the smaller sample,
and

23. The Wilcoxon Rank Sum Test
Performing a Wilcoxon Rank Sum Test
In Words In Symbols
Identify the claim. State the null and alternative hypotheses.
Specify the level of significance.
Determine the critical value.
Determine the sample sizes.
State H0 and Ha.
Identify .
Use Table 4 in Appendix B.
n1  n2
Continued.

24. The Wilcoxon Rank Sum Test
Performing a Wilcoxon Rank Sum Test
In Words In Symbols
Find the sum of the ranks for the smaller sample.
List the combined data in ascending order.
Rank the combined data.
Add the sum of the ranks for the smaller sample.
Calculate the test statistic. R
Continued.

25. The Wilcoxon Rank Sum Test
Performing a Wilcoxon Rank Sum Test
In Words In Symbols
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If z is in the rejection region, reject H0. Otherwise, fail to reject H0.

26. The Wilcoxon Rank Sum Test
Example:
An industry analyst claims that there is no difference in the salaries earned by workers in the manufacturing and construction industries. A random sample of 10 manufacturing and 10 construction workers and their salaries is shown below. At  = 0.10, can you reject the analyst’s claim? (Adapted from US Bureau of Labor Statistics)
Industry
Salary (in thousands of dollars)
Manufacturing
Construction
H0: There is no difference between the salaries. (Claim)
Ha: There is a difference between the salaries.
Continued.

27. The Wilcoxon Rank Sum Test
Example continued:
Because the test is a two-tailed test with  = 0.10, the critical values are
z0 =  and z0 =
The rejection regions are
z   and z 
To find the values of R, μR, andR, construct a table that shows the combined data in ascending order and the corresponding ranks.
Continued.

28. The Wilcoxon Rank Sum Test
Example continued:
Ordered data
Sample
Rank 26 C 1 27 2 28 3 29 M 4 30
5.5 31
7.5 32 9 33
11.5
Ordered data
Sample
Rank 33 M
11.5 34 C 14 35
15.5 38
17.5 45 19 47 20
Continued.

29. The Wilcoxon Rank Sum Test
Example continued:
Because the samples are the same size, n1 can be associated with either sample. If we let n1 be the sample of the construction workers, then R is the sum of the construction rankings.
R =
= 74.5
Using n1 = 10 and n2 = 10, we can find μR, andR.
Continued.

30. The Wilcoxon Rank Sum Test
Example continued:
When R = 74.5, μR = 105 andR = 13.23, the test statistic is
Since 2.31 is less than the critical value of 1.645,
H0 is rejected.
There is enough evidence at the 10% level to reject the claim that there is no difference in the salaries earned by workers in the manufacturing and construction industries.

31. The Kruskal-Wallis Test
§ 11.3
The Kruskal-Wallis Test

32. The Kruskal-Wallis Test
The Kruskal-Wallis test is a nonparametric test that can be used to determine whether three or more independent samples were selected from populations having the same distribution.
The null and alternative hypotheses for the Kruskal-Wallis test are as follows.
H0: There is no difference in the distribution of the populations.
Ha: There is a difference in the distribution of the populations.
Two conditions for using the Kruskal-Wallis test are that each sample must be randomly selected and the size of each sample must be at least 5. If these conditions are met, the test is approximated by a chi-square distribution with k – 1 degrees of freedom where k is the number of samples.

33. The Kruskal-Wallis Test
Test Statistic for the Kruskal-Wallis Test
Given three or more independent samples, the test statistic H for the Kruskal-Wallis test is
where
k represent the number of samples,
ni is the size of the ith sample,
N is the sum of the sample sizes,
and
Ri is the sum of the ranks of the ith sample.

34. The Kruskal-Wallis Test
Performing a Kruskal-Wallis Test
In Words In Symbols
Identify the claim. State the null and alternative hypotheses.
Specify the level of significance.
Identify the degrees of freedom
Determine the critical value and the rejection region.
State H0 and Ha.
Identify .
d.f. = k – 1
Use Table 6 in Appendix B.
Continued.

35. The Kruskal-Wallis Test
Performing a Kruskal-Wallis Test
In Words In Symbols
Find the sum of the ranks for each sample.
List the combined data in ascending order.
Rank the combined data.
Calculate the test statistic.
Continued.

36. The Kruskal-Wallis Test
Performing a Kruskal-Wallis Test
In Words In Symbols
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If H is in the rejection region, reject H0. Otherwise, fail to reject H0.

37. The Kruskal-Wallis Test
Example:
An insurance agent want to determine whether there is a difference in the annual premiums for home insurance in three states. He randomly selects homes from each state and records the annual premium for each state as shown below. At  = 0.05, can he conclude that the distributions of the annul premiums are different?
State
Annual Premium (in dollars)
New Jersey
New York
Pennsylvania
H0: There is no difference in the premiums in the three states.
Ha: There is a difference in the premiums in the three states (Claim)
Continued.

38. The Kruskal-Wallis Test
Example continued:
This is a right-tailed test with  = 0.05 and d.f. = k – 1 = 3 – 1 = 2.
From Table 6, the critical value is χ02 =
Ordered data
Sample
Rank
371 NJ 1
380 PA 2
382 3
383 4
387 5
405 6
411 7
420 8
441 9
470 10
Ordered data
Sample
Rank
474 NJ 11
484 PA 12
613 NY 13
653 14
663 15
684 16
719 17
753 18
869 19
1036 20
The table shows the order and rank of the data.
Continued.

39. The Kruskal-Wallis Test
Example continued:
The sum of the ranks for each sample is as follows.
R1 = = 46
R2 = = 118
R3 = = 46
The test statistic is
Because is greater than the critical value of 5.991, reject H0.
There is enough evidence at the 5% level to support the claim that the annual premiums are different in the three states.

40. § 11.4
Rank Correlation

41. The Spearman Rank Correlation Coefficient
The Spearman rank correlation coefficient rs is a measure of the strength of the relationship between two variables. The Spearman rank correlation coefficient is calculated using the ranks of paired sample data entries. The formula for the Spearman rank correlation coefficient is
where
n is the number of paired data entries,
and
d is the difference between the ranks of a paired data entry.

42. The Spearman Rank Correlation Coefficient
The values of rs range from 1 to 1, inclusive. If the ranks of corresponding data pairs are identical, rs is equal to +1. If the ranks are in “reverse” order, rs is equal to 1. If there is no relationship, rs is equal to 0.
To determine whether the correlation between variables is significant, you can perform a hypothesis test for the population correlation coefficient ρs.
The null and alternative hypotheses for this test are as follows.
H0: ρs = 0 (There is no correlation between the variables.)
Ha: ρs  0 (There is a significant correlation between the variables.)

43. The Spearman Rank Correlation Coefficient
Testing the Significance of the Correlation Coefficient
In Words In Symbols
State the null and alternative hypotheses.
Specify the level of significance.
Determine the critical value.
State H0 and Ha.
Identify .
Use Table 10 in Appendix B.
Continued.

44. The Spearman Rank Correlation Coefficient
Testing the Significance of the Correlation Coefficient
In Words In Symbols
Find the test statistic.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If is greater than the critical value, reject H0. Otherwise, fail to reject H0.

45. The Spearman Rank Correlation Coefficient
Example:
A Consumer Report article claims that the price of a portable CD player is related to its quality. To test this claim, you randomly select 11 portable CD players and determine the overall score and price of each. The overall score represents the error correction, locate speed, battery life, and headphone quality of a CD player. The results are in the table below. At  = 0.01, can you conclude that there is a correlation between the overall score and the price? (Adapted from Consumer Reports)
Overall Score
Price (in dollars) 82
150 78
100 68
120 67
140 61
145 60
Overall Score
Price (in dollars) 60
150 58 80 57
200 55 49 75
Continued.

46. The Spearman Rank Correlation Coefficient
Example continued:
H0: ρs = 0 (There is no correlation between score and price.)
Ha: ρs  0 (There is significant correlation between score and price.) (Claim)
Overall Score
Rank
Price (in dollars) d d2 82 11
150
9.5
1.5
2.25 78 10
100
4.5
5.5
30.25 68 9
120 6 3 67 8
140 7 1 61
145 -1 60 -4 16 58 4 80
2.5 57
200 -8 64 55 2
-0.5
0.25 49 75
Continued.

47. The Spearman Rank Correlation Coefficient
Example continued:
From Table 10 with  = 0.01 and n = 11, the critical value is
When n = 11 and ∑d 2 = 127, the test statistic is
Because < 0.818, we fail to reject H0.
At the 1% level, there is not enough evidence to conclude that there is a significant correlation between the overall score of a CD player and its price.

48. § 11.5
The Runs Test

49. The Runs Test for Randomness
A run is a sequence of data having the same characteristic. Each run is preceded by and followed by data with a different characteristic or by no data at all. The number of data in a run is called the length of the run.
Example:
The gender of babies born in a hospital in one month was recorded in order of birth, where F represents a female and M represents a male. Determine the number of runs and the length of each run.
F F F M M F F M F M M M F F F M M M M
There are 8 runs.
Length of runs:
F F F M M F F M F M M M F F F M M M M 3 2 2 1 1 3 3 4

50. The Runs Test for Randomness
The runs test for randomness is a nonparametric test that can be used to determine whether a sequence of sample data is random.
Test Statistic for the Runs Test
When n1  20 and n2  20, the test statistic for the runs test is G, the number of runs.
When n1 > 20 or n2 > 20, the test statistic for the runs test is
where

51. The Runs Test for Randomness
Performing a Runs Test for Randomness
In Words In Symbols
Identify the claim. State the null and alternative hypotheses.
Specify the level of significance. (Use  = 0.05 for the runs test.)
Determine the number of data that have each characteristic and the number of runs.
Determine the critical value.
State H0 and Ha.
Identify .
Determine n1, n2, and G.
If n1  20 and n2  20, use Table 12. If n1 > 20 or n2 > 20, use Table 4.
Continued.

52. The Runs Test for Randomness
Performing a Runs Test for Randomness
In Words In Symbols
Calculate the test statistic.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If n1  20 and n2  20, use G.
If n1 > 20 or n2 > 20, use
If G  the lower critical value, or if G  the upper critical value, reject H0. Otherwise, fail to reject H0.

53. The Runs Test for Randomness
Example:
An English professor at Smithville College is usually late for class. Students in his morning class record whether he is late (L) or on time (T ) for class each day. The results are shown below. At  = 0.05, can you conclude that the sequence is not random?
T T L T T L T T L T T L T T L T T L T T T T T T
L T T L T T L T T L T T L T L T T L T T T T T L
T T T T T L T T T L
T T L T T L T T L T T L T T L T T L T T T T T T
L T T L T T L T T L T T L T L T T L T T T T T L
T T T T T L T T T L
H0: The sequence of arrivals is random.
Ha: The sequence of arrivals is not random. (Claim)
n1 = the number of Ts = 42
n2 = the number of Ls = 16
G = the number runs = 32
Because n1 > 20, use Table 4 to find the critical values of
z0 =  and z0 = 1.96.
Continued.

54. The Runs Test for Randomness
Example continued:
Find the test statistic by first calculating μG and G.
Because 2.61 > 1.96, reject H0.
At the 5% level, there is enough evidence to support the claim that the sequence of arrivals is not random.

55. The Runs Test for Randomness
Example:
Students in the English professor’s afternoon class also record whether he is late (L) or on time (T ) for class each day. The results are shown below. At  = 0.05, can you conclude that this sequence is not random?
T T L T L L L L L T T T L L L
L L L L T L T L L L T L L L
T T L T L L L L L T T T L L L
L L L L T L T L L L T L L L
H0: The sequence of arrivals is random.
Ha: The sequence of arrivals is not random. (Claim)
n1 = the number of Ts = 9
n2 = the number of Ls = 20
G = the number runs = 12
Continued.

56. The Runs Test for Randomness
Example continued:
Because n1  20 and n2  20, use Table 12 to find the lower critical value 8 and the upper critical value 18.
The test statistic is the number of runs G = 12.
Because 12 is between the critical values of 8 and 18, we fail to reject H0.
At the 5% level, there is not enough evidence to support the claim that the sequence of arrivals is not random.