Nonparametric Tests

1
Chapter 11

• Nonparametric Tests

2
Chapter Outline

• 11.1 The Sign Test
• 11.2 The Wilcoxon Tests
• 11.3 The Kruskal-Wallis Test
• 11.4 Rank Correlation
• 11.5 The Runs Test

3
Section 11.1

• The Sign Test

4
Section 11.1 Objectives

• Use the sign test to test a population median
• Use the paired-sample sign test to test the
  difference between two population medians
  (dependent samples)

5
Nonparametric Test

• Nonparametric test
• A hypothesis test that does not require any
  specific conditions concerning the shape of the
  population or the value of any population
  parameters.
• Generally easier to perform than parametric
  tests.
• Usually less efficient than parametric tests
  (stronger evidence is required to reject the null
  hypothesis).

6
Sign Test for a Population Median

• Sign Test
• A nonparametric test that can be used to test a
  population median against a hypothesized value k.
• Left-tailed test
• H0 median ? k and Ha median lt k
• Right-tailed test
• H0 median ? k and Ha median gt k
• Two-tailed test
• H0 median k and Ha median ? k

7
Sign Test for a Population Median

• To use the sign test, each entry is compared with
  the hypothesized median k.
• If the entry is below the median, a ? sign is
  assigned.
• If the entry is above the median, a sign is
  assigned.
• If the entry is equal to the median, 0 is
  assigned.
• Compare the number of and signs.

8
Sign Test for a Population Median

• 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 gt 25, the test statistic for the sign test
  is
• where x is the smaller number of or ? signs
  and n is the sample size (the total number of
  or ? signs).

9
Performing a Sign Test for a Population Median
In Words In Symbols

1. State the claim. Identify the null and
   alternative hypotheses.
2. Specify the level of significance.
3. Determine the sample size n by assigning signs
   and signs to the sample data.
4. Determine the critical value.

State H0 and Ha.
Identify ?.
n total number of and signs
If n ? 25, use Table 8. If n gt 25, use Table 4.
10
Performing a Sign Test for a Population Median
In Words In Symbols

1. Calculate the test statistic.

If n ? 25, use x.If n gt 25, use
If the test statistic is less than or equal to
the critical value, reject H0. Otherwise, fail
to reject H0.

1. Make a decision to reject or fail to reject the
   null hypothesis.
2. Interpret the decision in the context of the
   original claim.

11
Example Using the Sign Test

• A bank manager claims that the median number of
  customers per day is no more than 750. A teller
  doubts the accuracy of this claim. The number of
  bank customers per day for 16 randomly selected
  days are listed below. At a 0.05, can the
  teller reject the bank managers claim?
• 775 765 801 742
• 754 753 739 751
• 745 750 777 769
• 756 760 782 789

12
Solution Using the Sign Test
median 750

• H0
• Ha

median gt 750

• Compare each data entry with the hypothesized
  median 750

775 765 801 742 754 753 739 751 745 750 777 769 75
6 760 782 789
0

• There are 3 signs and 12 signs
• n 12 3 15

13
Solution Using the Sign Test
0.05

• a
• Critical Value

Use Table 8 (n 25)
Critical value is 3
14
Solution Using the Sign Test

• Test Statistic

x 3 (n 25 use smaller number of or signs)
Reject H0

• Decision

At the 5 level of significance, the teller can
reject the bank managers claim that the median
number of customers per day is no more than 750.
15
Example Using the Sign Test

• A car dealership claims to give customers a
  median trade-in offer of at least 6000. A random
  sample of 103 transactions revealed that the
  trade-in offer for 60 automobiles was less than
  6000 and the trade-in offer for 40 automobiles
  was more than 6000. At a 0.01, can you reject
  the dealerships claim?

16
Solution Using the Sign Test

• H0
• Ha
• a
• Critical value

median 6000

• Test Statistic
• Decision

There are 60 signs and 40 signs. n 60 40
100 x 40
median lt 6000
0.01
n gt 25
Fail to Reject H0
-2.33
At the 1 level of significance you cannot reject
the dealerships claim.
-1.9
17
The Paired-Sample Sign Test

• Paired-sample sign test
• 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.

18
Performing The Paired-Sample Sign Test
In Words In Symbols

1. State the claim. Identify the null and
   alternative hypotheses.
2. Specify the level of significance.
3. 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
19
Performing The Paired-Sample Sign Test
In Words In Symbols

1. Determine the critical value.
2. Find the test statistic.

Use Table 8 inAppendix B.
x lesser number of and signs

1. Make a decision to reject or fail to reject the
   null hypothesis.
2. Interpret the decision in the context of the
   original claim.

If the test statistic is less than or equal to
the critical value, reject H0. Otherwise, fail
to reject H0.
20
Example Paired-Sample Sign Test

• A psychologist claims that the number of repeat
  offenders will decrease if first-time offenders
  complete a particular rehabilitation course. You
  randomly select 10 prisons and record the number
  of repeat offenders during a two-year period.
  Then, after first-time offenders complete the
  course, you record the number of repeat offenders
  at each prison for another two-year period. The
  results are shown on the next slide. Ata
  0.025, can you support the psychologists claim?

21
Example Paired-Sample Sign Test
Prison 1 2 3 4 5 6 7 8 9 10
Before 21 34 9 45 30 54 37 36 33 40
After 19 22 16 31 21 30 22 18 17 21
Sign
Solution
The number of repeat offenders will not decrease.

• H0
• Ha

The number of repeat offenders will decrease.

• Determine the sign of the difference between the
  before and after data.

22
Solution Paired-Sample Sign Test
Sign
0.025 (one-tailed)

• a
• n
• Critical value

1 9 10
Critical value is 1
23
Solution Paired-Sample Sign Test
Sign

• Test Statistic
• Decision

x 1 (the smaller number of or signs)
Reject H0
At the 2.5 level of significance, you can
support the psychologists claim that the number
of repeat offenders will decrease.
24
Section 11.1 Summary

• Used the sign test to test a population median
• Used the paired-sample sign test to test the
  difference between two population medians
  (dependent samples)

25
Section 11.2

• The Wilcoxon Tests

26
Section 11.2 Objectives

• Use the Wilcoxon signed-rank test to determine if
  two dependent samples are selected from
  populations having the same distribution
• Use the Wilcoxon rank sum test to determine if
  two independent samples are selected from
  populations having the same distribution.

27
The Wilcoxon Signed-Rank Test

• Wilcoxon Signed-Rank Test
• A nonparametric test that can be used to
  determine whether two dependent samples were
  selected from populations having the same
  distribution.
• Unlike the sign test, it considers the magnitude,
  or size, of the data entries.

28
Performing The Wilcoxon Signed-Rank Test
In Words In Symbols

1. State the claim. Identify the null and
   alternative hypotheses.
2. Specify the level of significance.
3. Determine the sample size n, which is the number
   of pairs of data for which the difference is not
   0.
4. Determine the critical value.

State H0 and Ha.
Identify ?.
Use Table 9 in Appendix B.
29
Performing The Wilcoxon Signed-Rank Test
In Words In Symbols

• Calculate the test statistic ws.
• Complete a table using the headers listed at the
  right.
• Find the sum of the positive ranks and the sum of
  the negative ranks.
• Select the smaller of absolute values of the sums.

Headers Sample 1, Sample 2, Difference, Absolute
value, Rank, and Signed rank. Signed rank takes
on the same sign as its corresponding difference.
30
Performing The Wilcoxon Signed-Rank Test
In Words In Symbols

1. Make a decision to reject or fail to reject the
   null hypothesis.
2. 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.
31
Example Wilcoxon Signed-Rank Test

• A sports psychologist believes that listening to
  music affects the length of athletes workout
  sessions. The length of time (in minutes) of 10
  athletes workout sessions, while listening to
  music and while not listening to music, are
  shown in the table. At a 0.05, can you support
  the sports psychologists claim?

With music 45 38 28 39 41 47 62 54 33 44
Without music 38 40 33 36 42 41 54 47 28 35
32
Solution Wilcoxon Signed-Rank Test
There is no difference in the length of the
athletes workout sessions.

• H0
• Ha
• a
• n

There is a difference in the length of the
athletes workout sessions.
0.05 (two-tailed test)
10 (the difference between each data pair is not
0)
33
Solution Wilcoxon Signed-Rank Test

• Critical Value

Table 9
Critical value is 8
34
Solution Wilcoxon Signed-Rank Test

• Test Statistic

With music Without music Difference Absolute value Rank Signed rank
45 38
38 40
28 33
39 36
41 42
47 41
62 54
54 47
33 28
44 35
7
7.5
-2
2
4.5
-5
3
3
1
-1
6
6
8
9
7
7.5
4.5
5
9
10
35
Solution Wilcoxon Signed-Rank Test

• Test Statistic

The sum of the negative ranks is -1 (-2)
(-4.5) -7.5
The sum of the positive ranks is (3) (4.5)
(6) (7.5) (7.5) (9) (10) 47.5
ws 7.5 (the smaller of the absolute value of
these two sums -7.5 lt 47.5)
36
Solution Wilcoxon Signed-Rank Test

• Decision

Reject H0
At the 5 level of significance, you have enough
evidence to support the claim that music makes a
difference in the length of athletes workout
sessions.
37
The Wilcoxon Rank Sum Test

• Wilcoxon Rank Sum Test
• 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.

38
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
39
Performing The Wilcoxon Rank Sum Test
In Words In Symbols

1. State the claim. Identify the null and
   alternative hypotheses.
2. Specify the level of significance.
3. Determine the critical value(s).
4. Determine the sample sizes.

State H0 and Ha.
Identify ?.
Use Table 4 in Appendix B.
n1 n2
40
Performing The 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.

R
41
Performing The Wilcoxon Rank Sum Test
In Words In Symbols

1. Calculate the test statistic.
2. Make a decision to reject or fail to reject the
   null hypothesis.
3. Interpret the decision in the context of the
   original claim.

If z is in the rejection region, reject H0.
Otherwise, fail to reject H0.
42
Example Wilcoxon Rank Sum Test

• The table shows the earnings (in thousands of
  dollars) of a random sample of 10 male and 12
  female pharmaceutical sales representatives. At a
  0.10, can you conclude that there is a
  difference between the males and females
  earnings?

Male 58 73 94 81 78 74 66 75 97 79
Female 66 57 81 73 65 78 71 67 64 77 80 70
43
Solution Wilcoxon Rank Sum Test
There is no difference between the males and the
females earnings.

• H0
• Ha

There is a difference between the males and the
females earnings.
0.10 (two-tailed test)

• a
• Rejection Region

44
Solution Wilcoxon Rank Sum Test
To find the values of R, µR, and?R, construct a
table that shows the combined data in ascending
order and the corresponding ranks.
Ordered data Sample Rank
57 F 1
58 M 2
64 F 3
65 F 4
66 M 5.5
66 F 5.5
67 F 7
70 F 8
71 F 9
73 F 10.5
73 F 10.5
Ordered data Sample Rank
74 M 12
75 M 13
77 F 14
78 M 15.5
78 F 15.5
79 M 17
80 F 18
81 M 19.5
81 F 19.5
94 M 21
97 M 22
45
Solution Wilcoxon Rank Sum Test

• Because the smaller sample is the sample of
  males, R is the sum of the male rankings.

R 2 5.5 10.5 12 13 15.5 17 19.5
21 22 138

• Using n1 10 and n2 12, we can find µR, and?R.

46
Solution Wilcoxon Rank Sum Test

• H0
• Ha

no difference in earnings.

• Test Statistic

difference in earnings.

• a
• Rejection Region

0.10

• Decision

Fail to reject H0
At the 10 level of significance, you cannot
conclude that there is a difference between the
males and females earnings.
1.52
47
Section 11.2 Summary

• Used the Wilcoxon signed-rank test to determine
  if two dependent samples are selected from
  populations having the same distribution
• Used the Wilcoxon rank sum test to determine if
  two independent samples are selected from
  populations having the same distribution.

48
Section 11.3

• The Kruskal-Wallis Test

49
Section 11.3 Objectives

• Use the Kruskal-Wallis test to determine whether
  three or more samples were selected from
  populations having the same distribution.

50
The Kruskal-Wallis Test

• Kruskal-Wallis test
• 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.
51
The Kruskal-Wallis Test

• Two conditions for using the Kruskal-Wallis test
  are
• Each sample must be randomly selected
• 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.

52
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, Ri is the sum of the ranks of
the ith sample.
53
Performing a Kruskal-Wallis Test
In Words In Symbols

1. State the claim. Identify the null and
   alternative hypotheses.
2. Specify the level of significance.
3. Identify the degrees of freedom
4. Determine the critical value and the rejection
   region.

State H0 and Ha.
Identify ?.
d.f. k 1
Use Table 6 in Appendix B.
54
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.

55
Performing a Kruskal-Wallis Test
In Words In Symbols
If H is in the rejection region, reject H0.
Otherwise, fail to reject H0.

1. Make a decision to reject or fail to reject the
   null hypothesis.
2. Interpret the decision in the context of the
   original claim.

56
Example Kruskal-Wallis Test

• You want to compare the hourly pay rates of
  actuaries who work in California, Indiana, and
  Maryland. To do so, you randomly select several
  actuaries in each state and record their hourly
  pay rate. The hourly pay rates are shown on the
  next slide. At a 0.01, can you conclude that
  the distributions of actuaries hourly pay rates
  in these three states are different? (Adapted
  from U.S. Bureau of Labor Statistics)

57
Example Kruskal-Wallis Test
Sample Hourly Pay Rates Sample Hourly Pay Rates Sample Hourly Pay Rates
CA(Sample 1) IN(Sample 2) MD(Sample 3)
40.50 33.45 49.68
44.98 40.12 44.94
47.78 38.65 48.80
43.20 35.98 49.20
37.10 35.97 40.37
49.88 4570 48.79
42.05 42.05 53.82
52.94 35.97 45.35
41.70 38.25 53.25
43.85 43.57
58
Solution Kruskal-Wallis Test

• H0
• Ha

There is no difference in the hourly pay rates in
the three states.
There is a difference in the hourly pay rates in
the three states.

• a
• d.f.
• Rejection Region

0.01
k 1 3 1 2
59
Solution Kruskal-Wallis Test
The table shows the combined data listed in
ascending order and the corresponding ranks.
Ordered Data Sample Rank
33.45 IN 1
35.97 IN 2.5
35.97 IN 2.5
35.98 IN 4
37.10 CA 5
38.25 IN 6
38.65 IN 7
40.12 IN 8
40.37 MD 9
40.50 CA 10
Ordered Data Sample Rank
41.70 CA 11
42.05 IN 12.5
42.05 CA 12.5
43.20 CA 14
43.57 MD 15
43,85 CA 16
44.94 MD 17
44.98 CA 18
45.35 MD 19
45.70 IN 20
Ordered Data Sample Rank
47.78 CA 21
48.79 MD 22
48.80 MD 23
49.20 MD 24
49.68 MD 25
49.88 CA 26
52.94 CA 27
53.25 MD 28
53.82 MD 29
60
Solution Kruskal-Wallis Test
The sum of the ranks for each sample is as
follows. R1 5 10 11 12.5 14 16
18 21 26 27 160.5 R2 1
2.5 2.5 4 6 7 8 12.5 20
63.5 R3 9 15 17 19 22 23 24
25 28 29 211
61
Solution Kruskal-Wallis Test

• H0
• Ha

no difference in hourly pay rates

• Test Statistic

H 13.12
difference in hourly pay rates.

• Decision

Reject H0
0.01

• a
• d.f.
• Rejection Region

At the 1 level of significance, you can conclude
that there is a difference in actuaries hourly
pay rates in California, Indiana, and Maryland.
3 1 2
13.12
62
Section 11.3 Summary

• Used the Kruskal-Wallis test to determine whether
  three or more samples were selected from
  populations having the same distribution.

63
Section 11.4

• Rank Correlation

64
Section 11.4 Objectives

• Use the Spearman rank correlation coefficient to
  determine whether the correlation between two
  variables is significant.

65
The Spearman Rank Correlation Coefficient

• Spearman Rank Correlation Coefficient
• A measure of the strength of the relationship
  between two variables.
• Nonparametric equivalent to the Pearson
  correlation coefficient.
• Calculated using the ranks of paired sample data
  entries.
• Denoted rs

66
The Spearman Rank Correlation Coefficient

• Spearman rank correlation coefficient rs
• The formula for the Spearman rank correlation
  coefficient is
• where n is the number of paired data entries
• d is the difference between the ranks of
  a paired data entry.

67
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 ?s.

68
The Spearman Rank Correlation Coefficient

• The null and alternative hypotheses for this test
  are as follows.

H0 ?s 0 (There is no correlation between the
variables.) Ha ?s ? 0
(There is a significant correlation between
the variables.)
69
Testing the Significance of the Spearman Rank
Correlation Coefficient
In Words In Symbols

1. State the null and alternative hypotheses.
2. Specify the level of significance.
3. Determine the critical value.
4. Find the test statistic.

State H0 and Ha.
Identify ?.
Use Table 10 in Appendix B.
70
Testing the Significance of the Spearman Rank
Correlation Coefficient
In Words In Symbols
If rs is greater than the critical value,
reject H0. Otherwise, fail to reject H0.

1. Make a decision to reject or fail to reject the
   null hypothesis.
2. Interpret the decision in the context of the
   original claim.

71
Example The Spearman Rank Correlation Coefficient

• The table shows the prices (in dollars per 100
  pounds) received by U.S. farmers for beef and
  lamb from 1999 to 2005. At a 0.05, can you
  conclude that there is a correlation between the
  beef and lamb prices? (Source U.S. Department
  of Agriculture)

Year Beef Lamb
1999 63.4 74.5
2000 68.6 79.8
2001 71.3 66.9
2002 66.5 74.1
2003 79.7 94.4
2004 85.8 101.0
2005 89.7 110.0
72
Solution The Spearman Rank Correlation
Coefficient

• H0
• Ha

?s 0 (no correlation between beef and lamb
prices)
?s ? 0 (correlation between beef and lamb prices)

• a
• n
• Critical value

0.05
7
The critical value is 0.786
73
Solution The Spearman Rank Correlation
Coefficient
Beef Rank Lamb Rank d d2
63.4 74.5
68.6 79.8
71.3 66.9
66.5 74.1
79.7 94.4
85.8 101.0
89.7 110.0
1
3
-2
4
3
4
-1
1
4
1
3
9
2
2
0
0
5
5
0
0
6
6
0
0
7
7
0
0
Sd2 14
74
Solution The Spearman Rank Correlation
Coefficient
?s 0

• H0
• Ha

• Test Statistic

?s ? 0

• a
• n
• Critical value

0.05
7

• Decision

Fail to Reject H0
The critical value is 0.786
At the 5 level of significance, you cannot
conclude that there is a significant correlation
between beef and lamb prices between 1999 and
2005.
75
Section 11.4 Summary

• Used the Spearman rank correlation coefficient to
  determine whether the correlation between two
  variables is significant.

76
Section 11.5

• The Runs Test

77
Section 11.5 Objectives

• Use the runs test to determine whether a data set
  is random.

78
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.

79
Example Finding the Number of Runs

• A liquid-dispensing machine has been designed to
  fill one-liter bottles. A quality control
  inspector decides whether each bottle is filled
  to an acceptable level and passes inspection (P)
  or fails inspection (F). Determine the number of
  runs for the sequence and find the length of each
  run.
• P P F F F F P F F F P P P P P P

There are 5 runs.
Solution
P P F F F F P F F F P P P P P P
Length of run
2
4
1
3
6
80
Runs Test for Randomness

• Runs Test for Randomness
• A nonparametric test that can be used to
  determine whether a sequence of sample data is
  random.
• The null and alternative hypotheses for this test
  are as follows.

H0 The sequence of data is random. Ha The
sequence of data is not random.
81
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 gt 20 or n2 gt 20, the test statistic for
  the runs test is

where
82
Performing a Runs Test for Randomness
In Words In Symbols

1. State the claim. Identify null and alternative
   hypotheses.
2. Specify the level of significance. (Use a 0.05
   for the runs test.)
3. Determine the number of data that have each
   characteristic and the number of runs.

State H0 and Ha.
Identify ?.
Determine n1, n2, and G.
83
Performing a Runs Test for Randomness
In Words In Symbols

1. Determine the critical values.
2. Calculate the test statistic.

If n1 ? 20 and n2 ? 20, use Table 12. If n1 gt 20
or n2 gt 20, use Table 4.
84
Performing a Runs Test for Randomness
In Words In Symbols

1. Make a decision to reject or fail to reject the
   null hypothesis.
2. Interpret the decision in the context of the
   original claim.

If G ? the lower critical value, or if G ? the
upper critical value, reject H0. Otherwise, fail
to reject H0.
85
Example Using the Runs Test

• A foreman for a construction company records
  injuries reported by workers during his shift.
  The following sequence shows whether any injuries
  were reported during each month in a recent year.
  I represents a month in which at least one injury
  was reported and N represents a month in which no
  injuries were reported. At a 0.05, can you
  conclude that the occurrence of injuries each
  month is not random?
• I I N N N I N I I N N N

86
Solution Using the Runs Test

• H0
• Ha

The occurrence of injuries is random.
The occurrence of injuries is not random.

• I I N N N I N I I N N N

• n1 number of Is
• n2 number of Ns
• G number of runs
• a
• Critical value

5
7
6
0.05
Because n1 20, n2 20, use Table 12
87
Solution Using the Runs Test

• Critical value
• n1 5 n2 7 G 6

The lower critical value is 3 and the upper
critical value is 11.
88
Solution Using the Runs Test

• H0
• Ha

random

• Decision

Fail to Reject H0
At the 5 level of significance, you do not have
enough evidence to support the claim that the
occurrence of injuries is not random. So, it
appears that the injuries reported by workers
during the foremans shift occur randomly.
not random

• n1 n2
• G
• a
• Critical value

5
7
6
0.05
lower critical value 3 upper critical value 11

• Test Statistic

G 6
89
Example Using the Runs Test

• You want to determine whether the selection of
  recently hired employees in a large company is
  random with respect to gender. The genders of 36
  recently hired employees are shown below. At a
  0.05, can you conclude that the selection is not
  random?
• M M F F F F M M M M M M
• F F F F F M M M M M M M
• F F F M M M M F M M F M

90
Solution Using the Runs Test

• H0
• Ha

The selection of employees is random.
The selection of employees is not random.

• M M F F F F M M M M M M
• F F F F F M M M M M M M
• F F F M M M M F M M F M

• n1 number of Fs
• n2 number of Ms
• G number of runs

14
22
11
91
Solution Using the Runs Test

• H0
• Ha

random

• Test Statistic

not random

• n1 n2
• G
• a
• Critical value

14
22
11
0.05

• Decision

n2 gt 20, use Table 4
92
Solution Using the Runs Test
93
Solution Using the Runs Test

• H0
• Ha

random

• Test Statistic

z -2.53
not random

• n1 n2
• G
• a
• Critical value

14
22

• Decision

Reject H0
11
You have enough evidence at the 5 level of
significance to support the claim that the
selection of employees with respect to gender is
not random.
0.05
-2.53
94
Section 11.5 Summary

• Used the runs test to determine whether a data
  set is random.