Statistics for Business and Economics chapter 19 Nonparametric Methods

Chapter 19
Nonparametric Methods
Learning Objectives
1.

Learn the difference between parametric and nonparametric methods.

2.

Know the advantages of nonparametric methods and when they are applicable.

3.

Be able to use the sign test to conduct hypothesis tests about a median.

4.

Learn how to use the sign test to test a hypothesis with matched samples.

5.

Be able to use the Wilcoxon signed-rank test to test a hypothesis with matched samples and to test a
hypothesis about the median of a symmetric population.

6.

Be able to use the Mann-Whitney-Wilcoxon test for the comparison of two populations when using
independent random samples from each population.

7.

Be able to use the Kruskal-Wallis test for the comparison of k populations when using independent
random samples from each population.

8.

Be able to compute the Spearman rank-correlation coefficient and test for a significant rank
correlation for two variables that use ordinal or rank-ordered data.

Solutions:
19 - 1

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Chapter 19

1.

H0: Median < 150
Ha: Median > 150
There are 22 + 5 = 27 observations where the value is different from 150.
Use the normal approximation with µ = .5n = .5(27) = 13.5 and

σ = .25n = .25(27) = 2.5981
With the number of plus signs = 22 in the upper tail, we use the continuity correction factor and the
normal distribution approximation as follows
21.5 − 13.5 

P ( x ≥ 21.5) = P  z ≥

= P ( z ≥ 3.08)
2.5981 ÷



The upper-tail p-value = (1.0000 - .9990) = .0010
p-value < .01; reject H0
Conclusion: The population median is greater than 150.
2.

Let p = the probability of a preference for brand A
H0 : p = .50
Ha : p ≠ .50
Dropping the individual with no preference, the binomial probabilities for n = 9 and p = .50 are as
follows.
x
0
1
2
3
4
5
6
7
8
9

Probability
0.0020
0.0176

0.0703
0.1641
0.2461
0.2461
0.1641
0.0703
0.0176
0.0020

Number of plus signs = 7
P(x > 7) = P(7) + P(8) +P(9) = .0703 + .0176 + .0020 = .0899
Two-tailed p-value = 2(.0899) = .1798
p-value > .05, do not reject H0. There is no indication that a difference in preference exists. A
larger sample size should be considered.
3.

H0: Median = 18
Ha: Median ≠ 18
Dropping the restaurant with 18 part-time employees, the binomial probabilities for n = 8 and p
19 - 2

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods
= .50 are as follows.
x
0
1

2
3
4
5
6
7
8

Probabilit
y
0.0039
0.0313
0.1094
0.2188
0.2734
0.2188
0.1094
0.0313
0.0039

Number of plus signs = 7
P(x > 7) = P(7) + P(8) = .0313 + .0039 = .0352
Since this is an upper-tailed test, p-value = .0352
p-value < .05, reject H0. Conclude that the population median is greater than 18. There has been
an increase in the median number of part-time employees.
4.

a.

H0: Median > 15

Ha: Median < 15

b.

Dropping Vanguard GNMA with net assets of $15 billion, the binomial probabilities for n = 9 and
p = .50 are as follows.
x
Probability
0
0.0020
1
0.0176
2
0.0703
3
0.1641
4
0.2461
5
0.2461
6
0.1641
7
0.0703
8
0.0176
9
0.0020
Number of plus signs = 1 American Funds with net assets of $22.4 billion.
P(x < 1) =P(0) + P(1) = . 0020 + .0176 = .0196

p-value = .0196

5.

p-value <.05, reject H0. Conclude that Bond Mutual Funds have a significantly lower median net
assets than Stock Mutual Funds.
H0: Median = 75,000
Ha: Median ≠ 75,000
Use the normal approximation with µ = .5n = .5(300) = 150 and

σ = .25n = .25(300) = 8.6603
19 - 3

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Chapter 19

With the number of plus signs = 165 in the upper tail, we use the continuity correction factor and
the normal distribution approximation as follows
164.5 − 150 

P ( x ≥ 164.5) = P  z ≥
= P ( z ≥ 1.67)
8.6603 ÷



The two-tailed p-value =2(1.0000 -. 9525) = .0950

p-value > .05;do not reject H0. We are unable to conclude that the median annual income for
Popular Photography & Imagining subscribers differs from $75,000.
6.

H0: Median < 56.2
Ha: Median > 56.2
There are n = 31 + 17 = 48 observations where the value is different from 56.2.
Use the normal approximation with µ = .5n = .5(48) = 24 and

σ = .25n = .25(48) = 3.4641
With the number of plus signs = 31in the upper tail, we use the continuity correction factor and the
normal distribution approximation as follows
30.5 − 24 

P( x ≥ 30.5) = P  z ≥
= P( z ≥ 1.88)
3.4641 ÷



Upper-tail p-value = (1.0000 - .9699) = .0301
p-value < .05; reject H0. Conclude that the median annual income for families living in Chicago is
greater that $56.2 thousand.
7.

a.

Let p = probability the shares held will be worth more after the split
H0 : p ≤ .50
Ha : p > .50

b.

Let the number of plus signs be the number of increases in value.
Use the binomial probability tables with n = 18 (there were 2 ties in the 20 observations)
With x = 14 plus signs,
P(x ≥ 14) = P(14) + P(15) + P(16) + P(17) + P(18)
= .0117 + .0031 + .0006 + .0001 + .0000 = .0155
Upper-tailed p-value = .0155
p-value ≤ .05, reject H0. The results support the conclusion that stock splits are beneficial for
shareholders.

8.

a.

H0 : p = .50
Ha : p ≠ .50
Dropping the individual with no preference, selected binomial probabilities for n = 15 and p = .50
19 - 4

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Nonparametric Methods
are as follows.
x
Probability
0

.0000
1
.0002
2
.0018
3
.0085
4
.0278
Let the response faster pace be a plus sign. Number of plus signs = 4.
Since this is in the lower tail of binomial distribution, we compute
P(x < 4) = P(0) + P(1) + P(2) + P(3) + P4)
= .0000 + .0002 + .0018 + .0085 + .0278 = .0384
Two-tailed p-value = 2(.0384) = .0768
p-value > .05, do not reject H0. The sample does not allow the conclusion that a difference in
preference exists for the fast pace or slower pace of life.
b.

9.

Of the original 16 respondents, 4/16(100) = 25% favored a faster pace of life and 11/16(100) =
68.8% favored a slower pace of life. There is an almost 3 to 1 in favor of the slower pace of life.
In addition, the p-value .0768 is low, but not low enough to detect a difference in preference.
While we are unable to reject H0 and conclude a difference in preference exists from this sample
of 16 respondents, continuing the study and taking a larger sample would be recommended.
Let p = proportion of adults who feel children will have a better future.
H0 : p = .50
Ha : p ≠ .50
Eliminating the responses that said about the same, we have
n = 242 + 310 = 552

Using the normal distribution, we have
µ = .5 n = .5(552) = 276

σ = .25n = .25(552) = 11.7473

With the number of plus signs = 242 in the lower tail, we will use the continuity correction factor
and the normal distribution approximation as follows:
242.5 − 276 

P( x ≤ 242.5) = P  z ≤
= P ( z ≤ −2.85)
11.7473 ÷



The two-tailed p-value =2(.0022) = .0044
p-value ≤ .05, reject H0. Conclude that there is a significant difference between the adults projecting
a better future and adults projects a worse future. In 2008, more adults were projecting a worse
future for their children.
10.

Let p = proportion who favor America Idol
19 - 5

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Chapter 19

H0 : p = .50
Ha : p ≠ .50
Eliminating the responses that said about the same, we have
n = 330+ 270 = 600
Using the normal distribution, we have
µ = .5 n = .5(600) = 300

σ = .25n = .25(600) = 12.2474
With the number of plus signs = 330 in the upper tail, we will use the continuity correction factor
and the normal distribution approximation as follows
329.5 − 300 

P( x ≥ 330) = P  z ≥
= P ( z ≥ 2.41)
12.2474 ÷



The two-tailed p-value = 2(1.0000 – 9920) = .0160
p-value ≤ .05, reject H0. Conclude that there is a significant difference between the preference for
American Idol and Dancing with the Stars. Based on the data, American Idol is most preferred.
11.

Let p = proportion who purchase brand A computers
H0 : p = .50
Ha : p ≠ .50
Eliminating purchases of other computers, we have
n = 202+ 175 = 377
Using the normal distribution, we have
µ = .5 n = .5(377) = 188.5

σ = .25n = .25(377) = 9.7082
With the number of plus signs = 202 in the upper tail, we will use the continuity correction factor
and the normal distribution approximation as follows
201.5 − 188.5 

P ( x ≥ 201.5) = P  z ≥
= P ( z ≥ 1.34)
9.7082 ÷



The two-tailed p-value = 2(1.0000 – .9099) = .1802
p-value > .05, do not reject H0. We are unable to conclude that there is a difference between the
market shares for the two brands of computers.
12.

H0: Median for Additive 1 - Median for Additive 2 = 0

19 - 6

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods

Ha: Median for Additive 1 - Median for Additive 2 ≠ 0
Additive
Car

1
2
3
4
5
6
7
8
9
10
11
12

1
20.12
23.56
22.03
19.15
21.23
24.77
16.16
18.55
21.87
24.23
23.21
25.02

2
18.05
21.77

22.57
17.06
21.22
23.80
17.20
14.98
20.03
21.15
22.78
23.70

Differenc
e
2.07
1.79
-0.54
2.09
0.01
0.97
-1.04
3.57
1.84
3.08
0.43
1.32

Absolute
Difference
2.07
1.79

0.54
2.09
0.01
0.97
1.04
3.57
1.84
3.08
0.43
1.32

Signed Ranks
Rank
9
7
3
10
1
4
5
12
8
11
2
6

Negative

-3
10

1
4
-5

Sum of Positive Signed Ranks

µT =
+

σT =
+

Positive
9
7

12
8
11
2
6
T + = 70

n(n + 1) 12(13)
=
= 39
4
4
n(n + 1)(2n + 1)
12(13)(25)

=
= 12.7475
24
24

T + = 70 is in the upper tail of the sampling distribution. Using the continuity correction factor, we
have:
69.5 − 39 

P (T + ≥ 70) = P  z ≥
= P ( z ≥ 2.39)
12.7475 ÷



Using z = 2.39, the p-value = 2(1.0000 - .9916) = .0168

13.

p-value < .05, reject H0. Conclude that there is a significant difference in the median miles per
gallon for the two additives.
H0: Median time without Relaxant - Median time with Relaxant < 0
Ha: Median time without Relaxant - Median time with Relaxant > 0
Relaxant

Absolute
Signed Ranks
Differenc
Subject
No

Yes
e
Difference
Rank
Negative
Positive
15
10
1
5
5
9
9
12
10
2
2
2
3
3
22
12
3
10
10
10
10
8
11
4

-3
3
6.5
-6.5
10
9
5
1
1
1
1
7
5
6
2
2
3
3
8
10
7
-2
2
3
-3
10
7
8
3
3

6.5
6.5
14
11
19 - 7 3
9
3
6.5
6.5
Learning. All Rights Reserved.
9
6 © 2010 Cengage
10 May not be
3
3
6.5
6.5
scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Sum of Positive Signed Ranks

T += 45.5


Chapter 19

µT =
+

σT =
+

n(n + 1) 10(11)
=
= 27.5
4
4
n(n + 1)(2n + 1)
10(11)(21)
=
= 9.8107
24
24

T + = 45.5 is in the upper tail of the sampling distribution. Using the continuity correction factor,
we have:
45 − 27.5 

P (T + ≥ 45.5) = P  z ≥
= P ( z ≥ 1.78)
12.7475 ÷



Using z = 1.78, p-value = (1.0000 - .9925) = .0375
p-value < .05, reject H0. Conclude that there is a significant difference in the median times to fall
asleep. Without the relaxant has a significantly greater median time and with the relaxant.

14.

H0: Median percent on-time in 2006 - Median percent on time in 2007 = 0

Ha: Median percent on-time in 2006 - Median percent on time in 2007 ≠ 0
Year

Airport
1
2
3
4
5
6
7
8
9
10
11

2006
71.78%
68.23%
77.98%
78.71%
77.59%
77.67%
76.67%
76.29%
69.39%
79.91%
75.55%

2007

69.69%
65.88%
78.40%
75.78%
73.45%
78.68%
76.38%
70.98%
62.84%
76.49%
72.42%

Differenc
e
2.09%
2.35%
-0.42%
2.93%
4.14%
-1.01%
0.29%
5.31%
6.55%
3.42%
3.13%

Absolute
Difference
2.09%
2.35%

0.42%
2.93%
4.14%
1.01%
0.29%
5.31%
6.55%
3.42%
3.13%

Signed Ranks
Rank
4
5
2
6
9
3
1
10
11
8
7

Negative

-2
6
9
-3

Sum of Positive Signed Ranks
19 - 8

Positive
4
5

1
10
11
8
7
T + = 61

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods

µT =
+

σT =
+

n(n + 1) 11(12)
=
= 33

4
4
n(n + 1)(2n + 1)
11(12)(23)
=
= 11.2472
24
24

T + = 61 is in the upper tail of the sampling distribution. Using the continuity correction factor, we
have:
60.5 − 33 

P(T + ≥ 61) = P  z ≥
= P ( z ≥ 2.45)
11.2472 ÷



Using z = 2.45, p-value = 2(1.0000 - .9929) = .0142
p-value < .05, reject H0. Conclude that there is a significance difference between the median
percentage of on-time flights for the two years. Median percentage of on-time flights was better
in 2006.

15.

H0: Median time for Service 1 - Median time for Service 2 = 0
Ha: Median time for Service 1 - Median time for Service 2 ≠ 0
Service

Delivery
1
2
3
4
5
6
7
8
9
10
11

1
24.5
26.0
28.0
21.0
18.0
36.0
25.0
21.0
24.0
26.0
31.0

Absolute
2
28.0
25.5

32.0
20.0
19.5
28.0
29.0
22.0
23.5
29.5
30.0

Differenc
e
-3.50
0.50
-4.00
1.00
-1.50
8.00
-4.00
-1.00
0.50
-3.50
1.00

Difference
3.50
0.50
4.00
1.00
1.50

8.00
4.00
1.00
0.50
3.50
1.00

Signed Ranks
Rank
7.5
1.5
9.5
4
6
11
9.5
4
1.5
7.5
4

Negative
-7.5

1.5
-9.5
4
-6
11
-9.5

-4
1.5
-7.5

Sum of Positive Signed Ranks

µT =
+

Positive

4
T + = 22

n(n + 1) 11(12)
=
= 33
4
4
19 - 9

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 19

σT =
+

n(n + 1)(2n + 1)
11(12)(23)
=
= 11.2472
24
24

T + = 22 is in the lower tail of the sampling distribution. Using the continuity correction factor, we
have:
22.5 − 33 

P(T + ≤ 22) = P  z ≤
= P ( z ≤ −.93)
11.2472 ÷


Using z = -.93, p-value = .1762
p-value > .05, do not reject H0. There is no significant difference between the median delivery
times for the two services.

19 - 10

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods
16.

H0: Median score for Round 1 - Median score for Round 2 = 0

Ha: Median score for Round 1 - Median score for Round 2 ≠ 0
Round

Golfer
1
2
3
4
5
6
7
8
9
10
11

10.

1
63
70
72
65
70
69
72
68
70
71
72

2
74
73
70
71
74
73
71
70
68
71
69

Absolute

Differenc
e
-11
-3
2
-6
-4
-4
1
-2
2
0
3

Signed Ranks

Difference
11
3
2
6
4
4
1
2
2

Rank
10
5.5
3
9
7.5
7.5
1
3
3

3

5.5

Negative
-10

-5.5

Positive

3
-9
-7.5
-7.5
1
-3
3
5.5

Sum of Positive Signed Ranks
T += 12.5
Carlos Franco (10) had the same score on both rounds as is removed from the sample. Thus, n =

µT =
+

σT =
+

n(n + 1) 10(11)
=
= 27.5
4
4
n(n + 1)(2n + 1)
10(11)(21)

=
= 9.8107
24
24

T + = 12.5 is in the lower tail of the sampling distribution. Using the continuity correction factor,
we have:
13 − 27.5 

P(T + ≤ 12.5) = P  z ≤
= P ( z ≤ −1.48)
9.8107 ÷



Using z = -1.48, p-value = 2(.0694) = .1388
p-value > .05, do not reject H0. There is no significant difference between the median scores for
the two rounds of golf.

19 - 11

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Chapter 19
17.

H0: Median = 500
Ha: Median ≠ 500

Student
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Score
635
502
447
701
405
590
439
453
337
447
471

387
464
476
514

Median
500
500
500
500
500
500
500
500
500
500
500
500
500
500
500

Absolute

Differenc
e
135
2
-53
201

-95
90
-61
-47
-163
-53
-29
-113
-36
-24
14

Difference
135
2
53
201
95
90
61
47
163
53
29
113
36
24
14

Signed Ranks

Rank
13
1
7.5
15
11
10
9
6
14
7.5
4
12
5
3
2

Negative

-7.5
15
-11
10
-9
-6
-14
-7.5
-4
-12
-5

-3

Sum of Positive Signed Ranks

µT ==

n(n + 1) 15(15 + 1)
=
= 60
4
4

σT =

n(n + 1)(2n + 1)
15(15 + 1)(30 + 1)
7440
=
=
= 17.6068
24
24
24

+

+

Positive
13

1

2
T += 41

T + = 41 is in the lower tail of the sampling distribution. Using the continuity correction factor, we
have:
41.5 − 60 

P(T + ≤ 41) = P  z ≤
= P ( z ≤ −1.05)
17.6068 ÷



Using z = -1.05, p-value = 2(.1469) = .2938
p-value > .05, do not reject H0. We cannot reject the hypothesis that the population median
writing test score is 500.

18.

H0: The two populations of additives are identical
19 - 12

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Nonparametric Methods
Ha: The two populations of additives are not identical

Additive 1
17.3
18.4
19.1
16.7
18.2
18.6
17.5

Rank
2
6
10
1
5
7
3

W =

34

Additive 2
18.7
17.8
21.3
21.0
22.1
18.7
19.8

20.7
20.2

Rank
8.5
4
15
14
16
8.5
11
13
12

µW = 1 n1 (n1 + n2 + 1) = 1 7(7 + 9 + 1) = 59.5
2
2
σW =

1 n n (n + n + 1) =
12 1 2 1 2

1 7(9)(7 + 9 + 1) = 9.4472
12

With W = 34 in the lower tail, we will use the continuity correction factor and the normal
distribution approximation as follows
34.5 − 59.5 

P(W ≤ 34) = P  z ≤

= P ( z ≤ −2.65)
9.4472 ÷



Using z = -2.65, the two-tailed p-value =2(.0040) = .0080
p-value < .05; reject H0
Conclusion: The two populations of fuel additives are not identical. The population with additive 2
tends to provide higher miles per gallon.

19 - 13

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Chapter 19
19. a.

H0: The two populations of salaries are identical
Ha: The two populations of salaries are not identical
Public
Accountan
t
50.2
58.8
56.3
58.2
54.2
55.0

50.9
59.5
57.0
51.9
W =

Financial
Rank
5
19
16
18
13
14
6
20
17
8.5
136.5

Planner
49.0
49.2
53.1
55.9
51.9
53.6
49.7
53.9
51.8

48.9

Rank
2
3
10
15
8.5
11
4
12
7
1

µW = 1 n1 (n1 + n2 + 1) = 1 10(10 + 10 + 1) = 105
2
2

σW =

1 n n (n + n + 1) =
12 1 2 1 2

1 10(10)(10 + 10 + 1) = 13.2288
12

With W = 136.5 in the upper tail, we will use the continuity correction factor and the normal
distribution approximation as follows
136 − 105 


P (W ≥ 136.5) = P  z ≥
= P ( z ≥ 2.34)
13.2288 ÷



Using z = 2.34, the two-tailed p-value =2(1.0000 - .9904) = .0192
p-value < .05; reject H0
Conclusion: The two populations of salaries are not identical. The population of public accountants
tends to have the higher salaries.
b.

Public Accountant

Median = (55.0+56.3)/2 = 55.65

$55,650

Financial Planner

Median = (51.8+51.9)/2 = 51.85

$51,850

19 - 14

© 2010 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Nonparametric Methods
20. a.

Median salary for men $54,900

(4 th position when ranked)

Median salary for women $40,400
b.

(4 th position when ranked)

H0: The two populations of salaries are identical
Ha: The two populations are of salaries and not identical
Men
35.6
80.5
50.2
67.2
43.2
54.9
60.3
W =

Rank
4
14
9
13
6

11
12
69

Women
49.5
40.4
32.9
45.5
30.8
52.5
29.8

Rank
8
5
3
7
2
10
1

µW = 1 n1 (n1 + n2 + 1) = 1 7(7 + 7 + 1) = 52.5
2
2
σ W = 1 n1n2 (n1 + n2 + 1) =
12

1 7(7)(7 + 7 + 1) = 7.8262
12

With W = 69 in the upper tail, we will use the continuity correction factor and the normal
distribution approximation as follows
68.5 − 52.5 

P(W ≥ 69) = P  z ≥
= P ( z ≥ 2.04)
7.8262 ÷



Using z = 2.04, the two-tailed p-value =2(1.000-.9793) = .0414
p-value < .05; reject H0
Conclusion: The two populations of salaries are not identical. The population of men tends to have
the higher salaries.

19 - 15

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 19
21.

H0: The two populations of hurricane wind speeds are identical
Ha: The two populations of hurricane wind speeds are not identical
Atlantic
Wind Speed
125

110
65
135
80
150
150
65
80
105
145
60
W =

Rank
19
16.5
3.5
21
10.5
23.5
23.5
3.5
10.5
14.5
22
1
169

Pacific
Wind Speed

105
75
65
90
70
110
130
95
65
120
75
70

Rank
14.5
8.5
3.5
12
6.5
16.5
20
13
3.5
18
8.5
6.5

µW = 1 n1 (n1 + n2 + 1) = 1 12(12 + 12 + 1) = 150
2
2

σW =

1 n n (n + n + 1) =
12 1 2 1 2

1 12(12)(12 + 12 + 1) = 17.3205
12

With W = 169 in the upper tail, we will use the continuity correction factor and the normal
distribution approximation as follows
168.5 − 150 

P (W ≥ 169) = P  z ≥
= P ( z ≥ 1.07)
17.3205 ÷



Using z = 1.07, the two-tailed p-value =2(1.0000 - .8577) = .2846
p-value > .05; do not reject H0
Conclusion: We cannot reject the null hypothesis that the two populations of hurricane wind speeds
are identical. There is no indication that there is a difference between the populations of hurricane
wind speeds in the Atlantic/Caribbean/Gulf of Mexico and the Eastern Pacific Ocean.

19 - 16

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Nonparametric Methods
22.

H0: The two populations of P/E ratios are identical
Ha: The two populations of P/E ratios are not identical
Japan
P/E Ratio
153
21
18
125
31
213
64
666
33
68

Rank
20
8
5
19
13
21
17
22
14
18

W =

157

U.S.
P/E Ratio
19
24
24
43
22
14
21
14
21
38
15
14

Rank
6
11.5
11.5
16
10
2
8
2
8
15

4
2

µW = 1 n1 (n1 + n2 + 1) = 1 10(10 + 12 + 1) = 115
2
2
σW =

1 n n (n + n + 1) =
12 1 2 1 2

1 10(12)(10 + 12 + 1) = 15.1658
12

With W = 157 in the upper tail, we will use the continuity correction factor and the normal
distribution approximation as follows
156.5 − 115 

P (W ≥ 157) = P  z ≥
= P ( z ≥ 2.74)
15.1658 ÷



Using z = 2.74, the two-tailed p-value =2(1.0000 - .9969) = .0062
p-value < .01; reject H0
Conclusion: The two populations of P/E ratios are not identical. The population of Japanese
companies tends to have higher P/E ratios than the population of United States companies.

19 - 17

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 19
23.

H0: The two populations of daily crimes are identical
Ha: The two populations of daily crimes are not identical
Winter
18
20
15
16
21
20
12
16
19
20
W =

Rank
6.5
11
2
3.5
13
11

1
3.5
9
11
71.5

Summer
28
18
24
32
18
29
23
38
28
18

Rank
16.5
6.5
15
19
6.5
18
14
20
16.5
6.5

µW = 1 n1 (n1 + n2 + 1) = 1 10(10 + 10 + 1) = 105
2
2
σW =

1 n n (n + n + 1) =
12 1 2 1 2

1 10(10)(10 + 10 + 1) = 13.2288
12

With W = 71.5 in the lower tail, we will use the continuity correction factor and the normal
distribution approximation as follows
72 − 105 

P(W ≤ 71.5) = P  z ≤
÷ = P( z ≤ −2.49)
13.2288



Using z = -.2.49, the two-tailed p-value =2(.0064) = .0128
p-value < .05; reject H0
Conclusion: The two populations of daily crimes are not identical. The population of daily crimes
in winter months tends to be less than the population of daily crimes in the summer months.

19 - 18

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Nonparametric Methods
24.

H0: The two populations of microwave prices are identical
Ha: The two populations of microwave prices are not identical
Dallas
445
489
405
485
439
449
436
420
430
405

W =

Rank
14.5
23
1.5
22
13
16
12
4

8.5
1.5

San Antonio
460
451
435
479
475
445
429
434
410
422
425
459
430

Rank
19
17
11
21
20
14.5
7
10
3
5
6

18
8.5

116

µW = 1 n1 (n1 + n2 + 1) = 1 10(10 + 13 + 1) = 120
2
2
σ W = 1 n1n2 (n1 + n2 + 1) =
12

1 10(13)(10 + 13 + 1) = 16.1245
12

With W = 116 in the lower tail, we will use the continuity correction factor and the normal
distribution approximation as follows
116.5 − 120 

P (W ≤ 116) = P  z ≤
= P ( z ≤ −.22)
16.1245 ÷



Using z = -.22, the two-tailed p-value =2(.4129) = .8258
p-value > .05; do not reject H0
Conclusion: We cannot reject the null hypothesis that the two populations of microware prices are
identical. There is no indication that there is a difference between the populations of microwave
prices for the two cities.

19 - 19

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 19
25.

H0: The two populations of draft positions are identical
Ha: The two populations of draft positions are not identical
Southeastern Conference

Player's
College
Georgia
Alabama
Vanderbilt
Florida
Mississippi
Mississippi
Auburn

Projected
Draft Position
1
2
14
18
20

24
27
W=

Atlantic Coast Conference
Player's
College
Georgia Tech
Wake Forrest
Virginia
Wake Forrest
Florida State
Maryland
Virginia

Rank
1
2
6
7
8
10
13
47

Projected
Draft Position
3
6
8

23
25
26
29

Rank
3
4
5
9
11
12
14

µW = 1 n1 (n1 + n2 + 1) = 1 7(7 + 7 + 1) = 52.5
2
2
σ W = 1 n1n2 (n1 + n2 + 1) =
12

1 7(7)(7 + 7 + 1) = 7.8262
12

With W = 47 in the lower tail, we will use the continuity correction factor and the normal
distribution approximation as follows
47.5 − 52.5 

P (W ≤ 47) = P  z ≤
= P ( z ≤ −.64)
7.8262 ÷




Using z = -.64, the two-tailed p-value =2(.2611) = .5211
p-value > .05; do not reject H0
Conclusion: We cannot reject the null hypothesis that the two populations of draft positions are
identical. There is no indication that there is a difference between the populations of draft position
preferences for the players from the two conferences.
26.

H0: All populations of product ratings are identical
Ha: Not all populations of product ratings are identical
A
50
62
75
48

Rank
4
8
10
3

B
80
95
98
87
19 - 20

Rank
11
14
15
12

C
60
45
30
58

Rank
7
2
1
6

© 2010 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods
65
Sum of
Ranks

9

90

13

34

57

65

5
21

k
 12  34 2 652 212  

Ri2 
12
H =
−
3(
n
+
1)
=
+
+


÷ − 3(16) = 10.22

∑  T
5
5  
 nT ( nT + 1) i =1 ni 
 15(16)  5
Using the χ2 table with df = 2, χ2 = 10.22 shows the p-value is between .005 and .01

Using Excel or Minitab, the p-value for χ2 = 10.22 is .0060.
p-value < .01, reject H0. Conclude that the populations of product ratings are not identical.
27.

H0: All populations of test scores are identical
Ha: Not all populations of test score are identical
A
540
400
490
530
490
610

Rank
11.5
2.5
8
10
8
18

Sum of

Ranks

58

B
450
540
400
410
480
370
550

Rank
5
11.5
2.5
4
6
1
13

C
600
630
580
490
590
620
570

43

k
 12  582 432 1092

Ri2 
12
H =
+
+

 − 3(nT + 1) = 
∑
7
7
 20(21)  6
 nT ( nT + 1) i =1 ni 

Rank
17
20
15
8
16
19
14
109



÷ − 3(21) = 9.06
 

Using the χ2 table with df = 2, χ2 = 9.06 shows the p-value is between .025 and .01
Using Excel or Minitab, the p-value for χ2 = 9.06 is .0108.
p-value < .05, reject H0. Conclude that the populations of test scores are not identical. In
particular, program C appears to provide the higher test scores.
28.

H0: All populations of calories burned are identical
Ha: Not all populations of calories burned are identical
Swimming
408
380
425
400
427
Sum of
Ranks

Rank
8
4
11
6
12

Tennis
415
485

450
420
530

41

Rank
9
14
13
10
15
61

19 - 21

Cycling
385
250
295
402
268

Rank
5
1
3
7
2
18

© 2010 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 19

k
 12  412 612 182  

Ri2 
12
H =
+
+

÷ − 3(16) = 9.26
 − 3(nT + 1) = 
∑
5
5  
 nT (nT + 1) i =1 ni 
 15(16)  5

Using the χ2 table with df = 2, χ2 = 9.26 shows the p-value is between .005 and .01
Using Excel or Minitab, the p-value for χ2 = 9.26 is .0098.
p-value < .05, reject H0. Conclude that the populations of calories burned by the three activities
are not identical. Cycling tends to have the lowest calories burned.
29.

H0: All populations of cruise ship ratings are identical
Ha: Not all populations of cruise ship ratings are identical
Holland
America
84.5
81.4
84.0
78.5
80.9
Sum of
Ranks

Rank
11
5
9.5
1
3

Princess
85.1
79.0
83.9
81.1
83.7

Rank
13
2
8

4
7

29.5

Royal
Caribbean
84.8
81.8
84.0
85.9
87.4

34

Rank
12
6
9.5
14
15
56.5

k
 12  29.52 34 2 56.52

Ri2 
12
H =
+

+

 − 3(nT + 1) = 
∑
5
5
 nT (nT + 1) i =1 ni 
 15(16)  5


÷ − 3(16) = 4.19
 

Using the χ2 table with df = 2, χ2 = 4.19 shows the p-value is greater than .10.
Using Excel or Minitab, the p-value for χ2 = 4.19 is .1231.
p-value > .05, do not reject H0.
We cannot reject the null hypothesis that the three populations of cruise ship rating are identical.
There is no indication that there is a difference among the populations of rating for the three cruise
ship lines.
30.

H0: All populations of training courses are identical
Ha: Not all populations of training courses are identical
Course

Sum of Ranks

A
3
14

10
12
13
52

B
2
7
1
5
11
26

19 - 22

C
19
16
9
18
17
79

D
20
4
15
6
8
53

© 2010 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods

k
 12  52 2 26 2 79 2 532  

Ri2 
12
H =
+
+
+

÷ − 3(21) = 8.03
 − 3(nT + 1) = 
∑
5
5
5  
 nT ( nT + 1) i =1 ni 
 20(21)  5

Using the χ2 table with df = 3, χ2 = 8.03 shows the p-value is between .025 and .05.
Using Excel or Minitab, the p-value for χ2 = 8.03 is .0454.
p-value < .05, reject H0. Conclude that the populations are not identical. There is a significant
difference in the quality of the courses offered at the four management development centers.

31.

H0: All populations of calories are identical
Ha: Not all populations of calories are identical
M&Ms
230
210
240
250
230
Sum of
Ranks

Rank
10
7
13
15
11

Kit Kat
225
205
245
235
220

56

Rank

9
5
14
12
8

Milky Way II
200
208
202
190
180

Rank
3
6
4
2
1

48

16

k
 12  562 482 152  

Ri2 
12
H =

−
3(
n
+
1)
=
+
+


÷ − 3(16) = 8.96
∑  T
5
5  
 nT (nT + 1) i =1 ni 
 15(16)  5

Using the χ2 table with df = 2, χ2 = 8.96 shows the p-value is between .01 and .025.
Using Excel or Minitab, the p-value for χ2 = 8.96 is .0113.
p-value < .05, reject H0. Conclude that the populations of calories are not identical for the three
candies.
Milky Way II appears to have the lowest number of calories.
32. a.

b.

Σdi2 = 52

rs = 1 −

6Σd i2
6(52)
= 1−
= .685
2
10(99)
n(n − 1)

σr =

1
=
n −1

s

z=

1
= .3333
9

rs − 0 .685
=
= 2.05
σ rs
.3333

p-value = 2(1.0000 - .9798) = .0404
p-value ≤ .05, reject H0. Conclude that significant positive rank correlation exists.

33.

Case 1:

Σdi2 = 0

19 - 23

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Chapter 19

rs = 1 −

6Σd i2
6(0)
= 1−
=1
2
6(36 − 1)
n(n − 1)

Case 2: Σdi2 = 70
rs = 1 −

6Σd i2
6(70)
= 1−

= −1
2
6(36 − 1)
n(n − 1)

With perfect agreement, rank correlation coefficient rs = 1.
With exact opposite ranking, rank correlation coefficient rs = -1.
Σdi2 = 250

34.

rs = 1 −

6Σd i2
6(250)
= 1−
= −.136
11(120)
n(n 2 − 1)

σr =

1
1
=
= .3162
n −1
10

s

z=

rs − 0 −.136
=
= −.43
σ rs
.3162

p-value = 2(.3336) = .6672
p-value > .05, do not reject H0. We cannot conclude that there is a significant relationship between
ranking based on the expenditure per student and the ranking based on the student-teacher ratio.
35. a.

b.

Σdi2 = 54

r = 1−

6Σd i2
6(54)
= 1−
= .673
2
n(n − 1)
10(102 − 1)

σr =

1
1
=
= .3333
n −1
10 − 1

s

z=

rs − µrs

σr

s

=

.673 − 0
= 2.02
.3333

p-value = 2(1.0000 - .9783) = .0434
c.

p-value ≤ .05, reject H0. Conclude that there is a significant positive rank correlation between a
company’s reputation and having stock that is desirable to purchase.

19 - 24

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Nonparametric Methods

36.
Driving Distance

Putting

di

1
5
4
9
6
10
2
3
7
8

5
6
10
2
7

3
8
9
4
1

-4
-1
-6
7
-1
7
-6
-6
3
7

rs = 1 −

6Σd i2
6(282)
= 1−
= −.709
10(100 − 1)
n(n 2 − 1)

σr =

1
=

n −1

s

z=

rs − µrs

σr

s

=

d i2
16
1
36
49
1
49
36
36
9
49
Σdi2 = 282

1
= .3333
9

−.709 − 0
= −2.13
.3333

p-value = 2(.0166) = .0332
p-value ≤ .10, reject H0. There is a significant negative rank correlation between driving distance
and putting. Professional golfers who rank high in driving distance tend to rank low in putting.
37.

Σdi2 = 38

rs = 1 −

6Σdi2
6(38)
= 1−
= .77
10(99)
n(n 2 − 1)

σr =

1
=
n −1

s

z=

1
= .3333
9

rs − 0
.77
=
= 2.31
σ rs
.3333

p-value = 2(1.0000 - .9896) = .0208
p-value ≤ .10, reject H0. Conclude that there is a significant positive rank correlation between
current students and recent graduates in terms of professor teaching ability.
38.

Let p = proportion who favor the proposal
19 - 25

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