Student’s Solutions Manual and Study Guide: Chapter 17
Page 1
Chapter 17
Nonparametric Statistics
LEARNING OBJECTIVES
This chapter presents several nonparametric statistics that can be used to
analyze data specifically. The key questions of the chapter are:
1. What are categorical data?
2. What are chi-square tests?
3. When can you use the chi-square goodness-of-fi t test and when can you use the chisquare test of independence?
4. What minimum sample size will you need in each cell to use chi-square tests?
Student’s Solutions Manual and Study Guide: Chapter 17
Page 2
CHAPTER OUTLINE
17.1
Runs Test
Small-Sample Runs Test
Large-Sample Runs Test
17.2
Mann-Whitney U Test
Small-Sample Case
Large-Sample Case
17.3
Wilcoxon Matched-Pairs Signed Rank Test
Small-Sample Case (n < 15)
Large-Sample Case (n > 15)
17.4
Kruskal-Wallis Test
17.5
Friedman Test
17.6
Spearman's Rank Correlation
KEY TERMS
Friedman Test
Kruskal-Wallis Test
Mann-Whitney U Test
Nonparametric Statistics
Parametric Statistics
Runs Test
Spearman’s Rank Correlation
Wilcoxon Matched-Pairs Signed Rank Test
Student’s Solutions Manual and Study Guide: Chapter 17
Page 3
STUDY QUESTIONS
1. Statistical techniques based on assumptions about the population from which the sample data
are selected are called _______________ statistics.
2. Statistical techniques based on fewer assumptions about the population and the parameters
are called _______________ statistics.
3. The nonparametric test for the randomness of a sequence of numbers is called the
________ test.
4.
Suppose the following sequence of outcomes is obtained using a random sampling process:
XXXYYYYXXXXXX
Testing to determine if this pattern is random, we obtain a critical value of ________ for
the lower tail at  = .025 (for each tail). Based on this critical value and the observed
sequence, we would conclude that the data are (random, nonrandom) ______________?
5.
Suppose we are attempting to randomly sample males and females in a study. The sequence
shown below is obtained:
MMFFFFFMFMMFMMFFFMMMMMMFFFFMFMFFMFFMMMFMFMMFF
The value of µR for these data is _______________. The value of R is _______________.
The value of R is __________. Suppose  = .05 for this test. The critical value of z is
_________. The observed value of z is ___________. The decision is to ______________
the null hypothesis. The conclusion is that the data are (random, not random)
__________________.
6.
The nonparametric counterpart of the t test to compare the means of two independent
populations is called _________________________.
Student’s Solutions Manual and Study Guide: Chapter 17
7.
8.
Page 4
A researcher wants to determine if the values obtained for population two are greater than
those for population one. To test this, the researcher randomly samples six scores from
population one (designated as group one) and six scores from population two (designated as
group two) shown below. The sampled groups are independent. Using the Mann-Whitney U
test, the values of W1 and W2 are _________ and ___________. The values of U1 and U2
are __________ and ___________. This is a ________ tailed test. The p-value obtained
from the table in the appendix of the text for this problem is _______. Assuming that
 = .05, the decision for this problem is to _______________ the null hypothesis.
Group 1
Group 2
124
131
128
126
132
141
129
138
136
137
142
139
Suppose a Mann-Whitney U test is being used to test to determine if two populations are
different or not. A value of alpha = .05 is used. Random samples of size 16 are gathered
from each population and the resulting data are given below:
Sample 1: 19, 27, 23, 29, 22, 20, 29, 31, 25, 17, 26, 23, 18, 28, 28, 33
Sample 2: 27, 35, 33, 28, 24, 33, 30, 31, 38, 39, 29, 41, 33, 34, 36, 34
The critical table value of z for this problem is _______________.
9.
In the problem presented in question 8 if sample one is designated as group one, the value
of W is ________.
10. In the problem presented in question 8, the value of U is __________.
11. For the problem presented in question 8, the observed value of z is _______________.
Based on this value and the critical value determined in question 8, the decision should be
to _______________ the null hypothesis.
12. The nonparametric alternative to the t test for two related samples or matched-pairs is
_______________.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 5
13. A researcher conducts a before/after study in which she believes that scores will decrease
after the treatment. A sample of seven people are used in the study. The resulting scores are
shown below. Using  = .05, the critical table T value for this problem is ________. The
observed value of T is ________. The decision is to _________ the null hypothesis.
Before
29
34
17
39
51
64
21
After
18
32
21
30
46
40
18
14. The Wilcoxon matched-pairs signed rank test for large samples uses a _______________
statistic to analyze the data.
15. Suppose the following samples of paired data are gathered and the Wilcoxon matched-pairs
signed is used to determine if there is a significant difference in the two populations from
which the samples are were gathered.
Sample 1
109
103
111
98
105
108
101
102
100
97
106
112
104
105
109
Sample 2
98
100
107
102
99
96
100
104
93
98
100
108
105
101
101
Let alpha be .10. The critical value of z for this problem is _______________.
16. The value of T for the problem presented in question 15 is _______________.
17. The mean value of T for the sample size of the problem presented in question 15 is
_______________.
The standard deviation of T for this sample size is _______________.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 6
18. The observed z value for the problem presented in question 15 is _______________.
Based on the critical value determined in question 15 and the observed value of z, a
statistician would decide to _______________ the null hypothesis.
19. The nonparametric alternative to the one-way analysis of variance is the
_________________________ test.
20. The value of K computed in the Kruskal-Wallis test is distributed approximately as a
_______________ value.
21. Suppose a researcher desires to analyze the data below using a Kruskal-Wallis test to
determine if there is a significant difference in the populations from which the four samples
were taken.
Sample 1
113
124
117
118
122
Sample 2
97
99
98
101
Sample 3
105
108
100
98
103
Sample 4
109
106
105
108
110
The degrees of freedom associated with the Kruskal-Wallis test of this data are _________.
22. The critical value of chi-square used to test the data for the problem in question 21 using
 = .05 is _______________.
23. The observed value of chi-square for the data of the problem in question 21 is __________.
Using this value and the critical value determined in question 22, a researcher would decide
to _______________ the null hypothesis.
24. The nonparametric equivalent to the randomized block design is the __________________.
25. A randomized block design is set up for a research study. However, the data are considered
to be only ordinal measurements. Therefore, the data are analyzed using a Friedman test.
The design and data are shown below. In analyzing these data, the sum of the squared ranks
is ________. Using  = .05, the critical value of 2 is _____________. The value of
2r is __________. Based on these statistics, the decision is to __________________ the
null hypothesis.
Block
1
2
3
4
A
105
106
110
108
B
97
98
105
104
Treatment
C
99
96
94
93
D
112
109
104
106
E
101
105
98
105
Student’s Solutions Manual and Study Guide: Chapter 17
Page 7
26. A nonparametric alternative to Pearson's product-moment correlation coefficient, r, is
_______________________.
27. The data below have been gathered in pairs from two variables:
x: 12, 19, 21, 34, 50, 69, 70
y: 8, 17, 15, 29, 31, 45, 39
The value of Spearman's rank correlation for this data is _______________.
28. The data below have been gathered in pairs from two variables:
x: 29, 13, 11, 9, 5, -1, -3, -4, -11
y: -12, -13, -16, -25, -26, -18, -20, -44, -90
The value of Spearman's correlation for this data is _______________.
29. The data below have been gathered in pairs from two variables:
x: 230, 221, 190, 130, 124, 109
y: 107, 150, 134, 138, 166, 178
The value of Spearman's rank correlation for this data is _______________.
30. If the value of Spearman's rank correlation is near -1, the two variables have
_______________ correlation. If the value of a Spearman's rank correlation is near 0, the
two variables have _______________ correlation.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 8
ANSWERS TO STUDY QUESTIONS
1.
Parametric Statistics
16. 15.5
2.
Nonparametric Statistics
17. 60, 17.61
3.
Runs
18.
4.
3, nonrandom
19. Kruskal-Wallis test
5.
23.489, 3.314, 22, + 1.96, -0.45
Fail to Reject, Random
20. Chi-square
6.
Mann-Whitney U Test
7.
28, 50, 29, 7, One, .0465, Reject
– 2.53, Reject
21. 3
22. 7.81473
23. 15.11, Reject
8.
+ 1.96
9.
164.5
24. Friedman Test
25. 840, 9.4877, 12, Reject
10. 227.5
26. Spearman’s Rank Correlation
11. 3.75, Reject
27. .929
12. Wilcoxon Matched-Pairs Signed Rank Test
28. .867
13. 4, 3, Reject
29.
-.829
14. z
30. High Negative, Little or No
15. + 1.645
Student’s Solutions Manual and Study Guide: Chapter 17
Page 9
SOLUTIONS TO CHAPTER 17
17.1
Ho: The observations in the sample are randomly generated.
Ha: The observations in the sample are not randomly generated.
This is a small sample runs test since n1, n2 < 20
 = .05, The lower tail critical value is 6 and the upper tail critical value is 16
n1 = 10
n2 = 10
R = 11
Since R = 11 is between the two critical values, the decision is to fail to reject the
null hypothesis.
The data are random.
17.3
n1 = 8
n2 = 52
 = .05
This is a two-tailed test and /2 = .025. The p-value from the printout is .0264.
Since the p-value is the lowest value of “alpha” for which the null hypothesis can
be rejected, the decision is to fail to reject the null hypothesis
(p-value = .0264 > .025). There is not enough evidence to reject that the data are
randomly generated.
17.5
Ho: The observations in the sample are randomly generated.
Ha: The observations in the sample are not randomly generated.
Since n1, n2 > 20, use large sample runs test
 = .05 Since this is a two-tailed test, /2 = .025, z.025 = + 1.96. If the
observed value of z is greater than 1.96 or less than -1.96, the decision is to reject
the null hypothesis.
R = 27 n1 = 40 n2 = 24
R 
2n1n2
2(40)( 24)
1 
 1 = 31
n1  n2
64
Student’s Solutions Manual and Study Guide: Chapter 17
R 
z
2n1n2 (2n1n2  n1  n2 )

(n1  n2 )2 (n1  n2  1)
R  R
R

Page 10
2(40)(24)2(40)( 24)  40  24
= 3.716
(64)2 (63)
27  31
= -1.08
3.716
Since the observed z of -1.08 is greater than the critical lower tail z value
of -1.96, the decision is to fail to reject the null hypothesis. The data are
randomly generated.
17.7
Ho: Group 1 is identical to Group 2
Ha: Group 1 is not identical to Group 2
Use the small sample Mann-Whitney U test since both n1, n2 < 10,  = .05. Since
this is a two-tailed test, /2 = .025. The p-value is obtained using Table A.13.
Value
11
13
13
14
15
17
18
18
21
21
22
23
23
24
26
29
Rank
1
2.5
2.5
4
5
6
7.5
7.5
9.5
9.5
11
12.5
12.5
14
15
16
Group
1
1
2
2
1
1
1
2
1
2
1
2
2
2
1
1
n1 = 8
n2 = 8
W1 = 1 + 2.5 + 5 + 6 + 7.5 + 9.5 + 15 + 16 = 62.5
U  n1  n2 
n1 (n1  1)
(8)(9)
 W1  (8)(8) 
 62.5 = 37.5
2
2
U '  n1  n2  U = 64 – 37.5 = 26.5
Student’s Solutions Manual and Study Guide: Chapter 17
Page 11
We use the small U which is 26.5
From Table A.13, the p-value for U = 27 is .3227(2) = .6454
Since this p-value is greater than /2 = .025, the decision is to fail to reject the
null hypothesis.
17.9
H0: The population of people 75 and older will have no greater contact with
physicians than do the population of people between 65 and 74.
Ha: The population of people 75 and older will have greater contact with
physicians than do the population of people between 65 and 74.
Contacts
6
8
9
9
10
11
11
12
12
13
13
13
14
15
16
17
W1 = 39
U1  n1  n2 
Rank
1
2
3.5
3.5
5
6.5
6.5
8.5
8.5
11
11
11
13
14
15
16
Group
1
1
1
2
2
1
1
1
2
1
2
2
2
2
2
2
n1 (n1  1)
(7)(8)
 W1  (7)(9) 
 39 = 52
2
2
U2  n1  n2  U1 = (7)(9) – 52 = 11
U = 11
From Table A.13, the p-value = .0156. Since this p-value is greater than  = .01,
the decision is to fail to reject the null hypothesis.
17.11
Ho: Males do not earn more than females
Ha: Males do earn more than females
Earnings
Rank
Gender
Student’s Solutions Manual and Study Guide: Chapter 17
$28,900
31,400
36,600
40,000
40,500
41,200
42,300
42,500
44,500
45,000
47,500
47,800
47,800
48,000
50,100
51,000
51,500
51,500
53,850
55,000
57,800
61,100
63,900
Page 12
1
2
3
4
5
6
7
8
9
10
11
12.5
12.5
14
15
16
17.5
17.5
19
20
21
22
23
n1 = 11
F
F
F
F
F
F
F
F
F
M
F
F
M
F
M
M
M
M
M
M
M
M
M
n2 = 12
W1 = 10 + 12.5 + 15 + 16 + 17.5 + 17.5 + 19 + 20 + 21 + 22 + 23 = 193.5

n1  n2 (11)(12)

= 66 and  
2
2
U 

 = .01,
(11)(12)(24)
= 16.25
12
n1 (n1  1)
(11)(12)
 W1  (11)(12) 
 193.5 = 4.5
2
2
U  n1  n2 
z
n1  n2 (n1  n2  1)

12

4.5  66
= -3.78
16.25
z.01 = -2.33
Since the observed z = -3.78 < z.01 = 2.33, the decision is to reject the null
hypothesis.
17.13 Ho: The population differences = 0
Ha: The population differences  0
1
2
d
Rank
Student’s Solutions Manual and Study Guide: Chapter 17
212
234
219
199
194
206
234
225
220
218
234
212
219
196
178
213
179
184
213
167
189
200
212
221
223
217
208
215
187
198
189
201
Page 13
33
50
6
32
5
6
22
4
-3
1
26
-3
32
-2
-11
12
15
16
7.5
13.5
6
7.5
11
5
- 3.5
1
12
-3.5
13.5
-2
-9
10
n = 16
T = min (T+ , T-) = T- = 3.5 + 3.5 + 2 + 9 = 18


z
(n)( n  1) (16)(17)

= 68
4
4
n(n  1)( 2n  1)
16(17)(33)
= 19.34

24
24
T 

 = .10

18  68
= -2.59
19.34
/2 = .05
z.05 = ±1.645
Since the observed z = -2.59 < z.05 = -1.645, the decision is to reject the null
hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 14
17.15 Ho: The population differences = 0
Ha: The population differences < 0
Before
10,500
8,870
12,300
10,510
5,570
9,150
11,980
6,740
7,340
13,400
12,200
10,570
9,880
12,100
9,000
11,800
10,500
After
12,600
10,660
11,890
14,630
8,580
10,115
14,320
6,900
8,890
16,540
11,300
13,330
9,990
14,050
9,500
12,450
13,450
d
-2,100
-1,790
410
-4,120
-3,010
-965
-2,370
-160
-1,550
-3,140
900
-2,760
-110
-1,950
-500
-650
-2,950
Rank
-11
-9
3
-17
-15
-7
-12
-2
-8
-16
6
-13
-1
-10
-4
-5
-14
Since n = 17, use the large sample test
T+ = 3 + 6 = 9
T =9


z
(n)( n  1) (17)(18)

= 76.5
4
4
n(n  1)( 2n  1)
17(18)(35)
= 21.12

24
24
T 

 = .05

9  76.5
= -3.20
21.12
z.05 = -1.645
Since the observed z = -3.20 < z.05 = -1.645, the decision is to reject the null
hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 15
17.17 Ho: The population differences = 0
Ha: The population differences < 0
1999
49
27
39
75
59
67
22
61
58
60
72
62
49
48
19
32
60
80
55
68
2009
54
38
38
80
53
68
43
67
73
55
58
57
63
49
39
34
66
90
57
58
d
-5
-11
1
-5
6
-1
-21
-6
-15
5
14
5
-14
-1
-20
-2
-6
-10
-2
10
Rank
-7.5
-15
2
-7.5
11
-2
-20
-11
-18
7.5
16.5
7.5
-16.5
-2
-19
-4.5
-11
-13.5
-4.5
13.5
n = 20
T+ = 2 + 11 + 7.5 + 16.5 + 7.5 + 13.5 = 58
T = 58


z
(n)( n  1) (20)( 21)

= 105
4
4
n(n  1)( 2n  1)

24
T 


For  = .10,
20(21)( 41)
= 26.79
24
58  105
= -1.75
26.79
z.10 = -1.28
Since the observed z = -1.75 < z.10 = -1.28, the decision is to reject the null
hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 16
17.19 Ho: The 5 populations are identical
Ha: At least one of the 5 populations is different
1
157
188
175
174
201
203
2
165
197
204
214
183
1
1
6
4
3
9
10.5
_
Tj 33.5
nj
2
2
7.5
12
14
5
_
40.5
6

5
3
219
257
243
231
217
203
BY RANKS
3
18
26
23.5
19
16.5
10.5
__
113.5
6
4
286
243
259
250
279
5__
197
215
235
217
240
233
213
4
29
23.5
27
25
28
5__
7.5
15
21
16.5
22
20
13_
115
__
132.5
5
7
2
(33.5) 2 (40.5) 2 (113.5) 2 (132.5) 2 (115) 2





= 8,062.67
nj
6
5
6
5
7
Tj
n = 29
2
Tj
12
12
K
 3(n  1) 
(8,062.67)  3(30) = 21.21

n(n  1)
nj
29(30)
 = .01
df = c - 1 = 5 - 1 = 4
2.01,4 = 13.2767
Since the observed K = 21.21 > 2.01,4 = 13.2767, the decision is to reject the null
hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 17
17.21 Ho: The 4 populations are identical
Ha: At least one of the 4 populations is different
Region 1
$1,200
450
110
800
375
200
Region 2
$225
950
100
350
275
Region 3
$ 675
500
1,100
310
660
Region 4
$1,075
1,050
750
180
330
680
425
By Ranks
Region 1
23
12
2
18
10
4
_
Tj
69
nj
Tj
n
2

j
6
Region 2
5
19
1
9
6
Region 3
15
13
22
7
14
_
40
_
71
Region 4
21
20
17
3
8
16
11
96
5
5
7
(69) 2 (40) 2 (71) 2 (96) 2



= 3,438.27
6
5
5
7
n = 23
2
Tj
12
12
K
 3(n  1) 
(3,438.27)  3(24) = 2.75

n(n  1)
nj
23(24)
 = .05
df = c - 1 = 4 - 1 = 3
2.05,3 = 7.8147
Since the observed K = 2.75 < 2.05,3 = 7.8147, the decision is to fail to reject
the null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 18
17.23 Ho: The 4 populations are identical
Ha: At least one of the 4 populations is different
Amusement Parks
0
1
1
0
2
1
0
Amusement Parks
2
5.5
5.5
2
11.5
5.5
2
Tj
nj
Tj
n
2

j
__
34
7
Lake Area
3
2
3
5
4
4
3
5
2
City
2
2
3
2
3
2
3
3
1
3
By Ranks
Lake Area
20.5
11.5
20.5
33
28.5
28.5
20.5
33
11.5
__
207.5
9
City
11.5
11.5
20.5
11.5
20.5
11.5
20.5
20.5
5.5
20.5
154.0
10
National Park
2
4
3
4
3
5
4
4
National Park
11.5
28.5
20.5
28.5
20.5
33
28.5
28.5
____
199.5
8
(34) 2 (207.5) 2 (154) 2 (199.5) 2



= 12,295.80
7
9
10
8
n = 34
2
Tj
12
12
K
 3(n  1) 
(12,295.80)  3(35) = 18.99

n(n  1)
nj
34(35)
 = .05
df = c - 1 = 4 - 1 = 3
 .05,3 = 7.8147
2
Since the observed K = 18.99 > 2.05,3 = 7.8147, the decision is to reject the
null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 19
17.25 Ho: The treatment populations are equal
Ha: At least one of the treatment populations yields larger values than at least one
other treatment population.
Use the Friedman test with  = .05
c = 5, b = 5, df = c - 1 = 4, 2.05,4 = 9.4877
If the observed value of 2 > 9.4877, then the decision will be to reject the null
hypothesis.
Shown below are the data ranked by blocks:
1
2
3
4
5
1
1
2.5
3
4
4
3
1
2
2
3
4
4
4
3
5
5
5
5
5
2
2
2.5
1
1
11.5
12
18
25
8.5
132.25
144
324
625
72.25
1
2
3
4
5
Rj
Rj2
Rj2 = 1,297.5
r 2 
12
12
2
R j  3b(c  1) 
(1,297.5)  3(5)(6) = 13.8

bc(c  1)
(5)(5)(6)
Since the observed value of r2 = 13.8 > 4,.052 = 9.4877, the decision is to
reject the null hypothesis. At least one treatment population yields larger values
than at least one other treatment population.
17.27 Ho: The treatment populations are equal
Ha: At least one of the treatment populations yields larger values than at least one
other treatment population.
Use the Friedman test with  = .01
c = 4, b = 6, df = c - 1 = 3, 2.01,3 = 11.3449
If the observed value of 2 > 11.3449, then the decision will be to reject the null
hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 20
Shown below are the data ranked by blocks:
1
2
3
4
5
6
1
1
2
1
1
1
2
2
4
3
4
3
3
3
3
3
4
3
4
4
4
4
2
1
2
2
2
1
Rj
8
20
22
10
64
400
484
100
Rj2
Rj2 = 1,048
r 2 
12
12
2
R j  3b(c  1) 
(1,048)  3(6)(5) = 14.8

bc(c  1)
(6)( 4)(5)
Since the observed value of r2 = 14.8 > 23,.01 = 11.3449, the decision is to
reject the null hypothesis. At least one treatment population yields larger values
than at least one other treatment population.
17.29 c = 4 treatments
b = 5 blocks
S = r2 = 2.04 with a p-value of .564.
Since the p-value of .564 >  = .10, .05, or .01, the decision is to fail to reject
the null hypothesis. There is no significant difference in treatments.
17.31
x
23
41
37
29
25
17
33
41
40
28
19
y
201
259
234
240
231
209
229
246
248
227
200
x Ranked
3
10.5
8
6
4
1
7
10.5
9
5
2
y Ranked
2
11
7
8
6
3
5
9
10
4
1
d
d2
1
1
-.5
0.25
1
1
-2
4
-2
4
-2
4
2
4
1.5
2.25
-1
1
1
1
1
1
2
d = 23.5
Student’s Solutions Manual and Study Guide: Chapter 17
Page 21
n = 11
rs  1 
17.33
x
99
67
82
46
80
57
49
91
6 d 2
n(n  1)
2
y
108
139
117
168
124
162
145
102
 1
6(23.5)
= .893
11(120)
x Ranked
8
4
6
1
5
3
2
7
y Ranked
2
5
3
8
4
7
6
1
d
d2
6
36
-1
1
3
9
-7
49
1
1
-4
16
-4
16
6
36
2
d = 164
n =8
rs  1 
6 d 2
n(n  1)
2
 1
6(164)
= -.95
8(63)
17.35
Year
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Bank Credit
Card (x)
2.51
2.86
2.33
2.54
2.54
2.18
3.34
2.86
2.74
2.54
3.18
3.53
3.51
3.11
Home Equity
Loan (y)
2.07
1.95
1.66
1.77
1.51
1.47
1.75
1.73
1.48
1.51
1.25
1.44
1.38
1.3
Ranked x
3
8.5
2
5
5
1
12
8.5
7
5
11
14
13
10
Ranked y
14
13
9
12
7.5
5
11
10
6
7.5
1
4
3
2
d
-11
-4.5
-7
-7
-2.5
-4
1
-1.5
1
-2.5
10
10
10
8
 d2 =
d2
121
20.25
49
49
6.25
16
1
2.25
1
6.25
100
100
100
64
636
Student’s Solutions Manual and Study Guide: Chapter 17
Page 22
n = 14
rs  1 
6 d 2
n(n  1)
2
 1
6(636)
14(14 2  1)
= -.398
There is a very modest negative correlation between overdue payments for bank
credit cards and home equity loans.
17.37
Number of Companies
Year
on NYSE (x)
1
1,774
2
1,885
3
2,088
4
2,361
5
2,570
6
2,675
7
2,907
8
3,047
9
3,114
10
3,025
11
2,862
Number of Equity
Issues on AMEX (y)
1,063
1,055
943
1,005
981
936
896
893
862
769
765
Ranked x Ranked y
1
11
2
10
3
7
4
9
5
8
6
6
8
5
10
4
11
3
9
2
7
1
d
-10
-8
-4
-5
-3
0
3
6
8
7
6
d 2 =
d2
100
64
16
25
9
0
9
36
64
49
36
408
n = 11
rs  1 
6 d 2
n(n  1)
2
 1
6(408)
= -0.855
11(112  1)
There is a strong negative correlation between the number of companies listed
on the NYSE and the number of equity issues on the American Stock Exchange.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 23
17.39 Ho: Population 1 has values that are the same as those in population 2
Ha: Population 1 has values that are different than those in population 2
Sample 1
573
532
544
565
540
548
536
523
Sample 2
547
566
551
538
557
560
557
547
 = .01 Since n1 = 8, n2 = 8 < 10, use the small sample Mann-Whitney U test.
x
523
532
536
538
540
544
547
547
548
551
557
557
560
565
566
573
Rank
1
2
3
4
5
6
7.5
7.5
9
10
11.5
11.5
13
14
15
16
Group
1
1
1
2
1
1
2
2
1
2
2
2
2
1
2
1
W1 = 1 + 2 + 3 + 5 + 6 + 9 + 14 + 16 = 56
U1  n1  n2 
n1 (n1  1)
(8)(9)
 W1  (8)(8) 
 56 = 44
2
2
U2  n1  n2  U1 = 8(8) - 44 = 20
Take the smaller value of U1, U2 = 20
From Table A.13, the p-value (1-tailed) is .1172, for 2-tailed, the p-value is .2344.
Since the p-value is >  = .05, the decision is to fail to reject the null
hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 24
17.41 nj = 7, n = 28, c = 4, df = 3
Group 1
6
11
8
10
13
7
10
Group 2
4
13
6
8
12
9
8
Group 3
3
7
7
5
10
8
5
Group 4
1
4
5
6
9
6
7
By Ranks:
Group 1
9.5
25
17.5
23
27.5
13.5
23
139
Tj
Tj
n
Group 2
3.5
27.5
9.5
17.5
26
20.5
17.5
122
Group 3
2
13.5
13.5
6
23
17.5
6
81.5
Group 4
1
3.5
6
9.5
20.5
9.5
13.5
63.5
2
= 2760.14 + 2126.29 + 948.89 + 576.04 = 6411.36
j
2
Tj
12
12
K
 3(n  1) 
(6411.36)  3(29) = 7.75

n(n  1)
nj
28(29)
The critical value of chi-square is: 23,.01 = 11.3449.
Since K = 7.75 < 23,.01 = 11.3449, the decision is to fail to reject the null
hypothesis.
17.43
Ranks
1
101
129
133
147
156
179
183
190
2
87
89
84
79
70
64
67
71
1
1
2
3
4
5
6
7
8
2
7
8
6
5
3
1
2
4
d
d2
-6
36
-6
36
-3
9
-1
1
2
4
5
25
5
25
4
16
2
d = 152
Student’s Solutions Manual and Study Guide: Chapter 17
Page 25
n=8
rs  1 
6 d 2
n(n  1)
2
 1
17.45 N = 40 n1 = 24 n2 = 16
6(152)
= -.81
8(63)
 = .05
Use the large sample runs test since both n1, n2 are not less than 20.
H0: The observations are randomly generated
Ha: The observations are not randomly generated
With a two-tailed test, /2 = .025, z.025 = + 1.96. If the observed z > .196
or < -1.96, the decision will be to reject the null hypothesis.
R = 19
R 
R 
z
2n1n2
2(24)(16)
1 
 1 = 20.2
n1  n2
24  16
2n1n2 (2n1n2  n1  n2 )

(n1  n2 )2 (n1  n2  1)
R  R
R

2(24)(16)2(24)(16)  24  16
= 2.993
(40)2 (39)
19  20.2
= -0.40
2.993
Since z = -0.40 > z.025 = -1.96, the decision is to fail to reject the null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 26
17.47 Ho: EMS workers are not older
Ha: EMS workers are older
Age
21
23
24
25
27
27
27
28
28
28
29
30
30
30
32
33
33
36
36
37
39
41
Rank
1
2
3
4
6
6
6
9
9
9
11
13
13
13
15
16.5
16.5
18.5
18.5
20
21
22
n1 = 10
Group
1
1
1
1
1
2
2
1
2
2
2
2
2
2
1
2
2
1
2
1
2
1
n2 = 12
W1 = 1 + 2 + 3 + 4 + 6 + 9 + 15 + 18.5 + 20 + 22 = 100.5

n1  n2 (10)(12)

= 60
2
2

n1  n2 (n1  n2  1)

12
U  n1  n2 
z
U 


(10)(12)( 23)
= 15.17
12
n1 (n1  1)
(10)(11)
 W1  (10)(12) 
 100.5 = 74.5
2
2
74.5  60
= 0.96
15.17
with  = .05, z.05 = -1.645
Since the observed z = 0.96 < z.05 =  1.645 , the decision is to fail to reject the
null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 27
17.49 H0: There is no difference between March and June
Ha: There is a difference between March and June
GMAT
350
430
460
470
490
500
510
520
530
530
540
550
550
560
570
570
590
600
610
630
n1 = 10
Rank
1
2
3
4
5
6
7
8
9.5
9.5
11
12.5
12.5
14
15.5
15.5
17
18
19
20
Month
J
M
J
J
M
M
M
J
M
J
M
M
J
M
M
J
J
M
J
J
n2 = 10
W1 = 1 + 3 + 4 + 8 + 9.5 + 12.5 + 15.5 + 17 + 19 + 20 = 109.5
U1  n1  n2 
n1 (n1  1)
(10)(11)
 W1  (10)(10) 
 109.5 = 45.5
2
2
U2  n1  n2  U1 = (10)(10) - 45.5 = 54.5
From Table A.13, the p-value for U = 46 is .3980 and for 45 is .3697. For a
two-tailed test, double the p-value to at least .739. Using  = .10, the decision is
to fail to reject the null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 28
17.51 Ho: The population differences = 0
Ha: The population differences  0
Box
185
109
92
105
60
45
25
58
161
108
89
123
34
68
59
78
No Box
170
112
90
87
51
49
11
40
165
82
94
139
21
55
60
52
d
15
-3
2
18
9
-4
14
18
-4
26
-5
-16
13
13
-1
26
Rank
11
-3
2
13.5
7
-4.5
10
13.5
-4.5
15.5
-6
-12
8.5
8.5
-1
15.5
n = 16
T- = 3 + 4.5 + 4.5 + 6 + 12 + 1 = 31
T = 31


z
(n)( n  1) (16)(17)

= 68
4
4
n(n  1)( 2n  1)
16(17)(33)

24
24
T 


= 19.34
31  68
= -1.91
19.34
 = .05, /2 = .025
z.025 = ±1.96
Since the observed z = -1.91 > z.025 = -1.96, the decision is to fail to reject the
null hypothesis.
Student’s Solutions Manual and Study Guide: Chapter 17
Page 29
17.53 n1 = 15, n2 = 15 Use the small sample Runs test
 = .05, /.025
H0: The observations in the sample were randomly generated.
Ha: The observations in the sample were not randomly generated
From Table A.11, lower tail critical value = 10
From Table A.12, upper tail critical value = 22
R = 21
Since R = 21 between the two critical values, the decision is to fail to reject the
null hypothesis. The observations were randomly generated.
17.55 Ho: With ties have no higher scores
Ha: With ties have higher scores
Rating
16
17
19
19
20
21
21
22
22
22
23
23
24
25
25
25
25
26
26
26
27
28
n1 = 11
Rank
1
2
3.5
3.5
5
6.5
6.5
9
9
9
11.5
11.5
13
15.5
15.5
15.5
15.5
19
19
19
21
22
Group
2
2
2
2
2
2
1
1
1
2
1
2
2
1
1
1
2
1
1
2
1
1
n2 = 11
W1 = 6.5 + 9 + 9 + 11.5 + 15.5 + 15.5 + 15.5 + 19 + 19 + 21 + 22 = 163.5
Student’s Solutions Manual and Study Guide: Chapter 17
n1  n2 (11)(11)

= 60.5
2
2

n1  n2 (n1  n2  1)
(11)(11)(23)
= 15.23

12
12

U  n1  n2 
z
Page 30
U 


n1 (n1  1)
(11)(12)
 W1  (11)(11) 
 163.5 = 23.5
2
2
23.5  60.5
= -2.43
15.23
For  = .05,
z.05 = 1.645
Since the observed z = -2.43 < z.05 = -1.645, the decision is to reject the null
hypothesis.
17.57 Ho: The 4 populations are identical
Ha: At least one of the 4 populations is different
45
216
215
218
216
219
214
55
228
224
225
222
226
225
70
219
220
221
223
224
85
218
216
217
221
218
217
55
23
18.5
20.5
16
22
20.5
120.5
70
11.5
13
14.5
17
18.5
74.5
85
9
4
6.5
14.5
9
6.5
49.5
6
5
6
By Ranks:
Tj
45
4
2
9
4
11.5
1
31.5
nj
6
Tj
n
2

j
(31.5) 2 (120.5) 2 (74.5) 2 (49.5)



6
6
5
6
= 4,103.84
Student’s Solutions Manual and Study Guide: Chapter 17
Page 31
n = 23
2
Tj
12
12
K
 3(n  1) 
(4,103.84)  3(24) = 17.21

n(n  1)
nj
23(24)
 = .01
df = c - 1 = 4 - 1 = 3
2.01,3 = 11.3449
Since the observed K = 17.21 > 2.01,3 = 11.3449, the decision is to reject the
null hypothesis.
17.59 Ho: The 3 populations are identical
Ha: At least one of the 3 populations is different
3-day
9
11
17
10
22
15
6
Quality
27
38
25
40
31
19
35
Mgmt. Inv.
16
21
18
28
29
20
31
By Ranks:
3-day
2
4
7
3
12
5
1
Tj
34
nj
Tj
n
7
2

j
Quality
14
20
13
21
17.5
9
19
113.5
Mgmt. Inv.
6
11
8
15
16
10
17.5
83.5
7
7
(34) 2 (113.5) 2 (83.5) 2


= 3,001.5
7
7
7
Student’s Solutions Manual and Study Guide: Chapter 17
Page 32
n = 21
2
Tj
12
12
K
 3(n  1) 
(3,001.5)  3(22) = 11.96

n(n  1)
nj
21(22)
 = .10
df = c - 1 = 3 - 1 = 2
2.10,2 = 4.6052
Since the observed K = 11.96 > 2.10,2 = 4.6052, the decision is to reject the
null hypothesis.
17.61 This problem uses a random block design which is analyzed by the Friedman
nonparametric test. There are 4 treatments and 10 blocks. The value of the
observed r2 (shown as S) is 12.16 (adjusted for ties) and has an associated
p-value of .007 that is significant at  = .01. At least one treatment population
yields larger values than at least one other treatment population. Examining the
treatment medians, treatment one has an estimated median of 20.125 and
treatment two has a treatment median of 25.875. These two are the farthest apart.
17.63 A large sample Mann-Whitney U test is being computed. There are 16
observations in each group. The null hypothesis is that the two populations are
identical. The alternate hypothesis is that the two populations are not identical.
The value of W is 191.5. The p-value for the test is .0066. The test is significant
at  = .01. The decision is to reject the null hypothesis. The two populations are
not identical. An examination of medians shows that the median for group two
(46.5) is larger than the median for group one (37.0).