Nonparametric Methods
Chapter 19
Nonparametric Methods
Learning Objectives
1.Learn the difference between parametric and nonparametric methods.
2.Know the particular advantages of nonparametric methods and when they are and when they are not applicable.
3.Learn how to use the sign test for the analysis of paired comparisons.
4.Be able to use the sign test to conduct hypothesis tests about a median.
5.Be able to use the Wilcoxon signed-rank test and the Mann-Whitney-Wilcoxon test to determine whether or not two populations have the same distribution.
6.Be able to use the Kruskal-Wallis tests for the comparison of k populations.
7.Be able to compute the Spearman rank correlation coefficient and test for a significant correlation between two sets of rankings.
Solutions:
1.Binomial Probabilities for n = 10, p = .50.
x / Probability / x / Probability0 / .0010 / 6 / .2051
1 / .0098 / 7 / .1172
2 / .0439 / 8 / .0439
3 / .1172 / 9 / .0098
4 / .2051 / 10 / .0010
5 / .2461
Number of plus signs is 7.
P(x 7)= P(7) + P(8) + P(9) + P(10)
= .1172 + .0439 + .0098 + .0010
= .1719
Two-tailed p-value = 2(.1719) = .3438
p-value > .05, do not reject H0. There is no indication that a difference exists.
2.There are n = 27 cases in which a value different from 150 is obtained.
Use the normal approximation with µ = np = .5(27) = 13.5 and
Use x = 22 as the number of plus signs and obtain the following test statistic:
Last entry in normal distribution table is z > 3.09, Area = .4990
For z = 3.27, p-value is less than (.5000 - .4990) = .0010
p-value .01, reject H0
Conclusion: The median is greater than 150.
3.a.Let p = probability the shares held will be worth more after the split
H0 : p .50
Ha : p > .50
If H0 cannot be rejected, there is no evidence to conclude stock splits continue to add value to stock holdings.
b.Let x be the number of plus signs (increases in value).
Use the binomial probability tables with n = 18 (there were 2 ties in the 20 cases)
P(x 14)= P(14) + P(15) + P(16) + P(17) + P(18)
= .0117 + .0031 + .0006 + .0001 + .0000
= .0155
p-value = .0155
p-value .05, reject H0. The results support the conclusion that stock splits are beneficial for shareholders.
4.We need to determine the number who said better and the number who said worse. The sum of the two is the sample size used for the study.
n = .34(1253) + .29(1253) = 789.4
Use the large sample test using the normal distribution. This means the value of n (n = 789.4 above) need not be integer. Hence,
µ = .5 n = .5(789.4) = 394.7
Let p = proportion of adults who feel children will have a better future.
H0: p .50
Ha: p > .50
With x = .34(1253) = 426
p-value = (.5000 - .4871) = .0129
p-value .05, reject H0
Conclusion: More adults feel their children will have a better future.
5.n = 185 + 165 = 350
µ = 0.5 n = 0.5(350) = 175
Using ER, x = 185
Area in tail = .5000 - .3577 = .1423
p-value = 2(.1423) = .2846
p-value > .05, do not reject H0. Cannot conclude there is a difference in preference for the two shows.
6.n = 202 + 158 = 360
µ = 0.5 n = 0.5(360) = 180
Using Packard Bell, x = 202
Area in tail = .5000 - .4898 = .0102
p-value = 2(.0102) = .0204
p-value .05, reject H0. Conclude Packard Bell and Compaq have different market shares.
7.µ = 0.5 n = 0.5(300) = 150
p-value = 2(.5000 - .4582) = .0836
Do not reject H0; we are unable to conclude that the median annual income differs.
8.µ = .5 n = .5(150) = 75
For 98 + signs
For z = 3.76, area in tail is less than .001
With two tails, p-value is less than .002
Using Excel, p-value = .0002
p-value .05, reject H0. Conclude that a home team advantage exists.
9.H0: Median 15
Ha: Median > 15
Use binomial probabilities with n = 8 and p = .50:
P(x 7)= P(7) + P(8)
= .0313 + .0039 = .0352
p-value = .0352
p-value .05, reject H0. Data does enable us to conclude that there has been an increase in the median number of part-time employees.
10.H0: Median = 152
Ha: Median 152
µ = .5n = .5(225) = 112.5
For x = 122
Area in tail = .5000 - .3980 = .1020
p-value = 2(.1020) = .2040
p-value > .05, do not reject H0. We are unable to conclude that the median annual income needed differs from that reported in the survey.
11.n = 50
µ = 0.5n = 0.5(50) = 25
x = 33 had wages greater than $585
p-value = .5000 - .4881 = .0119
p-value .05, reject H0. Conclude that the median weekly wage is greater than $585.
12.H0: The populations are identical
Ha: The populations are not identical
Additive 1 / Additive 2 / Difference / Absolute Value / Rank / Signed Rank20.12 / 18.05 / 2.07 / 2.07 / 9 / +9
23.56 / 21.77 / 1.79 / 1.79 / 7 / +7
22.03 / 22.57 / -.54 / .54 / 3 / -3
19.15 / 17.06 / 2.09 / 2.09 / 10 / +10
21.23 / 21.22 / .01 / .01 / 1 / +1
24.77 / 23.80 / .97 / .97 / 4 / +4
16.16 / 17.20 / -1.04 / 1.04 / 5 / -5
18.55 / 14.98 / 3.57 / 3.57 / 12 / +12
21.87 / 20.03 / 1.84 / 1.84 / 8 / +8
24.23 / 21.15 / 3.08 / 3.08 / 11 / +11
23.21 / 22.78 / .43 / .43 / 2 / +2
25.02 / 23.70 / 1.32 / 1.32 / 6 / +6
Total 62
µT = 0
p-value = 2(.5000 - .4925) = .0150
p-value .05, reject H0.
Conclusion: There is a significant difference in the additives.
13.
Without Relaxant / With Relaxant / Difference / Rank of Absolute Difference / Signed Rank15 / 10 / 5 / 9 / 9
12 / 10 / 2 / 3 / 3
22 / 12 / 10 / 10 / 10
8 / 11 / -3 / 6.5 / -6.5
10 / 9 / 1 / 1 / 1
7 / 5 / 2 / 3 / 3
8 / 10 / -2 / 3 / -3
10 / 7 / 3 / 6.5 / 6.5
14 / 11 / 3 / 6.5 / 6.5
9 / 6 / 3 / 6.5 / 6.5
Total 36
µT = 0
One-tailed test.
p-value = .5000 - .4664 = .0336
p-value .05, reject H0.
Conclusion: There is a significant difference in favor of the relaxant.
14.
Airport / Difference / AbsoluteDifference / Signed Rank
Boston Logan / 0.19 / 0.19 / 10
Chicago Midway / -0.02 / 0.02 / -3.5
Chicago O'Hare / 0.05 / 0.05 / 6
Denver / 0.04 / 0.04 / 5
Fort Lauderdale / -0.01 / 0.01 / -1.5
Los Angeles / 0.06 / 0.06 / 7
Miami / 0.02 / 0.02 / 3.5
New York (JFK) / 0.09 / 0.09 / 8
Orange County (CA) / 0.16 / 0.16 / 9
Washington (Dulles) / 0.01 / 0.01 / 1.5
T = 45
p-value = 2(.5000 - .4890) = .0220
p-value .05, reject H0. Conclude a difference exists with Avis higher.
15.
Service #1 / Service #2 / Difference / Rank of Absolute Difference / Signed Rank24.5 / 28.0 / -3.5 / 7.5 / -7.5
26.0 / 25.5 / 0.5 / 1.5 / 1.5
28.0 / 32.0 / -4.0 / 9.5 / -9.5
21.0 / 20.0 / 1.0 / 4 / 4.0
18.0 / 19.5 / -1.5 / 6 / -6.0
36.0 / 28.0 / 8.0 / 11 / 11.0
25.0 / 29.0 / -4.0 / 9.5 / -9.5
21.0 / 22.0 / -1.0 / 4 / -4.0
24.0 / 23.5 / 0.5 / 1.5 / 1.5
26.0 / 29.5 / -3.5 / 7.5 / -7.5
31.0 / 30.0 / 1.0 / 4 / 4.0
T = -22.0
µT = 0
p-value = 2(.5000 - .3365) = .3270
p-value > .05, do not reject H0. There is no significant difference.
16.
1997 P/E Ratio / Est. 1998 P/E Ratio / Difference / Rank / Signed Rank40 / 32 / 8 / 9 / 9
24 / 22 / 2 / 2.5 / 2.5
21 / 23 / -2 / 2.5 / -2.5
30 / 23 / 7 / 8 / 8
25 / 19 / 6 / 6.5 / 6.5
19 / 19 / 0 / 0 / 0
20 / 17 / 3 / 4 / 4
29 / 19 / 10 / 10 / 10
35 / 20 / 15 / 11 / 11
17 / 18 / -1 / 1 / -1
33 / 27 / 6 / 6.5 / 6.5
20 / 16 / 4 / 5 / 5
T = 59
n = 11 (discarding the 0)
µT = 0
p-value = 2(.5000 - .4956) = .0088
p-value .05, reject H0. Conclude a difference exists between the 1997 P/E ratios and the estimated 1998 P/E ratios.
17.
Precampaign / Postcampaign / Difference / Rank of Absolute Difference / Signed Rank130 / 160 / -30 / 10 / -10
100 / 105 / -5 / 2.5 / -2.5
120 / 140 / -20 / 9 / -9
95 / 90 / 5 / 2.5 / 2.5
140 / 130 / 10 / 4.5 / 4.5
80 / 82 / -2 / 1 / -1
65 / 55 / 10 / 4.5 / 4.5
90 / 105 / -15 / 7.5 / -7.5
140 / 152 / -12 / 6 / -6
125 / 140 / -15 / 7.5 / -7.5
T = -32
µT = 0
One tailed test to see if there is a significant improvement.
p-value = .5000 - .4484 = .0516
p-value > .05, do not reject H0. The difference is not significant at the = .05 level.
18.Rank the combined samples and find the rank sum for each sample.
This is a small sample test since n1 = 7 and n2 = 9
Additive 1 / Additive 2MPG / Rank / MPG / Rank
17.3 / 2 / 18.7 / 8.5
18.4 / 6 / 17.8 / 4
19.1 / 10 / 21.3 / 15
16.7 / 1 / 21.0 / 14
18.2 / 5 / 22.1 / 16
18.6 / 7 / 18.7 / 8.5
17.5 / 3 / 19.8 / 11
34 / 20.7 / 13
20.2 / 12
102
T = 34
With = .05, n1 = 7 and n2 = 9
TL = 41 and TU = 7(7 + 9 + 1) - 41 = 78
Since T = 34 < 41, we reject H0
Conclusion: There is a significant difference in gasoline mileage
19.a.
Public Accountant / Rank / Financial Planner / Rank25.2 / 5 / 24.0 / 2
33.8 / 19 / 24.2 / 3
31.3 / 16 / 28.1 / 10
33.2 / 18 / 30.9 / 15
29.2 / 13 / 26.9 / 8.5
30.0 / 14 / 28.6 / 11
25.9 / 6 / 24.7 / 4
34.5 / 20 / 28.9 / 12
31.7 / 17 / 26.8 / 7
26.9 / 8.5 / 23.9 / 1
136.5 / 73.5
T = 136.5
p-value = 2(.5000 - .4913) = .0174
p-value .05, reject H0. Salaries differ significantly for the two professions.
b.Public Accountant:$30,200
Financial Planner:$26,700
20.a.Median 4th salary for each
Men 49.9 Women 35.4
b.
Men / Rank / Women / Rank30.6 / 4 / 44.5 / 8
75.5 / 14 / 35.4 / 5
45.2 / 9 / 27.9 / 3
62.2 / 13 / 40.5 / 7
38.2 / 6 / 25.8 / 2
49.9 / 11 / 47.5 / 10
55.3 / 12 / 24.8 / 1
T = 36
From Tables TL = 37
T < TL Reject H0; Conclude populations differ. Men show higher salaries.
21.Sum of ranks (Model 1) = 185.5
Sum of ranks (Model 2) = 114.5
Use T = 185.5
p-value = 2(.5000 - .4798) = .0404
p-value .10, reject H0.
Conclusion: There is a significant difference between the populations.
22.H0 : There is no difference in the distributions of P/E ratios
Ha: There is a difference between the distributions of P/E ratios
Japan / United StatesCompany / P/E Ratio / Rank / Company / P/E Ratio / Rank
Sumitomo Corp. / 153 / 20 / Gannet / 19 / 6
Kinden / 21 / 8 / Motorola / 24 / 11.5
Heiwa / 18 / 5 / Schlumberger / 24 / 11.5
NCR Japan / 125 / 19 / Oracle Systems / 43 / 16
Suzuki Motor / 31 / 13 / Gap / 22 / 10
Fuji Bank / 213 / 21 / Winn-dixie / 14 / 2
Sumitomo Chemical / 64 / 17 / Ingersoll-Rand / 21 / 8
Seibu Railway / 666 / 22 / Am. Elec. Power / 14 / 2
Shiseido / 33 / 14 / Hercules / 21 / 8
Toho Gas / 68 / 18 / Times Mirror / 38 / 15
Total / 157 / WellPoint Health / 15 / 4
No. States Power / 14 / 2
Total / 96
T = 157
p-value = 2(.5000 - .4972) = .0056
p-value .01, reject H0.
We conclude that there is a significant difference in P/E ratios for the two countries.
23.Sum of ranks (Winter) = 71.5
Sum of ranks (Summer) = 138.5
Use T = 71.5
p-value = 2(.5000 - .4943) = .0114
p-value .05, reject H0. There is a significant difference
24.Sum of ranks (Dallas) = 116Sum of ranks (San Antonio) = 160
Use T = 116
p-value = 2(.5000 - .0987) = .8026
p-value > .05, do not reject H0. There is no evidence to conclude that there is a difference.
25.
Kitchen / Rank / Master Bedroom / Rank25,200 / 16 / 18,000 / 4
17,400 / 2 / 22,900 / 11
22,800 / 10 / 26,400 / 17
21,900 / 9 / 24,800 / 15
19,700 / 5.5 / 26,900 / 18
23,000 / 12 / 17,800 / 3
19,700 / 5.5 / 24,600 / 14
16,900 / 1 / 21,000 / 7
21,800 / 8 / 89
23,600 / 13
82
From Appendix B,
TL = 73TU= n1 (n1 + n2 + 1) - TL
= 10 (10 + 8 + 1) - 73 = 117
Reject H0 if T < 73 or if T > 117
Since T = 82, do not reject
There is no significant difference between the costs.
26.
A / B / C4 / 11 / 7
8 / 14 / 2
10 / 15 / 1
3 / 12 / 6
9 / 13 / 5
34 / 65 / 21
Degrees of freedom = 2
Usingtable,= 10.22 shows p-value is between .005 and .01
Actual p-value = .006
p-value .05, reject H0. Conclude that the ratings for the products differ.
27.
A / B / C11.5 / 5.0 / 17.0
2.5 / 11.5 / 20.0
8.0 / 2.5 / 15.0
10.0 / 4.0 / 8.0
8.0 / 6.0 / 16.0
18.0 / 1.0 / 19.0
13.0 / 14.0
58.0 / 43.0 / 109.0
Degrees of freedom = 2
Usingtable,= 9.06 shows p-value is between .01 and .025
Actual p-value = .0108
p-value > .01, do not reject H0. We cannot conclude that there is a significant difference in test preparation programs.
28.
Swimming / Rank / Tennis / Rank / Cycling / Rank408 / 8 / 415 / 9 / 385 / 5
380 / 4 / 485 / 14 / 250 / 1
425 / 11 / 450 / 13 / 295 / 3
400 / 6 / 420 / 10 / 402 / 7
427 / 12 / 530 / 15 / 268 / 2
Sum / 41 / 61 / 18
Degrees of freedom = 2
Usingtable,= 9.26 shows p-value is between .005 and .01
Actual p-value = .0098
p-value .05, reject H0. Conclude that there is a significant difference in calories among the three activities.
29.
A / B / C2 / 2 / 12
7 / 4.5 / 14
4.5 / 9 / 10.5
2 / 7 / 13
7 / 10.5 / 15
22.5 / 33 / 64.5
Degrees of freedom = 2
Usingtable,= 9.56 shows p-value is between .005 and .01
Actual p-value = .0084
p-value .05, reject H0. Conclude that there is a significant difference in gas mileage among the three automobiles.
30.
Course 1 / Course 2 / Course 3 / Course 43 / 2 / 19 / 20
14 / 7 / 16 / 4
10 / 1 / 9 / 15
12 / 5 / 18 / 6
13 / 11 / 17 / 8
52 / 26 / 79 / 53
Degrees of freedom = 3
Usingtable,= 8.03 shows p-value is between .025 and .05
Actual p-value = .0454
p-value .05, reject H0. Conclude that there is a significant difference in the quality of courses offered by the four management development centers.
31.
M&Ms / Kit Kat / Milky Way II10.5 / 9 / 3
7 / 5 / 6
13 / 14 / 4
15 / 12 / 2
10.5 / 8 / 1
56 / 48 / 16
Degrees of freedom = 2
Usingtable,= 8.96 shows p-value is between .01 and .025
Actual p-value = .0113
p-value .05, reject H0. There are significant differences in calorie content among the three candies.
32.a.= 52
b.
p-value = 2(.5000 - .4798) = .0404
p-value .05, reject H0. Conclude that significant rank correlation exists.
33.Case 1:
= 0
Case 2:
= 70
With perfect agreement, rs = 1.
With exact opposite ranking, rs = -1.
34.= 250
p-value = 2(.5000 - .1664) = .6672
p-value > .05, do not reject H0. Conclude that there is not a significant relationship between the rankings.
35.a.= 54
b.H0 : pr 0
Ha : pr > 0
p-value = 2(.5000 - .4783) = .0434
c.p-value .05, reject H0. Conclude a significant positive rank correlation.
36.
Driving Distance / Putting / di /1 / 5 / -4 / 16
5 / 6 / -1 / 1
4 / 10 / -6 / 36
9 / 2 / 7 / 49
6 / 7 / -1 / 1
10 / 3 / 7 / 49
2 / 8 / -6 / 36
3 / 9 / -6 / 36
7 / 4 / 3 / 9
8 / 1 / 7 / 49
= 282
= 0
p-value = 2(.5000 - .4834) = .0332
p-value .10, reject H0. There is a significant negative rank correlation between driving distance and putting.
37.= 38
= 0
p-value = 2(.5000 - .4896) = .0208
p-value .10, reject H0. There is a significant rank correlation between current students and recent graduates.
38.n = 905 + 1045 = 1950
= .5n = .5(1950) = 975
p-value less than 2(.001) = .002
Actual p-value = .0016
p-value .05, reject H0. The difference in the favor-oppose opinion is significant.
39.a.n = 11 + 32 = 43
H0: Median $118,000
Ha: Median < $118,000
µ = .5n = .5(43) = 21.5
p-value less than .001
p-value .05, reject H0. We conclude that the median resale price for homes in Houston, Texas is below the national median.
b.n = 27 + 13 = 40
H0: Median $118,000
Ha: Median > $118,000
µ = .5 n = .5(40) = 20
p-value = .5000 - .4864 = .0136
p-value .05, reject H0. We conclude that the median resale price for homes in Boston, Massachusetts is above the national median.
40.Use the Wilcoxon Signed Rank Test
Homemaker / Difference / Signed Rank1 / -250 / -11
2 / 40 / 2
3 / 50 / 3
4 / -150 / -6
5 / -330 / -12
6 / -180 / -7
7 / -190 / -8.5
8 / -230 / -10
9 / -100 / -5
10 / -190 / -8.5
11 / -90 / -4
12 / 20 / 1
T = -66
p-value = 2(.5000 - .4952) = .0096
p-value .05, reject H0. Conclude that the models differ in terms of selling prices.
41.
Difference / Rank ofAbsolute Difference / Signed Rank
1.5 / 10 / 10.0
1.2 / 9 / 9.0
-.2 / 2.5 / -2.5
0 / /
.5 / 4 / 4.0
.7 / 6 / 6.0
.8 / 7 / 7.0
1.0 / 8 / 8.0
0 / /
.6 / 5 / 5.0
.2 / 2.5 / 2.5
-.01 / 1 / -1.0
T = 48
p-value = .5000 - .4929 = .0071
p-value .05, reject H0. Conclude that there is a significant weight gain.
42.Use the MWW test.
Sum of ranks (line 1) = 70
Sum of ranks (line 2) = 183
T = 70
p-value = 2(.5000 - .4985) = .0030
p-value .10, reject H0. Conclude that the weights differ for the two production lines.
43.
Method 1 / Method 2 / Method 38.5 / 4.5 / 2.0
15.0 / 14.0 / 7.0
6.0 / 16.0 / 10.0
17.0 / 8.5 / 1.0
18.0 / 12.5 / 3.0
12.5 / 11.0 / 4.5
77.0 / 66.5 / 27.5
Degrees of freedom = 3
Usingtable,= 7.96 shows p-value is between .025 and .05
Actual p-value = .0468
p-value .05, reject H0. Conclude that there is a significant difference among the methods.
44.
No Program / Company Program / Off Site Program16 / 12 / 7
9 / 20 / 1
10 / 17 / 4
15 / 19 / 2
11 / 6 / 3
13 / 18 / 8
14 / 5
74 / 106 / 30
Degrees of freedom = 2
Usingtable,= 12.61 shows p-value is between .01 and .025
Actual p-value = .0018
p-value .05, reject H0. There is a significant difference among the programs.
45.
Black / Jennings / Swanson / Wilson22.5 / 20.5 / 22.5 / 9.5
9.5 / 27.0 / 6.0 / 17.5
8.0 / 7.0 / 2.5 / 1.0
2.5 / 17.5 / 12.5 / 5.0
26.0 / 28.5 / 17.5 / 24.0
4.0 / 28.5 / 12.5 / 20.5
17.5 / 15.0
25.0 / 14.0
11.0
72.5 / 171.5 / 113.5 / 77.5
Degrees of freedom = 3
Usingtable,= 6.34 shows p-value is between .05 and .10
Actual p-value = .0962
p-value > .05, do not reject H0. We cannot conclude that there is a significant difference among the course evaluation ratings for the 4 instructors.
46.= 136
p-value = 2(.5000 - .4977) = .0046
p-value .10, reject H0. Conclude that there is a significant rank correlation between the two exams.
19 - 1