chap. 14: nonparametric statistics

Non-Parametric Statistics

Rupika Abeynayake, Ph.D (Statistics)

Professor in Applied Statistics

Content

Introduction: Distribution-Free Tests

Single Population Inferences

Comparing Two Populations:

Independent Samples

Comparing Two Populations: Paired

Difference Experiment

Comparing Three or More

Populations: Completely Randomized

Design

Content

Comparing Three or More

Populations: Randomized Block Design

Rank Correlation

Learning Objectives

• Develop the need for inferential

techniques that require fewer, or less

stringent, assumptions than parametric

methods

• Introduce nonparametric tests that are

based on ranks (i.e., on an ordering of

the sample measurements according to

their relative magnitudes)

Introduction:

Distribution-Free Tests

Parametric Test Procedures

• Involve population parameters— Example: population mean

• Require interval scale or ratio scale— Whole numbers or fractions

— Example: height in inches (72, 60.5, 54.7)

• Have stringent assumptions— Example: normal distribution

• Examples: z-test, t-test, F-test, 2-test

Nonparametric Test

Procedures

• Do not involve population parameters

— Example: probability distributions, independence

• Data measured on any scale

— Ratio or interval

— Ordinal

Example: good-better-best

— Nominal

Example: male-female

• Example: Wilcoxon rank sum test

Nonnormal Distributions -

t-Statistic is Invalid

Distribution-Free Tests

Distribution-free tests are statistical tests that do not rely on any underlying assumptions about the probability distribution of the sampled population.

The branch of inferential statistics devoted to distribution-free tests is called nonparametrics.

Nonparametric statistics (or tests) based on the ranks of measurements are called rank statistics (or rank tests).

Advantages of

Nonparametric Tests

• Used with all scales

• Easier to compute

• Make fewer assumptions

• Need not involve population parameters

• Results may be as exact as parametric procedures

Disadvantages of

Nonparametric Tests• May waste information

— If data permit using parametric procedures

— Example: converting data from ratio to ordinal scale

• Difficult to compute by hand for large samples

• Tables not widely available

• Power of the test is low

• It is merely for testing of hypothesis and no confidence limits could be calculated.

Frequently Used Nonparametric

Tests

• Sign Test

• Wilcoxon Signed Rank Test

• Wilcoxon Rank Sum Test – Mann-Whitny test

• Kruskal Wallis H-Test – Quade test

• Friedman test

• Spearman’s Rank Correlation Coefficient© 2011 Pearson Education, Inc

Parametric vs Nonparametric

• One Sample Sign Test

• Sign Rank Test

• Mann-Whitney Test

• Kruskal-Wallis Test

• Quade / Friedman Test

• Spearman Rank Test

• One-sample t-test

• Paired t-test

• Pooled t-test

• One-way ANOVA (CRD)

• Two-way ANOVA (RCBD)

• Correlation/Regression

Single Population Inferences

Sign Test

• Tests one population median, (eta)

• Corresponds to t-test for one mean

• Assumes population is continuous

• Small sample test statistic: Number of sample values above (or below) median

• Can use normal approximation if n 30

Sign Test Uses p-Value

to Make Decision

.031

.109

.219

.273

.219

.109

.031.004.004

0%

10%

20%

30%

0 1 2 3 4 5 6 7 8 X

P(X)

© 2011 Pearson Education, Inc

Binomial: n = 8 p = 0.5

P-value is the probability of getting an observation at least as

extreme as we got. If 7 of 8 observations ‘favor’ Ha, then

p-value = p(x 7) = .031 + .004 = .035.

If = .05, then reject H0 since p-value .

Sign Test for a Population

Median

One-Tailed Test

H0: = 0

Ha: > 0 [or Ha: < 0 ]

Test statistic:

S = Number of sample measurements greater than0 [or S = number of measurements less than 0]

Sign Test Example

You’re an analyst for a some fruit juice company. You’ve asked 7 people to rate a new fruit juice on a 5-point Likert scale (1 = terrible to 5 = excellent). The ratings are:

2 5 4 4 1 4 5

At the .05 level of significance, is there evidence that the median rating is greater than 3?

Sign Test Solution

• H0:

• Ha:

• =

• Test Statistic:


p-value:

Decision:

Conclusion:

Do not reject null

hypothesis at = .05

There is no evidence

median is greater than 3

P(x 5) = 1 – P(x 4)

= 1-0.7734

=0.2266

S = 5

(Ratings 1 & 2 are

less than = 3:

2, 5, 4, 4, 1, 4, 5)

= 3

> 3

.05


Median

Observed significance level:

p-value = P(x ≥ S)

where x has a binomial distribution with parameters n and p = .5

Rejection region: Reject H0 if p-value ≤ .05


Median

Two-Tailed Test

H0: = 0

Ha: ≠ 0

Test statistic:

S = Larger of S1 and S2, where S1 is the number of sample measurements less than0 and S2 is the number of measurements greater than 0


Median

Observed significance level:

p-value = 2P(x ≥ S)

where x has a binomial distribution with parameters n and p = .5

Rejection region: Reject H0 if p-value ≤ .05

Recent studies of the private practices of physicians

who saw no Medicaid patients suggested that the

median length of each patient visit was 22 minutes. It is

believed that the median visit length in practices with a

large Medicaid load is shorter than 22 minutes. A

random sample of 20 visits in practices with a large

Medicaid load yielded, in order, the following visit

lengths:

9.4 13.4 15.6 16.2 16.4 16.8

18.1 18.7 18.9 19.1 19.3 20.1

20.4 21.6 21.9 23.4 23.5 24.8

24.9 26.8

Based on these data, is there sufficient evidence to conclude that

the median visit length in practices with a large Medicaid load is

shorter than 22 minutes?

Conditions Required for Valid

Application of the Sign Test

The sample is selected randomly from a continuous probability distribution

[Note: No assumptions need to be made about the shape of the probability distribution.]

Large Sample Sign Test for a

Population Median

One-Tailed TestH0: = 0Ha: > 0 [or Ha: < 0 ]

Test statistic:

N is the number of sample not equal to the hypnotized valueS = Number of sample measurements greater than0 [or S = number of measurements less than 0] The “– .5” is the “correction for continuity.”

z

S .5 .5n

.5 n

Can use normal approximation if n 30


Population Median

The null hypothesized mean value is np = .5n, and the standard deviation is

Rejection region: z > z

npq n .5 .5 .5 n


Population Median

Two-Tailed Test

H0: = 0

Ha: ≠ 0

Test statistic:

S = Larger of S1 and S2, where S1 is the number of sample measurements less than0 and S2 is the number of measurements greater than 0


z

S .5 .5n

.5 n


Population Median

The null hypothesized mean value is np = .5n, and the standard deviation is

Rejection region: z > z

npq n .5 .5 .5 n

The Gordon Travel Agency claims that their median airfare for

all their clients to all destinations is $450.

This claim is being challenged by a competing agency, who

believe the median is different from $450.

A random sample of 300 tickets revealed

170 tickets were below $450. Use the 0.05 level of

significance.

Continued…

H0 is rejected if z is > than 1.96 or < than -1.96

… the value of z is 2.252

H0 is rejected.

We conclude that the median is not $450

450$median :H$450=median : 10H

252.23005.

)300(50.)5.170(

5.

50.)5.(

n

nxz

Comparing Two Populations:Related Samples

Wilcoxon signed rank test

To test difference between paired data

Example

Data

14

2

5

4

2

14

18

14

ExampleSorted

Data

2

2

4

5

14

14

14

18

Data

14

2

5

4

2

14

18

14

ExampleSorted

Data

2

2

4

5

14

14

14

18

Data

14

2

5

4

2

14

18

14

TIES

ExampleSorted

Data

2

2

4

5

14

14

14

18

Data

14

2

5

4

2

14

18

14

TIES

Rank them

anyway,

pretending they

were slightly

different

Example

Rank A

2

2

4

5

14

14

14

18

1

2

3

4

5

6

7

8

Sorted

DataData

14

2

5

4

2

14

18

14

Example

Rank A

2

2

4

5

14

14

14

18

1

2

3

4

5

6

7

8

Sorted

DataData

14

2

5

4

2

14

18

14

Find the

average of the

ranks for the

identical

values, and give

them all that

rank

Example

Rank A

2

2

4

5

14

14

14

18

1

2

3

4

5

6

7

8

Sorted

DataData

14

2

5

4

2

14

18

14

Average = 1.5

Average = 6

Example

Rank A

2

2

4

5

14

14

14

18

1

2

3

4

5

6

7

8

Sorted

DataData

14

2

5

4

2

14

18

14

1.5

1.5

3

4

6

6

6

8

Rank

Example

Rank A

2

2

4

5

14

14

14

18

1

2

3

4

5

6

7

8

Sorted

DataData

14

2

5

4

2

14

18

14

1.5

1.5

3

4

6

6

6

8

Rank

These can now be used for the Mann-Whitney U test

Patient

Hours of sleep

Drug Placebo

1 6.1 5.2

2 7.0 7.9

3 8.2 3.9

4 7.6 4.7

5 6.5 5.3

6 8.4 5.4

7 6.9 4.2

8 6.7 6.1

9 7.4 3.8

10 5.8 6.3

EXAMPLE

Null Hypothesis: Hours of sleep are the same using placebo & the drug

STEP 1• Exclude any differences which are zero

• Ignore their signs

• Put the rest of differences in ascending order

• Assign them ranks

• If any differences are equal, average their

ranks

STEP 2

• Count up the ranks of +ives as W+

• Count up the ranks of –ives as W-

STEP 3• If there is no difference between drug (W+)

and placebo (W-), then W+ & W- would be similar

• If there is a difference

one sum would be much smaller and

the other much larger than expected

• The smaller sum is denoted as W

• W = smaller of W+ and W-

STEP 4

• Compare the value obtained with the

critical values (5%, 2% and 1% ) in table

• N is the number of differences that

were ranked (not the total number of

differences)

• So the zero differences are excluded

Patient

Hours of sleep

Difference

Rank

Ignoring signDrug Placebo

1 6.1 5.2 0.9 3.5*

2 7.0 7.9 -0.9 -3.5*

3 8.2 3.9 4.3 10

4 7.6 4.7 2.9 7

5 6.5 5.3 1.2 5

6 8.4 5.4 3.0 8

7 6.9 4.2 2.7 6

8 6.7 6.1 0.6 2

9 7.4 3.8 3.6 9

10 5.8 6.3 -0.5 -1

3rd & 4th ranks are tied hence averaged; W = smaller of W+ (50.5) and W-

(4.5)

Here, calculated value of W-= 4.5 ; tabulated value of T= 8 (at 5%)

significant at 5% level indicating that the drug (hypnotic) is more effective

than placebo

Wilcoxon test Class Example:

In order to investigate whether adults report verbally presented

material more accurately from their right than from their left

ear, a dichotic listening task was carried out. The data were

found to be positively skewed.

1 25 32

2 29 30

3 10 8

4 31 32

5 27 20

6 24 32

7 26 27

8 29 30

9 30 32

10 32 32

11 20 30

12 5 32

Wilcoxon Signed-Rank Test

Use the Wilcoxon matched-pair signed-rank test

to determine if the R&D expenses as a percent of income

(EXAMPLE 1) have declined.

Use the .05 significance level

Step 1: H0: Expenditure same in two years

H1: Expenditure not same in two years

Step 2: H0 is rejected if the smaller of the rank

sums is greater than or equal critical value

ontinued…

Company 2000 2001 Difference ABS-Diff Rank R+ R-

Savoth Glass 20 16 4 4

Ruisi Glass 14 13 1 1

Rubin Inc.. 23 20 3 3

Vaught 24 17 7 7

Lambert Glass 31 22 9 9

Pimental 22 20 2 2

Olson Glass 14 20 - 6 6

Flynn Glass 18 11 7 7

ontinued…


The smaller rank sum is….., which is ……..than

critical value of T.


Savoth Glass 20 16 4 4

Ruisi Glass 14 13 1 1

Rubin Inc.. 23 20 3 3

Vaught 24 17 7 7

Lambert Glass 31 22 9 9

Pimental 22 20 2 2

Olson Glass 14 20 - 6 6

Flynn Glass 18 11 7 7


The smaller rank sum is 5, which is greater to the

critical value of 3.

H0 can't be rejected

There is no evidence to say that expenditure is differ in two

years .


Savoth Glass 20 16 4 4 4 4 *

Ruisi Glass 14 13 1 1 1 1 *

Rubin Inc.. 23 20 3 3 3 3 *

Vaught 24 17 7 7 6.5 6.5 *

Lambert Glass 31 22 9 9 8 8 *

Pimental 22 20 2 2 2 2 *

Olson Glass 14 20 - 6 6 5 * 5

Flynn Glass 18 11 7 7 6.5 6 .5 *


Example

Let Xi denote the length, in centimeters, of a

randomly selected pygmy sunfish, i = 1, 2, ...

10. If we obtain the following data set:

5.0 3.9 5.2 5.5 2.8

6.1 6.4 2.6 1.7 4.3

can we conclude that the median length of

pygmy sunfish differs significantly from 3.7

centimeters?

Solution.

We are interested in testing the null

hypothesis

H0: m = 3.7

against the alternative hypothesis

HA: m ≠ 3.7.

Solution. Recall that we are interested in testing

the null hypothesis H0: m = 3.7 against the

alternative hypothesis HA: m ≠ 3.7.

W = 15 (smaller value) for the given data set.

Tabulated value = 8

Tabulated value<calculated value

Can’t Reject H0

we cannot reject the null hypothesis.

There is no evidence at the 0.05 level to

conclude that the median length of

pygmy sunfish differs significantly

from 3.7 centimeters.

Normal Approximation when Sample

size is > 30


Homework

The median age of the onset of diabetes is thought

to be 45 years. The ages at onset of a random

sample of 30 people with diabetes are:

35.5 44.5 39.8 33.3 51.4

51.3 30.5 48.9 42.1 40.3

46.8 38.0 40.1 36.8 39.3

65.4 42.6 42.8 59.8 52.4

26.2 60.9 45.6 27.1 47.3

36.6 55.6 45.1 52.2 43.5

Solution. We are interested in testing the null

hypothesis

H0: m = 45

against the alternative hypothesis

HA: m ≠ 45.

Comparing Two Populations:Independent Samples

Mann-Whitney U test

• Tests two independent population probability distributions

• Corresponds to t-test for two independent means

• Assumptions

— Independent, random samples

— Populations are continuous

• Can use normal approximation if ni 10

Mann-Whitney test, step-by-step:

Does it make any difference to students'

comprehension of statistics whether the lectures

are in Theory base or in practical base?

Group 1: statistics lectures in Theory base.

Group 2: statistics lectures in Practical base.

theory group

(raw scores)

Practical group

(raw scores)

18 17

15 13

17 12

13 16

11 10

16 15

10 11

17 13

12

Step 1:

• Rank all the scores together, regardless of group

Step 2:

• Add up the ranks for group 1, to get R1.

• Add up the ranks for group 2, to get R2.

Step 3:

• n1 is the number of subjects in group 1; n2 is the number

of subjects in group 2. Here, n1 = 8 and n2 = 9.

Mann-Whitney U: CalculationStep 4:

Calculate two versions of the U statistic using:

U1 = (n1 x n2) + 2

(n1 + 1) x n1- ∑R1

AND…

U2 = (n1 x n2) + 2

(n2 + 1) x n2- ∑R2

Mann-Whitney U: Calculation

Calculate two versions of the U statistic using:

U1 = (n1 x n2) + 2

(n1 + 1) x n1- ∑R1

…OR to save time you can calculate U1 and then U2 as follows

U2 = (n1x n2) - U1

Mann-Whitney U: CalculationStep 5:

Select the smaller of the two U statistics (U1 & U2)

now consult a table of critical values for the

Mann-Whitney test

Calculated U must be less than critical U to

conclude a significant difference

theory group

(raw scores)

Ranks Practica group

(raw scores)

Ranks

18 17 17 15

15 10.5 13 8

17 15 12 5.5

13 8 16 12.5

11 3.5 10 1.5

16 12.5 15 10.5

10 1.5 11 3.5

17 15 13 8

12 5.5

Mean:

SD:

14.63

2.97

Mean:

SD:

13.22

2.33

Median: 15.5 Median: 13

Add up the ranks for group 1, to get R1.

Here, R1 = 83.

Add up the ranks for group 2, to get R2.

Here, R2 = 70.

In the present example, U1 = 25, and, U2 =

47

Therefore, use 25 as U.

Here, the critical value of U for N1 = 8 and N2 = 9 is 15.

Our obtained U of 25 is larger than this, and so we cant

reject the Ho (Null hypothesis)

Conclusion: There is no evidence to say that Students

score for statistics is depend on the teaching methods.

N 2

N 1 5 6 7 8

2 3 5 6 7 8

3 5 6 8 10 11

5 6 8 10 12 14

6 8 10 13 15 17

7 10 12 15 17 20

8 11 14 17 20 23

9

10

5

6

7

8

9 10

Non-smokers (n=15) Heavy smokers (n=14)Birth wt (Kg)

3.99

3.79

3.60*

3.73

3.21

3.60*

4.08

3.61

3.83

3.31

4.13

3.26

3.54

3.51

2.71

Birth wt (Kg)

3.18

2.84

2.90

3.27

3.85

3.52

3.23

2.76

3.60*

3.75

3.59

3.63

2.38

2.34

Null Hypothesis: Median birth weight is same between non-smokers & smokers

Non-smokers (n=15) Heavy smokers (n=14)Birth wt (Kg) Rank Birth wt (Kg) Rank

3.99 27 3.18 73.79 24 2.84 53.60* 18 2.90 63.73 22 3.27 113.21 8 3.85 263.60* 18 3.52 14

4.08 28 3.23 9

3.61 20 2.76 4

3.83 25 3.60* 183.31 12 3.75 234.13 29 3.59 163.26 10 3.63 213.54 15 2.38 23.51 13 2.34 12.71 3

Sum=272 Sum=163

* 17, 18 & 19are tied hence the ranks are

averaged

Hence calculated value of T = ; tabulated value of T

(14,15) =

median birth weights are not same for non-

smokers & smokers

they are significantly different

Mann-Whitney U testLarge-sample approximation:

Use this when n1& n2 are both > 10

Compare to the standard normal distribution

Mann-Whitney U test

Z 2U n1n2

n1n2 n1 n2 1 / 3

• The Kruskal-Wallis H Test is a nonparametric procedure that can be used to compare more than two populations in a completely randomized design.

• All n = n1+n2+…+nk measurements are jointly ranked (i.e. treat as one large sample).

• We use the sums of the ranks of the k samples to compare the distributions.

The Kruskal-Wallis H Test

Rank the total measurements in all k samples

from 1 to n. Tied observations are assigned average of

the ranks they would have gotten if not tied.

Calculate

Ti = rank sum for the ith sample i = 1, 2,…,k

And the test statistic


)1(3)1(

12 2

nn

T

nnH

i

i

H0: the k distributions are identical versus

Ha: at least one distribution is different

Test statistic: Kruskal-Wallis H

When H0 is true, the test statistic H has an

approximate chi-square distribution with df

= k-1.

Use a right-tailed rejection region or p-

value based on the Chi-square distribution.


Example

Four groups of students were randomly

assigned to be taught with four different

techniques, and their achievement test scores

were recorded. Are the distributions of test

scores the same, or do they differ in location?

88628179

67

78

59

3

83

69

75

2

73

87

65

1

80

89

94

4

Teaching Methods

H0: the distributions of scores are the same

Ha: the distributions differ in location

88628179

67

78

59

3

83

69

75

2

73

87

65

1

80

89

94

4

55153531Ti

(14)(2)(11)(9)

(4)

(8)

(1)

(12)

(5)

(7)

(6)

(13)

(3)

(10)

(15)

(16)

96.8)17(34

55153531

)17(16

12

)1(3)1(

12

2222

2

nn

T

nnH

i

i :statistic Test

Rank the 16

measurements

from 1 to 16,

and calculate

the four rank

sums.

Teaching MethodsH0: the distributions of scores are the same

Ha: the distributions differ in location

96.8)17(34

55153531

)17(16

12

)1(3)1(

12

2222

2

nn

T

nnH

i

i :statistic Test

Rejection region: For a right-

tailed chi-square test with =

.05 and df = 4-1 =3, reject H0 if

H 7.81.

Reject H0. There is sufficient

evidence to indicate that there

is a difference in test scores for

the four teaching techniques.

Kruskal-Wallis test Class example

Does it make any difference to students’ comprehension of

statistics whether the lectures are given in teacher centred,

student centred or self study?

Group A: teacher centred;

Group B: Student centred ;

Group C: Self study

DV: student rating of lecturer's intelligibility on 100-point

scale ("0" = "incomprehensible").

Ratings - so use a nonparametric test.

Group A

(raw score)

Group B

(raw score)

Group C

(raw score)

20 25 19

27 33 20

19 35 25

23 36 22

Group A

(raw score)

(rank) Group B

(raw score)

(rank) Group C

(raw score) (rank)

20 3.5 25 7.5 19 1.5

27 9 33 10 20 3.5

19 1.5 35 11 25 7.5

23 6 36 12 22 5

M = 22.25

SD = 3.59

M = 32.25

SD = 4.99

M = 21.50

SD = 2.65

Tc1 (the total for the 1st group) is 20.

Tc2 (for the 2nd group) is 40.5.

Tc3 (for the 3rd group) is 17.5.

)1(3)1(

12 2

nn

T

nnH

i

i

12.613362.58613*12

12

62.5864

5.17

4

5.40

4

202222

H

n

Tc

c

)(

Degrees of freedom are the number of groups minus one. Here, d.f. = 3 - 1 = 2.

Assessing the significance of H depends on the number of participants and the

number of groups.

(a) If you have 3 groups and N in each group is 5 or less:

Special tables exist for small sample sizes – but you really should run more

participants!

(b) If N in each group is larger than 5:

Treat H as Chi-Square.

H is statistically significant if it is larger than the critical value of Chi-Square for

these d.f.

For 2 d.f., a Chi-Square of 5.99 has a p = .05 of occurring by chance.

Our H is bigger than 5.692, and so even less likely to occur by chance!

Conclusion:

The three groups differ significantly;

Table of Chi-square p

n 0.99 0.95 0.50 0.30 0.20 0.10 0.05 0.01

1 0.0002 0.004 0.46 1.07 1.64 2.71 3.84 6.64

2 0.020 0.103 1.39 2.41 3.22 4.00 5.99 9.21

3 0.115 0.35 2.17 3.66 4.64 6.25 7.82 11.34

4 0.30 0.71 3.86 4.88 5.99 7.78 9.49 13.28

5 0.55 1.14 4.35 6.06 7.29 9.24 11.07 15.09

6 0.87 1.64 5.35 7.23 8.56 10.64 12.59 16.81

7 1.24 2.17 6.35 8.38 9.80 12.02 14.07 18.48

8 1.65 2.73 7.34 9.52 11.03 13.36 15.51 20.09

9 2.09 3.32 8.34 10.66 12.24 14.68 16.92 21.67

10 2.56 3.94 9.34 11. 78 13.44 15.99 18.31 23.21

11 3.05 4.58 10.34 12.90 14.63 17.28 19.68 24.72

12 3.37 5.23 11.34 14.01 15.81 18.55 21.03 26.22

13 4.11 5.89 12.34 15.12 16.98 19.81 22.36 27.69

14 4;66 6.57 13.34 16.22 18.15 21.06 23.68 29.14

15 5.23 7.26 14.34 17.32 19.31 22.31 25.00 30.58

16 5.81 7.96 15.34 18.42 20.46 23.54 26.30 32.06

17 6.41 8.67 16.34 19.51. 21.62 24.77 27.59 33.41

18 7.02 9.39 17.34 20.60 22.76 25.99 28.87 34.80

19 7.63 10.12 18.34 21.69 23.90 27.20 30.14 36.19

20 8.26 10.85 19.34 22.78 25.04 28.41 31.41 37.57

Friedman’s test

Effects on worker mood of different types of

music:

Five workers. Each is tested three times, once

under each of the following conditions:

condition 1: silence.

condition 2: "easy-listening‖ music.

condition 3: marching-band music.

DV: mood rating ("0" = unhappy, "10" = very

happy).

Step 1:

Rank each subject's scores individually.

Worker 1's scores are 4, 5, 6: these get ranks of 1, 2, 3.

Worker 4's scores are 3, 7, 5: these get ranks of 1, 3, 2 .

Silence

(raw

score)

Silence

(ranked

score)

Easy

(raw

score)

Easy

(ranked

score)

Band

(raw

score)

Band

(ranked

score)

Wkr 1: 4 1 5 2 6 3

Wkr 2: 2 1 7 2.5 7 2.5

Wkr 3: 6 1.5 6 1.5 8 3

Wkr 4: 3 1 7 3 5 2

Wkr 5: 3 1 8 2 9 3

M = 3.00

SD = 1.52

M = 6.00

SD = 1.14

M = 7.00

SD = 1.58

Silence

(raw

score)

Silence

(ranked

score)

Easy

(raw

score)

Easy

(ranked

score)

Band

(raw

score)

Band

(ranked

score)

Wkr 1: 4 1 5 2 6 3

Wkr 2: 2 1 7 2.5 7 2.5

Wkr 3: 6 1.5 6 1.5 8 3

Wkr 4: 3 1 7 3 5 2

Wkr 5: 3 1 8 2 9 3

rank

total:5.5 11 13.5

Step 2:

Find the rank total for each condition, using the ranks

from all subjects within that condition.

Rank total for ‖Silence" condition: 1+1+1.5+1+1 = 5.5.

Rank total for ―Easy Listening‖ condition = 11.

Rank total for ―Marching Band‖ condition = 13.5.

Step 3: Work out r2

13

1

12 22

CNTc

CCNr

C is the number of conditions.

N is the number of subjects.

Tc2 is the sum of the squared rank totals

for each condition.

To get Tc2 :

(a) square each rank total:

5.52 = 30.25. 112 = 121. 13.52 = 182.25.

(b) Add together these squared totals.

30.25 + 121 + 182.25 = 333.5.

13

1

12 22

CNTc

CCNr

7.64535.333435

122

r

13

1

12 22

÷

CNTc

CCNr

Step 5:

Assessing the statistical significance of r2 depends

on the number of participants and the number of

groups.

r2 = 6.7

Step 4:

Degrees of freedom = number of conditions minus

one. df = 3 - 1 = 2.

Use a Chi-Square table

Compare your obtained r2 value to the critical

value of 2 for your d.f.

If your obtained r2 is bigger than the critical 2

value, your conditions are significantly different.

The test only tells you that some kind of

difference exists; look at the median or mean

score for each condition to see where the

difference comes from.

Our obtained r2 is 6.7.

For 2 d.f., a 2 value of 5.99 would occur by

chance with a probability of .05.

Our obtained value is bigger than 5.99.

Therefore our obtained r2 is even less likely to

occur by chance: p < .05.

Conclusion: the conditions are significantly

different. Music does affect worker mood.

Term Description

j 1, 2, ..., k

k the number of treatment conditions

n the number of blocks

Rj the sum of ranks for treatment j

i 1, 2, ..., m

m the number of sets of ties

ti the number of tied scores in the ithset of ties

If the data have ties, the formula is:

where C is a correction factor that is equal to:

Notation

Exercise: Friedman Test

Six judges with different expertise were asked to

rank 5 different types of ready to drink milk

packets from 1 to 10, where 10 represents a most

prefer. The data were given in the table. Can you

select a best drink according to the rank given by

the judges.

Judge A B C D E1 3.9 4.1 4.2 4.1 3.32 9.4 9.5 9.4 9.0 8.63 9.7 9.3 9.3 9.2 8.44 8.3 8.0 7.9 8.6 7.45 9.8 8.9 9.0 9.0 8.36 9.9 10.0 9.7 9.6 9.1

Rank-Order Correlation

…Spearman’s coefficient of rank correlation reports the

association between two sets of ranked observations

Features

…it can range from –1.00 up to 1.00

…it is similar to Pearson’s coefficient of

correlation, but is based on ranked data

ontinued…

Spearman’s


Formula (to find the coefficient of rank correlation)

d is the difference in the ranks and n is the number of

observations.

rd

n ns

1

6

1

2

2

( )

ontinued…

Spearman’s


Testing the significance of rs

State the null hypothesis:

Rank correlation in population is 0

State the alternate hypothesis:

Rank correlation in population is not 0.

The value of the test statistic is computed from …

t rs

n

rs2

2

1

Spearman’s


The preseason football rankings for the Atlantic Coast Conference

by the coaches and sports writers are shown below. What is the

coefficient of rank correlation between the two groups?

School Coaches Writers

Maryland 2 3

NC State 3 4

NC 6 6

Virginia 5 5

Clemson 4 2

Wake Forest 7 8

Duke 8 7

Florida State 1 1

School Coaches Writers d d2

Maryland 2 3 -1 1

NC State 3 4 -1 1

NC 6 6 0 0

Virginia 5 5 0 0

Clemson 4 2 2 4

Wake Forest 7 8 -1 1

Duke 8 7 1 1

Florida State 1 1 0 0

= 0.905)18(8

)8(61

2

)1(

61

2

2

nn

drs

There is a strong correlation between the ranks of the coaches and

the sports writers!

The hypothesis tested is that prices should decrease with

distance from the key area of surrounding the Contemporary Art

Museum. Following data are from continuous sampling of the

price of a 50ml bottle water at every convenience store.

Convenience

Store

Distance from

CAM (m)

Price of 50ml

bottle (Rs)

1 50 28.0

2 175 22.0

3 270 30.0

4 375 20.0

5 425 20.0

6 580 22.0

7 710 18.0

8 790 16.0

9 890 20.0

10 980 18.5

Conveni

ence

Store

Distance

from CAM

(m)

Rank

distanc

e

Price of

50ml

bottle

(Rs)

Rank

price

Difference

between

ranks (d)

d²

1 50 10 28.0 2 8 64

2 175 9 22.0 3.5 5.5 30.25

3 270 8 30.0 1 7 49

4 375 7 20.0 6 1 1

5 425 6 20.0 6 0 0

6 580 5 22.0 3.5 1.5 2.25

7 710 4 18.0 9 -5 25

8 790 3 16.0 10 -7 49

9 890 2 20.0 6 -4 16

10 980 1 18.5 8 -7 49

d² = 285.5

chap. 14: nonparametric statistics

Documents