nonparametric statistical techniques

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Nonparametric StatisticalTechniquesNonparametric StatisticalTechniques

Chapter 17

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The statistical techniques introduced in this chapter deal with ordinal data.

We test to determine whether the population locations differ.

In testing the locations we will not refer to any parameter, thus the procedure’s name.

17.1 Introduction

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When comparing two populations the hypotheses generally are:

17.1 Introduction

H0: The population locations are the same

H1: (i) The locations differ, or(ii) Population 1 is located to the right (left) of

population 2

The random variable X1 is generally larger (smaller) than X2.

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17.2 Wilcoxon Rank Sum Test

The problem characteristics of this test are: The problem objective is to compare two populations. The data are either ordinal or interval (but not normal). The samples are independent.

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Wilcoxon Rank Sum Test – Example

Example 17.1 Based on the two samples shown below, can we infer

at 5% significance level that the location of population 1 is to the left of the location of population 2?

Sample 1: 22, 23, 20; Sample 2: 18, 27, 26;The hypotheses are:

H0: The two population locations are the same.H1: The location of population 1 is to the left of the

location of population 2.

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Graphical DemonstrationWhy use the sum of ranks to test locations?

Sum of ranks = 37Sum of ranks = 41

76 921 3 4 5 8 10 11 12

If the locations of the two populations are about the same, (the null hypothesis is true)we would expect the ranks to be evenly spread between the samples.

In this case the sum of ranks for the two samples will be close to one another.

Two hypothetical populations and their corresponding samples are presented, the GREEN population and the PURPLE population.

Populations

Let us rank the observations of the two samples together

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Allow the GREEN population to shift to the left of the PURPLE population.


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766 7 92


1 3 4 5 8 10 11 1292

The green sample is expected to shift to the left too.As a result, several observations exchange location.

What happens to the sum of ranks? Click.

Attentio

nAttention

Attentio

n




9

6 7


1 3 4 5 8 10 11 1292



The “green” sum decreases , and the “purple” sum increases.Changing the relative location of two populations affect the sum of ranks of the two samples combined.


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Example 17.1 – continued Test statistic

1. Rank all the six observations (1 for the smallest).

Sample 1 22 23 20

Sample 2 18 27 26

RankRank1

6

5

3

4

2

2. Calculate the sum of ranks: 9

2. Calculate the sum of ranks:12

3. Let T = 9 be the test statistic (We arbitrarily define the test statistic as the rank sum of sample 1.


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Example 17.1 - continued If T is sufficiently small then most of the smaller

observations are located in population 1. Reject the null hypothesis.

Question: How small is sufficiently small? We need to look at the distribution of T.

Wilcoxon Rank Sum Test – Rationale

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1,2,3

6 7 8 9 10 11 12 13 14 15

1,2,4 1,2,5 1,2,6

1,3,4

1,3,6

1,3,5 1,4,5

1,4,6 1,5,6

2,3,4 2,3,5

2,3,6

2,4,5

2,4,6

2,5,6

3,4,5

3,4,6

3,5,6 4,5,6

T

.05

.10

.15

T is the rank sum of a sample of size 3.

This sample received the ranks 3, 4, 5

If H0 is true (the two populations have the same location), each ranking is equally likely, and each possible value of T has the same probability = 1/20

This sample received the ranks 1, 2, 3

The distribution of T under H0 for two samples of size 3

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The distribution of T under H0 for two samples of size 3

1,2,3

6 7 8 9 10 11 12 13 14 15

1,2,4 1,2,5 1,2,6

1,3,4

1,3,6

1,3,5 1,4,5

1,4,6 1,5,6

2,3,4 2,3,5

2,3,6

2,4,5

2,4,6

2,5,6

3,4,5

3,4,6

3,5,6 4,5,6

T

.05

.10

.15

The significance level is 5%,and under H0 P(T 6) = .05.Thus, the critical value of T is 6.

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• Example 17.1 - continued

Conclusion

H0 is rejected if TSince T = 9, there is insufficient evidence to conclude that population 1 is located to the left of population 2, at the 5% significance level.


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Critical values of the Wilcoxon Rank Sum Test

n2 n13 4 5 . . . 10

4 6 18 11 25 17 33 . . . 61 895 6 21 12 28 18 37 . . . 64 96...

10 9 33 16 44 24 56 . . . 79 131

= .025 for two tail test, or = .05 for one tail test

Using the table: For given two samples of sizes n1 and n2, P(T<TL)=P(T>TU)=

For a two tail test: P(T<11) = P(T>25) = .025 if n1=4 and n2=4.For a one tail test: P(T<11) = P(T>25) = .05 if n1=4 and n2=4.

11 25

A similar table exists for = .05 (one tail test) and = .10 (two tail test)

TL TU TL TU TL TU TL TU

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Wilcoxon rank sum test for samples where n > 10

• The test statistic is approximately normally distributed with the following parameters:

n1(n1 + n2 + 1)2

E(T) =

12)1nn(nn 2121

T

Therefore,

Z =T - E(T)

T

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• Example 17.2 (using Wilcoxon rank sum test with ordinal data)

A pharmaceutical company is planning to introduce a new painkiller.

To determine the effectiveness of the drug, 30 people were randomly selected.

15 were given the tested drug (Sample 1). 15 were given aspirin (Sample 2).

Each participant was asked to indicate which one of five statements best represented the effectiveness of the drug they took.

Wilcoxon rank sum test for samples where n > 10, Example

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Example 17.2 – continued Summary of the experiment results.

The drug taken was… Painkiller Aspirinextremely effective (5) 6 1quite effective (4) 3 5somewhat effective (3) 4 3slightly effective (2) 1 4not at all effective (1) 1 2

SolutionThe objective is to compare two populations of ordinal data.The two samples are independent.Wilcoxon rank test is the appropriate technique to apply.

Wilcoxon test for samples where n > 10, Example

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The hypothesesH0: The locations of population 1 and 2 are the same

H1: The location of population 1 is to the right of the location

of population 2.Note: A high score selected from among the five possible scores 1, 2, 3, 4, 5, indicates high effectiveness.


Received the new painkiller Received Aspirin

Solving by hand To reject the null hypothesis, we need to show that z is

“large enough”. First we rank the observations, Secondly, we run a z-test, with rejection region of Z > Z.

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Ranking the raw data

Painkiller Rank Aspirin Rank1 2 1 22 6 1 23 12 2 63 12 2 63 12 2 63 12 2 64 19.5 3 124 19.5 3 124 19.5 3 125 27 4 19.55 27 4 19.55 27 4 19.55 27 4 19.55 27 4 19.5

There are three observationswith an effectiveness score of 1.

The original ranks for these observations are 1, 2 , and 3.This tie is broken by giving eachobservation the average rank of 2.

Sum of ranks: T1=276.5 T2=188.5

These are the effectiveness scores provided by the experiment participants for each drug.


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To standardize the test statistic we need:

E(T) = n1(n1+n2+1)/2= (15)(31)/2=232.5

1.2412

)1nn(nn 2121T

83.1)T(ET

zT

83.1)T(ET

zT


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For 5% significance level z=1.645.Since z = 1.83 > 1.645, there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At 5% significance level,

the new drugs is perceived as more effective than Aspirin.


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Excel solution (Xm17-02)


Wilcoxon Rank Sum Test

Rank sum ObservationsNew 276.5 15Aspirin 188.5 15z Stat 1.83P(Z<=z) one-tail 0.034z Critical one-Tail 1.6449P(Z<=z) two-tail 0.068z Critical two-Tail 1.96

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Wilcoxon rank sum test for non-normal interval data, Example

The human resource manager of a large company wanted to compare how long business and non-business graduates worked for the company before quitting.

Two samples of 25 business graduates and 20 non-business graduates were randomly selected.

The data representing their time with the company were recorded.

• Retaining Workers

25

Business Non-Bus60 2511 6018 2219 245 2325 36. .. .. .

Can the personnel manager conclude at 5% significance level that a difference in duration of employment exists between business and non-business graduates?

• Retaining workers - continued


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Solution The problem objective is to compare two populations of

interval data. The samples are independent. The non-normality of the two populations is apparent from the

sample histograms:

02468

101214

5 20 35 50 65 More

0

2

4

6

8

10

15 25 35 45 55 65 More

Non Business graduates Business graduates


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Solution – continued The Wilcoxon rank test is the correct procedure to run.

H0: The two population locations are the same

H1: The location of population 1(business graduates) is different from the location of population 2

(non- business graduates).


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Solution – continued Solving by hand

The rejection region is After the ranking process is completed, we have:

T = Tbusiness graduates = 463.

E(T) = n1(n1+n2+1)/2=575;

T=[n1n2(n1+n2+1)/12]1/2=43.8

96.1zzz 025.2/

56.28.43575463)T(ET

zT

Reject the null hypothesis


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Excel solution (Workers.xls)

There is a strong evidence to infer that the duration of employment is different for business and non-business graduates


Wilcoxon Rank Sum Test

Rank Sum ObservationsBusiness 463 25Non-Bus 572 20z Stat -2.56P(Z<=z) one-tail 0.0053z Critical one-tail 1.6449P(Z<=z) two-tail 0.0106z Critical two-tail 1.96

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Required conditions for nonparametric tests

A rejection of the null hypothesis when performing a nonparametric test can occur due to: different location different spread (variance) different shape (distribution).

Since we are interested in the location, we require that the two distributions are identical, except for location.

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17.3 Sign Test and Wilcoxon Signed Rank Sum Test

Two techniques for matched pairs experiment are introduced.

the objective is to compare two populations. the data are either ordinal or interval (but not

normal). The samples are matched by pairs.

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The Sign Test

This test is employed when: The problem objective is to compare two populations, and The data are ordinal, and The experimental design is matched pairs.

The hypothesesH0: The two population locations are the sameH1: The two population locations differ or population 1

is right (left) of population 2

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The Sign Test –Statistic and Sampling Distribution

A matched pair experiment calls for a test of matched pair differences.

The test statistic and sampling distribution Record the sign of all the matched-pair-differences. The number of positive (or negative) differences is the

test statistic.

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The number of positive or negative differences is binomial, with: n = the number of non-zero differences p = the probability that a difference is positive (negative)

If the two populations have the same locations (H0 is true), it is expected that

Thus, under H0: p = 0.5

Number of positive differences = Number of negative differences

The Sign Test - Rationale

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The test statistic and sampling distribution The hypotheses:

H0: The two population locations are the same

H1: The two population locations are different

The Sign Test - Rationale

H0: p .5H1: p .5

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The Test – continued The hypotheses tested

H0: p .5H1: p .5

The binomial variable can be approximated by a normal variable if np and n(1-p) > 5.The Z- statistic becomes

.10nwhere

n5.

n5.x

)5)(.5(.n

n5.x

)p1(np

npxz

.10nwhere

n5.

n5.x

)5)(.5(.n

n5.x

)p1(np

npxz

The Sign Test –Statistic and Sampling Distribution

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Example 17.3 (Xm17-03) In an experiment to determine which car is perceived

to have the more comfortable ride, 25 people took two rides:

One ride in a European model. One ride in a North American car.

Each person ranked the cars on a scale of 1 (ride is very uncomfortable) to 5 (ride is very comfortable).

The Sign Test – Example

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Respondent European American1 4 52 2 13 5 44 3 25 2 16 5 37 1 38 4 29 4 2. . .

Do these data allow us to conclude at 5% significance level that the European car is perceived to be more comfortable?


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• SolutionRespondent European American

1 4 52 2 13 5 44 3 25 2 16 5 37 1 38 4 29 4 2. . .

Difference-111112

-222.

• We compare two populations• The data are ordinal• A matched pair experiment


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• Solution– The hypotheses are:

H0: The two population location are the same. H1: The European car population is located to the

right of the American car population.– The test.

• There were 18 positive, 5 negatives, and 2 zero differences. Thus, X = 18, n = 23(!).

• Z = [x-np]/[np(1-p)].5 = [18-.5(23)]/[.5{23}.5] = 2.71• The rejection region is z > z. For = .05 we have z > 1.645.

The p-value = P(Z > 2.71) = .0034


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Using the computer: Tools > Data Analysis Plus > Sign Test

• Excel – Solution (Xm17-03)


123456789

101112

A B C D E FSign Test

Difference European - American

Positive Differences 18Negative Differences 5Zero Differences 2z Stat 2.71P(Z<=z) one-tail 0.0034z Critical one-tail 1.6449P(Z<=z) two-tail 0.0068z Critical two-tail 1.96

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Conclusion: Since the p-value < we reject the null hypothesis.At 5% significance level there is sufficient evidence to infer that the European car is perceived as more comfortable than the American car.


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Checking the required conditions Observe the sample histograms (Xm17-03)

The populations are similar in shape and spread

European cars

0

2

4

6

8

1 2 3 4 5 More

American cars

0

2

4

6

8

1 2 3 4 5 More


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This test is used when the problem objective is to compare two populations, the data are interval but not normal, the samples are matched pairs.

The test statistic and sampling distribution T is based on rank sum of the absolute values of the

positive and negative differences When n <=30, reject H0 if T>TU or T<TL(TL and TU tabulated

values related to n). When n > 30, T is approximately normally distributed.

Use a Z-test.

Wilcoxon Signed Rank Sum Test

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Example 17.4 Does “flextime” work-schedule help reduce the travel

time of workers to work? A random sample of 32 workers was selected, and

workers recorded their travel time before and after the program was implemented.

The hypotheses test are The two population locations are the same. The two population locations are different.

Wilcoxon Signed Rank Sum Test,Example

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Example 17.4 Does “flextime” work-schedule help reduce the travel time of

workers to work? A random sample of 32 workers was selected, and workers

recorded their travel time before and after the program was implemented.

The hypotheses are H0: The two population locations are the same. H1: The two population locations are different.

The rejection region:|z| > z

The rejection region:|z| > z

Wilcoxon Signed Rank Sum Test, Example

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Worker 8:00-Arr Flextime Difference ABS(Diff.) Ranks3 43 44 -1 1 4.55 16 15 1 1 4.58 38 39 -1 1 4.5

12 13 12 1 1 4.516 18 19 -1 1 4.523 19 18 1 1 4.527 51 50 1 1 4.530 20 19 1 1 4.5

4 46 44 2 2 136 26 28 -2 2 139 61 63 -2 2 13

10 52 54 -2 2 1313 69 71 -2 2 1315 53 55 -2 2 1318 25 23 2 2 1328 40 38 2 2 1331 19 21 -2 2 13

1 34 31 3 3 2111 68 65 3 3 2117 41 38 3 3 21

This data were sorted by the absolute valueof the differences.

12345678

Ties were broken by assigning the average rank to the tied observations

Average rank =(1 + 8)/2 = 4.5

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Worker 8:00-Arr Flextime Differtence ABS(Diff.) Ranks3 43 44 -1 1 4.55 16 15 1 1 4.58 38 39 -1 1 4.5

12 13 12 1 1 4.516 18 19 -1 1 4.523 19 18 1 1 4.527 51 50 1 1 4.530 20 19 1 1 4.5

4 46 44 2 2 136 26 28 -2 2 139 61 63 -2 2 13

10 52 54 -2 2 1313 69 71 -2 2 1315 53 55 -2 2 1318 25 23 2 2 1328 40 38 2 2 1331 19 21 -2 2 13

1 34 31 3 3 2111 68 65 3 3 2117 41 38 3 3 21

T is the rank sum of the positive differences. T = T+ = 367.5

E(T) = n(n+1)/4 = 32(33)/4 = 264T = [n(n+1)(2n+1)/24].5 = 53.48

The test statistic is:

Z = T

E(T)TT

E(T)T

53.48264= 1.94367.5 -

=

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Excel – solution (Xm17-04)


Wilcoxon Signed Rank Sum Test

Difference 8:00-Arr - Flextime

T+ 367.5T- 160.5Observations (for test) 32z Stat 1.94P(Z<=z) one-tail 0.0265z Critical 1.6449P(Z<=z) two-tail 0.053z Critical two-tail 1.96

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The rejection region for = .05 is |z| > z.025 = 1.96

Conclusion: Since |1.94| < 1.96, There is insufficient evidence to infer that the flextime program was effective at 5% significance level.

Solution – continued


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17.4 Kruskal-Wallis Test

The problem characteristics for this test are: The problem objective is to compare two or more populations. The data are either ordinal or interval but not normal. The samples are independent.

The hypotheses areH0: The location of all the k populations are the same.

H1: At least two population locations differ.

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Rank the data from 1(smallest) to n (largest). Calculate the rank sums T1, T2,…Tk for all the k

samples. Calculate the statistic H as follows:

)1(3)1(

12

1

2

nn

T

nnH

k

j j

j)1(3

)1(

12

1

2

nn

T

nnH

k

j j

j

Kruskal-Wallis Test Statistic

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Test Rationale and Rejection region

If all the populations have the same location (H0

is true)… The ranks should be evenly distributed among the k

samples. The statistic H will be small.

Uneven distribution of ranks1 4 72 5 83 6 9

T1=6 T2=15 T3=24H = 7.2

Even distribution of ranks1 2 34 5 69 8 7

T1=14 T2=15 T3=16H = .0888

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Sampling distribution When the sample sizes 5, H is approximately

chi-squared distributed with k-1 degrees of freedom. The rejection region:

Since a large value of H justifies the rejection of H0,

we have:2

1k,H 21k,H

Test Rationale and Rejection Region

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Example 17.5 How do customers rate three shifts with respect to

speed of service in a certain restaurant? Three samples of 10 customer response-cards were

randomly selected, one sample from each shift. Customer ratings were recorded.

The Kruskal-Wallis Test Example

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4:00-mid Mid-8:00 8:00-4:004 3 34 4 13 2 34 2 23 3 13 4 33 3 43 3 22 2 43 3 1

Can we conclude that customers perceive the speed of service to be different among the three shifts at 5% significance level?


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Solution The problem objective is to compare three

populations. The data are ordinal. The hypotheses:

H0: The locations of all three populations are the same.

H1: At least two population locations differ.


58

Solution - continued Test statistic:

4:00-mid Mid-8:00 8:00-4:004 3 34 4 13 2 34 2 23 3 13 4 33 3 43 3 22 2 43 3 1

27

27

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

16.5

27

27

27

27

27

2

2

2

6.5

6.5

6.5

6.56.5

6.5

1

3

2

3

4

7

5

6

8

T1 = 186.5 T2 = 156.0 T3 = 122.5

n = n1 + n2 + n3 = 10+10+10 = 30

Ranking

64.2

)130(3

10

5.122

10

0.156

10

5.186

)130(30

12

)1(3)1(

12

222

2

n

n

T

nnH

j

j


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For = .05, 2,k-1 = 2

.05,2 = 5.99147

Solution - continued The critical value


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Solution – Excel (Xm17-05)

123456789

1011

A B C DKruskal-Wallis Test

Group Rank Sum Observations4:00-mid 186.5 10Mid-8:00 156 108:00-4:00 122.5 10

H Stat 2.64df 2p-value 0.2665chi-squared Critical 5.9915

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Conclusion: Since H=2.64 < 5.99147, do not reject the null hypothesis. There is insufficient evidence to conclude at 5% significance level, that there is a difference in customers’ perception regarding service speed among the three shifts.


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17.5 Friedman Test The problem characteristics of this test are:

The problem objective is to compare two or more populations. The data are either ordinal or interval but not normal. (For

normal populations we use ANOVA). The data are generated from a blocked experiment (samples

are not independent). The hypotheses are

The location of all the k populations are the same. At least two population locations differ.

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Test Statistic and Rejection Region

The test statistic is

The rejection region is

)1k(b3T)1k(bk

12F

k

1j

2jr

)1k(b3T)1k(bk

12F

k

1j

2jr

21k,rF 21k,rF

b = the number of blocksK = the number of treatments

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The Friedman Test Example

Example 17.6 Four managers evaluate applicants for a job in an

accounting firm on several dimensions. Eight applicants were randomly selected, and their

evaluations by the four managers recorded. Manager

Applicant 1 2 3 41 2 1 2 22 4 2 3 23 2 2 2 34 3 1 3 25 3 2 3 56 2 2 3 47 4 1 5 58 3 2 5 3

Can we conclude at 5% significance level thatthere are differences inthe way managersevaluate candidates?

Can we conclude at 5% significance level thatthere are differences inthe way managersevaluate candidates?

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Solution The problem objective is to compare four populations Data are ordinal. This is a randomized block design experiment

because each applicant (block) was ranked four times. The appropriate procedure is the Friedman test


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Solution The hypotheses areH0: The locations of all four populations are the same.

H1: At least two population locations differ. The data

ManagerApplicant 1 2 3 4

1 2 1 2 22 4 2 3 23 2 2 2 34 3 1 3 25 3 2 3 56 2 2 3 47 4 1 5 58 3 2 5 3


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ManagerApplicant 1 2 3 4

1 2 1 2 22 4 2 3 23 2 2 2 34 3 1 3 25 3 2 3 56 2 2 3 47 4 1 5 58 3 2 5 3

T1 = 21 T2 = 10 T3 = 24.5 T4 = 24.5

34

23.52.51.522.5

11.5

211.5111

33

23.52.533.54

31.5

42443.52.5


How to rank, block by block.

Applicant 1:

Scores: 2 1 2 2

Actual ranks: 2 1 3 4

Averaged ranks: 3 1 3 3

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SolutionIn our problem:b = 8 (number of blocks) k = 4 (number of treatments, populations)


61.10)14(35.244.241021)14)(4(8

12

)1(3)1(

12

2222

1

2

kbTkbk

Fk

jjr

69

Solution

We have : Fr = 10.61; Let = .05, then 2.05, 4-1 = 7.8147


70


Solution – Excel (Xm17-06)

Friedman Test

Group Rank SumManager1 21Manager2 10Manager3 24.5Manager4 24.5

Fr Stat 10.61df 3p-value 0.0140chi-squared Critical 7.8147

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Conclusion: Since Fr =10.61> 7.8147, reject the null hypothesis. There is sufficient evidence to conclude at 5% significance level, that the managers’ evaluations differ.


nonparametric statistical techniques

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