introduction to nonparametric...

Introduction toNonparametric Statistics

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Chapter Goals

After completing this chapter, you should beable to:Recognize when and how to use the WilcoxonWilcoxonsigned rank testsigned rank test for a population medianRecognize the situations for which the Wilcoxonsigned rank test applies and be able to use it fordecision-makingKnow when and how to perform a Manna Mann--WhitneyWhitneyUU--testtestPerform nonparametric analysis of variance usingthe KruskalKruskal--Wallis oneWallis one--way ANOVAway ANOVA

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Nonparametric Statistics

Nonparametric StatisticsFewer restrictive assumptions aboutdata levels and underlying probabilitydistributions

Population distributions may be skewedThe level of data measurement may onlybe ordinal or nominal

Data measured on any scaled (nominal,ordinal, ratio, interval)

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Advantages of Nonparametric TestsAdvantages of Nonparametric Tests

1. Used With All Scales2. Easier to Compute

Developed Originally Before WideComputer Use

3. Make Fewer Assumptions4. Need Not Involve

Population Parameters5. Results May Be as Exact

as Parametric Procedures © 1984-1994 T/Maker Co.

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Disadvantages ofNonparametric Tests

1. May Waste InformationIf Data Permit Using ParametricProceduresExample: Converting Data FromRatio to Ordinal Scale

2. Difficult to Compute byHand for Large Samples

3. Tables Not Widely Available

© 1984-1994 T/Maker Co.

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Frequently Used NonparametricTests

Sign Test

MannMann--Whitney UWhitney U--TestTest

WilcoxonWilcoxon MatchedMatched--PairsPairs Signed Rank TestSigned Rank Test

KruskalKruskal WallisWallis HH--TestTest

Spearman’s Rank Correlation Coefficient

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Nonparametric Tests for TwoPopulation Centers

NonparametricTests for Two

Population Centers

WilcoxonMatched-Pairs

Signed Rank Test

Mann-WhitneyU-test

LargeSamples

SmallSamples

LargeSamples

SmallSamples

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Mann-Whitney U-Test

Used to compare two samples from two populations

Assumptions:The two samples are independent and randomThe value measured is a continuous variableThe measurement scale used is at least ordinalIf they differ, the distributions of the two populations willdiffer only with respect to the central location

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Consider two samplescombine into a single list, but keep track ofwhich sample each value came fromrank the values in the combined list from lowto high

For ties, assign each the average rank of the tied values

separate back into two samples, each valuekeeping its assigned rankingsum the rankings for each sample

Mann-Whitney U-Test(continued)

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If the sum of rankings from one samplediffers enough from the sum of rankingsfrom the other sample, we conclude thereis a difference in the population medians

Mann-Whitney U-Test(continued)

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(continued)

Mann-Whitney U-Test

Mann-Whitney U-Statistics

111

211 2)1( TnnnnU

222

212 2)1( TnnnnU

where:n1 and n2 are the two sample sizes

T1 and T2 = sum of ranks for samples 1 and 2

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(continued)

Mann-Whitney U-Test

Claim: Median class size for Math is largerthan the median class size for English

A random sample of 9 Math and 9 Englishclasses is selected (samples do not have tobe of equal size), alfa = 0.05

Rank the combined values and then splitthem back into the separate samples

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Suppose the results are:

Class size (Math, M) Class size (English, E)

234534783466629581

304718344461542840

(continued)

Mann-Whitney U-TestClic

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Size Rank18 123 228 330 434 634 634 640 844 9

Size Rank45 1047 1154 1261 1362 1466 1578 1681 1795 18

Ranking for combined samples

tied

(continued)


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Split back into the original samples:Class size(Math, M) Rank Class size

(English, E) Rank

234534783466629581

210616615141817

304718344461542840

411169131238

= 104 = 67

(continued)


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H0: MedianM = MedianE

HA: MedianM > MedianE

Claim: Median class size forMath is larger than themedian class size for English

221042

(9)(10)(9)(9)T2

1)(nnnnU 111

211

59672

(9)(10)(9)(9)T2

1)(nnnnU 222

212

Note: U1 + U2 = n1n2

(continued)

Mann-Whitney U-Test

Math:

English:

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• One side test

• For one-tailed tests like this one, check thealternative hypothesis to see if U1 or U2 should beused as the test statistic

• Since the alternative hypothesis indicates thatpopulation 1 (Math) has a higher median, use U1 asthe test statistic

• Use U1 as the test statistic: U = 22

• Compare U = 22 to the critical value U from theappropriate table

• If value for U is less than , where U=U1, Reject H0

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• n1 = 9, n2 = 9, U = 22

• So the table value is 0.0567 (> alfa 0.05)

• So the decision is do not reject Ho.

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Mann-Whitney U-Test forLarge Samples

The U statistic approaches a normal distributionas sample sizes increase

If samples (n1 and n2) are larger than 10, anormal approximation can be used

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The mean and standard deviation forMann-Whitney U Test Statistic:

(continued)

221

2

nnU

12)1)()(( 2121

2

nnnnU

Where n1 and n2 are sample sizes from populations 1 and 2

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Normal approximation for Mann-WhitneyU Test Statistic:

(continued)

12)1)()((

22121

212

nnnn

nnUz

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Large Sample Example

We wish to test

Suppose two samples are obtained:n1 = 40 , n2 = 50When rankings are completed, the sum of ranksfor sample 1 is T1 = 1475When rankings are completed, the sum of ranksfor sample 2 is T2 = 2620

H0: Median1 = Median2

HA: Median1 < Median2

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134514752

(40)(41)(40)(50)T2

1)(nnnnU 111

211

65526202

(50)(51)(40)(50)T2

1)(nnnnU 222

212

What’s the decision??

Reject Ho if Z < Z depends on the alternative hypothesis

Note: Normal approximation always use U2

Compute the U statistics:

Large Sample Example(continued)

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Since z = -2.80 < -1.645, we reject H0

645.1z

Reject H0

0MedianMedian:H0MedianMedian:H

21A

210

80.2

12)15040)(50)(40(

1000655

12)1nn)(n)(n(

2nnU

z2121

21

= .05

Do not reject H0

0

Large Sample Example(continued)

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Wilcoxon Matched-PairsSigned Rank Test

The Mann-Whitney U-Test is used whensamples from two populations are independent

If samples are paired, they are not independent

Use Wilcoxon Matched-Pairs Signed Rank Testwith paired samples

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The Wilcoxon T Test Statistic

Performing the Small-Sample WilcoxonMatched Pairs Test (for n < 15)

Calculate the test statistic T using these steps:

Step 1: collect sample data

Step 2: compute di = difference between thesample 1 value and its paired sample 2 value

Step 3: rank the differences, and give each rankthe same sign as the sign of the difference value

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The Wilcoxon T Test Statistic

Performing the Small-Sample WilcoxonMatched Pairs Test (for n < 15)

Step 4: The test statistic is the sum of the absolutevalues of the ranks for the group with the smallerexpected sum

Look at the alternative hypothesis to determinethe group with the smaller expected sumFor two tailed tests, just choose the smallersum

(continued)

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Small Sample Example

Paired samples, n = 9:

Value (before) Value (after)

384534583046425541

304718343431243840

baA

b0

MedianMedian:HMedianMedian:H a

Claim: Medianvalue is smallerafter than before

0MedianMedian:H0Median-Median:H

abA

a0 b

or can be stated as :

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Small Sample Example

Paired samples, n = 9:Value

(before)Value(after)

Differenced

Rankof d

Ranks with smallerexpected sum

364534583046425541

304718543831246240

6-2164-81518-71

4-283-679-51

2

6

5

= T = 13

(continued)

Dipilih jumlah T-

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The calculated T value is T = 13

Complete the test by comparing the calculatedT value to the critical T-value

For n = 9 and = .025 for a one-tailed test,T = 6

Since T > T , do not reject H0

T = 6

T = 13

do not reject H0reject H0

Small Sample Example(continued)

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Wilcoxon Matched Pairs Testfor Large Samples

The T statistic approaches a normaldistribution as sample size increases

If the number of paired values is larger than 15,a normal approximation can be used

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The mean and standard deviation forWilcoxon T :

(continued)

4)1(nn

T

24)12)(1)(( nnn

T

where n is the number of paired values


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Normal approximation for the Wilcoxon TTest Statistic:

(continued)

24)12)(1(

4)1(

nnn

nnTz


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Tests the equality of more than 2 populationmediansAssumptions:

variables have a continuous distribution.the data are at least ordinal.samples are independent.samples come from populations whose onlypossible difference is that at least one mayhave a different central location than theothers.

Kruskal-Wallis One-Way ANOVAClic

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Kruskal-Wallis Test Procedure

Obtain relative rankings for each value

In event of tie, each of the tied values getsthe average rank

Sum the rankings for data from each of the kgroups

Compute the H test statistic

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Kruskal-Wallis Test Procedure

The Kruskal-Wallis H test statistic:(with k – 1 degrees of freedom)

)1N(3nR

)1N(N12H

k

1i i

2i

where:N = Sum of sample sizes in all samplesk = Number of samplesRi = Sum of ranks in the ith sampleni = Size of the ith sample

(continued)

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Complete the test by comparing thecalculated H value to a critical 2 value fromthe chi-square distribution with k – 1degrees of freedom

(The chi-square distribution is Appendix G)Decision rule

Reject H0 if test statistic H > 2

Otherwise do not reject H0

(continued)

Kruskal-Wallis Test ProcedureClic

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Do different departments have different classsizes?

Kruskal-Wallis Example

Class size(Math, M)

Class size(English, E)

Class size(History, H)

2345547866

5560724570

3040183444

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Do different departments have different classsizes?


Class size(Math, M) Ranking Class size

(English, E) Ranking Class size(History, H) Ranking

2341547866

2691512

5560724570

101114813

3040183444

35147

= 44 = 56 = 20

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The H statistic is

(continued)


72.6)115(35

205

565

44)115(15

12

)1N(3nR

)1N(N12H

222

k

1i i

2i

equalareMedianspopulationallotN:H

MedianMedianMedian:H

A

HEM0

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Since H = 6.72 <do not reject H0

(continued)


4877.9205.

Compare H = 6.72 to the critical value from thechi-square distribution for 5 – 1 = 4 degrees offreedom and = .05:

4877.9205.

There is not sufficient evidence to rejectthat the population medians are all equal

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Kruskal-Wallis Correction

If tied rankings occur, give each observation themean rank for which it is tiedThe H statistic is influenced by ties, and shouldbe corrected

Correction for tied rankings:NN

)tt(1 3

g

1ii

3i

where:g = Number of different groups of tiesti = Number of tied observations in the ith tied group of scoresN = Total number of observations

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H Statistic Corrected forTied Rankings

Corrected H statistic:

NN

)tt(1

)1N(3nR

)1N(N12

H

3

g

1ii

3i

k

1i i

2i

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