Non Parametric Presentation

• Friedman Test&• Quade Test

Group Members

Iqra Tanveer (05)Anwar Ul Haq (28)Ameer Umar Khan (30)Irfan Hussain (39)Ammar Ahmad Khan (45)

Learning Objectives• History• Introduction• Assumptions• General Procedure• Applications• Advantages• Disadvantages• Example

History• Friedman,

Milton (December 1937). • Friedman, Milton (March

1940).• “A comparison of

alternative tests of significance for the problem of m rankings“.

• Kendall, M. G. Rank Correlation Methods. (1970) London.

History• Hollander, M., and Wolfe, D. (1973). New

York.• Siegel, Sidney, and Castellan, N. John

Nonparametric Statistics for the Behavioral Sciences. (1988).

Introduction

Introduction

Introduction

• The Friedman test is a non-parametric statisticaltest developed by the U.S. economist MiltonFriedman.

• Friedman test is a non-parametric randomizedblock analysis of variance.

• Similar to the parametric repeatedmeasures ANOVA. It is used to detect differencesin treatments across multiple test attempts.

Introduction• The procedure involves ranking each row

(or block) together, then considering the valuesof ranks by columns. Applicable to completeblock designs.

• It is thus a special case of the Durbin test.• The Friedman test is used for one-way repeated

measures analysis of variance by ranks. In its useof ranks it is similar to the Kruskal-Wallis one-way analysis of variance by ranks.

Introduction

• It is non-parametric test in which we use the f-table for critical region.

Assumptions• Data should consist of three or more than three

samples.• Data should be consist of random samples from

population.• All samples data should be independent. • Measurement scale should be at least ordinal.• Variable of interest should be continuous.• Data need not be normally distributed.• Within each block the observation may be rank

according to some criteria of interest.

General Procedure

1. Null & Alternative Hypothesis.Ho: The distributions (whatever they are)

are the same across repeated measures.H1: The distributions across repeated

measures are different.2. Level Of Significance.

α = 0.01, 0.05, 0.01…. (chose as required

for test)

General Procedure

3. Test StatisticsStep 1If no ties occur in data.

T1 =12

bk(k + 1)

j=1

k

Rj −b(k + 1)

2

2

General ProcedureIf ties occur in data

T1 =(k − 1) j=1

k Rj2 − bC1

A1 − C1Where

A1 = i=1b j=1

k R(xij)2

And

C1 =bk(k + 1)2

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General ProcedureStep 2

Put the value of 𝑇1 into this equation

𝑇2 =(𝑏−1)𝑇1

𝑏(𝑘−1)𝑇1

Where 𝑇2 ↝ 𝐹 𝑘1, 𝑘2 𝛼And𝑘1=k-1

𝑘2=(k-1)(b-1)

General Procedure4. Calculation5. Critical Region

if 𝑇2> 𝐹 𝑘1, 𝑘2 𝛼 Reject Ho6. DecisionFrom the provided evidence as our calculated value is …….. So we …… and conclude that…….

ApplicationsThis can be used to perform the testing in every field, where comparison between variables is required.1. Used to compare the effects of same fertilizer in

different patches of field having different fertility levels. (In agricultural Field)

2. Comparison between different companies cold drinks.

3. Test the equality of difference car’s engines performance. (In industries)

Applications4. Comparison between the average

performance of players. (Games)5. Comparison of different pain-killer tablets

average effect.So this is valuable test used as non-

parametric test of multiple comparison. Where data is not normally distributed. That is assumption of normality is violated.

Advantages

1. Since the Friedman test ranks the values in each row, it is not affected by sources of variability that equally affect all values in a row (since that factor won't change the ranks within the row).

2. The test controls experimental variability between subjects, thus increasing the power of the test.

Disadvantage

• Since this test does not make a distribution assumption, it is not as powerful as the ANOVA.

ExampleA B C D4 3 2 14 2 3 13 1.5 1.5 43 1 2 44 2 1 32 2 2 41 3 2 42 4 1 3

3.5 1 2 3.54 1 3 24 2 3 1

3.5 1 2 3.538 23.5 34.5 30

Ranked Data

Solution1. Null & Alternative Hypothesis.

Ho: The distributions (whatever they are) are the same across repeated measures.H1: The distributions across repeated measures are different.

2. Level Of Significance.α=0.05

Solution3. Test StatisticsBecause ties occur in data so….

𝑇1 =(𝑘 − 1) 𝑗=1

𝑘 𝑅𝑗2 − 𝑏𝐶1

𝐴1 − 𝐶1

4. Calculationcalculate the 𝑨𝟏 & 𝑪𝟏

SolutionAs we know that:

𝐴1 =

𝑖=1

𝑏

𝑗=1

𝑘

𝑅(𝑥𝑖𝑗)2

And

𝐶1 =𝑏𝑘(𝑘+1)2

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By using these formulas we calculate that𝐴1=456.5 𝐶1=300

Solution

𝑇1 =(𝑘 − 1) 𝑗=1

𝑘 𝑅𝑗2 − 𝑏𝐶1

𝐴1 − 𝐶1

𝑇1 = 8.097

Put 𝑇1 in 𝑇2 formula

𝑇2 =(𝑏−1)𝑇1

𝑏(𝑘−1)𝑇1=5.6006

SolutionCritical Value

𝑇2 ↝ 𝐹 𝑘1, 𝑘2 𝛼

And𝑘1=k-1

𝑘2=(k-1)(b-1)CR=2.8742

SolutionDecision:

From the provided evidence as our calculated is greater than our tabulated value. so, we will reject the H0 and hence concluded that the distributions across repeated measures are different.

Multiple Comparison Test

𝑅𝑗 − 𝑅𝑖 ≥ 𝑡1− 𝛼 2𝐴1−𝐶1

(𝑏−1)(𝑘−1)1 −

𝑇1

𝑏(𝑘−1)

1 2

with d.f 𝑡(𝑘−1)(𝑏−1)• If ties Occur

𝐴1 − 𝐶1 =𝑏𝑘(𝑘+1)(𝑘−1)

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