non parametric tests-marketing reasearch

8/4/2019 Non Parametric Tests-Marketing Reasearch

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ChapterNonparametric Methods

Sign Test

RANK _SUM -TESTS

1 Mann-Whitney U Test

2 Kruskal-Wallis Test

Rank Correlation


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Most of the statistical methods referred to as

parametric require the use of interval- or ratio-scaleddata.

Nonparametric methods are often the only way toanalyze nominal or ordinal data and draw statistical

conclusions. Nonparametric methods require no assumptions

about the population probability distributions.

Nonparametric methods are often called distribution-

free methods.

Nonparametric Methods


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Nonparametric Methods

In general, for a statistical method to be classified as

nonparametric, it must satisfy at least one of thefollowing conditions.

The method can be used with nominal data.

The method can be used with ordinal data.

The method can be used with interval or ratio datawhen no assumption can be made about thepopulation probability distribution.


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Sign Test

A common application of the sign test involves using

a sample of n potential customers to identify apreference for one of two brands of a product.

The objective is to determine whether there is adifference in preference between the two items being

compared. To record the preference data, we use a plus sign if

the individual prefers one brand and a minus sign ifthe individual prefers the other brand.

Because the data are recorded as plus and minussigns, this test is called the sign test.


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Example: Peanut Butter Taste Test

Sign Test: Large-Sample Case

As part of a market research study, a sample of 36

consumers were asked to taste two brands of peanut

butter and indicate a preference. Do the data shown

below indicate a significant difference in the consumerpreferences for the two brands?

18 preferred Hoppy Peanut Butter (+ sign recorded)

12 preferred Pokey Peanut Butter (_sign recorded)

6 had no preferenceThe analysis is based on a sample size of 18 + 12 = 30.


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When H0 is true

p=0.5 q=0.5

= np = 30( 0.5) =15 = npq = 30(0.5)(0.5) = 7.5 = 2.24

x =18 = number preferring Hoppy Peanut Butter In the sample z = (x - )/ follows standard normal distribution


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Rejection Rule

Using .05 level of significance,Reject H0 if z < -1.96 or z > 1.96

Test Statistic

z = (18 - 15)/2.74 = 3/2.74 = 1.095

ConclusionDo not reject H0. There is no difference in

preference for the two brands of peanut butter.

Fewer than 10 or more than 20 individuals

would have to have a preference for a particularbrand in order for us to reject H0.


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SIGN TEST EXAMPLE

Q In a sample of 300 shoppers, 160 indicated they

prefer fluoride toothpaste, 120 favored non fluoride,and 20 were indifferent. At a 0.05 level of significance,use sign test for testing a difference in the preferencefor the two kinds of toothpaste.

160 preferred Fluoride Toothpaste(+ sign recorded)120 preferred Non Fluiride Toothpaste(_sign recorded)

20 had no preference

The analysis is based on a sample size of 160 + 120 = 280.


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Hypotheses

H0: No preference for one brand over the other exists

Ha: A preference for one brand over the other exists

Sampling Distribution

2.74

Sampling distributionof the number of +values if there is nobrand preference

= 140 = .5(280)

Example: SIGN Test


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Example: SIGN Test

When H0 is true

p=0.5 q=0.5

= np = 280( 0.5) =140 = npq = 280(0.5)(0.5) = 70 = 8.36

x =160 = number preferring Fluoiride Toothpaste In the sample z = (x - )/ follows standard normal distribution


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Example: SIGN Test

Rejection Rule


Test Statistic

z = (160 - 140)/8.36 = 20/8.36 = 2.39

ConclusionReject H0. There is sufficient evidence in the

sample to conclude that a difference in preferenceexists for the two brands of Tooth Pastes.


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Mann-Whitney-U Test

This test is another nonparametric method for

determining whether there is a difference betweentwo populations.

This test does not require interval data or theassumption that both populations are normally

distributed. The only requirement is that the measurement scale

for the data is at least ordinal.


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14Slide

Mann-Whitney-U-Test

Instead of testing for the difference between the

means of two populations, this method tests todetermine whether the two populations are identical.

The hypotheses are:

H0: The two populations are identicalHa: The two populations are not identical


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15Slide

Example: Westin Freezers

Mann-Whitney-U- Test (Large-Sample Case)

Manufacturer labels indicate the annual energycost associated with operating home appliances suchas freezers.

The energy costs for a sample of 10 Westin

freezers and a sample of 10 Brand-X Freezers areshown on the next slide. Do the data indicate, usinga= .05, that a difference exists in the annual energycosts associated with the two brands of freezers?


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16Slide


Westin Freezers Brand-X Freezers

$55.10 $56.10

54.50 54.70

53.20 54.40

53.00 55.4055.50 54.10

54.90 56.00

55.80 55.50

54.00 55.0054.20 54.30

55.20 57.00


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HypothesesH0: There is no difference between the two

populations. So Annual energy costs for Westinfreezers and Brand-X freezers are the same.

Ha: There is difference between the twopopulations. In particular they have different means.Annual energy costs differ for the two brands offreezers.


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18Slide

First, rank the combined data from the lowest to

the highest values, with tied values being assignedthe average of the tied rankings.

Then, compute R1, the sum of the ranks for the firstsample.

Then, compute R2, the sum of the ranks for thesecond sample.

Let n 1 be number of observations in the first sample

Let n 2 be number of observations in the secondsample

Mann-Whitney-Wilcoxon Test:Large-Sample Case


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19Slide

Compute Test Statistics

U= {n1 n2 + n1 ( n1 + 1 )/ 2 } - R1

u = n1 n2 /2 u = n1 n2 ( n1 + n2 + 1 )/12

Test Statistics: Z= (U u )/ u


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20Slide

Sampling Distribution of Ufor Identical Populations

Mean

u

= n1n1 /2

Standard Deviation

Distribution Form

Approximately normal, providedn1 > 10 and n2 > 10

Mann-Whitney-Wilcoxon Test:Large-Sample Case

1 2 1 21 ( 1)12T n n n n


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21Slide


Westin Freezers Rank Brand-X Freezers Rank$55.10 12 $56.10 19

54.50 8 54.70 9

53.20 2 54.40 7

53.00 1 55.40 1455.50 15.5 54.10 4

54.90 10 56.00 18

55.80 17 55.50 15.5

54.00 3 55.00 1154.20 5 54.30 6

55.20 13 57.00 20

Sum of Ranks 86.5 Sum of Ranks 123.5


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22Slide

R 1= 86.5

R 2= 123.5

n1 = 10

n 2 = 10

U ={ (10) ( 10) + 10( 11)/2 ) -86.5 =(100+55) -86.5

=155-86.5

= 68.5


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Sampling Distribution

u13.23

Sampling distribution

of U if populationsare identical

u

= 50 =(1/2)(10)(10)T


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Rejection Rule

Using .05 level of significance,

Reject H0 if z < -1.96 or z > 1.96

Test Statistic

z = (U-

u

)/

u

= (68.5 - 50)/13.23 = 1.40 Conclusion

Do not reject H0. There is no difference in theannual energy cost associated with the two brands of

freezers.


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Kruskal Wallis Test


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26Slide

Kruskal-Wallis-Test

Kruskal-Wallis-Test

HypothesesH0: There is no difference between the K

populations..

Ha: There is a difference between the Kpopulations. In particular they have different means.

K k l W lli T


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27Slide

First, rank the combined data from the lowest to

the highest values, with tied values being assigned theaverage of the tied rankings.

Then, compute R1, the sum of the ranks for the firstsample.

Then, compute R2, the sum of the ranks for the secondsample. And so on upto p th sample

Then , compute Rp , the sum of the ranks for the

p th sample.

Let n 1 be number of observations in the first sample

Let n 2 be number of observations in the second sample

Let n p be number of observations in the pth sample

Kruskal-Wallis-TestLarge-Sample Case

Kruskal Wallis Test


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28Slide

Kruskal-Wallis-Test

Compute Test Statistics

Test Statistics is K= {12/n(n+1) } Rj2 /nj - 3(n+1) K follows a Chi Square Distribution with p-1 d.f

when all sample sizes are at least 5 and p is thenumber of groups

Kruskal Wallis Test


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29Slide

Kruskal-Wallis-Test

A sample of 20 pilot students were divided into 3

groups of 6, 5 and 9.For the purpose of their writtentests the first group was trained through use of videocassettes , the second through use of Audio cassettesand the third group through classroom teaching. The

scores of the students in the three groups are asfollows.

Video-74, 88,82,93,55,70

Audoo-78,80,65,57, 89

Classroom-68,83,50,91,84,77,94,81,92 Test the hypothesis that the mean scores of stundent

pilots trained by each of these three methods is equal


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30Slide

Video-74(7), 88(15),82(12),93(19),55(2),70(6)

Audoo-78(9),80(10),65(4),57(3), 89(16)

Classroom-68(5),83(13),50(1),91(17),84(14),77(8),94(20),81(11),92(18)

R1 = 61 n1 = 6

R2= 42 n2 = 5

R3 = 107 n3 = 9 n = 20 p= 3

K= {12/n(n+1) } Rj2 /nj - 3(n+1) Test Statistics : K= 1.143

Follows Chi Square distribution with 3-1=2 d.f

Table value is 4.605 .Since 1.143 less than 4.065 we accept thehypothesis & conclude that there is no difference in the threeteaching methods


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31Slide

Rank Correlation

The Pearson correlation coefficient, r, is a measure of

the linear association between two variables forwhich interval or ratio data are available.

The Spearman rank-correlation coefficient, rs , is ameasure of association between two variables when

only ordinal data are available. Values of rs can range from 1.0 to +1.0, where

values near 1.0 indicate a strong positiveassociation between the rankings, and

values near -1.0 indicate a strong negativeassociation between the rankings.


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32Slide

Rank Correlation

Spearman Rank-Correlation Coefficient, rs

where: n = number of items being rankedxi = rank of item i with respect to one variable

yi = rank of item i with respect to a secondvariable

di = xi - yi

2

2

61

( 1)

i

s

dr

n n


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33Slide

Test for Significant Rank Correlation

We may want to use sample results to make an

inference about the population rank correlationps. To do so, we must test the hypotheses:

H0: ps = 0

Ha: ps = 0


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34Slide

Sampling Distribution ofrswhenps = 0

Mean

Standard Deviation

Distribution Form

Approximately normal, provided n > 10

Rank Correlation

0sr

1

1sr

n


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35Slide

Example: Connor Investors

Rank Correlation

Connor Investors provides a portfoliomanagement service for its clients. Two of Connorsanalysts rated ten investments from high (6) to low(1) risk as shown below. Use rank correlation, with

a= .10, to comment on the agreement of the twoanalysts rankings.

Investment A B C D E F G H I J

Analyst #1 1 4 9 8 6 3 5 7 2 10Analyst #2 1 5 6 2 9 7 3 10 4 8


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36Slide

Analyst #1 Analyst #2

Investment Ranking Ranking Differ. (Differ.)2

A 1 1 0 0B 4 5 -1 1C 9 6 3 9

D 8 2 6 36E 6 9 -3 9F 3 7 -4 16G 5 3 2 4H 7 10 -3 9I 2 4 -2 4

J 10 8 2 4Sum = 92



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Rejection Rule


Test Statistic

z = (rs - r)/r= (.4424 - 0)/.3333 = 1.33

Conclusion

Do no reject H0. There is not a significant rankcorrelation. The two analysts are not showingagreement in their rating of the risk associated withthe different investments.

2

2

6 6(92)1 1 0.4424( 1) 10(100 1)

i

s

d

r n n

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