Post on 07-Apr-2018
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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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Slide
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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Slide
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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Slide
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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Slide
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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Example: Peanut Butter Taste Test
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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Example: Peanut Butter Taste Test
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
Using .05 level of significance,Reject H0 if z < -1.96 or z > 1.96
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
Example: Westin Freezers
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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17Slide
Example: Westin Freezers
Mann-Whitney-U- Test (Large-Sample Case)
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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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
Example: Westin Freezers
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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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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23Slide
Example: Westin Freezers
Mann-Whitney-U- Test (Large-Sample Case)
Sampling Distribution
u13.23
Sampling distribution
of U if populationsare identical
u
= 50 =(1/2)(10)(10)T
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24Slide
Example: Westin Freezers
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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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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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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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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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
Example: Connor Investors
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Example: Connor Investors
Rejection Rule
Using .10 level of significance,Reject H0 if z < -1.645 or z > 1.645
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