© 2005 Thomson/South-Western Slide 1 · 4 © 2005 Thomson/South-Western Slide 10 Hypotheses Ha:...

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1Slide© 2005 Thomson/South-Western

Slides Prepared byJOHN S. LOUCKS

St. Edward’s University


Chapter 19Nonparametric Methods

Sign TestWilcoxon Signed-Rank TestMann-Whitney-Wilcoxon TestKruskal-Wallis TestRank Correlation


Nonparametric Methods

Most of the statistical methods referred to as parametric require the use of interval- or ratio-scaled data.Nonparametric methods are often the only way to analyze 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.

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Nonparametric Methods

In general, for a statistical method to be classified as nonparametric, it must satisfy at least one of the following 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 data

when no assumption can be made about the population probability distribution.


Sign Test

A common application of the sign test involves using a sample of n potential customers to identify a preference for one of two brands of a product.The objective is to determine whether there is a difference 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 if the individual prefers the other brand.Because the data are recorded as plus and minus signs, this test is called the sign test.


Sign Test: Small-Sample Case

The small-sample case for the sign test should be used whenever n < 20.The hypotheses are

a : .50H p ≠a : .50H p ≠

0 : .50H p =0 : .50H p =

A preference for one brandover the other exists.

No preference for one brandover the other exists.

The number of plus signs is our test statistic.Assuming H0 is true, the sampling distribution for the test statistic is a binomial distribution with p = .5.H0 is rejected if the p-value < level of significance, α.

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Sign Test: Large-Sample Case

Using H0: p = .5 and n > 20, the sampling distribution for the number of plus signs can be approximated by a normal distribution.When no preference is stated (H0: p = .5), the sampling distribution will have:

The test statistic is:

H0 is rejected if the p-value < level of significance, α.

Mean: µ = .50nStandard Deviation: .25nσ = .25nσ =

xz µσ−

=xz µσ−

= (x is the numberof plus signs)


Example: Ketchup Taste Test


A B

As part of a market research study, asample of 36 consumers were asked to tastetwo brands of ketchup and indicate a preference. Do the data shown on the nextslide indicate a significant difference in theconsumer preferences for the two brands?


18 preferred Brand A Ketchup(+ sign recorded)

12 preferred Brand B Ketchup(_ sign recorded)

6 had no preference


Example: Ketchup Taste Test

A BThe analysis will be based ona sample size of 18 + 12 = 30.

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Hypotheses

a : .50H p ≠a : .50H p ≠

A B


0 : .50H p =0 : .50H p =

A preference for one brand over the other exists

No preference for one brand over the other exists


Sampling Distribution for Number of Plus Signs

µ = .5(30) = 15

A B


.25 .25(30) 2.74nσ = = =.25 .25(30) 2.74nσ = = =


Rejection Rule A B


p-Value = 2(.5000 - .3643) = .2714p-Value

z = (x – µ)/σ = (18 - 15)/2.74 = 3/2.74 = 1.10Test Statistic

Using .05 level of significance:Reject H0 if p-value < .05

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A B


ConclusionBecause the p-value > α, we cannot reject H0.

There is insufficient evidence in the sample to conclude that a difference in preference exists for the two brands of ketchup.


Hypothesis Test About a Median

We can apply the sign test by:• Using a plus sign whenever the data in the sample

are above the hypothesized value of the median• Using a minus sign whenever the data in the

sample are below the hypothesized value of the median

• Discarding any data exactly equal to the hypothesized median



34 years= 34 years=H0: Median Age34 years≠ 34 years≠Ha: Median Age

Example: Trim Fitness CenterA hypothesis test is being conducted

about the median age of female membersof the Trim Fitness Center.

In a sample of 40 female members, 25 are olderthan 34, 14 are younger than 34, and 1 is 34. Is theresufficient evidence to reject H0? Assume α = .05.

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p-Value = 2(.5000 − .4608) = .0784

µ = .5(39) = 19.5

.25 .25(39) 3.12nσ = = =.25 .25(39) 3.12nσ = = =


p-Value

z = (x – µ)/σ = (25 – 19.5)/3.12 = 1.76Test Statistic

Mean and Standard Deviation



Rejection Rule

ConclusionDo not reject H0. The p-value for this two-tail

test is .0784. There is insufficient evidence in the sample to conclude that the median age is not 34 for female members of Trim Fitness Center.

Using .05 level of significance:Reject H0 if p-value < .05


Wilcoxon Signed-Rank Test

This test is the nonparametric alternative to the parametric matched-sample test presented in Chapter 10.The methodology of the parametric matched-sample analysis requires:• interval data, and• the assumption that the population of differences

between the pairs of observations is normally distributed.

If the assumption of normally distributed differences is not appropriate, the Wilcoxon signed-rank test can be used.

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Example: Express Deliveries


A firm has decided to select oneof two express delivery services toprovide next-day deliveries to itsdistrict offices.

To test the delivery times of the two services, thefirm sends two reports to a sample of 10 district offices, with one report carried by one service and theother report carried by the second service. Do the dataon the next slide indicate a difference in the twoservices?



SeattleLos AngelesBostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver

32 hrs.301916151814107

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25 hrs.24151513151589

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District Office OverNight NiteFlite



Preliminary Steps of the Test• Compute the differences between the paired

observations.• Discard any differences of zero.• Rank the absolute value of the differences from

lowest to highest. Tied differences are assigned the average ranking of their positions.

• Give the ranks the sign of the original difference in the data.

• Sum the signed ranks.. . . next we will determine whether the

sum is significantly different from zero.

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SeattleLos AngelesBostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver

764123−1

2−2

5

District Office Differ. |Diff.| Rank Sign. Rank1097

1.546

1.5448

+10+9+7

+1.5+4+6−1.5+4−4+8

+44



HypothesesH0: The delivery times of the two services are the

same; neither offers faster service than the other.Ha: Delivery times differ between the two services;

recommend the one with the smaller times.


Sampling Distribution of T for Identical Populations

µT = 0

( 1)(2 1) 10(11)(21) 19.626 6T

n n nσ + += = =

( 1)(2 1) 10(11)(21) 19.626 6T

n n nσ + += = =


T

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Rejection RuleUsing .05 level of significance,

Reject H0 if p-value < .05Test Statistic

p-Valuez = (T - µT )/σT = (44 - 0)/19.62 = 2.24

p-Value = 2(.5000 - .4875) = .025


ConclusionReject H0. The p-value for this two-tail test is

.025. There is sufficient evidence in the sample to conclude that a difference exists in the delivery times provided by the two services.



Mann-Whitney-Wilcoxon Test

This test is another nonparametric method for determining whether there is a difference between two populations.This test, unlike the Wilcoxon signed-rank test, is notbased on a matched sample.This test does not require interval data or the assumption 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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Ha: The two populations are not identicalH0: The two populations are identical

Instead of testing for the difference between the means of two populations, this method tests to determine whether the two populations are identical.

The hypotheses are:



Example: Westin FreezersManufacturer labels indicate the

annual energy cost associated withoperating home appliances such asfreezers.

The energy costs for a sample of10 Westin freezers and a sample of 10Easton Freezers are shown on the next slide. Do thedata indicate, using α = .05, that a difference exists inthe annual energy costs for the two brands of freezers?



$55.10 54.5053.2053.0055.5054.9055.8054.0054.2055.20

$56.10 54.7054.4055.4054.1056.0055.5055.0054.3057.00

Westin Freezers Easton Freezers

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Hypotheses


Ha: Annual energy costs differ for the twobrands of freezers.

H0: Annual energy costs for Westin freezersand Easton freezers are the same.


Mann-Whitney-Wilcoxon Test:Large-Sample Case

First, rank the combined data from the lowest to the highest values, with tied values being assigned the average of the tied rankings.Then, compute T, the sum of the ranks for the first sample.Then, compare the observed value of T to the sampling distribution of T for identical populations. The value of the standardized test statistic z will provide the basis for deciding whether to reject H0.


Mann-Whitney-Wilcoxon Test:Large-Sample Case

1 2 1 21 ( 1)12T n n n nσ = + +1 2 1 21 ( 1)12T n n n nσ = + +

Approximately normal, providedn1 > 10 and n2 > 10

µT = n1(n1 + n2 + 1)

Sampling Distribution of T for Identical Populations• Mean

• Standard Deviation

• Distribution Form

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$55.10 54.5053.2053.0055.5054.9055.8054.0054.2055.20

$56.10 54.7054.4055.4054.1056.0055.5055.0054.3057.00

Westin Freezers Easton Freezers

Sum of Ranks Sum of Ranks

Rank Rank

86.5 123.5

12

128

15.5101735

13

1997

144

1815.5116

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Sampling Distribution of T for Identical Populations

µT = ½(10)(21) = 105


1 2 1 21 ( 1)121 (10)(10)(21)12

13.23

T n n n nσ = + +

=

=

1 2 1 21 ( 1)121 (10)(10)(21)12

13.23

T n n n nσ = + +

=

=

T


Rejection RuleUsing .05 level of significance,

Reject H0 if p-value < .05Test Statistic

p-Valuez = (T - µT )/σT = (86.5 − 105)/13.23 = -1.40

p-Value = 2(.5000 - .4192) = .1616


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ConclusionDo not reject H0. The p-value > α. There is

insufficient evidence in the sample data to conclude that there is a difference in the annual energy cost associated with the two brands of freezers.


Kruskal-Wallis Test

The Mann-Whitney-Wilcoxon test has been extended by Kruskal and Wallis for cases of three or more populations.

The Kruskal-Wallis test can be used with ordinal data as well as with interval or ratio data.

Also, the Kruskal-Wallis test does not require the assumption of normally distributed populations.

Ha: Not all populations are identicalH0: All populations are identical


Test Statistic

Kruskal-Wallis Test

=

⎡ ⎤= − +⎢ ⎥+⎣ ⎦

∑2

1

12 3( 1)( 1)

ki

TiT T i

RW nn n n=

⎡ ⎤= − +⎢ ⎥+⎣ ⎦

∑2

1

12 3( 1)( 1)

ki

TiT T i

RW nn n n

where: k = number of populationsni = number of items in sample i

nT = Σni = total number of items in all samplesRi = sum of the ranks for sample i

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Kruskal-Wallis Test

When the populations are identical, the sampling distribution of the test statistic W can be approximated by a chi-square distribution with k – 1 degrees of freedom.This approximation is acceptable if each of the sample sizes ni is > 5.

The rejection rule is: Reject H0 if p-value < α


Rank Correlation

The Pearson correlation coefficient, r, is a measure of the linear association between two variables for which interval or ratio data are available.The Spearman rank-correlation coefficient, rs , is a measure 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 positive

association between the rankings, and• values near -1.0 indicate a strong negative

association between the rankings.


Rank Correlation

Spearman Rank-Correlation Coefficient, rs

2

2

61

( 1)i

s

dr

n n= −

−∑ 2

2

61

( 1)i

s

dr

n n= −

−∑

where: n = number of items being rankedxi = rank of item i with respect to one variableyi = rank of item i with respect to a second variabledi = xi - yi

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Test for Significant Rank Correlation

0 : 0sH p =0 : 0sH p =

a : 0sH p ≠a : 0sH p ≠

We may want to use sample results to make an inference about the population rank correlation ps.

To do so, we must test the hypotheses:

(No rank correlation exists)(Rank correlation exists)


Rank Correlation

0sr

µ = 0sr

µ =

11sr n

σ =−1

1sr nσ =

−

Approximately normal, provided n > 10

Sampling Distribution of rs when ps = 0• Mean

• Standard Deviation

• Distribution Form


Rank Correlation

Example: Crennor InvestorsCrennor Investors provides

a portfolio management servicefor its clients. Two of Crennor’sanalysts ranked ten investmentsas shown on the next slide. Userank correlation, with α = .10, tocomment on the agreement of the two analysts’rankings.

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Rank Correlation

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

Investment A B C D E F G H I J

Example: Crennor Investors

0 : 0sH p =0 : 0sH p =

a : 0sH p ≠a : 0sH p ≠(No rank correlation exists)(Rank correlation exists)

• Analysts’ Rankings

• Hypotheses


Rank Correlation

ABCDEFGHIJ

149863572

10

15629731048

0-136-3-42-3-22

019

369

164944

Sum = 92

InvestmentAnalyst #1Ranking

Analyst #2Ranking Differ. (Differ.)2


Sampling Distribution of rsAssuming No Rank Correlation

Rank Correlation

1 .33310 1sr

σ = =−

1 .33310 1sr

σ = =−

µr = 0rs

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Test Statistic2

2

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

is

dr

n n= − = − =

− −∑ 2

2

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

is

dr

n n= − = − =

− −∑

Rank Correlation

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

Rejection RuleWith .10 level of significance:

Reject H0 if p-value < .10

p-Valuep-Value = 2(.5000 - .4082) = .1836


Do no reject H0. The p-value > α. There is not a significant rank correlation. The two analysts are not showing agreement in their ranking of the risk associated with the different investments.

Rank Correlation

Conclusion


End of Chapter 19

© 2005 Thomson/South-Western Slide 1 · 4 © 2005 Thomson/South-Western Slide 10 Hypotheses Ha:...

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