HT-A Chapter on Non-Parameteric Test-Others

7/31/2019 HT-A Chapter on Non-Parameteric Test-Others

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A Ch apt e r on A Ch apt e r on

NonNon--Par am et r i c Test s Par am et r i c Test s


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Nonp aram et r i c Test : Pr ocedur es Nonp aram et r i c Test : Pr ocedur es

When classical methods are not applicable, anappropriate nonparametric procedure can be selected.

One sample: Sign test and Wilcoxon (Signed rank) test

Two related samples: Sign test and Wilcoxon (Signed rank) test

Two independent samples: Mann-Whitney-U test


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Nonp aram et r i c Test : Pr ocedur es Nonp aram et r i c Test : Pr ocedur es

Nonparametric tests may be broadly defined as either:

Those whose test statistic does not depend upon the formof the underlying population distribution from which the

sample data were drawn - thus the nameDistribution Free

Those which are not concerned with the parameters of a

population.


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Nonp aram et r i c Test : Adv an t ages Nonp aram et r i c Test : Adv an t ages

Make fewer, less stringent assumptions (which are moreeasily met) than classical procedures.

May be used on all types of data

- Qualitative data (nominal scale),

- Data in rank form (ordinal scale),

- Truly quantitative data (interval and ratio scale).


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Nonp aram et r i c Test : Adv an t ages Nonp aram et r i c Test : Adv an t ages

Are generally easy to apply and quick to compute when

the sample size is small.

Permit the solution of problems that do not involve the

testing of population parameters.

May be equally (or almost) as powerful as the classical

procedure when the assumptions of the latter are met, andwhen they are not met may be quite a bit more powerful.


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Nonp aram et r i c Test : Disadvan t ages Nonp aram et r i c Test : Disadvan t ages

When all the assumptions of the classical procedures aremet it is disadvantageous to apply nonparametric methods,because the researcher will not take full advantage of data.

In aforementioned condition, nonparametric methods willwaste part of the sample information.

In such circumstances some very simple and quick

nonparametric tests (such as the sign test) haveconsiderable less power than the classical procedures andshould be avoided.


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Nonp aram et r i c Test : Disadvan t ages Nonp aram et r i c Test : Disadvan t ages

As the sample size gets larger, data manipulation requiredfor nonparametric procedures are sometimes laboriousunless appropriate computer software is available.

Special tables with critical values are needed and they arenot readily available as are the corresponding tables forclassical methods.


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Nonp aram et r i c Test : M iscon cep t ions Nonp aram et r i c Test : M iscon cep t ions

Nonparametric methods are not assumptions free.

Nonparametric methods are not always quick to perform

especially when a sample is large.

Distribution-free methods are available only for the

simplest experimental situations.


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Nonp aram et r i c Test : M iscon cep t ions Nonp aram et r i c Test : M iscon cep t ions

Nonparametric methods are inferior to classic methods andwaste information when applied to numerical observations.(For example Barbara Hazard [1997] ,Statistical Methods forHealth Care Research. p. 100: Other things being equal,

parametric techniques are more powerful... )

This leads us to the problem how to evaluate the

performance of a test.


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These tests use the Median and do not assumeanything about distribution, i.e. distribution free

Mathematically, value is ignored (i.e. the magnitude ofdifferences are not compared)

Instead, data is analysed simply according to rank.

NonNon--Param et r i c Test s: Rem ark s Param et r i c Test s: Rem ark s


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Is there any difference in the length of time that males and

females can sustain an isometric muscular contraction?

E n e r g y I n t a k e ( c a l o r i e s p e r d a y )

1 5 0 0 2 5 0 0 3 5 0 0 4 5 0 0 5 5 0 0

NumberofPeople

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

1 6 0

16 17 18 19 20

Sustained Isometric Torque (seconds)

t-test

Mean A Mean B

Par am et r i c Vs. NonPar am et r i c Vs. Non--P a r am e t r i c P a r am e t r i c


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E n e r g y I n t a k e ( c a l o r i e s p e r d a y )

1 5 0 0 2 5 0 0 3 5 0 0 4 5 0 0 5 5 0 0

NumberofPeople

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

1 6 0

16 17 18 19 20

Sustained Isometric Torque (seconds)

Mean

AMean

B

Is there any difference in the length of time that males and

females can sustain an isometric muscular contraction?

Par am et r i c Vs. NonPar am et r i c Vs. Non--P a r am e t r i c P a r am e t r i c

Comparingthemeans

doesnotgiveava

lid

reflectionofthegroupdiffere

nces.


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A.A. One sample:One sample:

Sign testWhen population is Heavily Skewed

Wilcoxon testWhen population is Not Normal, butapproximately Symmetric

NonNon--Param et r i c Test s: W h en t o Use Param et r i c Test s: W h en t o Use


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B. TwoB. Two relatedrelated samples:samples:

Sign test

Wilcoxon test

(Conditions are same as above)

C. TwoC. Two independentindependent samples:samples:

Mann-Whitney-U testWhen population is Not Normal

NonNon--Param et r i c Test s: W h en t o Use Param et r i c Test s: W h en t o Use


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Th e Sig n Test Th e Sig n Test


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It is a versatile and exceptionally easy-to-applynonparametric test.

It focuses on the median rather than the mean as a

measure of central tendency.

The only assumption is that the variables come from acontinuous distribution.

It can be applied even to the Nominal DataNominal Data.

Th e Sig n Test : Abo u t Th e Sig n Test : Abo u t


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It is useful in two situations:

When test a hypothesis concerning the median of apopulation (an alternative to one-sample t-test), and

To test equality of the medians in the case of twodependent samples (as an alternative to matched-pairst test).

Recall that the value of median is the same as for the mean when a distribution is symmetric

Both tests are carried out using the same procedure



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It is called the sign test because we convert original

observations into plus and minus signs. That is, we countthe number of observations greater than the hypotheticalmedian.

When dealing with 2 related samples we count thenumber of times one treatment has a higher value than theother.

Ties are not counted (theoretically, with continuousdistributions, there should be no ties, but with limitedmeasuring instruments ties do occur).



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If the null hypothesis is true we would expectapproximately equal number of + and - signs.

If either positive or negative signs predominate, there isevidence that the null hypothesis is false.

As a test statistic we can use number of positive signs.



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Used in situations with limited ability to assess ranking of differences:

can only assess if score for a subject is less than, greater than, orequal to the paired score

Test statistic depends only on the sign of the differences

Special case ofone-sample Binomial Test with p=0.5

Assumptions:Assumptions:

Random sample Ordinal measurement with continuous values

Th e Sign Test : I n Nu t sh e l l Th e Sign Test : I n Nu t sh e l l


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Th e Sig n Test : Ex am p le Th e Sig n Test : Ex am p le --11

One Sample Case:One Sample Case:

In an anesthetic used for major surgery, the meannumber of hours it takes for the anesthesia to wear off is 7.

A new agent has been suggested that supposedlyprovides relief much sooner.

In a series of 12 surgeries using the new anesthetic, thefollowing times for recovery were observed:

Recovery time: 4 4 5 5 5 6 6 6 6 7 9 11


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Course of Action:Course of Action:

(a) If it is known that the population is heavily skewed,which test would you recommend to assess.

(b) Whether the new agent significantly reduces recoverytime? Explain your answer.

(c) Carry out the test you recommended in (a).


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StepStep--0: Check for Normality0: Check for Normality

Average: 6.16667

StDev: 2.03753N: 12

Anderson-Darling Normality Test

A-Squared: 0.783P-Value: 0.030

4 5 6 7 8 9 10 11

.001

.01

.05

.20

.50

.80

.95

.99

.999

Probability

Recovery Tim

Normal Probability Plot


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StepStep--1: Convert the Data1: Convert the Data

Recovery time: 4 4 5 5 5 6 6 6 6 7 9 11

Coded to Sign: - - - - - - - - - 0 + +

H0 : The population median = 7

HA : The population median is 7

StepStep--2: State the Hypothesis2: State the Hypothesis


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significance level, = 0.05

StepStep--4: Test Statistics4: Test Statistics

Test statistic for Sign Test is either the observed numberof +Sign or the observed number ofSign

The Nature of the alternative hypothesis determines which

one of the possible hypotheses is appropriate

HA : P(+) > P(-); HA : P(+) < P(-); HA : P(+) P(-)

StepStep--3: State the Significance Level3: State the Significance Level


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If the Null Hypothesis true; i.e., the population median is equalto 7, then we can say that:

StepStep--5: Distribution of Test Statistics5: Distribution of Test Statistics

H0 : P(+) = P(-) = 0.5

We note that nine outcomes are less than the median and twoare larger

W e exc lude t he one t ha t i s equa l t o t he m ed ian , i .e . 0

According to test statistic, we expected 5 pluses under H0



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As per conditions, Binomial Distribution fit well

StepStep--5a: Distribution of Test Statistics5a: Distribution of Test Statistics

StepStep--6: Calculation of Test Statistics6: Calculation of Test Statistics

( | , )x k n kP k x n p c p q n kk 0

Where: k = The Test Statistics; n = sample sizex = # of success; p/q = Prob. of success/failure



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StepStep--7: Values of Test Statistics7: Values of Test Statistics

(11 C 0)*((0.5)^0)*((0.5)^(11-0))+(11 C 1)*((0.5)^1)*((0.5)^(11-1))

+(11 C 2)*((0.5)^2)*((0.5)^(11-2))

??? = 0.03271484??? = 0.03271484



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StepStep--7a: Values of Test Statistics7a: Values of Test Statistics

Sign test of median = 7.000 versus < 7.000

N Below Equal AboveAbove P MedianRecovery 12 9 1 2 0.0327 6.000

Test

statistic



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StepStep--8: Conclusion8: Conclusion

pp--value = 0.03271484value = 0.03271484

Significance level, = 0.05

critical T toconclude a significant difference

Conclusion

Median A > Median B

Wi lcoxon Wi l coxon Sign ed Ran k : Ex am ple Sign ed Ran k : Ex am ple --22


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Th e Man nTh e Man n--W h i t n ey U Test W h i t n ey U Test


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Test of ranks between two samples

Ranks the pooled observations in the two samples andthen total the ranks in each sample

If the medians are the same, the ranks will be similar

Assumptions: Independent, random samples

Ordinal scale and continuous values Any difference reflected in the medians

MannMann--W h i t n ey U Test : Abou t W h i t n ey U Test : Abou t


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Used when you have two conditions:Used when you have two conditions:

Each performed by a separate groupseparate group of subjects.

Each subject produces one score.

Tests whether there a statistically significant differencebetween the two groups.

MannMann--W h i t n ey U Test : Abou t W h i t n ey U Test : Abou t


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MannMann--W h i t n ey U Test : Ex am p le W h i t n ey U Test : Ex am p le --11

Does it make any difference to students comprehension ofstatistics whether the lectures are in Hindi or in English?

Group 1: statistics lectures in Hindi.Group 2: statistics lectures in English.

Lecturer intelligibility ratings by students:Lecturer intelligibility ratings by students:

0 = "unintelligible",100 = "highly intelligible"


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Hindi GroupHindi Group

RawRaw--ScoreScore

Hindi GroupHindi GroupRankRank

English GroupEnglish GroupRawRaw--ScoreScore

English GroupEnglish GroupRankRank

St ep 1 : St ep 1 : Rank all the scores together, regardless of group.



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MannMann--W h i t n ey U Test : On Rank W h i t n ey U Test : On Rank

Rev ision o f how t o Rank scor es: Rev ision o f how t o Rank scores :

Lowest score gets rank of 1; next lowest gets 2; and so on. Two or more scores with the same value are tied, in this case

- Give each tied score the rank it would have had,

- Add the ranks for the tied scores, and divide by the number oftied scores. Each of the ties gets this average rank.

- The next score after the set of ties gets the rank it would haveobtained, had there been no tied scores.

Example:Example: Raw score: 6 34 34 48Originalrank: 1 2 3 4Actualrank: 1 2.5 2.5 4


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St ep 2 : St ep 2 :

Add up the ranks for group 1, to get T1.Add up the ranks for group 2, to get T2.Here, T1 = 83 and T2 = 70.

St ep 3 : St ep 3 :

N1 is the number of subjects in group 1;N2 is the number of subjects in group 2.Here, N1 = 8 and N2 = 9.

St ep 4 : St ep 4 :

Call the larger of these two rank totals Tx. Here, Tx = 83.Nx is the number of subjects in this group; here, Nx = 8.



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Hindi GroupHindi Group

RawRaw--ScoreScoreHindi GroupHindi Group

RankRankEnglish GroupEnglish Group

RawRaw--ScoreScoreEnglish GroupEnglish Group

RankRank

St ep 2 St ep 2 --4 :4 :

Total RankTotal Rank

(T1)(T1)Total RankTotal Rank

(T2)(T2)83.083.0 70.070.0



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St ep 5 : St ep 5 : Find U:

Nx (Nx + 1)U = N1 * N2 + ---------------- - Tx

2

I n Ex a m p le: I n Ex a m p l e: 8 * (8 + 1)

U = 8 * 9 + ---------------- - 832

U = 72 + 36 - 83 = 25



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If there are unequal numbers of subjects:If there are unequal numbers of subjects:

As in the present example

Calculate U forCalculate U for both rankboth rank totals and then use thetotals and then use the smallersmaller U.U.

In the present example:In the present example:For T1, U = 25,For T2, U = 47.

Therefore, use 25 as U.Therefore, use 25 as U.



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St ep 6 : St ep 6 :

Check the critical value of U, taking into account N1 and N2.

If our obtained U is smaller than the critical value of U.

Reject the null hypothesis and conclude that our two groups dodiffer significantly.



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Conclusion:Conclusion: Ratings of lecturer intelligibility are unaffected bywhether the lectures are given in English or in Hindi.

From the table:From the table: The critical value ofU for N1 = 8 and N2 = 9 is

1515, while

obtained U of25, which25, which is larger.larger.

So conclude that there is nosignificant difference between our

two groups.



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St ep 1 : St ep 1 : Rank all the data from both groups in one series, thentotal each

Student

School A School B

StudentGrade GradeRank Rank

J. S.

L. D.H. L.M. J.T. M.T. S.

P. H.

T. J.

M. M.K. S.P. S.R. M.P. W.

A. F.

B-

B-A+D-B+A-

F

D

C+C+B-EC-

A-

9.0

9.014.03.0

11.012.5

1.0

4.0

6.56.59.02.05.0

12.5

RA = 59.5 RB = 45.5



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St ep 2 : St ep 2 : Calculate two versions of the U statistic using:

U1 = (nA xnB) +2

(nA + 1) xnA- RA

AND

U2 = (nA xnB) +

2

(nB + 1) xnB- RB



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St ep 3 : St ep 3 : U statistics:

Select the smallersmaller of the two, i.e., Min(UMin(U11 , U, U22))

Table of critical values for the MannTable of critical values for the Mann--Whitney test:Whitney test:

n

0.05

0.01

6

5

2

7

8

4

8

13

7

9

17

11

Calculated U must be > critical U toconclude a significant difference

Conclusion

Median A>Median B



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Th e Med ian Test Th e Med ian Test


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Used to test the hypothesis that two samples are from populations withequal medians

Calculates the proportions in each group above/below the commonmedian of the two groups

Uses a chi-square test to test the differences between these frequencies

Assumptions:Assumptions: Independent, random samples

Ordinal scale and continuous values Any difference reflected in median

Th e Med ian Test : Abou t Th e Med ian Test : Abou t


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Th e Med ian Test : Abou t Th e Med ian Test : Abou t

The Median Test is used to test for location differences

between two or more independent populations.

The Median Test does not take into account the distance

from the median.

It, like the Sign Test, only takes into account which side

of the median the observations lie on.


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Th e Med ian Test : Assu m pt ion Th e Med ian Test : Assu m pt ion

Two/K independent samples.

Need independence both within and among samples.

Two/K populations have the same shape, but notnecessarily the same distribution.

Random Samples from each population


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Th e Med ian Test : Hyp ot h esis Th e Med ian Test : Hyp ot h esis

Two Samples:

H0: 1 = 2, the medians are the same

H1: 1 2, the medians are not the same

K Samples:

H0: 1 = 2 = 3 =. . . = , the medians are the same

H1: At least one of the medians is different


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Th e Med ian Test : Pr ocedu r e Th e Med ian Test : Pr ocedu r e

Find common median (M) by combining all the samples,(m+n=N).

Under H0, half the observations from each sample shouldbe above M.

Create Contingency Table, where the rows are thepopulations and the columns are whether the observation isabove or below M.


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Th e Med ian Test : Pr ocedu r e Th e Med ian Test : Pr ocedu r e

Use the idea of permutation (Binomial with p=.5) forfinding the probability,

E(a under H0)=m/2 and E(b under H0)=n/2.

Ties:

- Observations that equal the median.

- Correcting ties:* Omit the observation and proceed as normal.

* If lots of ties, then split up the observations in away that makes rejecting H0 less likely.


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Th e Median Test : Con t in g en cy Tab l es

Two Samples:

21=N11=c+d10=a+b

11=n4=d7=bType 4

10=m7=c3=aType 1

Below median(M=6)

Above median(M=6)

* We have 3 tiesFour Samples:

48=N24=B24=A

12=n44=b48=a4Type 412=n34=b38=a3Type 3

12=n29=b23=a2Type 2

12=n17=b15=a1Type 1

Below median(M=5.5)

Above median(M=5.5)


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Th e Med ian Test : Test St a t i st i cs Th e Median Test : Test St a t i st i cs

Two Samples:

To find the actual probability of getting results like or moreextreme than ours is found by the following Hyper-GeometricDistribution:

+

ba

N

b

n

a

m

P*=P*=


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Four Samples:

The probability distribution for the k sample Median Test isfound by just slightly modifying the previous test statistic. Wesimply now account for all the populations.

A

N

ak

nk

a

n

a

n

2

2

1

1

P*=P*= =>=>( )

( )

!)!(

!

!

!!

biai

ni

N

BA


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UseUse

Ch iCh i--Squ are Test o f I n dependence Squ are Test o f I n dependence


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Th e Med ian Test : Ex am p le Th e Med ian Test : Ex am ple

Q. Do urban and rural male junior high school students differ

with respect to their level of mental health

Data:

Members of a random sample of12 male studentsfrom a rural and an independent sample of16 male studentsfrom urban, junior high school were collected.


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Observations:School

Urban Rural Urban Rural35 29 25 5026 50 27 3727 43 45 34

21 22 46 3127 42 3338 47 2623 42 4625 32 41


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Calculation of Test Statistics: First Step

Compute the common Median:

Combined the Two samples

Arrange the observations ascending orders

In this Example: # of Observation is even

So, The median is: (33+34)/2 = 33.5


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Calculation of Test Statistics: Second Step

Compute the 2x2 Contingency Table:

Urban Rural Total

# of scores above median 6 8 14

# of scores below median 10 4 14

Total 16 16 28


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Calculation of Test Statistics: Third Step

Test Statistics:

X Y Total

A a b a+ b

B c d c+ d Total a+ c b+ d n

2( )2

( )( )( )( )

n ad bcx

a c b d a b c d

=+ + + +


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Calculation of Test Statistics: Fourth Step

Computation:

228(6 4 8 10)2 2.33(16)(12)(14)(14)

x = =


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Th e Med ian Test : Nex t ???Th e Med ian Test : Nex t ???

What rest to do in computation ???

Why to use Median Test Not Chi-Square???


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Ch oi ce of St a t is t ica l Test s Ch o ice o f St a t is t i ca l Test s

ANOVAANOVAANOVA with

repeatedmeasures

IndependentIndependenttt--testtest

PairedPaired tt--testtesttt--testtestI n t e r v al or Rat io

Median TestMedian Test

Kruskal-Wallis

Friedman 2-way ANOVA

K-S;

Median Test;Median Test;

MannMann--Whitney UWhitney U

Sigh TestSigh Test &Wilcoxon

Signed Rank

K-SOrd ina l

Chi-squareCochran QChiChi--squaresquareMcNemarChiChi--squaresquareN o m i n a l

k - I n d e p e n d e n t Samples

K-Re la ted Samp les

2 - I n d e p en d e n t Samp les

2-Re la ted Samples1 -Samp le


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Documents

HT-A Chapter on Non-Parameteric Test-Others