Wilcoxon rank sum test Wilcoxon rank sum test (or the Mann-Whitney U (or the Mann-Whitney U

test)test)In In statisticsstatistics, the , the Mann-Whitney Mann-Whitney UU test (also test (also called the called the Mann-Whitney-WilcoxonMann-Whitney-Wilcoxon ( (MWWMWW), ), Wilcoxon rank-sum testWilcoxon rank-sum test, or , or Wilcoxon-Wilcoxon-Mann-WhitneyMann-Whitney test) is a test) is a non-parametricnon-parametric test test for assessing whether two for assessing whether two samplessamples of of observations come from the same distributionobservations come from the same distribution . .

It requires the two samples to be It requires the two samples to be independentindependent, , and the observations to be and the observations to be ordinalordinal or or continuous measurements, i.e. one can at continuous measurements, i.e. one can at least say, of any two observations, which is the least say, of any two observations, which is the greatergreater . .

It is one of the best-known non-parametric It is one of the best-known non-parametric significance testssignificance tests . .

It was proposed initially by It was proposed initially by WilcoxonWilcoxon (1945), (1945), for equal sample sizes, and extended to for equal sample sizes, and extended to arbitrary sample sizes and in other ways arbitrary sample sizes and in other ways by by Mann andMann and WhitneyWhitney (1947) (1947) . .

MWW is virtually identical in performing an MWW is virtually identical in performing an ordinary parametric two-sample ordinary parametric two-sample tt test on test on the data after ranking over the combined the data after ranking over the combined samplessamples..

Here, we do NOT have paired data, but Here, we do NOT have paired data, but rather nrather n11 values from group 1 and n values from group 1 and n22 values from group 2values from group 2 . .

Assumptions:Assumptions: Independence within and Independence within and between groups, same distributions in between groups, same distributions in both groupsboth groups

We want to test whether the values in the We want to test whether the values in the groups are samples from different groups are samples from different distributionsdistributions..

A- For small sample size n< 20A- For small sample size n< 20: :

--Rank in an ascending order all values for the two Rank in an ascending order all values for the two groups together as if they were from the same groups together as if they were from the same groupgroup..

**If ANY OF THE VALUES ARE EQUAL AVARAGE THEIR If ANY OF THE VALUES ARE EQUAL AVARAGE THEIR RANKSRANKS..

--Let Let TT be the sum of the ranks of the values from be the sum of the ranks of the values from group hasing smaller sum of ranksgroup hasing smaller sum of ranks . .

Under the assumption that the values come from the Under the assumption that the values come from the same distribution, the distribution of same distribution, the distribution of T T is knownis known . .

The expectation and variance under the null The expectation and variance under the null hypothesis are simple functions of hypothesis are simple functions of nn11 (smaller (smaller sample size )and sample size )and nn22..

--The calculated The calculated TT value represents the lower value represents the lower limit of the range of all tested values. limit of the range of all tested values. Where the sum of ranks in the large Where the sum of ranks in the large sample size group is the upper limitsample size group is the upper limit..

--By comparing the calculated By comparing the calculated TT with the with the tabulated one at tabulated one at n1,n2n1,n2 values, we can values, we can predict the value of predict the value of PP and the range of and the range of significance. significance. ((tabletable A8) A8)

--If the calculated If the calculated TT < tabulated or < tabulated or P P < < tabulated the difference is significanttabulated the difference is significant . .

Wilcoxon Rank Sum Test Wilcoxon Rank Sum Test ExampleExample

You’re a production planner. You want to see if You’re a production planner. You want to see if the operating rates for 2 factories is the samethe operating rates for 2 factories is the same . .

For factory 1, the For factory 1, the ratesrates (% of capacity) are (% of capacity) are 7171, , 8282, , 7777, , 9292, , 8888 . .

For factory 2, the For factory 2, the ratesrates are are 8585, , 8282, , 9494 & & 9797 . .

Do the factory rates Do the factory rates have the same have the same probability probability distributionsdistributions at the at the

00.10.10 level level ? ?

Wilcoxon Rank Sum Test Wilcoxon Rank Sum Test SolutionSolution

HH00: : Identical DistribIdentical Distrib..

HHaa: : Shifted Left or Shifted Left or RightRight

= =

nn11 = =nn22 = =

Critical Value(s)Critical Value(s)::

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

RanksRanks

Wilcoxon Rank Sum Test Wilcoxon Rank Sum Test Computation TableComputation Table

Factory 1 Factory 2Rate Rank Rate Rank

71 1 85 582 3 3.5 82 4 3.577 2 94 892 7 97 988 6 ... ...

Rank Sum 19.5 25.5

Wilcoxon Rank Sum Wilcoxon Rank Sum Table (Portion)Table (Portion)

n1

3 4 5 ..

TL TU TL TU TL TU ..

3 6 15 7 17 7 20 ..n2 4 7 17 12 24 13 27 ..

5 7 20 13 27 19 36 ..: : : : : : : :

n1

3 4 5 ..

TL TU TL TU TL TU ..

3 6 15 7 17 7 20 ..n2 4 7 17 12 24 13 27 ..

5 7 20 13 27 19 36 ..: : : : : : : :

= .05 one-tailed; = .05 one-tailed; = .10 two-tailed = .10 two-tailed

Wilcoxon Rank Sum Test Wilcoxon Rank Sum Test SolutionSolution

HH00: : Identical DistribIdentical Distrib..HHaa: : Shifted Left or Shifted Left or

RightRight = = .10.10nn11 = = 4 4 nn22 = = 55 Critical Value(s)Critical Value(s)::

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

Do Not Reject at Do Not Reject at = .10 = .10

There Is No Evidence There Is No Evidence Distrib. Are Not EqualDistrib. Are Not Equal

RejectReject RejectRejectDo Not Do Not RejectReject

1313 2727 RanksRanks

TT22 = 5 + 3.5 + 8+ 9 = 25.5 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)(Smallest Sample)

Since the sum of ranks from the Since the sum of ranks from the largest sample size is the upper limit largest sample size is the upper limit (19.5) is smaller than tabulated value (19.5) is smaller than tabulated value the difference us significantthe difference us significant . .

Ex 2 : In medical statistics P.347Ex 2 : In medical statistics P.347

MannMann--Whitney test assumptions are as Whitney test assumptions are as followsfollows: :

11--The two samples of size The two samples of size nn and and m m are are independentindependent

22--The measurement scale is at least The measurement scale is at least ordinalordinal..

33--The variable of interest is The variable of interest is continuouscontinuous . .

44--Populations difference is with respect to their Populations difference is with respect to their medianmedian..

H0: if M1=M2H0: if M1=M2

HAHA::

M1>M2 OR M1<M2----------- One sided testM1>M2 OR M1<M2----------- One sided test

M1 does not equal M2-----Two sided testM1 does not equal M2-----Two sided test

ProcedureProcedure

--The observations in both groups are The observations in both groups are arranged and ranked in a descending arranged and ranked in a descending manner as beforemanner as before..

--The smaller sum of ranks The smaller sum of ranks (S)(S) in in n n group is calculatedgroup is calculated..

--T T value is calculated from the value is calculated from the following formulafollowing formula::

TT==SS-n(n+1)/2-n(n+1)/2

ExampleExample

111.31.3

221.51.5

332.12.1

443.23.2

553.43.4

664.34.3

774.94.9

886.36.3

997.17.1

Rank Group 1 Group 2

Rank sum: 16 29

T=16- 5(5+1)/2 = 1T=16- 5(5+1)/2 = 1Form a special table (L table ) we can find Form a special table (L table ) we can find tabulated T at specified P value. Through tabulated T at specified P value. Through nn and and mm values. (n is n1 .; m is n2)values. (n is n1 .; m is n2)

--If the calculated T < or = tabulated TIf the calculated T < or = tabulated TOR OR if the calculated T at P value < or = the if the calculated T at P value < or = the specified P the difference is significantspecified P the difference is significant..

- -Tabulated T at p=.05 is 3Tabulated T at p=.05 is 3T = 1 at is at p =.01T = 1 at is at p =.01So, In this example H0 is rejected and the So, In this example H0 is rejected and the difference is significantdifference is significant..

B- For large samples B- For large samples nn or or mm >20 >20

The normal approximation is appliedThe normal approximation is applied: : Z Z = = T – m n / 2T – m n / 2

sq.root (nm (n + m +1)/ 12sq.root (nm (n + m +1)/ 12 The calculated Z value is compared with the The calculated Z value is compared with the tabulated one in normal distribution table tabulated one in normal distribution table

atat the specified P value . *Smaller p is a significant difference.

Ex : in handout P;679Ex : in handout P;679

Method II (the modified Wilcoxon-MannMethod II (the modified Wilcoxon-Mann--Whitney test)Whitney test)

Depends upon the calculation of a measure known as Depends upon the calculation of a measure known as UUwhich for either sample is equal to the difference between the which for either sample is equal to the difference between the maximum possible value of T for the sample versus the maximum possible value of T for the sample versus the

actually observed T valueactually observed T value. .

Thus for sample A it would beThus for sample A it would be: : UAUA= = TA TA [max] — [max] — TATA

= =n a n bn a n b +na (na+1) / 2— +na (na+1) / 2— TATA

and for sample Band for sample B: : UBUB= = TBTB[max] — [max] — TBTB

= =na nb +nb(nb+1na nb +nb(nb+1) /2— ) /2— TBTB

TATA =the sum of the na ranks in =the sum of the na ranks in group Agroup A

TBTB =the sum of the nb ranks in =the sum of the nb ranks in group Bgroup B

TA TA [max]= na nb +na(na+1) / 2[max]= na nb +na(na+1) / 2

TB TB [max]= na nb +nb(nb+1) / 2[max]= na nb +nb(nb+1) / 2

ExampleExample

Twenty-one persons seeking treatment for Twenty-one persons seeking treatment for claustrophobia are claustrophobia are independentlyindependently and and randomlyrandomly sorted into sorted into two groupstwo groups, the first , the first of size na=11 and the second of of size na=11 and the second of size nb=10. The members of the first group size nb=10. The members of the first group each individually receive Treatment A over each individually receive Treatment A over a period of 15 weeks, while those of the a period of 15 weeks, while those of the second group receive Treatment B. The second group receive Treatment B. The investigators' investigators' directional hypothesisdirectional hypothesis is is that Treatment A will prove the more that Treatment A will prove the more effectiveeffective

At the end of the experimental treatment At the end of the experimental treatment period, the subjects are individually placed period, the subjects are individually placed in a series of claustrophobia test in a series of claustrophobia test situations, knowing that their reactions to situations, knowing that their reactions to these situations are being recorded on these situations are being recorded on videotape. Subsequently three clinical videotape. Subsequently three clinical experts, uninvolved in the experimental experts, uninvolved in the experimental treatment and not knowing which subject treatment and not knowing which subject received which treatment, received which treatment, independentlyindependently view the videotapes and rate each subject view the videotapes and rate each subject according to the according to the degreedegree of claustrophobic of claustrophobic tendency shown in the test situations. tendency shown in the test situations. Each judge's Each judge's rating rating takes place along a 10-takes place along a 10-point scale, with 1="point scale, with 1="very lowvery low" and " and 10="10="very highvery high"; and the final measure for "; and the final measure for each subject is the simple average of the each subject is the simple average of the ratings of the three judges for that subject.ratings of the three judges for that subject.

The following table shows the average The following table shows the average ratings for each subject in each of the ratings for each subject in each of the two groupstwo groups..

Group AGroup B

  4.6   4.7   4.9    5.1   5.2   5.5    5.8   6.1   6.5    6.5   7.2

  5.2   5.3   5.4    5.6   6.2   6.3    6.8   7.7   8.0    8.1

mean5.66.5

It begins by assembling the measures It begins by assembling the measures from samples from samples AA and  and BB into a single set into a single set of size of size N=na +nbN=na +nb . .

These measures are then rank-ordered These measures are then rank-ordered from lowest (from lowest (rank#1rank#1) to highest ) to highest ((rank#Nrank#N), with tied ranks included ), with tied ranks included where appropriatewhere appropriate . .

For the present set of data, this process For the present set of data, this process would yield the following resultwould yield the following result . .

Raw rank from sample Raw rank from sample measuremeasure

4.64.6 4.74.74.94.95.15.15.25.25.25.25.35.35.45.45.55.55.65.65.85.86.16.16.26.26.36.36.56.56.56.56.86.87.27.27.77.78.08.08.18.1

11223344

5.55.55.55.5778899

10101111121213131414

15.515.515.515.517171818191920202121

AAAAAAAAAABBBBBBAABBAAAABBBBAAAABBAABBBBBB

Our final bit of mechanics is just a dab of Our final bit of mechanics is just a dab of symbolic notationsymbolic notation . .

TATA =the sum of the =the sum of the nana ranks in group A ranks in group A

TBTB =the sum of the =the sum of the nbnb ranks in group B ranks in group B

TA TA [max]= nanb +na(na+1) / 2[max]= nanb +na(na+1) / 2

TB TB [max]= nanb +nb(nb+1) / 2[max]= nanb +nb(nb+1) / 2

For the present exampleFor the present example : :TATA =96.5[with =96.5[with n n a a =11]=11]TBTB =134.5[with =134.5[with n n b b =10]=10]

TA TA [max]= (11)(10) +11(12) / 2= 176[max]= (11)(10) +11(12) / 2= 176

TB TB [max]= (11)(10) +10(11) / 2= 165[max]= (11)(10) +10(11) / 2= 165

UAUA = 176—96.5 = 79.5 = 176—96.5 = 79.5 or or   UBUB = 165—134.5 = 30.5 = 165—134.5 = 30.5

UA UA andand UB UB are assigned as the upper and are assigned as the upper and lower limits of the observed lower limits of the observed UU value value. .

N/B N/B : The group hasing larger n is the A : The group hasing larger n is the A group and its group and its UU represents the upper limitrepresents the upper limit..

By comparing the critical values of By comparing the critical values of observed observed UA and UBUA and UB at na and nb with the at na and nb with the tabulated intervals we can conclude thattabulated intervals we can conclude that: :

** **If the calculated U is deviated from the If the calculated U is deviated from the tabulated U (if UB is smaller and / or UA tabulated U (if UB is smaller and / or UA is larger ) , the difference is significant is larger ) , the difference is significant and vice versaand vice versa..

****We may also find the level of significance We may also find the level of significance PP that match the calculated U intervals that match the calculated U intervals

- -To be significant difference To be significant difference PP calculated calculated has to be < or = to the tabulated has to be < or = to the tabulated PP value value..For one or two tailed testsFor one or two tailed tests..

In our example ,the tabulated intervals In our example ,the tabulated intervals for U were 31 , 79 at na =11 , nb=10 at for U were 31 , 79 at na =11 , nb=10 at p=0.05 p=0.05 ( for directional test )( for directional test ). .

The calculated U intervals were The calculated U intervals were 30.5,79.5 which will be at P < 0.0530.5,79.5 which will be at P < 0.05. .

So, treatment So, treatment AA will significantly improve will significantly improve claustrophobia than treatment claustrophobia than treatment BB

Test for approximation can be Test for approximation can be done as followsdone as follows: :

--If want to measure approximation to normal If want to measure approximation to normal distribution at P = 0.05distribution at P = 0.05

ZZ= = TTobs - obs - UUTT ± 0.05± 0.05 ơ ơ TT

TTobsobs    as the observed value of either TA or TBas the observed value of either TA or TB

UUTT as the the mean of the corresponding sampling as the the mean of the corresponding sampling distribution of T distribution of T ơ ơ TT as the standard deviation of that sampling as the standard deviation of that sampling distributiondistribution

correction for continuity:correction for continuity:  —.5 when   —.5 when TobsTobs>>TT  +.5 when   +.5 when TobsTobs<<TT

ơ ơ TT = sqrt = sqrt [[nnaannbb(N+1) / 12(N+1) / 12]]

N= total sum of ranks in A and BN= total sum of ranks in A and B

In our exampleIn our example: : --The observed value of The observed value of TATA=96.5=96.5 belongs to a belongs to a

sampling distribution whose mean is equal to sampling distribution whose mean is equal to 121121    wherewhere

UTUTAA = =n an a (N+1) / 2 = 11(21+1) /2 =121 (N+1) / 2 = 11(21+1) /2 =121--The observed value of The observed value of TBTB=134.5 belongs to a sampling =134.5 belongs to a sampling

distribution whose mean is equal to 110distribution whose mean is equal to 110

UTUTBB = =n bn b (N+1) / 2 = 10 (21+1) /2 =110 (N+1) / 2 = 10 (21+1) /2 =110

ơ ơ TT = sqrt = sqrt [[10x11(21+1) / 1210x11(21+1) / 12] =] =± 14.2± 14.2

ZZA= A= (96.5-121)+0.05 (96.5-121)+0.05 = - = -24 24 = -1.69 = -1.69

14.214.2

ZZBB= = (134.5-110)-0.05 (134.5-110)-0.05 = + = +24 24 = +1.69 = +1.69

14.214.2 14.214.2

So, the CI of calculated Z are -1.69So, the CI of calculated Z are -1.69

These values are compared with tabulated These values are compared with tabulated ones at ones at P P =0.05 for directional test=0.05 for directional test..

If calculated critical If calculated critical zz values > or = values > or = tabulated the null hypothesis is not true and tabulated the null hypothesis is not true and there is no approximation because P value there is no approximation because P value will be <will be <tabulatedtabulated. .

In our example: The tabulated CI forIn our example: The tabulated CI for z z at 0.05 are at 0.05 are

± 1.645 ± 1.645 and matches a and matches a PP value =0. value =0. 04550455

So , Treatment A is more potent than BSo , Treatment A is more potent than B