Introduction Siegel-Tukey test named after Sidney Siegel and

John Tukey, is a non-parametric test which may be applied to the data measured at least on an ordinal scale. It tests for the differences in scale between two groups.

The test is used to determine if one of two groups of data tends to have more widely dispersed values than the other.

The test was published in 1980 by Sidney Siegel and John Wilder Tukey in the journal of the American Statistical Association in the article “A Non-parametric Sum Of Ranks Procedure For Relative Spread in Unpaired Samples “.

Assumptions Each sample has been randomly selected

from the population. The two samples are independent of one

another. The level of measurement the data

represent is at least ordinal. The two populations from which the

samples are derived have equal medians.

Procedure Null hypothesis

H0 : δ2A = δ

2B

When the sample sizes are equal , then sum of the ranks also will be equal.

ԐR1 = ԐR2 Alternate hypothesis

H1: δ2A ≠ δ

2B

Procedure (Continued)2) α=0.053) Test statistic Siegel-Tukey test4) CalculationsBy combining the groups. The ranking is done by alternate extremes (rank 1 is lowest, 2 and 3 are the two highest, 4 and 5 are the two next lowest etc)Sum the rank of first and second group, after this procedure apply Mann-Whitney U test to find out the U value.

Procedure(Continued) U1 = n1n2+n1(n1+1)/2‒ ԐR1 U2 = n1n2+n2(n2+1)‒ԐR2 U = Min( U1,U2)

5)Critical region Ucal ˂ Utab ˂ Ucal6) Conclusion Accept or reject H0

In case when the sample size is less than 8

EXAMPLETwo plastics each produced by a different process were tested for ultimate strength. The measurement shown below represent breaking load in units of 1000 pounds per square inch. Plastic 1: 15.3, 18.7, 22.3, 17.6, 15.1, 14.8Plastic 2: 21.2, 22.4, 18.3, 19.3, 17.1, 27.7 Use Siegel-Tukey test to test the hypothesis that : δ2

1 = δ2

2

Solution1) H0 : : δ2

1 = δ22

H1: δ21 ≠ δ

22

2) α=0.05

3) Test statisticSiegel-Tukey test4) CalculationsArrange combine samples in ascending order,

Solution ( Continued ) 14.8, 15.1, 15.3, 17.1, 17.6, 18.3, 18.7, 19.3, 1 4 5 8 9 12 11

10 21.2, 22.3, 22.4, 27.7 7 6 3 2 R1 = 1+4+5+9+11+6 = 36 R2 = 8+12+10+7+3+2 = 42 U1 = n1n2+n1(n1+1)/2‒ ԐR1 = 36+6( 6+1 ) /2 – 36 =21

Solution ( Continued ) U2 = n1n2- U1 = 36 – 21 U2 = 15 U = min ( U1 , U2 ) U = min ( 21 , 15 ) U = 15

5) Critical regionIf U cal lies between 5 and 31 we accept H0, otherwise we will reject

Solution ( Continued )6) Conclusion As the calculated value of U lies in the interval 5 and 31. So, we accept the null hypothesis.

In case of tied ranks and n less than 8

EXAMPLEX1 : 10, 10, 9, 1, 0, 0X2 : 6, 6, 5, 5, 4, 4

SOLUTION:1) H0 : : δ2

1 = δ2

2

H1: δ21 ≠ δ

22

2) α=0.05

3) Test statisticSiegel-Tukey test

Solution ( Continued )4) CalculationsArrange combine samples in ascending order, Arrange data set 0, 0, 1, 4, 4, 5, 5, 6, 6, 9, 10,

10 Ranks 1 4 5 8 9 12 11 10 7 6

3 2 Tied adjusted rank 2.5, 2.5, 5, 8.5, 8.5, 11.5,

11.5, 8.5, 8.5, 6, 2.5, 2.5 R1 = 2.5+2.5+6+5+2.5+2.5 = 21

Solution ( Continued ) R2 = 8.5 +8.5+11.5+11.5+8.5+8.5 = 57 U1 = n1n2+n1(n1+1)/2‒ ԐR1 U1 = 36+6( 6+1)/2 – 21 U1 = 36+21 – 21 U1 = 36 U2 = n1n2 – U1 U2 = 36 – 36 U2 = 0

Solution ( Continued ) U = min ( U1,U2 ) U = min (36,0) U = 0

5) Critical regionIf U cal lies between 5 and 31 we will accept, otherwise we will reject6) ConclusionAs the calculated value of U does not lie in the interval 5 and 31. So, we will reject the null hypothesis.

In case of tied ranks and n greater than 8

If the sample size is greater than 8 and there are tied ranks in the data. We will use the formula Where s denotes the no of pair of ties and t denotes no of tied ranks

Incase of n greater than 8EXAMPLE

Group 1: 3.1, 5.3, 6.4, 6.2, 3.8, 7.5, 5.8, 4.3, 5.9, 4.9 Group 2: 9, 5.6, 6.3, 8.5, 4.6, 7.1, 5.5, 7.9, 6.8, 5.7,

8.9Solution:

1) H0 : : δ21 = δ

22

H1: δ21 ≠ δ

22

2) α=0.05

3) Test statisticSiegel-Tukey test

Solution ( Continued )4) CalculationsArrange combine samples in ascending order 3.1, 3.8, 4.3, 4.6, 4.9, 5.3, 5.5, 5.6, 5.7, 5.8, 1 4 5 8 9 12 13 16 17

20 5.9, 6.2, 6.3, 6.4, 6,8, 7.1, 7.5, 7.9, 8.5, 8.9,

9.0 21 19 18 15 14 11 10 7 6 3

2R1 = 1+4+5+9+12+20+21+19+15+10 = 116

Solution ( C0ntinued )R2 = 8+13+16+17+18+14+11+7+6+3+2 = 115U1 = n1n2+n1(n1+1)/2‒ ԐR1U1 = 10(11)+10(10+1)/2 – 116U1 = 149U2 = n1n2+n2(n2+1)‒ԐR2

U2 = 10(11)+11(11+1)/2 – 115U2 = 61

Solution ( Continued ) U=min( U1,U2 ) U=min( 149,61 ) U=61

Solution ( Continued ) Z = 0.42

5) Critical regionIf Zcal ≥ Ztab we reject our null hypothesis6) ConclusionSince calculated value is less than tabulated value. So, we accept the null hypothesis.

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