Javier Garcia - Verdugo Sanchez Six Sigma Training. Week 3. Median Tests.

1. Median Tests
99
Mean 66 4
Probability Plot of M1 - M4
Normal
95
90
80
70
60
50
cent
Mean 66,4
StDev 63,72
N 40
AD 2,311
P-Value <0,005
50
40
30
20
10
5
Per
250200150100500-50-100
1
M1 - M4
Week 3
No normal distribution – What now?
Knorr-Bremse Group
When do We Test the Median
Your analysis of variance is unvalid.
Wh h ll b d ?What shall be done now?
− the evaluated residuals are not normal distributed
− no homogeneity of variance
Within statistics there are many test methods available. There is not
enough time to discus them all within the BB training.g g
For the above mentioned cases, you can test for differences in
medians instead of means. We will discuss 2 of the most common
median tests.
Mood’s Median Test
Kruskal-Wallis Test
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 2/12
Kruskal Wallis Test

2. Median Tests
One problem statement and two solutions?
• Mood’s Median Test
– Tests equality of medians for 2 and more factor levels.
Th d t f thi t t i th t it i b t i t tli– The advantage of this test is that it is robust against outliers.
– Assumption: the distributions under investigations have the same
shape.shape.
– The power of this test decreases while the distributions get more
unequal. In this case we use the Kruskal-Wallis Test will have more
power.
• Kruskal Wallis Median Test• Kruskal-Wallis Median Test
– Tests equality of medians for 2 and more factor levels
– Comparison with the Mood’s Median Test:Comparison with the Mood s Median Test:
• More robust against unequal distributions
• Less robust against outliers
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 3/12
Suggestion How to Proceed…
You conduct an analysis of variance (ANOVA) and the diagnostic
tools around it. You will not trust the results if:
• the variances are not homogeneous
• the data and/or (more important) the residuals are not normal
distributed
In these cases you will make a better decision based on additionalIn these cases you will make a better decision based on additional
information coming from median tests.
If the results (p-values) confirm the conclusion of the ANOVA theIf the results (p values) confirm the conclusion of the ANOVA the
trust in the ANOVA evaluation is re-established.
If you get different results you may focus on not previouslyIf you get different results, you may focus on not previously
considered factors or on a modification of the sample plan.
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 4/12

3. Example
A high reject rate for a pneumatic module shall be reduced. During a
functional test, the module will be rejected if a specified pressure limit (pa)
will be exceededwill be exceeded.
The first improvement action, an operator training, was conducted within
th i d f D 2002 th h A 2003 (M1 M4) S b tlthe period from Dec. 2002 through Apr. 2003 (M1– M4). Subsequently
further improvements, a change of the assembly process and a new
material, were introduced. Each month samples of the reject rate were
taken to monitor and evaluated the effect of the improvement actions.
Some of the samples don’t follow a normal distribution, some showp ,
different variances.
1 Analyze if the Training efforts within M1 – M4 show a significant1. Analyze if the Training efforts within M1 – M4 show a significant
improvement.
2 Analyze the overall period M1 M82. Analyze the overall period M1 – M8
File: Pneum-test.mtw
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 5/12
Mood’s Median Test
File: Pneum-test.mtw
M1 12/2002 M2 01/2003 M3 03/2003 M4 04/2003 M5 06/2003 M6 07/2003 M7 08/2003 M8 09/2003 M1 - M4 M1 - M 8
16 89 195 8 37 31 23 36 16 M1 12/2002 16 M1 12/2002
146 147 157 29 21 41 30 31 146 M1 12/2002 146 M1 12/2002
151 77 26 4 16 23 24 30 151 M1 12/2002 151 M1 12/2002
1 67 27 145 20 31 25 32 1 M1 12/2002 1 M1 12/2002
2 163 10 87 25 31 22 28 2 M1 12/2002 2 M1 12/20022 163 10 87 25 31 22 28 2 M1 12/2002 2 M1 12/2002
41 29 135 112 16 30 30 31 41 M1 12/2002 41 M1 12/2002
1 115 1 27 26 27 28 33 1 M1 12/2002 1 M1 12/2002
8 8 18 164 17 34 26 31 8 M1 12/2002 8 M1 12/2002
105 6 16 103 17 29 22 31 105 M1 12/2002 105 M1 12/2002
1 181 27 11 22 28 30 31 1 M1 12/2002 1 M1 12/2002
89 M2 01/2003 89 M2 01/2003
147 M2 01/2003 147 M2 01/2003
Check if a significant difference in the reject rate within M1 - M4 cang j
be detected. The ANOVA test shows p-value of 0,556, no significant
improvement! Due to the fact that the data don’t show a normal
distribution a Mood‘s Median Test was performed to support thisdistribution a Mood‘s Median Test was performed to support this
assumption,.
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 6/12

4. Mood’s Median Test
Stat
>Nonparametrics
>Mood’s Median Test…
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 7/12
Mood’s Median Test
Mood Median Test: M1 - M4 versus C10
Mood median test for M1 - M4
Chi-Square = 3 51 DF = 3 P = 0 320Chi Square = 3,51 DF = 3 P = 0,320
Individual 95,0% CIs
C10 N<= N> Median Q3-Q1 +---------+---------+---------+------
M1 12/2002 6 4 12 114 (-*---------------------)
M2 01/2003 3 7 83 127 (------------*------------)
M3 03/2003 7 3 27 126 (-*-----------------------)
M4 04/2003 5 5 58 110 (---------*------------)M4 04/2003 5 5 58 110 ( )
+---------+---------+---------+------
0 50 100 150
Overall median = 29
Interpretation of the results:
The Mood‘s Median test compares the median values and performs a Chip p
Square test. The p – value is clearly below the critical Chi Square value for 3
degrees of freedom.
In addition we receive the information how many individual values are above
or below the overall median.
Wh t i d i i ?
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 8/12
What is our decision ?

5. Kruskal-Wallis Test
File: Pneum-test.mtw
M1 12/2002 M2 01/2003 M3 03/2003 M4 04/2003 M5 06/2003 M6 07/2003 M7 08/2003 M8 09/2003 M1 - M4 M1 - M 8
16 89 195 8 37 31 23 36 16 M1 12/2002 16 M1 12/2002
146 147 157 29 21 41 30 31 146 M1 12/2002 146 M1 12/2002
151 77 26 4 16 23 24 30 151 M1 12/2002 151 M1 12/2002
1 67 27 145 20 31 25 32 1 M1 12/2002 1 M1 12/2002
2 163 10 87 25 31 22 28 2 M1 12/2002 2 M1 12/20022 163 10 87 25 31 22 28 2 M1 12/2002 2 M1 12/2002
41 29 135 112 16 30 30 31 41 M1 12/2002 41 M1 12/2002
1 115 1 27 26 27 28 33 1 M1 12/2002 1 M1 12/2002
8 8 18 164 17 34 26 31 8 M1 12/2002 8 M1 12/2002
105 6 16 103 17 29 22 31 105 M1 12/2002 105 M1 12/2002
1 181 27 11 22 28 30 31 1 M1 12/2002 1 M1 12/2002
89 M2 01/2003 89 M2 01/2003
147 M2 01/2003 147 M2 01/2003
Now we evaluate if a significant difference exists within the timeg
frame M1 - M8. The ANOVA test shows a p-value of 0,015,
significant difference! Because the data have different distributions
the Kruskal Wallis Test will be performed in addition to support thisthe Kruskal-Wallis Test will be performed in addition to support this
assumption.
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 9/12
Kruskal-Wallis Test
Stat
>Nonparametrics
> Kruskal-Wallis…
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 10/12

6. K k l W lli T M1 M 8 C12
Kruskal-Wallis Test
Kruskal-Wallis Test: M1 - M 8 versus C12
Kruskal-Wallis Test on M1 - M 8
C12 N Median Ave Rank Z
M1 12/20 10 12,00 31,4 -1,33
M2 01/20 10 83,00 55,0 2,11
M3 03/20 10 26 50 37 4 0 45M3 03/20 10 26,50 37,4 -0,45
M4 04/20 10 58,00 45,4 0,71
M5 06/20 10 20,50 24,8 -2,29
M6 07/20 10 30,50 45,8 0,77M6 07/20 10 30,50 45,8 0,77
M7 08/20 10 25,50 33,1 -1,08
M8 09/20 10 31,00 51,3 1,56
Overall 80 40,5
H = 14,33 DF = 7 P = 0,046
H = 14,35 DF = 7 P = 0,045 (adjusted for ties)
At this evaluation an average rank of the numbers will be calculated. The
test statistic H is a derivation from the Chi Square distribution. In this case
the calculated value is slightly above the critical valuethe calculated value is slightly above the critical value.
Details of the evaluation e.g.: Werner Voß; “Taschenbuch der Statistik”
Wh t i d i i ?
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 11/12
What is our decision ?
Example: Yield Improvement
In the 2000 the yield of a
production process should be
1999 2000 2000 Phase Yield Phase
90,8 95,79 1 90,8 1
92,2 95,1 1 92,2 1
86 2 93 35 1 86 2 1
production process should be
improved.
How you proceed in applying
86,2 93,35 1 86,2 1
93,7 94,49 1 93,7 1
95 97,11 2 95 1
95,8 97,08 2 95,8 1
96,78 95,11 2 96,78 1
96 95 32 2 96 1How you proceed in applying
statistical test methods?
96 95,32 2 96 1
93,68 95,12 2 93,68 1
96,46 95,7 2 96,46 1
92,73 94,59 2 92,73 1
93,95 95,6 2 93,95 1
95 79 195,79 1
95,1 1
93,35 1
94,49 1
97,11 2
97 08 2
File: Median Test.mtw
This project was started in February. After collection of
the inputs and outputs improvement actions were
97,08 2
95,11 2
95,32 2
95,12 2
95,7 2
the inputs and outputs improvement actions were
determined based on results from C&E matrix and
FMEA. These improvements were implemented
94,59 2
95,6 2
between April 20 and May 6. For the baseline definition
data from 1999 and 2000 Jan. - April were used.
Knorr-Bremse Group 09 BB W3 median tests 08, D. Szemkus/H. Winkler Page 12/12