Median Tests

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Mean 66 4

Probability Plot of M1 - M4Normal

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6050ce

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Mean 66,4StDev 63,72N 40AD 2,311P-Value <0,005

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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

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Kruskal Wallis Test

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.

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ExampleA 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 show p ,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

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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 816 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 12/2002151 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 12/20022 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 12/200241 29 135 112 16 30 30 31 41 M1 12/2002 41 M1 12/20021 115 1 27 26 27 28 33 1 M1 12/2002 1 M1 12/20028 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 12/20021 181 27 11 22 28 30 31 1 M1 12/2002 1 M1 12/2002

89 M2 01/2003 89 M2 01/2003147 M2 01/2003 147 M2 01/2003

Check if a significant difference in the reject rate within M1 - M4 can g jbe 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,.

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Mood’s Median TestStat

>Nonparametrics

>Mood’s Median Test…

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Mood’s Median TestMood Median Test: M1 - M4 versus C10

Mood median test for M1 - M4Chi-Square = 3 51 DF = 3 P = 0 320Chi Square = 3,51 DF = 3 P = 0,320

Individual 95,0% CIsC10 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 Chi p pSquare 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 ?

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What is our decision ?

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 816 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 12/2002151 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 12/20022 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 12/200241 29 135 112 16 30 30 31 41 M1 12/2002 41 M1 12/20021 115 1 27 26 27 28 33 1 M1 12/2002 1 M1 12/20028 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 12/20021 181 27 11 22 28 30 31 1 M1 12/2002 1 M1 12/2002

89 M2 01/2003 89 M2 01/2003147 M2 01/2003 147 M2 01/2003

Now we evaluate if a significant difference exists within the time gframe 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.

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Kruskal-Wallis Test Stat

>Nonparametrics

> Kruskal-Wallis…

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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 ZM1 12/20 10 12,00 31,4 -1,33M2 01/20 10 83,00 55,0 2,11M3 03/20 10 26 50 37 4 0 45M3 03/20 10 26,50 37,4 -0,45M4 04/20 10 58,00 45,4 0,71M5 06/20 10 20,50 24,8 -2,29M6 07/20 10 30,50 45,8 0,77M6 07/20 10 30,50 45,8 0,77M7 08/20 10 25,50 33,1 -1,08M8 09/20 10 31,00 51,3 1,56Overall 80 40,5

H = 14,33 DF = 7 P = 0,046H = 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 ?

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What is our decision ?

Example: Yield Improvement

In the 2000 the yield of a production process should be

1999 2000 2000 Phase Yield Phase90,8 95,79 1 90,8 192,2 95,1 1 92,2 186 2 93 35 1 86 2 1production process should be

improved.

How you proceed in applying

86,2 93,35 1 86,2 193,7 94,49 1 93,7 195 97,11 2 95 1

95,8 97,08 2 95,8 196,78 95,11 2 96,78 1

96 95 32 2 96 1How you proceed in applying statistical test methods?

96 95,32 2 96 193,68 95,12 2 93,68 196,46 95,7 2 96,46 192,73 94,59 2 92,73 193,95 95,6 2 93,95 1

95 79 195,79 195,1 1

93,35 194,49 197,11 297 08 2

File: Median Test.mtw

This project was started in February. After collection of the inputs and outputs improvement actions were

97,08 295,11 295,32 295,12 295,7 2the inputs and outputs improvement actions were

determined based on results from C&E matrix and FMEA. These improvements were implemented

94,59 295,6 2

between April 20 and May 6. For the baseline definition data from 1999 and 2000 Jan. - April were used.

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