javier garcia - verdugo sanchez - six sigma training - w1 statistical methods

Statistical Methods for Process Improvements

xn

∑n

x=x 1=i

i∑n

n 2) i(X∑ − X

1-n1=i=s

Week 1

Knorr-Bremse Group

Why do We Need Statistics?• Variability

– Does a process hit the target with a minimum of variability?

Th M V l d i if i Th St d d D i ti– The Mean Value determines if a process is on target. The Standard Deviationdescribes the variability of the process.

• StabilityStability– How does the process behave over time?

– A stable process has a consistent mean and a predictable variability over time.

13 UCL=13,116

Xbar Chart of Process A

351

Xbar Chart of Process B

le M

ea

n

12

11

10__X=9,959

le M

ea

n

30

25

20UCL=18 19

1

Sa

mp

9

8

7 LCL=6,803

Sa

mp

l

15

10

__X=12,09

UCL=18,19

LCL=5 98

Sample24222018161412108642

6

,

Sample24222018161412108642

5LCL=5,98

1

Knorr-Bremse Group 08 BB W1 Statistical Methods 07, D. Szemkus/H. Winkler Page 2/37

Which of these two processes would you prefer?

Interpretation of Variation• Every process varies with time. Some processes show controlled variation,

while other processes have uncontrolled variation. (Walter Shewhart).

• A controlled variation is characterized by a stable and consistent pattern• A controlled variation is characterized by a stable and consistent pattern of variation with time. Reasons for this type of variation are common causes.

• An uncontrolled variation is characterized by unpredictable variation. Reasons here are special or assignable causes.

• Process A runs with controlled variationSpecial CProcess A runs with controlled variation.

• Process B shows uncontrolled variation.

Xb Ch t f P A Xb Ch t f P B

Causes!

13

12

UCL=13,116

Xbar Chart of Process A

35

30

1

Xbar Chart of Process B

Sa

mp

le M

ea

n

11

10

9

__X=9,959

Sa

mp

le M

ea

n

25

20

15

UCL=18,19

1

24222018161412108642

8

7

6

LCL=6,803

24222018161412108642

10

5

__X=12,09

LCL=5,98

1


Sample24222018161412108642

Sample24222018161412108642

Can We Accept Variation?

• Every process shows variation

• We accept variation if:p

– The total variation of the output is relativley small compared to the process specification and the process is on target.

– The process is stable over time.

ost

LSL USLNom

Co

The traditional view of variation

Accepted Variation

LSL USLNom

tC

ost

The new view of variation


Probabilities

Value Comb. Probability

2 1 ,0278,

3 2 ,0556

4 3 ,0833

5 4 11115 4 ,1111

6 5 ,1389

7 6 ,1667,

8 5 ,1389

9 4 ,1111

10 3 083310 3 ,0833

11 2 ,0556

12 1 ,0278,

Total 36 1,0000

Probability of a value for dice 1 = 1/6 = .1667

Probability of a value for dice 2 = 1/6 = .1667


Probability for any combination = 1/6 x 1/6 = 1/36 = .0278

Graphic of a Probability Function

Customer requirement: values between 3 and 11

.18

16

Value Comb. Probability

2 1 0278.16

.142.8%2.8% LSL USL

2 1 ,0278

3 2 ,0556

4 3 ,0833

.12

.10

5 4 ,1111

6 5 ,1389

7 6 1667

.08

.06

7 6 ,1667

8 5 ,1389

9 4 ,1111.06

.04

02

10 3 ,0833

11 2 ,0556

12 1 0278

Performance: 94 4%

.02

2 1210864 140Sum of the die results

12 1 ,0278

Total 36 1,0000


Performance: 94.4%

The Normal Distribution Curve

Units

µ


The Normal Distribution Curve

Specification Limitp(x > a) = 1

σ 2π e-(1/2)[(x - µ )/σ ]2

a

∞

dx

Area of the YieldProbabilityProbabilityof defects

+ infinite- infinite


Analytical Approach • Determine if the process is stable.

• If the process is not stable:

• Identify and eliminate the causes for instability

• If the process is stable:

• Determine/estimate the total amount of variation

• Identify the sources of variationy

• Reduce the variation

• We will now discuss statistical tools which will help us to do so.e o d scuss s a s ca oo s c e p us o do so


Overview Data Types

• Attribute Data (Qualitative)

– CategoriesCategories

– Yes, No

– Go, No goGo, No go

– Machine 1, Machine 2, Machine 3

– Pass/FailPass/Fail

• Variable Data (Quantitative)

– Discrete (Count) DataDiscrete (Count) Data

• Maintenance Equipment Failures, Number of HV-Arcs

• Number of Customer ReturnsNumber of Customer Returns

• Defects per Unit

– Continuous DataContinuous Data

• Decimal Subdivisions are meaningful

• Time, Pressure, Conveyor Speed


Time, Pressure, Conveyor Speed

A Selection of Statistical Techniques

Factor X = Input

Discrete / Attributive Continuous / VariableDiscrete / Attributive Continuous / Variable

ut

te ve

L i ti

= O

utp

u

Dis

cret

Attr

ibut

ivChi - Square

Logistic

Regression

nse

Y = A

s

Res

po

n

ntin

uou

aria

ble T - Test

ANOVA ( F - Test) Regression

R

Con Va

Median Tests

Statistical techniques for all combination of data types are available


y

Basic Definitions in Statistics

• Types of data

• Measuring scaleMeasuring scale

• Measures for the center of the data

M• Mean

• Median

• Measures for the variation of data

• RangeRange

• Variance

S d d d i i• Standard deviation

• Normal distribution and normal probabilities

• Standard (Z) transformation

• Process capability metrics


Process capability metrics

Sample vs. Population

X = Mean value of a sample µ = Mean value of the population

= Standard Deviation of a sample

σ = Standard Deviation of the population

S

Statistics Estimation Parameter


Two Important Statistical Equations

Xi

N

∑Mean of the

µ =N

i=1

)(XN

2∑

population

N

)(X= 1=i

2i∑ −µ

σStandard deviation of the population

n

∑

Nthe population

Mean of the sample

x=x 1=i

i∑sample

St d d d i ti f

n n 2) i(X∑ − XStandard deviation of

the sample1-n

1=i)i(

=s

∑


1n

Description of the Center

• Mean: Arithmetic average of the data

– Reflects the influence of all data

– Strongly influenced by extreme values

xx

ni

i

n

==∑

1

• Median: Reflects the 50% rank – the center of d t ft ti f l t hi h

i=1

data after sorting from low to high

– Does not include all values in the calculation

– Is “robust” to extreme values

Two successive steps influence the mean additively


Description of the Variation• The Range is the distance between the extreme values of a

data set.

• The Variance is the sum of the average squared deviation of each data point from the mean divided by the degrees of freedomfreedom.

• The Standard Deviation is the square root of the variance.

The most common and useful measure of variation is theThe most common and useful measure of variation is the standard deviation

You can calculate the variance of two steps by

Important!2A Variance of step AIf σ =

You can calculate the variance of two steps by adding the variances.

Important!

222222

2B

A

Variance of step Band

Variance of step AIf

σσ

=

2B

2ATotal

2B

2A

2Difference

2B

2A

2Total andthan

σσσ

σσσσσσ

+=

−=+=


BATotal

The Calculation of the Standard Deviation

2)X-(X X-X X11234

n 2

4567

11=i

2)i(X∑ − X78910 1-n10Σ

MeanVariance n

1=i

2) i(X∑ − XVariance

s

1-nAssignment: Calculate the Standard Deviation of the following numbers: 2, 1, 3, 5, 4, 3 Use the form sheet above


the form sheet above.

The Effect of the Quadratic Deviation

By squaring the deviations extreme values heavily effect the meaneffect the mean.

(x - x)2(x x)

100

50

Sq-

De

v

0

S

1050

0

Deviates


Descriptive Statistics for 3 DistributionsStat

>Basic Statistics

>Display Des>Display Des...

File: Distribution mtw

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

File: Distribution.mtw

Q Q

Symmetric 500 0 70,000 0,447 10,000 29,824 63,412 69,977 76,653

Pos Asym 500 0 70,000 0,447 10,000 62,921 63,647 65,695 72,821


Neg Asym 500 0 70,000 0,447 10,000 1,866 67,891 73,783 76,290

The Histogram of these Distributions

140

120

Histogram of Pos AsymStat

>Basic Statistics

>Display Des

Fre

qu

en

cy

100

80

60

>Display Des...

>Graphs Positive Asymmetry

P A130120110100908070

40

20

0

Pos Asym

Symmetric

70

60

Histogram of Symmetric

250

200

Histogram of Neg Asym

Symmetric Distribution

Fre

qu

en

cy

50

40

30

Fre

qu

en

cy

200

150

100

Negative Asymmetry

10090807060504030

20

10

07260483624120

50

0


Symmetric Neg Asym

Mean & MedianMean, Median

70 Mean 70 00

Histogram (with Normal Curve) of Symmetric

Histogram with Normal Curve

en

cy

60

50

40

Mean 70,00StDev 10,00N 500

Fre

qu

e

30

20

10

Symmetric10090807060504030

0

140

120

Mean 70,00StDev 10,00N 500

Histogram (with Normal Curve) of Pos Asym

250

200

Mean 70,00StDev 10,00N 500

Histogram (with Normal Curve) of Neg Asym

MeanMean

Fre

qu

en

cy

100

80

60 Fre

qu

en

cy

200

150

100

Median

Mean

Median

125,0112,5100,087,575,062,550,0

40

20

0847260483624120

50

0


Pos Asym,,,,,,,

Neg Asym

Different Forms of Distributions

Distribution 1

Distribution 2Distribution 2

Distribution 3

How do you interpret the differences?


y p

Areas under the Normal Curve

0.4

68 %

0.3

95 %

equ

ency

0.2

Fre

0.1

99,73 %

0

0 1 2 3 4-1-2-3-4Output

0.0

The shape of the curve is determined by the Standard Deviation and the Mean.


Deviation and the Mean.

Areas under the Normal Curve

Rules of thumb for the normal distribution

Rule 1• Roughly 60-75% of all data are in the area of

+/- 1 standard deviation from the mean.

Rule 2• Usually 90-98% of the data are in the area ofUsually 90 98% of the data are in the area of

+/- 2 standard deviation from the mean.

Rule 3Rule 3• About 99-100% of the data are in the area of

+/- 3 standard deviation from the mean+/- 3 standard deviation from the mean.

File. Distribution.mtw


File. Distribution.mtw

Testing for Normal Distributions

• Diagrams to describe test results for normal distribution are useful. We get information about the behavior of the distribution. If h d f ll l di ib i h l b biliIf the data follows a normal distribution the normal probability diagram displays a straight line.

Mi i b d hi d hi di (• Minitab does this test and generates this diagram (see next page). Additional to the graph Minitab displays „A square“ and a p value“„p value .

• A square is a calculated test value after Anderson/Darling. Its value shows the summed squared distances of the single datavalue shows the summed squared distances of the single data points from the straight line. Big A square values indicate that the data don’t follow a normal distribution.

• The p value helps to decide whether the data are normally distributed or not.

p values <0,05: Data are non normal

p values >0,05: Normal distributed Data


Testing for Normal DistributionsStat

>Basic Statistics

>Normality Test>Normality Test

70

60

Mean 70,00StDev 10,00N 500

Histogram (with Normal Curve) of Symmetric

99,9

99

Mean 70,00StDev 10,00N 500AD 0 418

Probability Plot of SymmetricNormal

Fre

qu

en

cy

50

40

30

Pe

rce

nt

9590

807060504030

0,328AD 0,418P-Value

20

10

0

20

10

5

1

0,1


Symmetric10090807060504030

Symmetric11010090807060504030


140

120

Mean 70,00StDev 10,00N 500

Histogram (with Normal Curve) of Pos Asym

qu

en

cy

120

100

80

Fre

q

60

40

20

Pos Asym125,0112,5100,087,575,062,550,0

0

99,9Mean 70,00

Probability Plot of Pos AsymNormal

en

t

99

9590

807060

<0,005

StDev 10,00N 500AD 46,489P-Value

Pe

rce

50403020

10

5

1

Pos Asym130120110100908070605040

1

0,1



250 Mean 70,00StDev 10,00N 500

Histogram (with Normal Curve) of Neg Asym

qu

en

cy

200

150

Fre

q

100

50

Neg Asym847260483624120

0

99,9Mean 70,00

Probability Plot of Neg AsymNormal

en

t

99

9590

807060

<0,005

StDev 10,00N 500AD 44,491P-Value

Pe

rce

50403020

10

5

1

Neg Asym100806040200

1

0,1


Analyze a Mystery Distribution

Generate a normal distribution diagram for the mystery data set C4 What is your conclusion?

Probability Plot of Mystery

mystery data set C4. What is your conclusion?

99,9

99

Mean 100,0StDev 32,38

Probability Plot of MysteryNormal

99

9590

80

<0,005

N 500AD 27,108P-Value

Pe

rce

nt 70

6050403020

10

5

1

Mystery200150100500

0,1


Descriptive Statistics with MinitabStat

>Basic Statistics

>Graphical Summary>Graphical Summary…

A nderson-Darling Normality Test

A -Squared 0,42

Summary for Symmetric

V ariance 100,000Skewness -0,050008Kurtosis 0,423256

A Squared 0,42P-V alue 0,328

Mean 70,000StDev 10,000

10090807060504030

Kurtosis 0,423256N 500

Minimum 29,8241st Q uartile 63,412Median 69,9773rd Q uartile 76,653Maximum 103,301,

95% C onfidence Interv al for Mean

69,121 70,879

95% C onfidence Interv al for Median

69,021 70,737

95% C onfidence Interv al for StDev95% Confidence Intervals

Median

Mean

71,070,570,069,569,0

9,416 10,66295% Confidence Intervals


Descriptive Statistics with Minitab


A -Squared 0,42

Summary for Symmetric

V ariance 100,000Skewness 0 050008

A Squared 0,42P-V alue 0,328

Mean 70,000StDev 10,000

Skewness -0,050008Kurtosis 0,423256N 500

Minimum 29,8241st Q uartile 63,412

10090807060504030

Median 69,9773rd Q uartile 76,653Maximum 103,301


69,121 70,879, ,


69,021 70,737

95% C onfidence Interv al for StDev


Median

Mean

71,070,570,069,569,0

Skewness und Kurtosis are related on asymmetry and flatness of the di t ib ti Th l t 0 th l di t ib t d


distribution. The closer to 0, the more normal distributed.



A -Squared 46,49

Summary for Pos Asym


A Squared 46,49P-V alue < 0,005

Mean 70,000StDev 10,000

Skewness 2,41707Kurtosis 6,93041N 500


130120110100908070



69,121 70,879, ,


65,260 66,501



Median

Mean

71706968676665

A positive skewness number shows a positive distortion. A kurtosis number h hi h k f th di t ib ti


shows a high peak of the distribution.



A -Squared 44,49

Summary for Neg Asym


A Squared 44,49P-V alue < 0,005

Mean 70,000StDev 10,000

Skewness -2,8688Kurtosis 11,5897N 500


7260483624120



69,121 70,879, ,


73,162 74,326



Median

Mean

75747372717069

A negative skewness number shows a negative distortion. The kurtosis b i iti i d h hi h k


number is positive again and shows a high peak.


Summary for MysteryA nderson-Darling Normality Test

A -Squared 27,11P-V alue < 0,005

Mean 100,00StDev 32 38V ariance 1048,78Skewness 0,00716Kurtosis -1,63184N 500

StDev 32,38

160140120100806040

Minimum 41,771st Q uartile 68,69Median 104,203rd Q uartile 130,81Maximum 162,82

f d l f95% C onfidence Interv al for Mean

97,15 102,85


82,78 117,66


Median

Mean



1201101009080


Probability Plots Graph

>Probability Plot…

99 9

Probability Plot of SymmetricNormal - 95% CI

99,9

99

95

Mean 70,00StDev 10,00N 500AD 0,418P-Value 0,328

90

8070605040e

rce

nt

,

403020

10

5

Pe

11010090807060504030

1

0,1

S t i


Symmetric

Some Exercises

• Analyze the variable Y in the file Delivery Time.mtw.y y

Are the data normal distributed?

I hi h t 95% f th l ?• In which area you expect 95% of the values?

• Analyze the variable Y, days / receiving, in the file Late Payment.mtw

• What kind of distribution?

• In which area you expect 99% of the values?y p


Summary

• Types of data

• Scale of measurements

• Measure of the center of the data• Measure of the center of the data

• Mean

• Median

• Measure of the spread of dataMeasure of the spread of data

• Range

• Variance

• Standard deviation

• Normal distribution and normal probabilities


javier garcia - verdugo sanchez - six sigma training - w1 statistical methods

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