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Evaluation of Non fNormal Distributions
Histogram (with Normal Curve) of C3
18
16
14
Mean 4,924StDev 3,092N 100
Fre
qu
en
cy
12
10
8
F
6
4
2
C315129630
0 Week 2
Knorr-Bremse Group
About this Module
How can we handle non normal data?
This module will show you possibilities how to process non normal distributed data setsnon normal distributed data sets.
Content
• The Box Cox transformation of dataThe Box Cox transformation of data
• The Johnson transformation of data
• Capability Analysis of Non Normal Data
• Examination of the mean• Examination of the mean
• Central limit theorem
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 2/39
When do we Need Normal Distribution?
For the following evaluations we need normal distributed data:
• Description of the process spread with mean and standard deviation
• Calculation of confidence intervals
C ti f t l h t (SPC)• Creation of control charts (SPC)
• Calculation of process capability metrics
Recommendation:Recommendation:
At first try to transfer the data. This is relatively fast and simple If that does not show success than use thesimple. If that does not show success than use the central limit theorem. That requires an increased number of samples which costs time and money.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 3/39
p y
Data Transformation
Which type of transformation?
Fil T f ti t (Th ifi ti li it i 12)File: Transformation.mtw (The upper specification limit is 12)
First review the raw data, which transformation will be successful?
Try different transformation methods:
− Square root
− Natural logarithm, ln
Logarithm base 10 log− Logarithm base 10, log 10
− Square
Review each transformation with histograms and normal distribution curves.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 4/39
Which of the transformation is close to a normal distribution?
Transformation with Minitab
With the transfer function of Minitab you can search for the best exponent to reach normal distribution of the data set The exponentexponent, to reach normal distribution of the data set. The exponent
is named lambda.
λ= Y*Y = YY
Here some examples of
Lamda Werte Transformation2y2=λ
Lambda Values
Here some examples of lambda values for the transfer function.
yyln0=λ
5,0=λ
y/1y/1
5,0−=λ1−=λ
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 5/39
The Box Cox TransformationStat>Control Charts>Box-Cox Transformation…
7
Lower CL Upper CL
Estimate 0,23
(using 95,0% confidence)
Lambda
Box-Cox Plot of Raw data
6
5
De
v
Estimate 0,23
Lower CL 0,13Upper CL 0,32
Rounded Value 0,23
4
3
StD
3210-1
2
Lambda
Limit
Minitab calculates the lambda value for the best transformation and the 95
% confidence interval
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 6/39
Process Capability with Lambda ValuesStat>Quality Tools>Capability Analysis
N lUSL*
Process Capability of Raw dataUsing Box-Cox Transformation With Lambda = 0,23
>Normal…>Box Cox
transformed dataLSL *Target *USL 12Sample Mean 2,83266Sample N 500StDev (O v erall) 2,23467
Process DataPp *PPL *PPU 0,80Ppk 0,80C pm *
O v erall C apability
Sample Mean* 1,19844StDev (O v erall)* 0,239086
LSL* *Target* *USL* 1,77097
A fter Transformation
1,81,61,41,21,00,80,6
O bserv ed Performance Exp. O v erall PerformancePPM < LSL *PPM > USL 4000,00PPM Total 4000,00
O bserv ed PerformancePPM < LSL* *PPM > USL* 8317,96PPM Total 8317,96
Exp. O v erall Performance
Subsequent you evaluate the data of the capability analysis. Activate under
Options - Box-Cox power p ptransformation. Enter the before calculated value of the exponent.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 7/39
Control Chart with Lambda Values15
10UCL=9 39
1
1
11
11
I Chart of Raw dataStat>Control Charts>Variable Charts for Individuals
Ind
ivid
ua
l Va
lue
5_X=2,83
UCL=9,39>Individuals… I Chart Options>Box-Cox
Observation500450400350300250200150100501
0
-5LCL=-3,72
2,00UCL 1 918
I Chart of Raw dataUsing Box-Cox Transformation With Lambda = 0,23
Observation
alu
e
1,75
1,50
UCL=1,918
Ind
ivid
ua
l Va
1,25
1,00
_X=1,193
500450400350300250200150100501
0,75
0,50 LCL=0,469
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 8/39
Observation
Comparison Box Cox / Johnson Transformation
With the Box Cox Transformation we try to transform not normal distributed data into normal distributed data for a simplified analysis.y
Johnson used the reverse way:Johnson used the reverse way:
He was aware, that the normal distribution has a lot of advantages (easy understandable well known to use with many good and know(easy understandable, well known, to use with many good and know tools)
He tried to “distort” a normal distribution with a combination of shape keeping and shape changing transformations. He evaluated which shapes could be later transformed back into a normal distribution.normal distribution.
With MINITAB 14 this transformation is now available.
Lets have a detailed look on this transformation:
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 9/39
Johnson Transformation in General
Johnson differentiates between 3 in principle different transformation approaches (MINITAB determines the most suitable one):
− Johnson distribution SBSB stands for system two sided limited data, this is e.g. a U-distribution. Therefore we can use data which even the Box Cox transformation can notTherefore we can use data, which even the Box Cox transformation can not use.
− Johnson distribution SLSL stands for system one sided limited, this is the case with e.g. a log.-normal distribution or also a F-distribution. These data can be limited transformed with Box Cox.
− Johnson distribution SUSU stands here for two side unlimited data. These data should be transformed with the Box Cox transformation without problems toowith the Box Cox transformation without problems too.
Which shape these three types of distribution may have and which equations result out of these distributions you can see on the next pagesequations result out of these distributions you can see on the next pages.
Here we can see, that up to four unknowns have to be determined. This can be solved only with computer support
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 10/39
can be solved only with computer support.
Johnson Distribution SB
SB => System two sided limited
Function: x‘= λ + δ ∗ ln[(x ε) / (γ + ε x)]
f(x) λ 1 0
Function: x = λ + δ ∗ ln[(x − ε) / (γ + ε − x)]
Where:(x) λ = 1,0
δ = 0,5x => Original value &
x‘ => transformed valueλ = 0δ = 1,0 λ ≈ 0
δ = 0,6
λ = 0
δ 0,6
Hint: δ = 0,5Hint:In MINITAB use for „ln“ -„Natural log“! Thi lt i th f l i
γεx
This results in the formula in „LOGE()“
In Excel: LN()
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 11/39
γεIn Excel: LN()
Johnson Distribution SL
SL => System one sided limited (log-normal distributed data)
Function: x‘= λ + δ ∗ ln(x ε)
f(x)
Function: x = λ + δ ∗ ln(x − ε)
Where:(x)
x => Original value &
x‘ => transformed value
Hint:Hint:In MINITAB use for „ln“ -„Natural log“! Thi lt i th f l i
εx
This results in the formula in „LOGE()“
In Excel: LN()
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 12/39
εIn Excel: LN()
Johnson Distribution SU
SU => System two sided unlimited
Function: x‘= λ + δ ∗ Sinh-1[(x ε) / λ]
f(x)
Function: x = λ + δ ∗ Sinh 1[(x − ε) / λ]
Where: (x)
x => Original value &
x‘ => transformed value
λ = 1,0δ = 0,5
λ = 2δ = 0 5
x
δ = 0,5
Hint:In Excel: ARCSINHYP()
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 13/39
In Excel: ARCSINHYP()
Example Johnson 1 (1)
Lets start with a transformation of a uniform distributed set of data.
In the file Johnson1 mtw the Output1“ has 700 valuesIn the file Johnson1.mtw the „Output1 has 700 values.
The histogram and the normal distribution test shows a significant deviation from a normal distributiondeviation from a normal distribution.
File: JOHNSON1. MTW
Histogram of Output1Normal
Probability Plot of Output1Normal - 95% CI
50
40
Mean 3,010StDev 0,5707N 700
Normal
99,99
99
95
Mean
<0,005
3,010StDev 0,5707N 700AD 8,033P-Value
Normal - 95% CI
Fre
qu
en
cy 30
20 Pe
rce
nt 80
50
20
5
Output14,03,63,22,82,42,0
10
0
Output154321
5
1
0,01
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 14/39
Output1 Output1
Example Johnson 1 (2)Lest try a first Johnson Transformation. From the theory MINITAB should use the System SB. After the transformation of Output1 we store the results in the column trans“store the results in the column „trans .
Stat>Quality Tools>Johnson Transformation…
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 15/39
Example Johnson 1 (3)
The graphical analysis:
99,99N 700 t 0 8
0,76Probability Plot for Original Data Select a T ransformation
Johnson Transformation for Output1Normal distribution test without
Transformation
Per
cent
99
90
50
N 700AD 8,033P-Value <0,005
alue
for
AD
tes
t 0,8
0,6
0,4
0,2
Transformation
P
5,54,02,51,0
10
1
0,01 Z Value
P-V
a
1,21,00,80,60,4
,
0,0Ref P
(P-Value = 0.005 means <= 0.005)Normal distribution
test with
t
99,99
99
90
N 700AD 0,241P-Value 0,774 P-V alue for Best F it: 0,773999
Z for Best F it: 0,76Best Transformation Ty pe: SB
Probability Plot for T ransformed Datatest with
transformation Equation for the transformation
Per
cent
50
10
1
Best Transformation Ty pe: SBTransformation function equals-0,0223806 + 0,645292 * Log( ( X - 1,96884 ) / ( 4,04224 - X ) )
The transformation was successful
5,02,50,0-2,50,01
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 16/39
The transformation was successful
Example Johnson 1 (4)
If we want to determine the portion outside the specification limit we use the capability analysis for non normal distribution.
Stat>Quality Tools>Capability Analysis>Nonnormal…
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 17/39
Example Johnson 1 (5)We receive:
Process Capability of Output1
Equation for the transformation
USL*f
Process Capability of Output1Johnson Transformation with SB Distribution Type
-0,022 + 0,645 * Log( ( X - 1,969 ) / ( 4,042 - X ) )
transformed dataProcess Data
Sample N 700
LSL *Target *USL 3,5Sample Mean 3,0096
O v erall C apabilityPp *PPL *PPU 0,22Ppk 0,22
Sample N 700StDev 0,570739Shape1 -0,0223806Shape2 0,645292Location 1,96884Scale 2,0734
Exp. O v erall PerformancePPM < LSL *PPM > USL 251344PPM Total 251344
A fter Transformation
LSL* *Target* *USL* 0,647477Sample Mean* -0,0166779StDev * 0 990882StDev * 0,990882
O bserv ed PerformancePPM < LSL *PPM > USL 258571PPM Total 258571
2,41,60,80,0-0,8-1,6-2,4
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 18/39
Exercise Johnson 2
How does the Johnson transformation react onHow does the Johnson transformation react on outliers?
Try the transformation of the data „Output2“ in the same worksheet.same worksheet.
Does it work without the outliers?
What is the big difference in both equations?
Can you understand the difference?
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 19/39
Example Johnson 3 (1)
Lets try with Output3:
500
400
Mean 21,16StDev 104,8N 500
Histogram of Output3Normal
Fre
qu
en
cy 300
200
100
0
Probability Plot of Output3Normal - 95% CI
Output321001800150012009006003000 99,9
99
9590
80
Mean
<0,005
21,16StDev 104,8N 500AD 136,954P-Value
No normal distribution!
It seems that the data can not be below 0
Pe
rce
nt
80706050403020
10
be below 0 (System SL?).
Output32000150010005000
5
1
0,1
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 20/39
Output3
Example Johnson 3 (2)
Stat>Quality Tools>J h T f ti>Johnson Transformation…
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 21/39
Example Johnson 3 (3)
We receive:
Johnson Transformation for Output3
99,9
99
90
N 500AD 136,954P-Value <0,005
AD
tes
t 0,60
0,45
0,68Probability P lot for Original Data Select a T ransformation
Johnson Transformation for Output3
Per
cent
90
50
10
1
P-V
alue
for
A
1,21,00,80,60,40,2
0,30
0,15
0,00Ref P
2000100000,1
99,9
99N 500AD 0,314
Z Value
P V l f B t F it 0 544750
Probability P lot for T ransformed Data
(P-Value = 0.005 means <= 0.005)
Per
cent
90
50
10
P-Value 0,545 P-V alue for Best F it: 0,544750Z for Best F it: 0,68Best Transformation Ty pe: SLTransformation function equals-0,532048 + 0,502955 * Log( X - 0,986507 )
The transformation was successfully! Minitab used the System SL
420-2
1
0,1
The transformation was successfully! Minitab used the System SL. The equation is shorter that with the system SB. Calculate the proportion above the limit of 200.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 22/39
p p
Example Johnson 4 (1)
In the last example we use both sided unlimited data
Th l O t t i l d 1000 l Fil JOHNSON2SU MTWThe column Output includes 1000 values.
Since we have positive and negative values a Box Cox-T f ti ld t k ith t ti
File: JOHNSON2SU. MTW
Transformation would not work without preparation.
Normality check results in non normal distribution!
Probability Plot of OutputNormal - 95% CI
Histogram of OutputNormal
99,99
99
95
Mean
<0,005
38,06StDev 1011N 1000AD 7,298P-Value
Normal 95% CI
140
120
100
Mean 38,06StDev 1011N 1000
Normal
Pe
rce
nt 80
50
20Fre
qu
en
cy
100
80
60
500025000-2500-5000
5
1
0,0140003000200010000-1000-2000
40
20
0
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 23/39
OutputOutput
Example Johnson 4 (2)
The result of the transformation:
99,99N 1000 t 0 8
1Probability Plot for Or iginal Data Select a T ransformation
Johnson Transformation for Output
rcen
t
99
90
50
N 1000AD 7,298P-Value <0,005
e fo
r A
D t
est 0,8
0,6
0,4
Per
500005000
10
1
0,01 Z Value
P-V
alu
1,21,00,80,60,40,2
0,2
0,0Ref P
50000-5000
99,99
99
N 1000AD 0,238
P V l f B F i 0 783934
Probability P lot for T ransformed Data
(P-Value = 0.005 means <= 0.005)
Per
cent
99
90
50
10
,P-Value 0,784 P-V alue for Best F it: 0,783934
Z for Best F it: 1Best Transformation Ty pe: SUTransformation function equals-4,33230 + 3,22900 * A sinh( ( X + 2788,42 ) / 1519,24 )
5,02,50,0-2,5
10
1
0,01
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 24/39
Example Johnson 4 (3)
What is the change, if we shift the mean by addition of 2352?
P-V alue for Best F it: 0,783934Z for Best F it: 1Best Transformation Ty pe: SUTransformation function equals
P-V alue for Best F it: 0,783934Z for Best F it: 1Best Transformation Ty pe: SUTransformation function equals
Only one parameter of the transformation equation has changed.
Transformation function equals-4,33230 + 3,22900 * A sinh( ( X + 436,415 ) / 1519,24 )
Transformation function equals-4,33230 + 3,22900 * A sinh( ( X + 2788,42 ) / 1519,24 )
y p q g
99 99
Probability Plot of transoutput2352; transNormal - 95% CI
99,99
99
95Mean
0,78447,78 10,38 1000 0,613 0,111
StDev N AD P-0,01120 0,9922 1000 0,238
Variabletransoutput2352transIn fact, the Box Cox
transformationk l B t it th t
Pe
rce
nt 80
50
20
47,78 10,38 1000 0,613 0,111
works also. But it seems, that the model fit in this example with Johnson Transformation
5
1
0,01
with Johnson Transformation is more effective. (higher P-value)
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 25/39
Data9080706050403020100
Identify the Individual DistributionStat>Quality Tools>Individual Distribution Identification…
Goodness of F it Test
Probability Plot for Raw dataNormal - 95% C I Exponential - 95% C I
Pe
rce
nt
99,9
99
90
50
Pe
rce
nt
99,99
90
50
10
NormalA D = 14,076 P-V alue < 0,005
ExponentialA D = 10,394
Raw data
P
1680
10
1
0,1
Raw data
P
100,00010,0001,0000,1000,0100,001
1
P-V alue = 0,195
Gamma
P-V alue < 0,003
WeibullA D = 0,529
Minitab runs the goodness of fit test for all theoretical distributions
en
t
99,99
90
50
10 en
t
99,99
99
90
50
GammaA D = 0,349 P-V alue > 0,250
Weibull - 95% C I Gamma - 95% C I theoretical distributions.
Together with the Anderson Darling Value
R d t
Pe
rce
10,001,000,100,01
10
1
R d t
Pe
rce
10,001,000,100,01
10
1
Anderson Darling Value Minitab is calculating the
p-Value.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 26/39
Raw data Raw datap
Capability Analysis of Non Normal DataBased on the best fit (lowest AD value and highest p-value) Minitab is able to run a capability analysis. In our example we get the best fit
with a Gamma distributionwith a Gamma distribution.
Process Capability of Raw dataCalculations Based on Gamma Distribution Model
USL
LSL *Target *
Process DataPp *PPL *
O v erall C apability
Stat
USL 12Sample Mean 2,83266Sample N 500Shape 1,57919Scale 1,79374
PPU 0,81Ppk 0,81
*O bserv ed Performance
PPM < LSL *PPM > USL 4537,96PPM Total 4537,96
Exp. O v erall Performance
>Quality Tools>Capability Analysis>Nonnormal…
PPM < LSL *PPM > USL 4000,00PPM Total 4000,00
14121086420
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 27/39
14121086420
The Quality of MeansHow precise will a sample average predict the center of the population?
Also a mean value has a standard deviation which is called the standard error of the mean (SEMean Standard Error).
sSE x= s
s x=n
SEMean
U i th t d d f th ti t th lit
ns
x
Using the standard error of the average we can estimate the quality of the mean (see confidence interval).
The equation indicates that a mean is more robust than the sample by the factor of the square root of the sample size.
Transferred to the praxis this means that e.g.: it diminishes the measurement error by a half using an average out of 4 readings.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 28/39
Think about the Following Exercise
• Let´s assume you have a bucket filled with a large number of hit h t O h h t t b tiwhite sheets. On each sheet you wrote a number representing a
normal distribution with a mean and a standard deviation.
P ll d l 9 h t l l t th f th b d• Pull randomly 9 sheets, calculate the mean of these numbers and note this number on a green paper. Subsequently you put the white papers back in the bucket.white papers back in the bucket.
• Put the green paper in another bucket.
R t th t til th th b k t i f ll f h t• Repeat that until the other bucket is full of green sheets.
• The contents of the bucket with the white papers represents the l ti f th i di id l lpopulation of the individual values.
• The contents of the bucket with the green papers represents di t ib ti f th th ldistribution of the mean the samples.
• Lets make this exercise with Minitab.
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 29/39
Lets Proof the Theory We generate simulated data to proof the theory.
Generate 9 columns with numbers from a normal distribution with a mean = 70 and standard deviation of 9.
The columns C1 to C9 represents the white sheets, column C10 the green sheets.
What standard deviation do we expect in column C10? What standard de iation do e act all ha e?deviation do we actually have?
Calc>Random Data
Calc>Row Statistics>Random Data
>Normal…>Row Statistics…
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 30/39
The Evaluation in Minitab
Variable N Mean Median TrMean StDev SE Mean
C1 250 70,022 69,755 70,004 9,214 0,583C1 250 70,022 69,755 70,004 9,214 0,583
C2 250 69,683 69,743 69,672 8,998 0,569
C3 250 71,287 71,928 71,324 9,072 0,574
C4 250 69,158 68,719 69,072 8,807 0,557C4 250 69,158 68,719 69,072 8,807 0,557
C5 250 71,088 70,894 71,167 9,381 0,593
C6 250 70,343 70,585 70,416 9,068 0,574
C7 250 70 630 70 991 70 697 9 208 0 582C7 250 70,630 70,991 70,697 9,208 0,582
C8 250 70,222 70,057 70,203 8,650 0,547
C9 250 69,990 70,076 69,953 8,776 0,555
C10 250 70 269 70 219 70 267 3 055 0 193C10 250 70,269 70,219 70,267 3,055 0,193
9sEnter the equation:
How can we describe
99
3 =n
ss x
x=
How can we describe these values?
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 31/39
The Graphical EvaluationSingle values vs. Mean values
I Chart of Values by Subs
1001 Mean
1
I Chart of Values by Subs
lue
90
80 UCL=79,411
div
idu
al V
al
70_X=69,91
UCL 79,41
Ind
60
50
LCL=60,421
Spread of single values:
Observation500450400350300250200150100501
40 45 – 95
Spread of means (n=9):Spread of means (n=9):
60 - 80
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 32/39
View of Mean Values
St d d e o le i e d Std
10Stdv 10Stdv 5
Variable
Standard error vs. sample size n and Stdv
8
or
Stdv 1
6
4an
da
rd E
rro
4
2
Sta
5040302010
0
5040302010Sample Size
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 33/39
The Central Limit Theorem
The central limit theorem gives the condition for the convergence of a sequence of distributions to a normal distribution.
Lets look at any distributed population with the mean value µ and the variance σ2. Then we can make the following statement: The gdistribution of mean values approximates with a growing sample size n against a normal distribution with the expected mean = µ and the
i 2/variance = σ2/n
Z transformation of mean values: Z – transformation of individual Z - transformation of mean values:values of a sample:
xZ
n
n σµ−
=σ
µ-x=Z
nσ
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 34/39
Exercise
Lets use the central limit theorem for the investigation of non normal distributed data.
File: Central Limit.mtw
H h 10 l ith d d t hi h f ll ChiHere we have 10 columns with random data which follow a Chi –square distribution. From the columns 1 – 5 and 1 – 10 we have
calculated means values.
Investigate the distribution of the columns. How many observations do we need until a near normal distribution appears.
Rule:
pp
Rule:
The stronger the distribution of the single values d i t f l di t ib ti th l th ldeviates from a normal distribution the larger the sample
size has to be. (minimal = 3; maximal = 30)
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 35/39
The Individual Values
20
15
1
1
I Chart of 1
18
16
Mean 5,255StDev 3,394N 100
Histogram of 1Normal
vid
ua
l Va
lue
15
10
_X 5 25
UCL=14,75
qu
en
cy
16
14
12
10
N 100
Ind
iv 5
0
X=5,25
LCL= 4 24
Fre
q 8
6
4
2
Observation1009080706050403020101
-5LCL=-4,24
115129630-3
0
Probability Plot of 1
99,9
99
95
Mean
<0 005
5,255StDev 3,394N 100AD 3,235P-Value
Probability Plot of 1Normal
Pe
rce
nt
90
80706050403020
<0,005P Value
20151050-5
10
5
1
0,1
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 36/39
1201510505
Mean Values; n = 5
18 Mean 4,886StDev 1 426
Histogram of n=5Normal
cy
16
14
12
StDev 1,426N 100
Fre
qu
en
c
10
8
6
9,07,56,04,53,01,5
4
2
0 99,9Mean 4,886StD 1 426
Probability Plot of n=5Normal
n=5
t
99
9590
8070
0,030
StDev 1,426N 100AD 0,839P-Value
Pe
rce
nt 70
6050403020
10
5
1086420
5
1
0,1
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 37/39
n=5
Mean Values; n = 10
20Mean 4,870StDev 1 078
Histogram of n=10Normal
cy
20
15
StDev 1,078N 100
Fre
qu
en
c
10
876543
5
0 99,9Mean 4,870StD 1 078
Probability Plot of n=10Normal
n=10
t
99
9590
8070
0,332
StDev 1,078N 100AD 0,413P-Value
Pe
rce
nt 70
6050403020
10
5
987654321
5
1
0,1
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 38/39
n=10
Example with Uniform Distributed Data
16
14
Mean 94,62StDev 2,663N 100
Histogram of 1Normal
99,9
99
Mean 94,62StDev 2,663N 100
Probability Plot of 1Normal
qu
en
cy
14
12
10
8
N 100
rce
nt
99
9590
8070605040
0,098
N 100AD 0,631P-Value
Fre
q
6
4
2
Pe 40
3020
10
5
1
11009896949290
0
1105100959085
0,1
18
16
14
Mean 94,83StDev 1,301N 100
Histogram of Mean 5Normal
99,9
99
Mean 94,83StDev 1,301N 100AD 0,299
Probability Plot of Mean 5Normal
Fre
qu
en
cy
14
12
10
8 Pe
rce
nt
9590
80706050403020
0,579P-Value
98996999392
6
4
2
09998996999392990
20
10
5
1
0,1
Knorr-Bremse Group 08 BB W2 non normal data 08, D. Szemkus/H. Winkler Page 39/39
Mean 598979695949392
Mean 599989796959493929190