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CSSBB Tutorial Series : Lesson 9

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lesson-9/)

July 18, 2014July 19, 2014 Lean Six SigmaLesson 9 : Measure Phase Part 5 of 5

Topics Covered :

Process Capability Analysis

o Process Capability Indices : Cp/Cpk/Pp/Ppk

o Application to non- normal & aribute data

o Six Sigma Metrics : DPMO / PPM /RTY


The only man who behaved sensibly was my tailor: he took my measure anew every time

improveandinnovate | Thoughts and ideas on Quality , Improvement and In... http://improveandinnovate.wordpress.com/?goback=.gde_3151110_memb...

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he saw me, whilst all the rest went on with their old measurements and expected them to fitme.

George Bernard Shaw


Process Capability refers to the ability of a process to meet customersspecifications. Depending on the process and the quality characteristic of interest,several methods are available for computing process capabilities.

Process Capabilities are generally expressed in terms of unitless numbers calledProcess Capability Indices or Process Performance Indices. These are ratios ofprocess spread to tolerance (i.e customers specifications)

Key requirements for computing Process Capability are :

The process should be stable i.e , the process operates within the Upper ControlLimit and the Lower Control Limit . This means the corresponding control chartshould be studied to assess process stability. ( Will be discussed in a later lessonon Control Charts ).The data follows a normal or near normal distribution . If the data is not normal ,the process capabilities can still be computed but a2er transforming the data .

Process Capability Indices ( C , C )

Figure 32(a and b) shows two processes with same set of customer specifications (Lower Specification Limit LSL and Upper Specification Limit USL ) . Whichprocess has a be6er capability of meeting customers requirements ?

p pk


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(h6ps://improveandinnovate.files.wordpress.com/2014/07/slide232.jpg)

Quite obviously , the process in 32(a). Part of the process in Fg 32(b) is outside thecustomers specification limits and hence will produce more non conforming productthan the process in Fig. 32(a). The only information one cannot obtain from theabove figures is how much non conforming product will each process produce? Suchinformation can be obtained with the help of process capability analysis.

Process Capability Indices are helpful when comparing two processes . For example:Vendor evaluation and rating can be done using Process Capability Indices.

Another advantage is that the Process Capability Indices provide a universallanguage that can be used to communicate process performance across industries &processes .

(UCL LCL) is called the Process Width and is given by : UCL LCL = 6, hence


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The concept of Cp is explained in the Fig. 33


ii) Cpk : The assumption while computing Cp using the earlier formula is that the

process mean is centered between the customer specification , which may not bealways true . Fig. 34 shows two process with the same Cp but with differentcapabilities. In such cases , the Cp fails to provide a correct estimate of the processperformance.


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Hence, we use a second process performance index called Cpk which is given by :


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Note : Cp Vs. Cpk : While the Cp is a good indicator of the potential of the processto perform , the Cpk is a realistic measure of the ability of the process to meetcustomers specifications.

Typical Values of Cp & Cpk :

Cp = Cpk implies the process mean is equal to the customers target.Cpk = 1 implies 99.72% of the process is within customer specification limits (Refer fig. 17 under normal distribution curve). This means the process is justcapable. This is also called a 3 sigma level processCpk = 1.33 implies a 4 sigma level process . Quite o2en customers specify thisas the minimum requirement for their vendors.Cpk = 1.67 implies a 5 sigma level processCpk = 2 implies a 6 sigma level process

Note : Refer Six Sigma Metrics in the following sections for details on Process Sigma

Level

Example : A filling machine in a boling process is expected to fill an average of 300ml.

Specifications for this process are :

LSL = 295ml , USL = 305ml

A sample of 72( 24 subgroups of size n= 3) boles from the process were taken and the


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control chart indicated the process was stable with Lower Control limit ( LCL ) = 298 andUpper Control Limit ( UCL) = 306 with a mean of 302ml . Compute Cp & Cpk for thisprocess.

Solution :

Cp = ( USL LSL ) / (UCL LCL)

= (305 295) / (306 298)

= 10/8 = 1.25


Note : UCL LCL= 6* = 8 ; Hence , 3* = 4

The Cp indicates the process is capable. However, since the process is not centered , theCpk will not be equal to the Cp . The actual capability of the process is much lower than theCp suggests !

Cp Vs. Cpk : The following set of figures ( Fig. 36) show how the Cp and Cpk arerelated.


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Process Performance Indices ( Pp , Ppk, Cpm)

The Cp & Cpk discussed earlier are called short term capability indices. This isbecause the estimates of variation ( or standard deviation) are based on short termsamples . The short term standard deviation used for Cp / Cpk calculations isestimated by :

= (R(bar) / d2 , where

R(bar) is the average range of subgroups in the X(bar) R control chart and d is aconstant ( refer Xbar R Control Charts in subsequent lessons ! )

On the other hand , the Process Performance Indicators ( Pp & Ppk) use the overall (long term ) standard deviation computed by the formula :

(h6ps://improveandinnovate.files.wordpress.com

/2014/07/slide142.jpg)

st

2


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The Process Performance Indicators ( Pp , Ppk ) were introduced by the AutomotiveIndustry Action Group ( AIAG) .

The formulae for Pp and Ppk are very much similar to those of Cp & Cpk except thatthe standard deviation is computed using different methods as discussed above.Hence ,

Note : It is a common practice to report Pp and Ppk for processes that are not incontrol. Many experts consider the use of Pp and Ppk unnecessary as it ismeaningless to estimate capability of a process that is not stable.

Cpm : The Cpm is another important Capability Ratio , and is given by :


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Process Capability for Non Normal Data

As discussed earlier, the formula for the process capability indices assume that thedata comes from a normally distributed process. However ,If one were to computethe capability of processes assuming normality when it is actually not , the resultscould be in error .

Hence , it is important to first assess if the data is normally distributed . This can bedone with methods such as probability plo6ing or more easily by hypothesis testingmethods such as the Anderson Darling test.

If the data is not normally distributed , one can try

o transforming the data to obtain a normal distribution Or

o Try fi6ing other known distributions to the data such as Exponential , Weibull,Lognormal etc. This can be done with standard statistical so2ware packages such asthe MINITAB.

o If no known distribution fits the data , one should work with non- parametricmethods and the above capability indices will not be valid.

The Box Cox Transformation Method

One approach to making non-normal data resemble normal data is by using atransformation. Among the many methods available for transforming non- normaldata to normal , the Box Cox is one of the most popular. All transformationmethods use transformation functions that convert a non- normal distribution to anormal distribution . The Box-Cox transformation function is defined as

Y = Y

where , Y is the transformed value of Y( the response variable ) and is thetransformation parameter. For = 0, the natural log of the data is taken instead ofusing the above formula. For example :

t

t


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For = -1 , the transformed value of Y will be 1 /Y ,For = 2 , the transformed value of Y will be YFor = 1/2 , the transformed value of Y will be = 0 , the transformed value of Y will be ln(y)

Example : The following table ( Fig 37) shows data of time taken to resolve customercomplaints ( TAT in hrs. ) An Anderson Darling Test using MINITAB indicates that thedata is not normally distributed. Refer Fig. 38 : Minitab Output . The p-value for the test is0.032 ( < of 0.05)


Note :

i) 10 data points may not be a statistically significant sample size to establish normality .

ii) Concept of hypothesis testing and p-value is discussed in detail in subsequent lessons :The Analyze Phase.

The data was transformed using the Box Cox Transformation method . Figure 39 isthe Minitab Output for the Box Cox plot . It shows the most likely value of , thetransformation parameter . In this case , the value of = 0 . This means the originaldata can be transformed using ln(y) as the transformation function. Thetransformed data is shown in the table in Figure 40 .

2


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A normality test of the transformed data , using Anderson Darling Test confirmsthat the transformed data is normally distributed. The corresponding p value is0.079 ( > ) . Refer Figure 41 .




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Six Sigma Metrics & Capability Analysis for A;ribute Data

The Six Sigma Methodology uses a set of metrics to measure process performancethat are similar to the Process Capability Indices , These metrics are o2en used todemonstrate the amount of improvement achieved by six sigma improvement teamsby estimating process capabilities before and a2er the project . The Six Sigmametrics can be used for variable as well as for a6ribute data ( count and proportions). Some of the commonly used metrics are :

Defects Per Million Opportunities ( DPMO)Process Yields or PPM DefectiveRolled Throughput Yields ( RTY)

Each of the above results can be translated into a common metric called ProcessSigma Level Or Sigma Capability . These metrics are derived from the normalprobability distribution .

The sigma capability ( also called z value) is a metric expressed as a single numberthat indicates defect rate of a process. This means , higher the sigma capability, thelower will be the defect rate and vice- versa . For example , A 6 sigma level processproduces only 3.4 Defects in a Million Opportunities ( 3.4 DPMO ) , whereas a 3sigmalevel produces 66,807 defects in a Million Opportunities. The following table (Fig.42 )shows process sigma levels for various defect rates


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Note : The above DPMO values include a 1.5 shiP to factor the long term processvariability a concept discussed later in the chapter.

Cpk & Process Sigma Level

Fig. 43 relates to the example discussed for computing Cp & Cpk. With reference tothis figure, it is easy to correlate Cpk with Process Sigma Level.

With reference to the normal probability distribution , the sigma level of a processcan be defined as the no. of standard deviations that can be fied between theprocess mean and the CLOSEST specification limit .


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Applying this definition of process sigma level we have :

The closest specification limit to the mean is the USL , Hence , the distance betweenprocess mean and the CLOSEST specific limit is :

USL- = 305 302 = 3 ( Refer fig. 43 )

Standard deviation = ( UCL LCL) /6

= ( 306 -298) / 6 = 8/6 = 1.33

Process sigma level ( z value ) = (USL )/

= 3/1.33 = 2.25

One would now observe that , Process sigma Level = 3* Cpk !

Defects Per Million Opportunities ( DPMO)

The DPMO is a Six Sigma metric that is used when a team needs to monitor processperformance with respect to defects.

Defect Vs. Defective : A defect is a non conformity in the product or processwith respect to a single quality characteristic that does not affect the functioningof the product / process, whereas a defective is an entire unit that isunacceptable to the customer. For example : A dent or a scratch on the body of car isone defect whereas a defective car is one that fails to start.

An Opportunity is a Critical to Quality Characteristic ( CTQ) specified by thecustomer. Thus , any failure to meet a CTQ requirement is termed a defect .

To compute DPMOs , consider the following example :

Example : A component has 4 defined opportunities /CTQs . 30 samples of the componentare inspected for defects against the 4 CTQs . Total defect count is 44 .

The Defects Per Opportunity( DPO ) is given by :

DPO = Observed Defects / Total Possible Defects

= 44/( 4 x 30)


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= 0.366666

Note : Each component provides 4 opportunities to produce a defect. Hence Total PossibleDefects = 4 x 30

Hence , DPMO = DPO x 10

= 366666

To translate this into a process sigma level we can look up the DPMO & Yieldconversion tables ( refer standard tables in Six Sigma Handbooks ) which gives a processsigma level of 1.84

We can also use the following formula in MS Excel:

NORMSINV( 1- defects/volume) + 1.5

Proportion Defective ( Yield or PPM)

This metric is used when monitoring rejections in processes / products . Commonlyused terms in the industry are Scrap% , First Time Thru , First Pass Yield , PPMetc

Estimating Process Sigma Level in this case is fairly simple : Convert the data in % yield =( 1 proportion defective) X100

Example : 92 out of 950 fasteners manufactured are defective.

Yield = ( 1 92/950) * 100

= 90.31%

From the DPMO& Yield conversion tables , this translates into a process sigma level of2.79

We can also use the following formula in MS Excel :

NORMSINV( yield in fraction) + 1.5

Rolled Throughput Yield ( RTY)

6


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This is an important six sigma metric and is useful when multiple processes areconnected in series ( Fig. 44 )


Example : Consider the 4 processes shown in Fig 44 above .Following are yields of each ofthe processes

o Process A : 97.5%

o Process B : 98%

o Process C : 99%

o Process D : 97.5%

The Rolled Thruput Yield is given as the product of all four yields i.e

RTY = (0.975 x 0.98 x 0.99 x 0.975) x 100 %

= 92.23 % !

From the conversion tables , this translates into a process sigma level of 1.42 !

Note : The RTY is a very important metric to establish the fact that even when each of the4 processes performs at 6sigma levels , the customer will receive an output that will belower than a 6sigma level.


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RTY & DPU : The RTY and Defects / Unit ( DPU ) can be related through a specialform of Poisson distribution as given below :

RTY = e

Short Term and Long Term Capability : The 1.5 ShiE

In the Six Sigma methodology , one comes across a term called the 1.5 shiE .This is an interesting theory that has been a subject of much debate. The inventors ofthe Six Methodology argued that in the long term a process is likely to show highervariation than in the short term. Hence , when computing process sigma levels usingshort term data ( samples) , one can obtain only short term process capabilities. Tofactor in the long term variation , a process shi2 of 1.5 from the mean isconsidered .

Example : If sample data indicates the process at 4.5 , then the long term sigma isconsidered as 4.5 1.5 = 3 . This can be generalized as :

Long Term Sigma Level = Short Term Sigma Level 1.5

Or

Z = Z 1.5

DPU

lt st


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The Table in Fig . 42 shows long term sigma levels. We can add another column toshow corresponding short term sigma levels. Refer Fig. 45.


Using the normal probability distribution concepts learnt earlier in the chapter , it iseasy to establish that a long term 6 sigma process actually produces only 0.002DPMO This is also called 2 parts per billion ( 2ppb) . This is equivalent to a short termprocess sigma level of 7.5 !

Suggested Reading :

Statistics For Management by Levin & Rubin1.Jurans Quality Handbook2.Introduction to Statistical Quality Control by D C Montgomery3.

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lesson-8/)

July 16, 2014July 17, 2014 Lean Six Sigma, UncategorizedLesson 8 : The Measure Phase Part 4

Topics Covered

Measurement Systems

Measurement Methods & Gauges

Measurement System Analysis Gage R & R Studies

Metrology-Basics

Measurement Systems

The only man who behaved sensibly was my tailor: he took my measure anew every timehe saw me, whilst all the rest went on with their old measurements and expected them to fitme.

George Bernard

Shaw

Introduction

Validating measurement systems is vital to successful data collection and analysis.Quite o2en , Six Sigma teams end up making wrong decisions due to poor quality ofdata . Data accuracy and precision is a function of the operator or inspectorresponsible for collecting the data as well as the gauges ( measuring instruments)used to collect data. An inadequate Measurement System might adversely impactthe process and (or) the improvement project in the following ways :

o Inaccurate data analysis might impact further decision making


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o The problem might get solved by simply fixing the measurement system and theproject might not be required at all.

o The team might tamper with process in an a6empt to reduce variation withoutactually realizing that the root cause(s) is elsewhere.

Measurement Methods & Gauges : Data collected with the help of anymeasuring instrument can be of two types : i) a6ribute or ii) variable. Variousmeasurement methods and gauges are used depending on the processrequirements and quality characteristic to be measured. Some of the commonlyused measurement methods are :

o Mechanical Systems : Example vernier calipers , micrometers , ring gages etc.

o Electronic Systems : Example Co-ordinate measuring machines ( CMMs)

o Optical Systems : Example Infrared Thermometers

o Pneumatic Systems : Example air calipers and air ring gauges

o Electron based systems : Example hot cathode ionization gauge

Measurement Systems Analysis ( MSA)

A Measurement System Analysis, is a designed experiment to identify thecomponents of variation in the process . The objective of any MSA is to ensure thatvariation due to measurement system is under control and does not adversely impactanalysis of the observed process variation. The MSA is an important part of any SixSigma improvement project.

The flow chart in Fig. 27 explains the concept of measurement system analysis.


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Measurement System Analysis : Terminologies

Measurement System Variation and Error can be classified into two categories :Accuracy and Precision

Accuracy : The ability of a measurement system to provide the correct results.Accuracy of a measurement system has three components :

o Bias : The absolute difference between the observed value and the true ( actual )value.

o Linearity A measure of the consistency of the accuracy of gage across the entirerange of the measurement system.

o Stability Ameasure of the accuracy of the system over a period of time.

Precision : Precision is a measure of the variation obtained from repeatedreadings with the same gauge.Precision has two components :

o Repeatability : Variation when one operator repeatedly measures the same unitwith the same measuring equipment.

o Reproducibility : Variation when multiple operators measure the same unit withthe same measuring equipment.

Discrimination : The ability of the measurement system to detect small changesin the value of the quality characteristic. As a rule of thumb , gauge selection


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should be such that it can detect atleast 1/10 th of the tolerance i.e ( USL LSL)specified by the customer.

Measurement System Analysis Types

Depending on the quality characteristic to be monitored , MSA can be classified intotwo types : i) Variable and ii) A;ribute

i) Variable MSA : The measurement system analysis used for variable data is calledvariable Gauge Repeatability and Reproducibility ( Gauge R & R ). It is typically usedin manufacturing environment for inspection and measuring tools such asmicrometers , Vernier calipers , height gauges etc. Two methods are commonly usedin the variable MSA :

o Analysis of Variance ( ANOVA) Method and

o X(bar) R Method.

The ANOVA method is a more robust method as we can compute the Operator*Partvariation with this method( Refer Fig. 27). This is not possible with the X(bar) Rmethod. MSA computations can be quite cumbersome and hence require use of aso2ware.

Note : Readers are advised to familiarize themselves with the ANOVA method (will bediscussed later !) for a good understanding of the variable MSA method.

Gauge R & R : Concept

The total observed variation in a process ( ) can be represented as :

= +

and the variation due to measurement system can be represented as :

= = +

Hence , the Gauge R&R , which is a % of measurement variation over the totalvariation is given by :

2Tota

2Total

2measurement process

2Process

2 Measurement Process

2 gage

2 (repeatability )

2 (reproducibility)


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(h6ps://improveandinnovate.files.wordpress.com/2014/07/slide104.jpg)Note : TypicalGR&R studies are done with 10 parts , 3 operators and each operator taking twomeasurements (trials) per part.

Example :The following data ( Fig. 28) relates to heights of aluminum fins ( mm) producedby a fin mill , measured by a height gauge. 10 fins were drawn at random from the process.Three operators measured heights of the 10 fins , twice , in random order. The objectivewas to compute Gauge R & R %


Solution :

The Gauge R & R can be worked out easily with the help of Minitab Statistical SoPware ,using the following commands :

Stat > Quality Tools > Gage Study > Gage R & R Study ( Crossed). Following are theresults : ( MINITAB OUTPUT in Fig. 29)


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Acceptance Criteria for Gauge R & R Study

For % contribution = ( )/ ( )*100 , the following is the acceptancecriteria :

1.

If % contribution is less than 1 % Measurement System is ExcellentIf % contribution is between 1% 10% Measurement System is acceptable ,but should be used with cautionIf % contribution is greater than 10% Measurement System is unacceptable needs to be replaced.

Note : The corresponding value is circled in the MINITAB Output Table ( Fig. 29)

2 .For % Study Variation = Precision to Total Variation ( P /TV )% , the following isthe acceptance criteria :

% Study Variation is Given By : 6* (Measurement SD / Total SD)

If % Study Variation is less than 10 % Measurement System is AcceptableIf % Study Variation is between 10% 30% Measurement System is acceptable ,but use with cautionIf % Study Variation is greater than 30% Measurement System is unacceptable needs to be replaced

Note : The corresponding value is circled in the MINITAB Output Table (Fig. 29)

3. No. of Distinct Categories ( NDC) is the no. of different groups in the data that

2gage

2 total


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the measurement system can discern . This should be greater than 5 . It iscomputed using the following formula :

NDC = 1.41x(Standard Deviation of Parts ) / Standard Deviation of Gauge

% Tolerance : Optionally, MINITAB will also return another important

information called SV/ Tolerance which is given by :

1.

% Tolerance = (6 * Measurement SD) / ( USL LSL )

where, USL & LSL are the upper and lower specification limits provided by thecustomer.

The operator * part interaction can be shown separately by the ANOVA method.Minitab Output omits this from the table if the p value for Op* Part interaction >0.25 . In the example discussed , there is no significant Operator * Part interaction.

Gauge R & R : Graphical Results : Minitab Output

The Minitab Output also provides a graphical summary of the values shown in thetables earlier. ( Fig. 30)

Components of variation graph displays Repeatability , Reproducibility andGauge R & R as a % of Total VariationAn important part of the graphical summary is the X(bar) & Range chart Thisconcept will be discussed in detail in subsequent lessons : The Control Phase .Note : For an acceptable MSA , one would expect most points in the X barChart to be outside the control limits indicating that the variation is primarily dueto differences between the parts and not due to the measurement system . Onthe other hand , if most points on the X-bar chart fall within control limits itindicates the variation is primarily due to measurement system


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NOTE : All acceptance guidelines are as defined by the Automotive Industry ActionGroup ( AIAG) , MSA Reference Manual , 3 edition. The AIAG was formed by the bigthree automakers i.e Ford , GM & Chrysler in the year 1991.Readers are advised to referto the latest AIAG manuals for any changes in the guidelines.

The AIAG method uses the following terminologies and definitions :

Gauge R & R , GRR =


/2014/07/slide73.jpg)

% Gauge R & R = %GRR = (GRR/TV)*100

EV = Equipment Variation = Repeatability

Hence , % EV = (EV/TV) *100

AV = Appraiser Variation = Reproducibility

Hence , % AV = (AV/TV) *100

Total Variation ( TV ) =

rd


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where , PV = Part- to Part Variation

The values of EV , AV and PV are computed using constants from the Xbar RControl Chart tables ( can be found in any book on SPC)

Gauge R & R : X(bar) R Method :

This method is similar to the ANOVA method and can be performed easily withMINITAB. One can expect minor differences in the results due to the fact that thismethod approximates standard deviation with range values using the control chartmethod. Refer Chapter 8 for detailed information on the Xbar R Control Chart. Asdiscussed earlier , this method does not compute Operator*Part interactionseparately.

Measurement System Analysis : A;ribute DataMSA for a6ribute data ( also called A6ribute Gauge R & R ) is used when the qualitycharacteristic to be monitored is a6ribute in nature , for example , ratings or rankings, pass/ fail in an inspection , accepting / rejecting an application form. etc . Followingare the steps for conducting an A6ribute Gauge R & R:

1. Identify the sample ,usually more than 30 ( some good ,some bad and someborderline cases)2. Have the items rated / graded first by an expert ,3. Select inspectors who would be rating the items4. Pass the items in a random order to each inspector and record the ratings5. Repeat the process to obtain ratings for a second trial ( in random order again ! ).Note : This method is also known as kappa method

Example : The Manager of a placement firm is concerned about the consistency ofher executives in short listing resumes for various positions . The Manager would liketo validate her concerns by using an A6ribute Gauge R & R method . 20 resumes ofapplicants to a position advertised were identified by the manager ( a mix of goodfits , poor fits and borderline cases) . Three executives were asked to grade theresumes ( Accept or Reject) with reference to the Job Description. Each executivewas given two trials , in random order . The resumes were also graded separately bythe manager regarded as the expert ( reference). The results are displayed in Fig. 30


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Solution :The following information can be gathered from the table : No. of times an executive agrees with herself In Sl.no. 3 ( shaded), Exec. A is notconsistent on both trials for the same resume No. of times an executive agrees with herself and also with other executives In Sl.No. 8(shaded) , all executives agree with themselves but executive B does not agreewith executives A & C No. of times all executives agree with each other and also with the expert. In Sl.No. 20 (shaded) , all executives agree with each other , but not with the expert. Thisis an estimate of A6ribute Gauge R & R %

In the above example, we have 14 occasions ( marked * ) where all executives agreed( accept or reject) with each other and also with the expert . Hence , the A6ributeGage R & R % is given as : (14/20 ) x 100 = 70 % . The target is 100% for all cases witha lower limit of 80%.

The same result can also obtained with MINITAB using the following command :Stat > Quality Tools > A6ribute Agreement AnalysisFig. 31 shows the Minitab output and Fig. 32 shows a graphical analysis of theappraisers performance .


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Conclusion :

Since the agreement is only 70% , the measurement system is not adequate . Figure31 also shows the 95% confidence interval for the agreement % and the Fleisskappa & Cohens kappa statistics . The corresponding p-values will indicate whetheror not to accept the null hypothesis.

Subsequent lessons will deal with confidence intervals , hypothesis testing andp-values.

Fig. 32 is a graphical analysis of the appraisers performance. Appraisers A and Bhave only 85% agreement with the expert ( standard) while the within appraiseragreement is low for appraiser A ( 90%)



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Metrology Basics

Metrology is defined as the science of measurement, embracing both experimentaland theoretical determinations at any level of uncertainty in any field of science andtechnology ( Source : The International Bureau of Weights and Measures ).

Metrology deals with the following subjects : Development and establishment of units of measurements , traceability standards. Application of the science of measurement in various manufacturing processeswhich include selection of the right measuring instruments and their calibration. Compliance to regulatory standards such as weights and measures , safety of theconsumers etc.

Traceability : A key concept in metrology is traceability which is the capability toverify the history, location, or application of an item by means of documented orrecorded identification.Traceability refers to an unbroken chain of comparisons relating an instrumentsmeasurements to a known standard. Every country maintains its own metrologysystem which define various standards.

Calibration : Calibration is a comparison between measurements one of knownmagnitude or correctness made or set with one device ( known as the standard) andanother measurement made in as similar a way as possible with a second device. Thisis done to ensure an instruments accuracy ( bias , stability and linearity) during itsuseful life.Calibration is a periodic activity and follows a pre defined schedule that dependson the measuring instrument and external factors. Calibration details need to berecorded and are o2en a subject of quality audits.

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lesson-6-part-3/)

July 15, 2014July 17, 2014 Lean Six SigmaLesson 7 : The Measure Phase Part 3

Topics Covered :

o Probability Distributions Discrete and Continuous

Probability Distributions


George Bernard Shaw

I . Probability Distributions

A probability distribution is a statistical model that describes characteristics of apopulation. Probability distributions can be used in Six Sigma for :

o Predicting probabilities of occurrence of future events

o Baselining process performance i.e understand how a process is currentlyperforming

o Comparing performance of multiple vendors ( processes ) etc.

Probability Distributions are primarily of two categories :

Discrete Probability Distributions, and


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Continuous Probability Distributions

Probability Distributions Terminologies

Random Variable : A variable can be called random if it takes unique numericalvalues with every outcome of an experiment. The value of the random variablewill vary from trial to trial as the experiment is repeated. Random Variables canbe discrete or continuous . For example :

o A coin is tossed several times. The outcome x is a discrete random variable as it cantake only values 0,1,2 etc.

o A process is run several times . The time taken to complete the process each time ( calledthe cycle time) is an example of a continuous random variable as it can take any positivevalue and not just integer values..

o Probability Density Function (pdf) is the probability of the random variabletaking a value equal to x, i.e

F(x) = P(X=x)

Cumulative Distribution Function (cdf) is denoted by F(x) and represents theprobability of the random variable ,X such that ,

F(x) = P(X x)

o The expected value (or population mean) of a random variable indicates itsaverage or central value. It is a summary value of the variables distribution. Thus

o For a discrete random variable the expected value is given by :

= E(X) = x p(x )

o For a continuous random variable , the expected value is given by :

= E(X) = x f(x)dx

Discrete Probability Distributions : Some situations call for discrete data, such as;the no. of applications rejected , the no. of abandoned callsetc. Such data can be represented by discrete probability distribution i.e a probabilitydistribution that can take only discrete values. The following are some commonly

i i


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used discrete probability distribution functions :

Poisson1.Binomial2.Hypergeometric3.

1. A Poisson distribution describes the count of the number of events that occur ina certain time interval or space. For example, the number of customers arriving everyhour , the number of calls received by a switchboard during a given time period etc. ThePoisson probability density function is given by :

is the mean ,

x is the Poisson distributed random variable and

x! , called n factorial = n*(n-1)*(n-2).x1

Note :

i) The mean of the Poisson process is ,

ii) The variance of the process is also , that is , = , so that the standarddeviation is =

iii) The Poisson distribution can be used when:

o The no. of possible occurrences is large

o The average no. of occurrences is constant

o The probability of the event is small

2


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Example : A soPware averages 8 defects in 8000 lines of code. What is the probability ofexactly 2 errors in 4000 lines of code ? What is the probability of less than 3 errors?

Solution :

The average defects in 4000 line of code (= ) = 4,

Probability of exactly 2 errors is P(x=2) = (e x 4 )/2! = 0.1465

Use of Poisson Tables : An easier way to arrive at the solution is to refer to thePoisson Tables and read off the value relating to = 4 & x = 2.

This problem can also be easily solved with the help of MS Excel using the formula :

Poisson ( 2,4,0) = 0.146525.

Note : The 0 at the end of the formula gives the probability density function . Forcumulative density function , the zero should be replaced with 1

o Probability of less than 3 errors

P( x


where,

P(x,n,p) is the probability of exactly x successes in n trials with a probability ofsuccess equal to p on each trial and

the total number of trials is fixed ;1.there are only two possible outcomes for each trial : success and failure2.the outcomes of all the trials are statistically independent;3.all the trials have an equal probability of success.4.

Note :The mean and variance of the binomial distribution are

= np and = np(1-p)

Example : The success rate of a Black Belt Certification test is 65% ! In a class of 10participants , what is the probability of 7 participants clearing the test. ? What is theprobability of more than 8 participants clearing the test?

Solution :

1.

2


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p = 0.65 , n = 10 , substituting ,

i) P( x=7) = 0.2522

Using the Binomial Tables : An easier way to arrive at the solution is to use theBinomial Table and read off the value relating to p=0.65 , n= 10 & x = 7 .

Note : This can also be solved with the help of MS Excel using the formula

BINOMDIST(7,10,0.65,0) = 0.25222

The binomial probability density function for n >8 is given by :P ( x 8) = 1- P(x

Example : A consignment of 25 parts contains 4 defectives. What is the probability that asample of 8 drawn at random will contain 2 defectives ? What is the probability that thesample will contain less than 2 defectives ?

Solution :

n=8 , N = 25, m = 4 ,

This problem can be solved with MS Excel using the formula HYPGEOMDIST()

P (x=2 ) = HYPGEOMDIST(2,8,4,25) =0.30P( x


where

x is the random variable such that

is the population mean

is the population standard deviation

= 3.14159

e = 2.71828

The normal probability distribution has certain properties that are very useful in ourunderstanding of the characteristics of the underlying process.

The distribution is symmetric i.e its skewness is zeroFor the normal probability distribution , the mean = median = mode.The curve extends from - to + along the x- axis.The width ( spread) of the curve is a function of the standard deviation . Higherthe standard deviation , wider the curve and vice- versa. Refer to Fig. 18


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Fig. 18 : Normal Distribution Curve

The areas under the curve represent the probabilities of the distribution forvarious values of the random variable. The mean () divides the curve into twoequal halves i.e 50% of the area under the curve is to the le2 of the mean and theother 50% is to the right of the mean. The areas under the curve between any twofinite limits can be obtained by integrating the density function between the twolimits. Some important areas to remember are :

o 68.26 % of the area under the curve is within the +/- 1 limit. This means 68.26%of the values of a normal random variable fall within the +/- 1 limit. Similarly,

o 95.44% of the area under the curve is within the +/- 2 limit., and

o 99.72% of the area under the curve is within the +/- 3 limit.

Areas under the normal curve between any finite limits , such as the above, can becomputed easily either with the help of tables or with statistical so2ware . Refer toFig 19a , 19b & 19c.


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The Standard Normality Probability Distribution

The Standard Normal Probability Distribution is a normal probability distributionof a random variable ( called z ) with a mean of 0 and a standard deviation of 1 .The random variable z is given by:


Thus , the z value can be seen as the number of standard deviations that a pointx is from

Example :The cycle time of an assembly process is known to be normally distributed with amean of 28 mins. and a standard deviation of 8 mins. i) What % of assemblies have a


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cycle time of less than 16 mins ? ii) What % of assemblies have a cycle time of more than42 mins ? iii) What % of assemblies have a cycle time between 16mins. and 42 mins.

Solution :

(hps://improveandinnovate.files.wordpress.com/2014/07/slide151.jpg)

Fig. 20 : Areas under the normal Curve

The areas under the normal curve that are of interest are marked in Figure 20. Firstthe x values need to be converted to their corresponding z-values .

i) z = (16-28)/8 = 1.5 ,

Hence , P(x 1.75)

1

1

2

2


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From the standard normal table , the area under the curve for a z- value of 1.75 =0.9599.

Note : This value can also be obtained with MS Excel using the formula

NORMSDIST(1.75)

Conclusion : As indicated in the table in Appendix I , this value ( 0.9599) is the areaunder the normal curve from - to z ( = 1.75 ) . Hence P(z>1,75) is the area to theright of z , which is equal to (1 0.9599) = 0.04. This means approximately 4% ofassemblies have a cycle time of more than 42 mins.

iii) z is the area between z & z ( Refer Fig. 20 ) . It can be easily computed as :

1 (area under z + area under z )

Hence , P ( 16


Fig. 21 : t- distribution

3. The F- Distribution :The F distribution is used in hypothesis testing for comparingtwo variances. Like the t distributions, the F distribution is actually a family ofdistributions .However , unlike the t- distribution , the F distribution ischaracterized by a pair of degrees of freedom i.e a numerator degrees of freedomand a denominator degrees of freedom. The random variable is the F ratio , whichis a ratio of two variances. Fig . 22 shows four F Distributions for various numeratorand denominator degrees of freedom. Use of F- Distribution is discussed in detail in The Analyze Phase.


Fig. 22 : The F Distribution


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4. The Exponential Distribution : The exponential distribution is used commonly inreliability engineering.It is used to model items with a constant failure rate such aselectronic & mechanical components and other applications such as wait times atcustomer service counters. The exponential distribution & the Poisson distributionare inversely related. That is , if a random variable, x, is exponentially distributed,then the reciprocal of x, y=1/x follows a Poisson distribution. Likewise, if x is Poissondistributed, then y=1/x is exponentially distributed. Refer Fig . 23 .

The exponential probability density function is given by :


The cumulative distribution function is given by :

(h6ps://improveandinnovate.files.wordpress.com/2014/07

/slide25.jpg)

= constant failure rate , for example : Number of failures / hr Or No. of failure / cycle

Thus , can also be expressed as , 1/

where , is the Mean Time Between Failures (MTBF)


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The standard deviation of the exponential distribution is given by



Fig 23 : The exponential distribution

Example :

The average life of a component is 1000 hrs. It is known that the failure rate of thecomponent is exponentially distributed . i) What is the probability that the component willlast 800 hrs ? ii) What is the probability that the component will last atleast 800 hrs ?

Solution :

Since the average life is 1000 hrs . ,

= 1/1000 = 0.001

i) Probability of exactly 800 hrs. = P(x=800) = 0.001*e^(-0.001*800)

= 0.000449

Note : The same value can be obtained with MS Excel by using the formula

EXPONDIST(800,0.001,0)


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ii ) Probability of at least 800 hrs. = P ( x > 800) = 1- P ( x 0

is the scale parameter, >0 , and

is the location parameter.

Figure 24 shows probability density functions of Weibull distribution for = 1 , = 0and for several values of


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Fig. 24 : The Weibull Distribution

The shape parameter gives the Weibull Distribution its flexibility. For example at = 1 , the Weibull is identical to the exponential distribution . if is between 3and 4 the Weibull distribution approximates the normal distribution. Refer Fig.24.The scale parameter, ,determines the range of the distribution.The location parameter , , indicates the location of the distribution along the x-axis.

The cumulative distribution function of a Weibull random variable is given by :


/slide72.jpg)

Example : A typical application of the Weibull Distribution is to describe the time to

failure of electronic components . The time to failure of an electronic component in aTelevision Set is known to follow a Weibull Distribution with a shape parameter of 0.5 , a


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scale parameter of 70 hrs. and a location parameter of 0. What is the probability of thecomponent lasting at least 200 hrs ?

Solution

= 0.5 , = 70 and = 0

P ( at least 200 hrs. ) = P ( x > 200 )

= 1- P ( x < = 200)

= 1- 0.8155

= 0.1844

Note :The same value can be obtained with MS Excel using the formula :

1 Weibull ( 200, 0.5 , 70,1)

6. The Lognormal Distribution : The log-normal distribution is the single-tailedprobability distribution of any random variable whose logarithm is normallydistributed.If a data set is known to follow a lognormal distribution, transforming thedata by taking a logarithm yields a data set that is normally distributed. While it iscommon to use natural logarithm ( denoted as ln),any base logarithm, such as base10 or base 2, can also be used to yield a normal distribution. The lognormaldistribution is commonly used to model the time to failure of mechanicalcomponents , where the failure is fatigue or stress related. Its probability densityfunction is given by :


/2014/07/slide63.jpg)

is called the location parameter which is also the log mean . It is the mean of


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the transformed data transformed data. is called the scale parameter which is alsothe log SD. It is the standard deviation of the . Fig 25 shows lognormal distributionsfor several values of .


Fig. 25 : The Lognormal Distribution

Example : The time to failure of a mechanical component is known to follow a lognormaldistribution . The following is the time to failure data in hours for 6 samples tested .

221, 365 , 420 , 310 , 396 , 289

i) What is the probability that a component will last upto 360 hrs. ? ii) What is theprobability that a component will last more than 440 hrs ?

Solution :

The table below shows the transformed data using natural logarithm as the base


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i) The probability of a component lasting upto 360 hrs. is given by :

p(x < =360) = p( x

1 NORMDIST( ln(440) , 5.7871 , 0.2379 , 1)

= 0.1039

7. The Chi- Square ( ) Distribution

If n random values z , z , , z are drawn from a standard normal distribution,squared, and summed, (z + z + z ) the resulting statistic is said to have achi-squared distribution with n -1 degrees of freedom This is a one-parameter familyof distributions, and the parameter, n, is the degrees of freedom of the distribution (Refer Fig. 26)


Fig. 26 : The Chi Square Distribution

The Chi- Square statistic is given by :


, where ,

s = sample variance of sample size n

2 = estimated population variance

2

1 2 n1

222

n2

2


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The Chi Square distribution has several applications in inferential statistics such asinterval estimates of variances and standard deviations , hypothesis testing such forGoodness of Fit etc . Application of the Chi-Square distribution is discussed in TheAnalyze Phase.

8. Bivariate Distribution :

In all the probability distributions discussed so far , we have seen distributionsrelating to only one variable . These are called univariate distributions. If multiplevariables need to be studied simultaneously, the resulting distribution will be called amultivariate distribution. A bivariate distribution is a from of multivariatedistribution with two variables .Bivariatedata arises from populations in which twovariables are associated with each observation .For example : A nutritionist may beinterested in studying how a particular diet influences both height and weight of children.The two variables of interest are height and weight. If both the variables are normallydistributed , we have a bivariate normal distribution.

Approximations to Probability Distributions

In many situations , one may find it necessary to approximate the real probabilitydistribution with a simpler distribution. For example , It is known that an underlyingdistribution is binomial but the analyst feels it would be easier to assume a normaldistribution and compute the probabilities . This can be done provided the data meetscertain conditions. Some of the commonly used approximations are described below:

The Normal approximation to the Binomial : The normal distribution may beused to approximate the binomial if :

np 5 and n(1-p) 5

The Poisson approximation to the Binomial : The Poisson distribution may beused to approximate the binomial if :

n is large and p is small ( < 0.1) such that np < 5

In this case , , the mean of the Poisson distribution = np

The Normal approximation to the Poisson : The normal distribution may beused to approximate the Poisson if :


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The mean > 10

In this case , the mean () and variance (2 ) of the normal distribution will be equalto

The Binomial approximation of the Hypergeometric : The binomial distributionmay be used to approximate the hypergeotmetric if :

n/N is small ( < 0.1)

In this case , probability of the binomial distribution, p = m/N ,

where , m is the number of successes in the population N

3 Comments




lesson-6/)


Topics Covered :

Handling Data

o Sampling techniques

o Data collection

o Basic Statistics & Probability

The only man who behaved sensibly was my tailor: he took my measure anew every time


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he saw me, whilst all the rest went on with their old measurements and expected them to fitme.

George Bernard Shaw

I. Handling Data

Collecting and analyzing data is one of the key requirements in the Measure Phase .The Six Sigma professional is expected to be an expert in data analysis. This sectionwill deal with some basic concepts of data management and statistics & probability.

Types of Data : Processes produce data of various types. The following chart (Fig. 1) shows data types with examples.


Fig. 1 : Data Types

Variable Data : Data that has units of measure ( ratio or interval type) . Example :Temperature , Pressure. Variable Data is of two types

Continuous Data : Data that can take fractional values. Example : Height , Weight, Time etc.

Discrete Data : Data that can take only integer values . Also called count data .Example : No. of guests in a hotel , no. of calls received in a day etc.

A;ribute Data : Data that is non- numeric and categorical . It is used to measurea6ributes of a product or process. Example : customer satisfaction ratings ( good ,


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satisfactory , poor ) , shades of colour in a fabric etc. A6ribute Data can be of two types :

Nominal Data : Categorical data without a specific order is called nominal data .Example : red , blue, black , green etc.Ordinal Data : Data that has ordered categories but no meaningful intervalsbetween the measurement is called ordinal data. Example : Customer satisfactionratings on a scale of 1-5 etc.

Note : The more continuous the data , the easier it is to analyse.

Levels of Measurement : Measurements are categorized into several levels . Aparticular level or scale of measurement will define how the data should betreated mathematically. Following are the scales of measurement :

o Nominal data have no order . They have names / labels to various categories.

o Ordinal data will have an order, but the interval between measurements is notmeaningful.

o Interval data have meaningful intervals between measurements, but there is notrue reference point (zero).

o Ratio data have the highest level of measurement. Ratios between measurementsas well as intervals are meaningful because there is a reference point (zero).

Sampling Methods :

For large populations , the time and cost involved in gathering data may beinfeasible. Sampling makes it easier to study a limited amount of data and drawinferences about the underlying population

Sampling is generally of two types :

Judgemental sampling or non random sampling where sampling is done basedon ones expertise and opinion.

1.

Random or probability sampling where all items / data points in the populationhave an equal chance of being chosen. Statistical analysis can be done with thisdata and inferences can be made as it is representative of the population.

2.

Random Sampling : Methods


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The following are the four commonly used methods of random sampling :

i) Simple Random Sampling : In this method each item in the population has equalprobability of ge6ing picked. This removes the element of bias in sampling. This isthe simplest form of random sampling and can be used to estimate populationparameters based on summary statistics. The easiest way to do a simple randomsampling is with the help of random numbers. Using MS Excel , one can generate atable of random numbers and use it to select the sample.

i) Systematic Sampling : In this method items are selected from the population atpre- defined intervals of time or space . For example : samples drawn from a processevery 30 mins or every 5 component from the assembly line etc. The disadvantages ofsuch a sampling method are quite apparent. There could be a bias introduced due toa pre- defined interval. However , it requires less time and consumes lesser resourcescompared to the simple random sampling method.

ii) Stratified Sampling : This method involves dividing the originalpopulation into homogenous groups ( strata) and then drawing a sample at randomfrom each group ( stratum) . For example : A market survey team would like to knowhow well a new brand of shampoo will be received . For this purpose , the team may firstdivide the population into various age groups ( teens , middle aged and senior citizens etc.) . Random sampling from each of the groups will ensure that each group is represented inthe sample. This method helps in reducing sampling effort for populations with largevariances.

iii) Cluster Sampling :This method is similar to Stratified Sampling . Thedifference being, stratified sampling is used when variation within groups is small ,but variation between groups is large. The opposite is the case for cluster samplingi.e the groups are essentially similar but within groups variation is large.

Data Collection

A formal data collection process should be established by the Six Sigma team. Thisprocess will ensure that the data collected is NOT time or person dependent. In casethe six sigma team has delegated the data collection activity to others , they shouldmonitor such activities by :

Questioning collectors for the understanding of operations definitions.Verifying collected data with source, through sampling or ad hoc QCEnsuring that pre-defined procedures are followed during the data collection

th


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process.

The following are some important properties of data that one needs to ensure whilecollecting data :

Data integrity : Is the data genuine ?Data precision & accuracy : Using the right operational definitions &appropriate gauges for collecting dataData consistency : Comparing apples with apples . The Six Sigma team shouldensure there are no changes in measurement systems , operational definitions ,metrics etc during the course of the project.Time traceability : Data must be traceable to the time it was collected.

Teams may use different templates ( formats ) for data collection depending uponspecific processes or organizational needs . A check sheet is a good example of adata collection format Refer Fig. 2


Fig 2 Check Sheet

Other techniques of data collection include

Data codingAutomating the data collection system


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Both the techniques will ensure that the data is free from human & gauge errors .

II. Basics Statistics

Terminologies and their Definitions : The table in Fig. 3 shows commonly usedterminologies in statistics and their definitions.


Summary Statistics :Summary Statistics are single numbers to describecharacteristics of a data set. Some of the most important summary statistics are :

Measures of central tendency : The three measure of central tendency are mean, median and mode.

1.

Mean ( arithmetic, weighted & geometric) : The arithmetic mean of a set of datais the average of all the values and is given by formula

2.

where , xbar is the sample mean,

is the population mean,

n is size of the sample and

N is the size of the population


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Median : Median is the middle value in a set of numbers , when the numbers are

arranged in an ascending or descending order. If the data set contains even no. ofobservations , the median is the average of the two middle values . The advantage ofmedian over the mean is that extreme values in a data set do not affect the value ofthe median.

Mode : Mode of a data set is the value that occurs with the highest frequency.

Though not used frequently by Six Sigma experts , this is an important measure ofcentral tendency. A data set can have one mode , two modes ( bi-modal) or multiplemodes ( multi modal).

2. Measures of dispersion : The three measures of dispersion are Range , Varianceand Standard Deviation.

Range : The range is the difference between the largest and the smallest value in

the data set . The range is easy to compute and hence can sometimes be a goodapproximation of the standard deviation. However , when the data contains extremevalues , this can lead to inaccurate estimates of variation in the underlying process.. Variance : The variance is a measure of dispersion that indicates how the data isspread about the mean of the data set . A high value of variance indicates highvariation in the process i.e the individual values are far away from the mean , andvice versa. Variance is given by the formula :

(h6ps://improveandinnovate.files.wordpress.com/2014/07/slide24.jpg)where , s2 is the sample variance ,and2 is the population variancen is size of the sample andN is the size of the population

Note : n-1 is the no. of degrees of freedom , a conceptthat will be discussed in subsequent chapters.

Standard deviation : The standard deviation is the square root of the variance and

is given by the formula :

(h6ps://improveandinnovate.files.wordpress.com/2014/07/slide33.jpg)Where , s is


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the sample standard deviation , and is the population standard deviationExample : Computing Standard DeviationGiven five values : 1, 2,3 4 & 5

The standard deviation can be computed using the following table :

(h6p://improveandinnovate.files.wordpress.com/2014/07/slide71.jpg)

Computing Standard Deviation

Standard Deviation = 10/5 = 1.414

Note :i) Standard deviation ( SD ) is the most commonly used measure of dispersion , as itcarries the same unit of measure as the individual values or the mean.ii) However , standard deviation values cannot be added up to compute a total SD .Instead , first the variances should be added and the corresponding square root gives thetotal SD, i.e


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/slide43.jpg)

The Central Limit TheoremThe central limit theorem ( CLT ) provides a relationship between the shape of thepopulation distribution & the shape of the sampling distribution of the mean. TheCLT is one of the most important theorems in statistics. The central limit theoremstates that :

Regardless of the shape of the population distribution and the size of sample, the mean of

the sampling distribution of the means will equal the population mean

Regardless of the shape of the population distribution , the sampling distribution will

approach normality as the sample size increases.

Mean of the Sampling distribution of the MeansGiven five values : 1 ,2, 3, 4, & 5 , the mean of sampling distribution of the meansfor a randomly drawn sample of size n = 3 would include the means of all possiblesamples of size n = 3 . ( Fig. 5)

(h6p://improveandinnovate.files.wordpress.com/2014/07/slide9.jpg)Fig 5 : Mean of Sampling distribution of means

Thus according to the Central Limit Theorem ,


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The population mean will be approximately 2.987 A probability distribution of the means of all possible samples is a distribution ofthe sample means Thus, instead of 5 values , if we could consider a larger population and take allpossible samples of size n = 2 ,3,4 20, 40 etc , the distribution of the means willapproach normality as the sample size increases. Fig. 6 shows such a distribution forsample size n = 3 for the data displayed in Fig. 5 The standard deviation of the sampling distribution of means is also called thestandard error of the mean and is given by :

(h6ps://improveandinnovate.files.wordpress.com/2014/07/slide53.jpg)Where , sigma is the population standard deviation.

(h6p://improveandinnovate.files.wordpress.com/2014/07/slide111.jpg)Fig 6: Sampling Distribution of Means for n=3

Note : The Central Limit Theorem allows us to make inferences about the populationparameters based upon sample statistics and hence is a very powerful theorem.

3. Skewness : A distribution is said to be skewed when the curve is not symmetricalabout its central tendency . Distributions can be right skewed ( positively skewed) orle2 skewed ( negatively skewed). Refer to Fig. 7 for examples of skeweddistributions.4. Kurtosis : Kurtosis is a measure of the peaked-ness of a curve. Two curves may


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have the same central tendency and dispersion properties , but one may be morepeaked than the other. The two curves are said to be having different degrees ofkurtosis.

Graphical Methods : Graphical methods are an an easy to understand approach

to data analysis and hence are commonly used for displaying descriptive statistics.However , making some of the graphs manually can be quite cumbersome and timeconsuming. Hence, use of so2ware is recommended.

1.Histograms : A histogram is a bar graph of raw system data. It displays basicinformation about the data such as central location, width of spread, and shape (Fig7) . The X- axis of the histogram shows the scale of measures which are divided intoseveral intervals. The Y- Axis shows frequency of occurrence of data . Each bar of thehistogram indicates frequency of data in the corresponding class interval. Thefollowing figure ( Fig.7) gives several possible shapes of a histogram.


Each histogram represents a unique data distribution. For example, A symmetrical histogram indicates data is distributed equally about the mean. Asymmetrical histogram also indicates normal distribution ( mean = median = mode) A bi-modal histogram indicates that the data may have come from two differentsources or populations ( ex.: data from two machines / two locations etc. ) A positive skew occurs when the mean > median (ex. : cycle times , surface finishetc) A negative skew occurs when the mean < median ( ex. : yields , cash flows etc) A random histogram indicates an unpredictable / uncontrolled process .


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2. Sca;er Diagram : The sca6er diagram is a graph showing pairs of plo6ed values oftwo factors, to examine whether the factors are related. The sca6er diagram is abasic form of regression analysis which is used to establish a y = mx + crelationship. A sca6er diagram may show : A positive correlation : As x increases , Y increases ( Fig 8a) A negative Correlation : A x increases , Y decreases ( Fig. 8b) A zero correlation : As x increases , Y remains constant ( Fig. 8c)


Note : The Sca6er Diagram does not indicate a cause and effect relationship . It onlyshows the direction of relationship between two variables.

Fig 8 : Sca6er Diagrams

3. Box and Whisker Plot : A box and whisker plot is used to understand datadistribution based on the median ( as compared to the histogram that shows howdata is distributed about the mean.)

Example : The following is the data for time taken (in days) to process 20 loanapplications ( also called Turnaround Time or TAT)

14 12 13 26 13 22 16 13 14 30 33 21 33 28 26 38 20 30 26 23


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A box and whisker plot for the data is shown in Fig 9.

(h6p://improveandinnovate.files.wordpress.com/2014/07/slide14.jpg)Fig. 9. : Box and Whisker Plot

Lower Whisker : It extends to the lowest value within the lower limit. Lower limit

= Q1- 1.5 (Q3 Q1) First Quartile ( Q1) : 25% of the data values are less than or equal to this value. The Median 50% of the observations are less than or equal to it. Third Quartile (Q3) : 75% of the data values are less than or equal to this value. Upper whisker : the upper whisker extends to the highest data value within theupper limit. Upper limit = Q3 + 1.5 (Q3 Q1) Note : A * beyond the whiskers indicates outliers.

5. Run Charts : Run charts are simple trend charts useful in monitoring processvariation with respect to time. Fig. 10 below is an example of a run chart.


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(h6p://improveandinnovate.files.wordpress.com/2014/07/slide15.jpg)Fig. 10 : Example of a Run Chart

6. Normal Probability Plot : The normal probability plot is a graphical technique forassessing whether or not a data set is approximately normally distributed. The dataare plo6ed against a theoretical normal distribution in such a way that the pointsshould form an approximate straight line. Values lying away from this line areindications of departures from normality . Refer Fig. 11 . These plots can be madeeasily with the help of standard statistical so2ware.

(h6p://improveandinnovate.files.wordpress.com/2014/07/slide16.jpg)Fig. 11 : Normal Probability Plot

The straight line in blue is the best fit line connecting the points. If most points fallon or around this line , we can conclude that the data represents a normaldistribution. The curved lines on either side of the straight line represent a 95%confidence interval for the best fit . The concept of normality assumptions andconfidence interval are discussed in more detail in Chapter 6 The Analyze Phase.

Basic ProbabilityProbability is the chance that an event will happen. Probabilities are expressed infractions such as 1/4, 2/5 or 0.25, 0.1 etc. Probability of an event occurring is given bythe formula :P(event) = No. of successful outcomes of the event /Total no. of possible outcomes

Probability Terminologies & Rules

o Events are said to be mutually exclusive if only one of the events can happen at atime, else the events will be called not mutually exclusive. Refer Figs. 12a & 12b. For


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example , the toss of a coin will result in either a head or a tail , not both. Hence ,

P(Head) or P(Tail) = 0.5 .

For mutually exclusive events : P( A or B) = P(A) + P(B)For events that are not mutually exclusive : P( A or B) = P(A) + P(B) P( A and B )


Fig. 12a : Mutually exclusive events

Fig. 12b : Mutually non- exclusive events

Events are said to be independent if the occurrence of one event does not affectthe probability of the occurrence of the other. Independent events can have threetypes of probabilities :

Marginal probability i.e is the simple probability of occurrence of an event. Forexample : probability of obtaining a head or a tail for every toss of a fair coin isalways 0.5

Joint probability is the probability of two or more events occurring together and isgiven by the product of their marginal probabilities , i.e P( AB) = P(A) x P(B)

Conditional Probability is the probability of an event occurring if another event


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has occurred preceding it i.e P(B/A) read as probability of event B given A hasoccurred = P(B) since A & B are independent events.

If the events are not independent ( dependent)

Conditional probability is given by P(B/A) =P(BA)/P(A)

Hence joint probability is given by P(BA) =P(B/A)*P(A)

Example : A box contains a total of 266 balls of four different colours ( pink, yellow,black and blue) and each colour is also of three different pa6erns ( do6ed , striped &plain) . A contingency table ( Refer Fig. 13) can be made to describe the compositionof the box.

(h6p://improveandinnovate.files.wordpress.com/2014/07/slide18.jpg)Fig. 13 : Contingency Table

What is the probability that a ball drawn at random will be pink ?

o Solution : P(pink) = 66/266 =0.248

What is the probability that a ball drawn at random will be striped ?

o Solution : P (striped) =57 /266 = 0.2142

What is the probability that a ball drawn at random is yellow and plain?


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o Solution : P ( yellow and plain) = 34/ 266=0.128

What is the probability that a ball drawn at random will be black or blue ?

o Solution : This is a typical case of mutually exclusive events. Hence, P(black orblue) = P (black) + P(blue) = 69/266 + 54/266 =0.462

What is the probability that a ball drawn at random is yellow or plain ?

o Solution : This is not a mutually exclusive event

o P(yellow or plain) : P(yellow) + P(plain) P(yellow and plain)= 77/266 + 104/266 34/266 = 0.552

A ball drawn at random is found to be black. What is the probability that the ball is

do6ed ?

o Solution : This is a case of conditional probability. Hence, P (do6ed, given the ball isblack) = P( do6ed/black) = P(do6ed and black) / P(black) = 36/266 69/266 = 0.133

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lesson-5/)



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Topics Covered:

Process Characteristics

Introduction to Measure PhaseMapping Process Flows : The SIPOC & Value Stream MappingAnalyzing Business Processes


George Bernard Shaw

Process Characteristics 1.

Introduction to Measure Phase : The Measure Phase of the DMAIC is requiredfor validating the data that was used in the Define Phase to establish theimprovement opportunity. Key activities in the Measure Phase are :

o Establish & Measure Project Ys

o Plan for Data Collection

o Validate Measurement System

o Establish Baseline Sigma

To begin with , key process elements and their relationships need to be identified .

In Chapter 2 : Business Process Management & Metrics we learnt the basicconcepts of business processes. In this section , we will discuss how to identify keyprocesses variables such as inputs and outputs and how they impact the processoutcome.

The SIPOC ( Supplier Input Process Output Customer) is a simple tool that isuseful in identifying process variables and their relationships. The SIPOC is a high level process map and is used by Six Sigma project teams to obtain a macro picture of


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the process as it flows from the suppliers to customers. The benefits of using SIPOCare :

It provides a birds-eye view of the process that can be easily communicated toall stakeholders of the projectHelps in identifying key process input variables ( KPIVs)and key process outputvariables ( KPOVs) and their linkages.The SIPOC helps in clearly defining the project scope i.e boundary limits.

Steps in making a SIPOC

The following is the sequence of steps in constructing a SIPOC :

Refer to the Project CharterIdentify the process (es) that are to be improvedWrite down the first and last steps of the process ( i.e the project scope)Fill in the intermediate steps ( remember the SIPOC is only a macro picture of theprocess , hence it is enough to describe the process in 6-10 high level steps. )For each process step :

o Identify the Output (s)

o Identify the all the Customers that will receive the Outputs o

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