2555571 six sigma and minitab 13

8/3/2019 2555571 Six Sigma and Minitab 13

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QSM 754SIX SIGMA APPLICATIONS

AGENDA


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Research and Benchmarking Web Sites

CommercialQuality Progresshttp://www.qualityprogress.asq.org

Minitabhttp://www.minitab.com

General Electric Annual Report 1997http://www.ge.com/annual97/sixsigma/

W. Edwards Deming Institute

http://www.deming.org/deminghtml/wedi.html

AcademicUniversity of Edinburgh

http://www.ucs.ed.ac.uk/usd/socsci/stats/minitab.html
http://www.qualityprogress.asq.org/http://www.minitab.com/http://www.ge.com/annual97/sixsigma/http://www.deming.org/deminghtml/wedi.htmlhttp://www.ucs.ed.ac.uk/usd/socsci/stats/minitab.htmlhttp://www.ucs.ed.ac.uk/usd/socsci/stats/minitab.htmlhttp://www.deming.org/deminghtml/wedi.htmlhttp://www.ge.com/annual97/sixsigma/http://www.minitab.com/http://www.qualityprogress.asq.org/


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Day 1 Agenda

Welcome and Introductions Course Structure

Meeting Guidelines/Course Agenda/Report Out Criteria

Group Expectations

Introduction to Six Sigma Applications

Red Bead Experiment

Introduction to Probability Distributions

Common Probability Distributions and Their Uses

Correlation Analysis


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Day 2 Agenda

Team Report Outs on Day 1 Material

Central Limit Theorem

Process Capability

Multi-Vari Analysis

Sample Size Considerations


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Day 3 Agenda

Team Report Outs on Day 2 Material

Confidence Intervals

Control Charts

Hypothesis Testing

ANOVA (Analysis of Variation)

Contingency Tables


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Day 4 Agenda

Team Report Outs on Practicum Application

Design of Experiments

Wrap Up - Positives and Deltas


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Class Guidelines

Q&A as we go Breaks Hourly

Homework Readings As assigned in Syllabus

Grading Class Preparation 30%

Team Classroom Exercises 30%

Team Presentations 40%

10 Minute Daily Presentation (Day 2 and 3) on Application of previous days work

20 minute final Practicum application (Last day)

Copy on Floppy as well as hard copy

Powerpoint preferred Rotate Presenters

Q&A from the class


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INTRODUCTION TO SIXSIGMA APPLICATIONS


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Learning Objectives

Have a broad understanding of statistical concepts andtools.

Understand how statistical concepts can be used toimprove business processes.

Understand the relationship between the curriculum and

the four step six sigma problem solving process(Measure, Analyze, Improve and Control).


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What is Six Sigma?

A Philosophy

A Quality Level

A Structured Problem-Solving Approach

A Program

Customer Critical To Quality (CTQ) Criteria Breakthrough Improvements Fact-driven, Measurement-based, Statistically Analyzed

Prioritization Controlling the Input & Process Variations Yields a Predictable

Product

6s = 3.4 Defects per Million Opportunities

Phased Project: Measure, Analyze, Improve, Control

Dedicated, Trained BlackBelts Prioritized Projects

Teams - Process Participants & Owners


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POSITIONING SIX SIGMATHE FRUIT OF SIX SIGMA

Ground FruitLogic and Intuition

Low Hanging FruitSeven Basic Tools

Bulk of FruitProcess Characterizationand Optimization

ProcessEntitlement

Sweet FruitDesign for Manufacturability


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UNLOCKING THE HIDDEN FACTORY

VALUESTREAM TO

THECUSTOMER

PROCESSES WHICHPROVIDE PRODUCT VALUEIN THE CUSTOMERS EYES

FEATURES ORCHARACTERISTICS THECUSTOMER WOULD PAYFOR.

WASTE DUE TOINCAPABLEPROCESSES

WASTE SCATTERED THROUGHOUTTHE VALUE STREAM

EXCESS INVENTORY

REWORK WAIT TIME EXCESS HANDLING EXCESS TRAVEL DISTANCES TEST AND INSPECTION

Waste is a significant cost driver and has a majorimpact on the bottom line...


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The Focus of Six Sigma..

Y = f(x)

All critical characteristics (Y)are driven by factors (x) whichare upstream from the

results.

Attempting to manage results(Y) only causes increasedcosts due to rework, test andinspection

Understanding and controllingthe causative factors (x) is thereal key to high quality at lowcost...


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INSPECTION EXERCISE

The necessity of training farm hands for first classfarms in the fatherly handling of farm livestock isforemost in the minds of farm owners. Since the

forefathers of the farm owners trained the farm handsfor first class farms in the fatherly handling of farmlivestock, the farm owners feel they should carry onwith the family tradition of training farm hands of first

class farms in the fatherly handling of farm livestockbecause they believe it is the basis of goodfundamental farm management.

How many fs can you identify in 1 minute of inspection.


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INSPECTION EXERCISE

The necessity of* training f*arm hands f*or f*irst classf*arms in the f*atherly handling of* f*arm livestock isf*oremost in the minds of* f*arm owners. Since the

f*oref*athers of* the f*arm owners trained the f*armhands f*or f*irst class f*arms in the f*atherly handlingof* f*arm livestock, the f*arm owners f*eel they shouldcarry on with the f*amily tradition of* training f*arm

hands of* f*irst class f*arms in the f*atherly handlingof* f*arm livestock because they believe it is the basisof* good f*undamental f*arm management.

How many fs can you identify in 1 minute of inspection.36 total are available.


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IMPROVEMENT ROADMAP

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Define the problem andverify the primary andsecondary measurementsystems.

Identify the few factorswhich are directlyinfluencing the problem.

Determine values for thefew contributing factors

which resolve theproblem.

Determine long termcontrol measures whichwill ensure that thecontributing factors

remain controlled.

Objective


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Measurements are critical...

If we cant accurately measure

something, we really dont know much

about it.

If we dont know much about it, wecant control it.

If we cant control it, we are at the

mercy of chance.


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WHY STATISTICS?THE ROLE OF STATISTICS IN SIX SIGMA..

WE DONT KNOW WHAT WE DONT KNOW IF WE DONT HAVE DATA, WE DONT KNOW

IF WE DONT KNOW, WE CAN NOT ACT

IF WE CAN NOT ACT, THE RISK IS HIGH

IF WE DO KNOW AND ACT, THE RISK IS MANAGED

IF WE DO KNOW AND DO NOT ACT, WE DESERVE THE LOSS.

DR. Mikel J. Harry

TO GET DATA WE MUST MEASURE

DATA MUST BE CONVERTED TO INFORMATION

INFORMATION IS DERIVED FROM DATA THROUGHSTATISTICS

USLT

LSL


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WHY STATISTICS?THE ROLE OF STATISTICS IN SIX SIGMA..

Ignorance is not bliss, it is the food of failure andthe breeding ground for loss.

DR. Mikel J. Harry

Years ago a statistician might have claimed that

statistics dealt with the processing of data. Todays statistician will be more likely to say thatstatistics is concerned with decision making in theface of uncertainty.

Bartlett

USLT

LSL


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Sales Receipts

On Time Delivery

Process Capacity

Order Fulfillment Time

Reduction of WasteProduct Development Time

Process Yields

Scrap Reduction

Inventory Reduction

Floor Space Utilization

WHAT DOES IT MEAN?

Random Chance or Certainty.

Which would you choose.?


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Learning Objectives

Have a broad understanding of statistical concepts andtools.

Understand how statistical concepts can be used toimprove business processes.

Understand the relationship between the curriculum and

the four step six sigma problem solving process(Measure, Analyze, Improve and Control).


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RED BEAD EXPERIMENT


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Learning Objectives

Have an understanding of the difference betweenrandom variation and a statistically significant event.

Understand the difference between attempting tomanage an outcome (Y) as opposed to managingupstream effects (xs).

Understand how the concept of statistical significancecan be used to improve business processes.


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WELCOME TO THE WHITE BEADFACTORY

HIRING NEEDS

BEADS ARE OUR BUSINESS

PRODUCTION SUPERVISOR

4 PRODUCTION WORKERS

2 INSPECTORS

1 INSPECTION SUPERVISOR

1 TALLY KEEPER


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STANDING ORDERS

Follow the process exactly.

Do not improvise or vary from the documented process.

Your performance will be based solely on your ability to producewhite beads.

No questions will be allowed after the initial training period.

Your defect quota is no more than 5 off color beads allowed perpaddle.


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WHITE BEAD MANUFACTURING PROCESSPROCEDURES The operator will take the bead paddle in the right hand.

Insert the bead paddle at a 45 degree angle into the bead bowl. Agitate the bead paddle gently in the bead bowl until all spaces are

filled.

Gently withdraw the bead paddle from the bowl at a 45 degree angleand allow the free beads to run off.

Without touching the beads, show the paddle to inspector #1 and waituntil the off color beads are tallied.

Move to inspector #2 and wait until the off color beads are tallied.

Inspector #1 and #2 show their tallies to the inspection supervisor. Ifthey agree, the inspection supervisor announces the count and thetally keeper will record the result. If they do not agree, the inspectionsupervisor will direct the inspectors to recount the paddle.

When the count is complete, the operator will return all the beads tothe bowl and hand the paddle to the next operator.


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INCENTIVE PROGRAM

Low bead counts will be rewarded with a bonus.

High bead counts will be punished with a reprimand.

Your performance will be based solely on your abilityto produce white beads.

Your defect quota is no more than 7 off color beadsallowed per paddle.


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PLANT RESTRUCTURE

Defect counts remain too high for the plant to beprofitable.

The two best performing production workers will beretained and the two worst performing productionworkers will be laid off.

Your performance will be based solely on your abilityto produce white beads.

Your defect quota is no more than 10 off color beadsallowed per paddle.


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OBSERVATIONS.

WHAT OBSERVATIONS DID YOU

MAKE ABOUT THIS PROCESS.?


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Y = f(x)

All critical characteristics (Y)are driven by factors (x) whichare downstream from the

results.




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Learning Objectives

Have an understanding of the difference betweenrandom variation and a statistically significant event.

Understand the difference between attempting tomanage an outcome (Y) as opposed to managingupstream effects (xs).

Understand how the concept of statistical significancecan be used to improve business processes.


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INTRODUCTION TO

PROBABILITYDISTRIBUTIONS


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Learning Objectives

Have a broad understanding of what probabilitydistributions are and why they are important.

Understand the role that probability distributions play indetermining whether an event is a random occurrence orsignificantly different.

Understand the common measures used to characterize apopulation central tendency and dispersion.

Understand the concept of Shift & Drift.


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Why do we Care?

An understanding ofProbability Distributions isnecessary to:

Understand the concept anduse of statistical tools.

Understand the significance ofrandom variation in everydaymeasures.

Understand the impact ofsignificance on the successfulresolution of a project.


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IMPROVEMENT ROADMAPUses of Probability Distributions

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Establish baseline datacharacteristics.

Project Uses

Identify and isolatesources of variation.

Use the concept of shift &drift to establish projectexpectations.

Demonstrate before andafter results are not

random chance.


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Focus on understanding the concepts

Visualize the concept

Dont get lost in the math.

KEYS TO SUCCESS


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Measurements are critical...

If we cant accurately measure

something, we really dont know much

about it.

If we dont know much about it, wecant control it.

If we cant control it, we are at the

mercy of chance.


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Types of Measures

Measures where the metric is composed of aclassification in one of two (or more) categories is calledAttribute data. This data is usually presented as acount or percent.

Good/Bad Yes/No

Hit/Miss etc.

Measures where the metric consists of a number whichindicates a precise value is called Variable data. Time

Miles/Hr


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COIN TOSS EXAMPLE

Take a coin from your pocket and toss it 200 times.

Keep track of the number of times the coin falls asheads.

When complete, the instructor will ask you for yourhead count.


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COIN TOSS EXAMPLE

130120110100908070

10000

5000

0

Cum

ulativeFrequency

Results from 10,000 people doing a coin toss 200 times.Cumulative Count

130120110100908070

600

500

400

300

200

100

0

"Head Count"

Frequency

Results from 10,000 people doing a coin toss 200 times.Count Frequency

130120110100908070

100

50

0

"Head Count"

Cumu

lativePercent

Results from 10,000 people doing a coin toss 200 times.Cumulative Percent

Cumulative Frequency

CumulativePercent

Cumulative count is simply the total frequencycount accumulated as you move from left to

right until we account for the total population of10,000 people.

Since we know how many people were in thispopulation (ie 10,000), we can divide each of thecumulative counts by 10,000 to give us a curvewith the cumulative percent of population.


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COIN TOSS PROBABILITY EXAMPLE

130120110100908070

100

50

0

Cumu

lativePercent

Results from 1 0,000 peop le doing a coin toss 200 timesCumulative Percent

This means that we can nowpredict the change thatcertain values can occur

based on these percentages.Note here that 50% of thevalues are less than ourexpected value of 100.

This means that in a futureexperiment set up the sameway, we would expect 50%of the values to be less than100.


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COIN TOSS EXAMPLE

130120110100908070

600

500

400

300

200

100

0

"Head Count"

Frequency

Results from 10,000 people doing a coin toss 200 times.Count Frequency

130120110100908070

100

50

0

"Head Count"

Cumu

lativePercent

Results from 10,000 people doing a coin toss 200 times.Cumulative Percent

We can now equate a probability to theoccurrence of specific values or groups ofvalues.

For example, we can see that the

occurrence of a Head count of less than74 or greater than 124 out of 200 tossesis so rare that a single occurrence wasnot registered out of 10,000 tries.

On the other hand, we can see that thechance of getting a count near (or at) 100

is much higher. With the data that wenow have, we can actually predict each ofthese values.


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COIN TOSS PROBABILITY DISTRIBUTION

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

NUMBER OF HEADS

PROCESS CENTEREDON EXPECTED VALUE

s

SIGMA (s ) IS A MEASUREOF SCATTER FROM THE

EXPECTED VALUE THAT

CAN BE USED TOCALCULATE A PROBABILITY

OF OCCURRENCE

SIGMA VALUE (Z)

CUM % OF POPULATION

58 65 72 79 86 93 100 107 114 121 128 135 142

.003 .135 2.275 15.87 50.0 84.1 97.7 99.86 99.997

130120110100908070

600

500

400

300

200

100

0

Frequency

If we know wherewe are in the

population we canequate that to aprobability value.This is the purposeof the sigma value(normal data).

% of population = probability of occurrence


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Probability and Statistics

the odds of Colorado University winning the nationaltitle are 3 to 1

Drew Bledsoes pass completion percentage for the last6 games is .58% versus .78% for the first 5 games

The Senator will win the election with 54% of the popularvote with a margin of +/- 3%

Probability and Statistics influence our lives daily Statistics is the universal lanuage for science Statistics is the art of collecting, classifying,presenting, interpreting and analyzing numericaldata, as well as making conclusions about thesystem from which the data was obtained.


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Population Vs. Sample (Certainty Vs. Uncertainty)

A sample is just a subset of all possible values

population

sample

Since the sample does not contain all the possible values,

there is some uncertainty about the population. Hence any

statistics, such as mean and standard deviation, are just

estimates of the true population parameters.


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Inferential Statistics

Inferential Statistics is the branch of statistics that deals withdrawing conclusions about a population based on informationobtained from a sample drawn from that population.

While descriptive statistics has been taught for centuries,inferential statistics is a relatively new phenomenon havingits roots in the 20th century.

We infer something about a population when only informationfrom a sample is known.

Probability is the link betweenDescriptive and Inferential Statistics


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WHAT DOES IT MEAN?

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

NUMBER OF HEADS

s

SIGMA VALUE (Z)

CUM % OF POPULATION

58 65 72 79 86 93 100 107 114 121 128 135 142

.003 .135 2.275 15.87 50.0 84.1 97.7 99.86 99.997

130120110100908070

600

500

400

300

200

100

0

Fre

quency

And the first 50trials showedHead Counts

greater than 130?

WHAT IF WE MADE A CHANGE TO THE PROCESS?

Chances are verygood that theprocess distributionhas changed. Infact, there is aprobability greater

than 99.999% thatit has changed.


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USES OF PROBABILITY DISTRIBUTIONS

CriticalValue

CriticalValue

CommonOccurrence

RareOccurrence

RareOccurrence

Primarily these distributions are used to test for significant differences in data sets.

To be classified as significant, the actual measured value must exceed a criticalvalue. The critical value is tabular value determined by the probability distributionand the risk of error. This risk of error is called a risk and indicates the probabilityof this value occurring naturally. So, an a risk of .05 (5%) means that this critical

value will be exceeded by a random occurrence less than 5% of the time.


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SO WHAT MAKES A DISTRIBUTION UNIQUE?

CENTRAL TENDENCY

Where a population is located.

DISPERSION

How wide a population is spread.

DISTRIBUTION FUNCTION

The mathematical formula thatbest describes the data (we willcover this in detail in the nextmodule).


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COIN TOSS CENTRAL TENDENCY

1 3 01 2 01 1 01 0 09 08 07 0

6 0 0

5 0 0

4 0 0

3 0 0

2 0 0

1 0 0

0

Num

berofoccurrences

What are some of the ways that we can easily indicatethe centering characteristic of the population?

Three measures have historically been used; the

mean, the median and the mode.


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WHAT IS THE MEAN?

ORDERED DATA SET

-5

-3

-1

-1

0

0

0

0

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

The mean has already been used in several earlier modulesand is the most common measure of central tendency for apopulation. The mean is simply the average value of thedata.

n=12

xi = - 2

mean xx

n

i= = =-

= - 212

17.

Mean


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WHAT IS THE MEDIAN?

ORDERED DATA SET

-5

-3

-1

-1

0

0

0

0

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

If we rank order (descending or ascending) the data set forthis distribution we could represent central tendency by theorder of the data points.

If we find the value half way (50%) through the data points, wehave another way of representing central tendency. This iscalled the median value.

Median

Value

Median

50% of datapoints


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WHAT IS THE MODE?

ORDERED DATA SET

-5

-3

-1

-1

0

0

0

0

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

If we rank order (descending or ascending) the data set forthis distribution we find several ways we can represent centraltendency.

We find that a single value occurs more often than any other.Since we know that there is a higher chance of this

occurrence in the middle of the distribution, we can use thisfeature as an indicator of central tendency. This is called themode.

Mode Mode


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MEASURES OF CENTRAL TENDENCY, SUMMARY

MEAN ( )(Otherwise known as the average)

XX

ni= =

-= 2

1217.

X

ORDERED DATA SET

-5

-3

-1

-1

0

0

00

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

ORDERED DATA SET

-5

-3

-1

-1

0

0

0

0

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

ORDERED DATA SET

-5

-3

-1

-1

0

0

00

0

1

3-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

MEDIAN

(50 percentile data point)

Here the median value falls between two zerovalues and therefore is zero. If the values were

say 2 and 3 instead, the median would be 2.5.MODE

(Most common value in the data set)

The mode in this case is 0 with 5 occurrenceswithin this data.

Mediann=12

n/2=6

n/2=6

} Mode = 0Mode = 0


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SO WHATS THE REAL DIFFERENCE?

MEAN

The mean is the mostconsistently accurate measure ofcentral tendency, but is more

difficult to calculate than theother measures.

MEDIAN AND MODE

The median and mode are both

very easy to determine. Thatsthe good news.The bad news

is that both are more susceptibleto bias than the mean.


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SO WHATS THE BOTTOM LINE?

MEAN

Use on all occasions unless acircumstance prohibits its use.

MEDIAN AND MODE

Only use if you cannot usemean.


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COIN TOSS POPULATION DISPERSION

1 3 01 2 01 1 01 0 09 08 07 0

6 0 0

5 0 0

4 0 0

3 0 0

2 0 0

1 0 0

0

Num

berofoccurrence

s

What are some of the ways that we can easily indicate the dispersion(spread) characteristic of the population?

Three measures have historically been used; the range, the standard

deviation and the variance.


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WHAT IS THE RANGE?

ORDERED DATA SET-5

-3

-1

-1

00

0

0

0

13

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

The range is a very common metric which is easilydetermined from any ordered sample. To calculate the rangesimply subtract the minimum value in the sample from themaximum value.

Range

RangeMaxMin

Range x xMAX MIN= - = - - =4 5 9( )


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WHAT IS THE VARIANCE/STANDARD DEVIATION?

The variance (s2) is a very robust metric which requires a fair amount of work todetermine. The standard deviation(s) is the square root of the variance and is themost commonly used measure of dispersion for larger sample sizes.

( )s

X X

n

i2

2

1

61 67

12 15 6=

-

-=

-=

..

DATA SET-5

-3-1-10000013

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

XX

n

i= =

-=

212

-.17X Xi -

-5-(-.17)=-4.83

-3-(-.17)=-2.83

-1-(-.17)=-.83

-1-(-.17)=-.83

0-(-.17)=.17

0-(-.17)=.17

0-(-.17)=.17

0-(-.17)=.17

0-(-.17)=.17

1-(-.17)=1.17

3-(-.17)=3.17

4-(-.17)=4.17

( )X Xi

-

2

(-4.83)2=23.32

(-2.83)2=8.01

(-.83)2=.69

(-.83)2=.69

(.17)2=.03

(.17)2=.03

(.17)2=.03

(.17)2=.03

(.17)2=.03

(1.17)2=1.37

(3.17)2=10.05

(4.17)2=17.39

61.67


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MEASURES OF DISPERSION

RANGE (R)(The maximum data value minus the minimum)

ORDERED DATA SET

-5-3

-1

-1

0

0

0

0

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

ORDERED DATA SET

-5

-3

-1

-1

0

0

00

0

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

VARIANCE (s2)

(Squared deviations around the center point)

STANDARD DEVIATION (s)

(Absolute deviation around the center point)

Min=-5

R X X= - = - - =max min ( )4 6 10

Max=4DATA SET

-5

-3

-1

-10

0

0

00

1

3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 4

XX

n

i= =

-= 2

12-.17

( )s

X X

ni2

2

1

61 67

12 15 6=

--

=-

= . .

X Xi --5-(-.17)=-4.83

-3-(-.17)=-2.83

-1-(-.17)=-.83

-1-(-.17)=-.83

0-(-.17)=.17

0-(-.17)=.17

0-(-.17)=.17

0-(-.17)=.17

0-(-.17)=.17

1-(-.17)=1.17

3-(-.17)=3.174-(-.17)=4.17

( )X Xi-

2

(-4.83)2=23.32

(-2.83)2=8.01

(-.83)2=.69

(-.83)2=.69

(.17)2=.03

(.17)2=.03

(.17)2=.03

(.17)2=.03

(.17)2=.03

(1.17)2=1.37

(3.17)2=10.05(4.17)2=17.39

61.67

s s= = =2 5 6 2 37. .


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SAMPLE MEAN AND VARIANCE EXAMPLE

$= =

XN

Xi

s ( )$22

1= =

-2s

n

-X Xi

Xi10151214

109

111210

12

1

2

3

4

5

6

7

8

910

SX

Xi

-X Xi ( )2-X Xi

2s


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SO WHATS THE REAL DIFFERENCE?

VARIANCE/ STANDARD DEVIATION

The standard deviation is the mostconsistently accurate measure ofcentral tendency for a single

population. The variance has theadded benefit of being additive overmultiple populations. Both are difficultand time consuming to calculate.

RANGEThe range is very easy to determine.Thats the good news.The bad news

is that it is very susceptible to bias.


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SO WHATS THE BOTTOM LINE?

VARIANCE/ STANDARDDEVIATION

Best used when you haveenough samples (>10).

RANGE

Good for small samples (10 orless).


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SO WHAT IS THIS SHIFT & DRIFT STUFF...

The project is progressing well and you wrap it up. 6 monthslater you are surprised to find that the population has taken a

shift.

-12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12

USLLSL


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SO WHAT HAPPENED?

All of our work was focused in a narrow time frame.Over time, other long term influences come and gowhich move the population and change some of itscharacteristics. This is called shift and drift.

Historically, this shift and driftprimarily impacts the position ofthe mean and shifts it 1.5 s fromits original position.

Original Study


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VARIATION FAMILIES

Variation is presentupon repeatmeasurements withinthe same sample.

Variation is presentupon measurements ofdifferent samples

collected within a shorttime frame.

Variation is presentupon measurementscollected with a

significant amount oftime between samples.

Sources ofVariation

Within IndividualSample

Piece to

Piece

Time to Time


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SO WHAT DOES IT MEAN?

To compensate for these longterm variations, we mustconsider two sets of metrics.

Short term metrics are thosewhich typically are associatedwith our work. Long term metricstake the short term metric dataand degrade it by an average of1.5s.


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IMPACT OF 1.5s SHIFT AND DRIFTZ PPM ST Cpk PPM LT (+1.5 s)

0.0 500,000 0.0 933,193

0.1 460,172 0.0 919,243

0.2 420,740 0.1 903,199

0.3 382,089 0.1 884,930

0.4 344,578 0.1 864,334

0.5 308,538 0.2 841,3450.6 274,253 0.2 815,940

0.7 241,964 0.2 788,145

0.8 211,855 0.3 758,036

0.9 184,060 0.3 725,747

1.0 158,655 0.3 691,462

1.1 135,666 0.4 655,4221.2 115,070 0.4 617,911

1.3 96,801 0.4 579,260

1.4 80,757 0.5 539,828

1.5 66,807 0.5 500,000

1.6 54,799 0.5 460,172

1.7 44,565 0.6 420,740

Here, you can see that theimpact of this concept is

potentially very significant. Inthe short term, we have driventhe defect rate down to 54,800

ppm and can expect to seeoccasional long term ppm tobe as bad as 460,000 ppm.


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SHIFT AND DRIFT EXERCISE

We have just completed a project and have presented thefollowing short term metrics:

Zst=3.5

PPMst=233

Cpkst=1.2

Calculate the long

term values for eachof these metrics.


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Learning Objectives

Have a broad understanding of what probabilitydistributions are and why they are important.

Understand the role that probability distributions play in

determining whether an event is a random occurrence orsignificantly different.

Understand the common measures used to characterize a

population central tendency and dispersion.

Understand the concept of Shift & Drift.


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COMMON PROBABILITY

DISTRIBUTIONS ANDTHEIR USES


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Learning Objectives

Have a broad understanding of how probabilitydistributions are used in improvement projects.

Review the origin and use of common probabilitydistributions.


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IMPROVEMENT ROADMAPUses of Probability Distributions

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Baselining Processes

Verifying Improvements

Common Uses


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Focus on understanding the use of the distributions

Practice with examples wherever possible

Focus on the use and context of the tool

KEYS TO SUCCESS

PROBABILITY DISTRIBUTIONS WHERE


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X

X X X

X

Data points vary, but as the data accumulates, it forms a distribution which occurs naturally.

Location Spread Shape

Distributions can vary in:

PROBABILITY DISTRIBUTIONS, WHEREDO THEY COME FROM?

COMMON PROBABILITY DISTRIBUTIONS


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-4

-3

-2

-1

0

1

2

3

4

0 1 2 3 4 5 6 7

-4

-3

-2

-1

0

1

2

3

4

0 1 2 3 4 5 6 7

0

1

2

3

4

0 1 2 3 4 5 6 7

Original Population

Subgroup Average

Subgroup Variance (s2)

Continuous Distribution

Normal Distribution

c2

Distribution

COMMON PROBABILITY DISTRIBUTIONS


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THE LANGUAGE OF MATH

Symbol Name Statistic Meaning Common Uses

a Alpha Significance level Hypothesis Testing,DOE

c2 Chi Square Probability Distribution Confidence Intervals, ContingencyTables, Hypothesis Testing

S Sum Sum of Individual values Variance Calculationst t, Student t Probability Distribution Hypothesis Testing, Confidence Interval

of the Mean

n Sample

Size

Total size of the Sample

Taken

Nearly all Functions

Nu Degree of Freedom Probability Distributions, HypothesisTesting, DOE

Beta Beta Risk Sample Size Determination Delta Difference between

population means

Sample Size Determination

SigmaValue

Number of Standard

Deviations a value Exists

from the Mean

Probability Distributions, Process

Capability, Sample Size Determinations

Population and Sample Symbology


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Population and Sample Symbology

Value Population Sample

Mean Variance s2 s2Standard Deviation s sProcess Capability Cp

Binomial Mean

x

P P

Cp


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THREE PROBABILITY DISTRIBUTIONS

tX

s

n

CALC = -

Significant t tCALC CRIT=

Significant F FCALC CRIT= F sscalc = 12

22

( )c

a,df

e a

e

f f

f

2

2

=-

SignificantCALC CRIT

= c c2 2

Note that in each case, a limit has been established to determine what israndom chance verses significant difference. This point is called the criticalvalue. If the calculated value exceeds this critical value, there is very lowprobability (P


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-1s +1s

68.26%

+/- 1s = 68%2 tail = 32%

1 tail = 16%

-2s +2s

95.46%

+/- 2s = 95%2 tail = 4.6%

1 tail = 2.3%

-3s +3s

99.73%

+/- 3s = 99.7%2 tail = 0.3%1 tail = .15%

Common Test Values

Z(1.6) = 5.5% (1 tail a=.05)Z(2.0) = 2.5% (2 tail a=.05)

Z TRANSFORM


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Y = f(x)

All critical characteristics (Y)are driven by factors (x) whichare downstream from the

results.



Probability distributions identify sourcesof causative factors (x). These can beidentified and verified by testing whichshows their significant effects againstthe backdrop of random noise.

BUT WHAT DISTRIBUTION


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SHOULD I USE?

CharacterizePopulation

PopulationAverage

PopulationVariance

DetermineConfidence

Interval forPoint Values

F Stat (n>30)

F Stat (n5)

Z Stat (n>30)

Z Stat (p)

t Stat (n30)

Z Stat (p)

t Stat (,n


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These interactions form a newpopulation which can now be

used to predict futureperformance.

HOW DO POPULATIONS INTERACT?


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HOW DO POPULATIONS INTERACT?ADDING TWO POPULATIONS

Population means interact in a simple intuitive manner.

1 2

Means Add

1 + 2 = new

Population dispersions interact in an additive manner

s1 s2

Variations Add

s12 + s22 = snew2

snew

new

HOW DO POPULATIONS INTERACT?


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HOW DO POPULATIONS INTERACT?SUBTRACTING TWO POPULATIONS

Population means interact in a simple intuitive manner.

1 2

Means Subtract

1 - 2 = new

Population dispersions interact in an additive manner

s1 s2

Variations Add

s12 + s22 = snew2

snew

new


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TRANSACTIONAL EXAMPLE

Orders are coming in with the following characteristics:

Shipments are going out with the followingcharacteristics:

Assuming nothing changes, what percent of the time willshipments exceed orders?

X = $53,000/week

s = $8,000

X = $60,000/weeks = $5,000



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To solve this problem, we must create a new distribution to model the situation posedin the problem. Since we are looking for shipments to exceed orders, the resulting

distribution is created as follows:

X X Xshipments orders shipments orders- = - = - =$60 , $53, $7 ,000 000 000

( ) ( )s s sshipments orders shipments orders- = + = + =2 22 2

5000 8000 $9434

The new distribution looks like this with a mean of $7000 and astandard deviation of $9434. This distribution represents theoccurrences of shipments exceeding orders. To answer the originalquestion (shipments>orders) we look for $0 on this new distribution.Any occurrence to the right of this point will represent shipments >orders. So, we need to calculate the percent of the curve that existsto the right of $0.

$7000

$0

Shipments > orders

X = $53,000 in orders/week

s = $8,000

X = $60,000 shipped/week

s = $5,000

Orders Shipments



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TRANSACTIONAL EXAMPLE, CONTINUED

X X Xshipments orders shipments orders- = - = - =$60 , $53, $7 ,000 000 000

( ) ( )s s sshipments orders shipments orders- = + = + =2 22 2

5000 8000 $9434

To calculate the percent of the curve to the right of $0 we need toconvert the difference between the $0 point and $7000 into sigma

intervals. Since we know every $9434 interval from the mean is onesigma, we can calculate this position as follows:

$7000

$0

Shipments > orders

074

-=

-=

X

ss

$0 $7000

$9434.

Look up .74s in the normal table and you will find .77. Therefore, the answer to the original

question is that 77% of the time, shipments will exceed orders.

Now, as a classroom exercise, what percent of the time willshipments exceed orders by $10,000?


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MANUFACTURING EXAMPLE

2 Blocks are being assembled end to end and significantvariation has been found in the overall assembly length.

The blocks have the following dimensions:

Determine the overall assembly length and standarddeviation.

X1 = 4.00 inches

s1 = .03 inches

X2 = 3.00 inches

s2 = .04 inches


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CORRELATION ANALYSIS


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Learning Objectives

Understand how correlation can be used to demonstratea relationship between two factors.

Know how to perform a correlation analysis and calculate

the coefficient of linear correlation (r).

Understand how a correlation analysis can be used in animprovement project.


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Why do we Care?

Correlation Analysis isnecessary to:

show a relationship betweentwo variables. This also sets the

stage for potential cause andeffect.

IMPROVEMENT ROADMAP


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IMPROVEMENT ROADMAPUses of Correlation Analysis

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Determine and quantifythe relationship betweenfactors (x) and outputcharacteristics (Y)..

Common Uses

KEYS TO SUCCESS


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Always plot the data

Remember: Correlation does not always imply cause & effect

Use correlation as a follow up to the Fishbone Diagram

Keep it simple and do not let the tool take on a life of its own

WHAT IS CORRELATION?


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Input or x variable (independent)

Output or yvariable

(dependent)

Correlation

Y= f(x)

As the input variable changes,

there is an influence or biason the output variable.



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A measurable relationship between two variable data characteristics.

Not necessarily Cause & Effect (Y=f(x))

Correlation requires paired data sets (ie (Y1,x1), (Y2,x2), etc)

The input variable is called the independent variable (x or KPIV) since it isindependent of any other constraints

The output variable is called the dependent variable (Y or KPOV) since itis (theoretically) dependent on the value of x.

The coefficient of linear correlation r is the measure of the strength ofthe relationship.

The square of r is the percent of the response (Y) which is related to theinput (x).



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CALCULATING r


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Coefficient of Linear Correlation

Calculate sample covariance( )

Calculate sx and sy for eachdata set

Use the calculated values tocompute rCALC.

Add a + for positive correlationand - for a negative correlation.

( )( )s

x x y y

nxyi i

=- -

-

1

rs sCALCsxy

x y

=

sxy

While this is the most precise method to calculatePearsons r, there is an easier way to come up with a fairlyclose approximation...

APPROXIMATING r


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Coefficient of Linear Correlation

Y=f(x)

x

rW

L -

1

r - - = -16 7

12 647.

..

+ = positive slope

- = negative slope

L| | | | | | | | | | | | | | |1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

6.7 12.6

Plot the data on orthogonal axisDraw an Oval around the data

Measure the length and widthof the Oval

Calculate the coefficient of

linear correlation (r) based onthe formulas below


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HOW DO I KNOW WHEN I HAVE CORRELATION?

OrderedPairs rCRIT

5 .88

6 .81

7 .75

8 .71

9 .67

10 .63

15 .51

20 .44

25 .40

30 .3650 .28

80 .22

100 .20

The answer should strike a familiar cord at this pointWe have confidence (95%) that we have correlationwhen |rCALC|> rCRIT.

Since sample size is a key determinate of rCRIT we needto use a table to determine the correct rCRIT given thenumber of ordered pairs which comprise the complete

data set.

So, in the preceding example we had 60 ordered pairsof data and we computed a rCALC of -.47. Using the tableat the left we determine that the rCRIT value for 60 is .26.

Comparing |rCALC

|> rCRIT

we get .47 > .26. Therefore thecalculated value exceeds the minimum critical valuerequired for significance.

Conclusion: We are 95% confident that the observedcorrelation is significant.

L i Obj i


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Learning Objectives

Understand how correlation can be used to demonstratea relationship between two factors.

Know how to perform a correlation analysis and calculate

the coefficient of linear correlation (r).

Understand how a correlation analysis can be used in ablackbelt story.


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CENTRAL LIMIT THEOREM

L i Obj ti


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Learning Objectives

Understand the concept of the Central Limit Theorem.

Understand the application of the Central Limit Theoremto increase the accuracy of measurements.


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IMPROVEMENT ROADMAP


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IMPROVEMENT ROADMAPUses of the Central Limit Theorem

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

The Central LimitTheorem underlies all

statistic techniques whichrely on normality as afundamental assumption

Common Uses

KEYS TO SUCCESS


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Focus on the practical application of the concept


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WHAT IS THE CENTRAL LIMIT THEOREM?

Central Limit Theorem

For almost all populations, the sampling distribution of themean can be approximated closely by a normal distribution,

provided the sample size is sufficiently large.

Normal

Why do we Care?


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What this means is that no matter what kindof distribution we sample, if the sample sizeis big enough, the distribution for the meanis approximately normal.

This is the key link that allows us to use

much of the inferential statistics we havebeen working with so far.

This is the reason that only a few probabilitydistributions (Z, t and c2) have such broadapplication.

If a random event happens a great many times,the average results are likely to be predictable.

Jacob Bernoulli

HOW DOES THIS WORK?


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ParentPopulation

n=2

n=5

n=10

As you average a larger

and larger number ofsamples, you can see howthe original sampledpopulation is transformed..

ANOTHER PRACTICAL ASPECT


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snxx=

sThis formula is for the standard error of the mean.

What that means in layman's terms is that thisformula is the prime driver of the error term of themean. Reducing this error term has a direct impacton improving the precision of our estimate of themean.

The practical aspect of all this is that if you want to improve the precision of anytest, increase the sample size.

So, if you want to reduce measurement error (for example) to determine a betterestimate of a true value, increase the sample size. The resulting error will bereduced by a factor of . The same goes for any significance testing.Increasing the sample size will reduce the error in a similar manner.

1n

DICE EXERCISE


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Break into 3 teamsTeam one will be using 2 diceTeam two will be using 4 diceTeam three will be using 6 dice

Each team will conduct 100 throws of their dice andrecord the average of each throw.

Plot a histogram of the resulting data.

Each team presents the results in a 10 min report out.

Learning Objectives


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Learning Objectives

Understand the concept of the Central Limit Theorem.

Understand the application of the Central Limit Theoremto increase the accuracy of measurements.


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PROCESS CAPABILITY

ANALYSIS

Learning Objectives


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Learning Objectives

Understand the role that process capability analysisplays in the successful completion of an improvementproject.

Know how to perform a process capability analysis.

Why do we Care?


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Why do we Care?

Process Capability Analysis isnecessary to:

determine the area of focuswhich will ensure successfulresolution of the project.

benchmark a process to enabledemonstrated levels ofimprovement after successfulresolution of the project.

demonstrate improvement aftersuccessful resolution of theproject.

IMPROVEMENT ROADMAP


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IMPROVEMENT ROADMAPUses of Process Capability Analysis

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Baselining a processprimary metric (Y) prior tostarting a project.

Common Uses

Characterizing thecapability of causitivefactors (x).

Characterizing a processprimary metric after

changes have beenimplemented todemonstrate the level ofimprovement.

KEYS TO SUCCESS


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Must have specification limits - Use process targets if no specs available

Dont get lost in the math

Relate to Z for comparisons (Cpk x 3 = Z)

For Attribute data use PPM conversion to Cpk and Z

WHAT IS PROCESS CAPABILITY?


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Process capability is simply a measure of how good a metric is performingagainst and established standard(s). Assuming we have a stable processgenerating the metric, it also allows us to predict the probability of the metricvalue being outside of the established standard(s).

Spec

Out of SpecIn Spec

Probability

Spec(Lower)

Spec(Upper)

In Spec Out of SpecOut of Spec

ProbabilityProbability

Upper and Lower Standards(Specifications)

Single Standard(Specification)

WHAT IS PROCESS CAPABILITY?


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Process capability (Cpk) is a function of how the population is centered (|-spec|) and the population spread (s).

Process Center

(|-spec|)

Spec(Lower)

Spec(Upper)

In SpecOut ofSpec

Out ofSpec

High Cpk Poor CpkSpec

(Lower)

Spec(Upper)

In Spec

Out ofSpec

Out ofSpec

Process Spread

(s)

Spec(Lower)

Spec(Upper)

In Spec

Out ofSpec

Out ofSpec

Spec(Lower)

Spec(Upper)

In Spec

Out ofSpec

Out ofSpec


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HOW IS PROCESS CAPABILITY CALCULATED

Spec(LSL)

Spec(USL)

Note:

LSL = Lower Spec Limit

USL = Upper Spec Limit

Distance between thepopulation mean andthe nearest spec limit(|-USL |). Thisdistance divided by3s is Cpk.

( )C

MIN LSL USL

PK=

- -

s

,

3

Expressed mathematically, this looks like:

PROCESS CAPABILITY EXAMPLE


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PROCESS CAPABILITY EXAMPLE

Calculation Values:

Upper Spec value = $200,000 maximumNo Lower Spec

= historical average = $250,000s = $20,000

Calculation:

Answer: Cpk= -.83

We want to calculate the process capability for our inventory. Thehistorical average monthly inventory is $250,000 with a standarddeviation of $20,000. Our inventory target is $200,000 maximum.

( ) ( )C

MIN LSL USL

PK =

- -

=

-

=

s

, $200, $250,

* $20, -.833

000 000

3 000


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Learning Objectives


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Learning Objectives

Understand the role that process capability analysisplays in the successful completion of an improvementproject.

Know how to perform a process capability analysis.


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MULTI-VARI ANALYSIS

Learning Objectives


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Learning Objectives

Understand how to use multi-vari charts in completing animprovment project.

Know how to properly gather data to construct multi-vari

charts.

Know how to construct a multi-vari chart.

Know how to interpret a multi-vari chart.

Why do we Care?


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Why do we Care?

Multi-Vari charts are a:

Simple, yet powerful way tosignificantly reduce the numberof potential factors which couldbe impacting your primary metric.

Quick and efficient method tosignificantly reduce the time andresources required to determinethe primary components of

variation.

IMPROVEMENT ROADMAPUses of Multi-Vari Charts


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Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Eliminate a large numberof factors from theuniverse of potentialfactors.

Common Uses

KEYS TO SUCCESS


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Gather data by systematically sampling the existing process

Perform ad hoc training on the tool for the team prior to use

Ensure your sampling plan is complete prior to gathering data

Have team members (or yourself) do the sampling to avoid bias

Careful planning before you start

A World of Possible Causes (KPIVs)..


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The Goal of the logical search

is to narrow down to 5-6 key variables !

technique

tooling

operator error

humidity

temperature

supplier hardness

lubrication

old machinery

wrong spec.tool wear

handling damagepressure heat treat

fatigue


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REDUCING THE POSSIBILITIES


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How many guesses do you think it will take to find asingle word in the text book?

Lets try and see.

REDUCING THE POSSIBILITIES


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How many guesses do you think it will take to find a singleword in the text book?

Statistically it should take no more than 17 guesses217=2X2X2X2X2X2X2X2X2X2X2X2X2X2X2X2X2= 131,072

Most Unabridged dictionaries have 127,000 words.

Reduction of possibilities can be an extremely powerfultechnique..

PLANNING A MULTI-VARI ANALYSIS


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Determine the possible families of variation.Determine how you will take the samples.

Take a stratified sample (in order of creation).

DO NOT take random samples.

Take a minimum of 3 samples per group and 3 groups.The samples must represent the full range of the process.

Does one sample or do just a few samples stand out?There could be a main effect or an interaction at the cause.

MULTI VARI ANALYSIS VARIATION FAMILIES


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MULTI-VARI ANALYSIS, VARIATION FAMILIES

Variation is presentupon repeatmeasurements within

the same sample.

Variation is presentupon measurements ofdifferent samples

collected within a shorttime frame.

Variation is presentupon measurementscollected with a

significant amount oftime between samples.

Sources ofVariation

Within IndividualSample

Piece toPiece

Time to Time

The key is reducing the number of possibilities to a manageable few.

MULTI-VARI ANALYSIS, VARIATION SOURCES


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Within Individual Sample

Measurement Accuracy

Out of Round

Irregularities in Part

Piece to Piece

Machine fixturing

Mold cavity differences

Time to Time

Material Changes

Setup Differences

Tool Wear

Calibration Drift

Operator Influence

Manufacturing

(Machining)

Transactional

(Order

Rate)

Piece to Piece

Customer Differences

Order Editor

Sales Office

Sales Rep

Within Individual Sample

Measurement Accuracy

Line Item Complexity

Time to Time

Seasonal Variation

Management Changes

Economic Shifts

Interest Rate

HOW TO DRAW THE CHART


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Range within asingle sample

Sample 1

Average within asingle sample

Plot the first sample range with a point for themaximum reading obtained, and a point for the

minimum reading. Connect the points and plota third point at the average of the within samplereadings

Step 1

Range betweentwo sampleaverages

Sample 1

Sample 2Sample 3

Plot the sample ranges for the remaining piece

to piece data. Connect the averages of the

within sample readings.

Step 2

Plot the time to time groups in the same

manner.

Time 1Time 2

Time 3

Step 3

READING THE TEA LEAVES.


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Common Patterns of Variation

Within Piece

Characterized by largevariation in readings

taken of the same singlesample, often fromdifferent positions withinthe sample.

Piece to Piece


taken between samplestaken within a short timeframe.

Time to Time


taken between samplestaken in groups with asignificant amount oftime elapsed betweengroups.

MULTI-VARI EXERCISE


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We have a part dimension which is considered to be impossible to manufacture. A

capability study seems to confirm that the process is operating with a Cpk=0(500,000 ppm). You and your team decide to use a Multi-Vari chart to localize thepotential sources of variation. You have gathered the following data:

Sample Day/Time

Beginningof Part

Middle ofPart

End ofPart

1 1/0900 .015 .017 .0182 1/0905 .010 .012 .015

3 1/0910 .013 .015 .016

4 2/1250 .014 .015 .018

5 2/1255 .009 .012 .0176 2/1300 .012 .014 .016

7 3/1600 .013 .014 .017

8 3/1605 .010 .013 .015

9 3/1610 .011 .014 .0179

Construct a multi-vari chart of the data

and interpret theresults.

Learning Objectives


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Understand how to use multi-vari charts in completing animprovment project.

Know how to properly gather data to construct multi-vari

charts.

Know how to construct a multi-vari chart.

Know how to interpret a multi-vari chart.


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SAMPLE SIZE

CONSIDERATIONS

Learning Objectives


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Understand the critical role having the right sample sizehas on an analysis or study.

Know how to determine the correct sample size for a

specific study.

Understand the limitations of different data types onsample size.

Why do we Care?


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The correct sample size isnecessary to:

ensure any tests you designhave a high probability of

success.properly utilize the type of datayou have chosen or are limited toworking with.

IMPROVEMENT ROADMAPUses of Sample Size Considerations


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Uses of Sample Size Considerations

Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Sample Sizeconsiderations are used inany situation where asample is being used to

infer a populationcharacteristic.

Common Uses

KEYS TO SUCCESS


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Use variable data wherever possible

Generally, more samples are better in any study

When there is any doubt, calculate the needed sample size

Use the provided excel spreadsheet to ease sample size calculations

CONFIDENCE INTERVALS


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The possibility of error exists in almost every system. This goes for point values aswell. While we report a specific value, that value only represents our best estimatefrom the data at hand. The best way to think about this is to use the form:

true value = point estimate +/- error

The error around the point value follows one of several common probability

distributions. As you have seen so far, we can increase our confidence is to gofurther and further out on the tails of this distribution.

PointValue

+/- 1s = 67% Confidence Band


This error band

which exists around

the point estimate iscalled the confidenceinterval.

BUT WHAT IF I MAKE THE WRONG DECISION?


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Not different (Ho)

Reality

TestD

ecision

Different (H1)

Not different (Ho)

Different (H1)

Correct Conclusion

Correct ConclusionaRisk Type I Error Producer Risk

Type II Error risk Consumer Risk

Test Reality = Different

Decision Point

Risk aRisk

WHY DO WE CARE IF WE HAVE THE TRUE VALUE?How confident do you want to be that you have made the right decision?


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How confident do you want to be that you have made the right decision?

Ho: Patient is not sick

H1: Patient is sick

Error Impact

Type I Error = Treating a patient who is not sick

Type II Error = Not treating a sick patient

A person does not feel well and checks into a hospital for tests.

Not different (Ho)

Reality

TestDecision

Different (H1)

Not different (Ho)

Different (H1)

Correct Conclusion

Correct ConclusionaRisk Type I Error Producer Risk

Type II Error risk Consumer Risk


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CONFIDENCE INTERVAL FORMULAS


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These individual formulas are not critical at this point, but notice that the onlyopportunity for decreasing the error band (confidence interval) without decreasing theconfidence factor, is to increase the sample size.

X t t na n a n + -/ , / ,2 1 2 1s sn XMean

sn

sn

a a

- --

1 1

1 22

22c s c/ /

Standard Deviation

Cpn

Cp Cpn

a n a nc1 2 12 2 121 1

- - -- -/ , / ,Process Capability

( ) ( )$

$ $

$

$ $

/ /p Z

p p

n p p Z

p p

na a-

+-

2 2

1 1

Percent Defective

SAMPLE SIZE EQUATIONS


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n ZX

a=-

/ 2

2

s

ns a

= +

sc

2

22 1/

Allowable error = -

(also known as )

X

Allowable error = s/s

( )n p pZ

E= - $ $/1 2

2

a Allowable error = E

Standard Deviation

Mean

Percent Defective

SAMPLE SIZE EXAMPLE


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Calculation Values:

Average tells you to use the mean formula

Significance: a = 5% (95% confident)Za/2 = Z.025 = 1.96

s=10 pounds

-x = error allowed = 2 pounds

Calculation:

Answer: n=97 Samples

We want to estimate the true average weight for a part within 2 pounds.

Historically, the part weight has had a standard deviation of 10 pounds.We would like to be 95% confident in the results.

nZ

X= -

= =

a

s

/

. *2

2 21 96 10

2 97

SAMPLE SIZE EXAMPLE


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Calculation Values:

Percent defective tells you to use the percent defect formula

Significance: a = 5% (95% confident)Za/2 = Z.025 = 1.96

p = 10% = .1

E = 1% = .01

Calculation:

Answer: n=3458 Samples

We want to estimate the true percent defective for a part within 1%.Historically, the part percent defective has been 10%. We would like tobe 95% confident in the results.

( ) ( )n p p ZE

= - = -

=

$ $ . ..

./

1 1 1 11 96

013458

2

2 2

a

Learning Objectives


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Understand the critical role having the right sample sizehas on an analysis or study.

Know how to determine the correct sample size for a

specific study.

Understand the limitations of different data types onsample size.


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CONFIDENCE INTERVALS

Learning Objectives


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Understand the concept of the confidence interval andhow it impacts an analysis or study.

Know how to determine the confidence interval for a

specific point value.

Know how to use the confidence interval to test futurepoint values for significant change.

Why do we Care?


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Understanding the confidenceinterval is key to:

understanding the limitations ofquotes in point estimate data.

being able to quickly andefficiently screen a series of pointestimate data for significance.

IMPROVEMENT ROADMAPUses of Confidence Intervals


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Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Used in any situationwhere data is beingevaluated for significance.

Common Uses

KEYS TO SUCCESS


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Use variable data wherever possible

Generally, more samples are better (limited only by cost)

Recalculate confidence intervals frequently

Use an excel spreadsheet to ease calculations


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So, what does this do for me?


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The confidence intervalestablishes a way to test whetheror not a significant change hasoccurred in the sampledpopulation. This concept iscalled significance or hypothesistesting.

Being able to tell when asignificant change has occurredhelps in preventing us from

interpreting a significant changefrom a random event andresponding accordingly.

REMEMBER OUR OLD FRIEND SHIFT & DRIFT?


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All of our work was focused in a narrow time frame.Over time, other long term influences come and gowhich move the population and change some of itscharacteristics.

Confidence Intervals give usthe tool to allow us to be able tosort the significant changes fromthe insignificant.

Original Study

USING CONFIDENCE INTERVALS TO SCREEN DATA


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2 3 4 5 6 7

TIME

95% ConfidenceInterval

Significant Change?

WHAT KIND OF PROBLEM DO YOU HAVE?

Analysis for a significant change asks the question What happened tomake this significantly different from the rest?

Analysis for a series of random events focuses on the process and asksthe question What is designed into this process which causes it to have

this characteristic?.

CONFIDENCE INTERVAL FORMULAS


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These individual formulas enable us to calculate the confidence interval for many ofthe most common metrics.

Mean

sn

sn

a a

- --

1 1

1 22

22c s c/ /

Standard Deviation

Cpn

Cp Cpn

a n a nc1 2 12 2 121 1

- - -- -/ , / ,Process Capability

( ) ( )$ $ $ $ $ $/ /p Z p pn p p Z p pna a- + -2 21 1Percent Defective

X t tn

a n a n + -/ , / ,2 1 2 1n

Xs s

CONFIDENCE INTERVAL EXAMPLE


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Calculation Values:

Average defect rate of 14,000 ppm = 14,000/1,000,000 = .014

Significance: a = 5% (95% confident) Za/2 = Z.025 = 1.96

n=10,000

Comparison defect rate of 23,000 ppm = .023

Calculation:

Answer: Yes, .023 is significantly outside of the expected 95% confidence interval of.012 to .016.

Over the past 6 months, we have received 10,000 parts from a vendor with

an average defect rate of 14,000 dpm. The most recent batch of partsproved to have 23,000 dpm. Should we be concerned? We would like tobe 95% confident in the results.

( ) ( )$

$ $

$

$ $

/ /p Z

p p

n p p Z

p p

na a- + -2 2

1 1

( ) ( ). .

. .

,. .

. .

,014 1 96

014 1 014

10 000014 1 96

014 1 014

10 000

- + -p . . . .014 0023 014 0023 +p. .012 016p

CONFIDENCE INTERVAL EXERCISE


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We are tracking the gas mileage of our late model ford and find thathistorically, we have averaged 28 MPG. After a tune up at Billy Bobsauto repair we find that we only got 24 MPG average with a standarddeviation of 3 MPG in the next 16 fillups. Should we be concerned? Wewould like to be 95% confident in the results.

What do you think?

Learning Objectives


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Understand the concept of the confidence interval andhow it impacts an analysis or study.

Know how to determine the confidence interval for a

specific point value.

Know how to use the confidence interval to test futurepoint values for significant change.


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Learning Objectives


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Understand how to select the correct control chart for anapplication.

Know how to fill out and maintain a control chart.

Know how to interpret a control chart to determine theoccurrence of special causes of variation.

Why do we Care?


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Control charts are useful to:

determine the occurrence ofspecial cause situations.

Utilize the opportunities

presented by special causesituations to identify and correct

the occurrence of the special

causes .

IMPROVEMENT ROADMAPUses of Control Charts


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Breakthrough

Strategy

Characterization

Phase1:

Measurement

Phase 2:

Analysis

Optimization

Phase 3:

Improvement

Phase4:

Control

Control charts can beeffectively used todetermine special cause

situations in theMeasurement andAnalysis phases

Common Uses


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What is a Special Cause?


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Remember our earlier work with confidence intervals? Any occurrence which falls

outside the confidence interval has a low probability of occurring by random chanceand therefore is significantly different. If we can identify and correct the cause, we

have an opportunity to significantly improve the stability of the process. Due to theamount of data involved, control charts have historically used 99% confidence fordetermining the occurrence of these special causes

PointValueSpecial cause

occurrence.

X


What is a Control Chart ?


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A control chart is simply a run chart with confidence intervals calculated and drawn in.These Statistical control limits form the trip wires which enable us to determine

when a process characteristic is operating under the influence of a Special cause.

+/- 3s =

99% ConfidenceInterval

So how do I construct a control chart?


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First things first:Select the metric to be evaluated

Select the right control chart for themetric

Gather enough data to calculate thecontrol limits

Plot the data on the chart

Draw the control limits (UCL & LCL)

onto the chart.Continue the run, investigating andcorrecting the cause of any out of

control occurrence.

How do I select the correct chart ?


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What type ofdata do I have?

Variable Attribute

Counting defects or

defectives?

X-s Chart IMR ChartX-R Chart

n > 10 1 < n < 10 n = 1Defectives Defects

What subgroup size

is available?

Constant

Sample Size?

Constant

Opportunity?

yes yesno no

np Chart u Chartp Chart c Chart

Note: A defective unit can have

more than one defect.

How do I calculate the control limits?

X R Chart-


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CL X=

UCL X RA= + 2LCL X RA= - 2

X = average of the subgroup averages

R = average of the subgroup range values

2A = a constant function of subgroup size (n)

For the averages chart:

CL R=

UCL RD= 4LCL RD= 3

For the range chart:

n D4 D3 A2

2 3.27 0 1.88

3 2.57 0 1.02

4 2.28 0 0.73

5 2.11 0 0.58

6 2.00 0 0.48

7 1.92 0.08 0.42

8 1.86 0.14 0.37

9 1.82 0.18 0.34

UCL = upper control limit

LCL = lower control limit


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How do I calculate the control limits?

c and u Charts


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UCL U U

nu = + 3

U= total number of nonconformities/total units evaluated

n = number evaluated in subgroup

For varied opportunity (u): For constant opportunity (c):

C= total number of nonconformities/total number of subgroups

Note: U charts have an individually calculated control limit for each point plotted

LCL U U

nu = - 3

UCL C C C = +3

LCL C CC = -3

How do I interpret the charts?


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The process is said to be out of control if: One or more points fall outside of the control limits

When you divide the chart into zones as shown and:

2 out of 3 points on the same side of the centerline in Zone A

4 out of 5 points on the same side of the centerline in Zone A or B

9 successive points on one side of the centerline

6 successive points successively increasing or decreasing 14 points successively alternating up and down

15 points in a row within Zone C (above and/or below centerline)

Zone A

Zone B

Zone C

Zone C

Zone B

Zone A

Upper Control Limit (UCL)

Lower Control Limit (LCL)

Centerline/Average


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Learning Objectives


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Understand how to select the correct control chart for anapplication.

Know how to fill out and maintain a control chart.

Know how to interpret a control chart to determine out ofcontrol situations.


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HYPOTHESIS TESTING ROADMAP...


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Test Used

F Stat (n>30)

F Stat (n5)

Test Used

Z Stat (n>30)

Z Stat (p)

t Stat (n


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Determine the hypothesis to be tested (Ho:=, < or >). Determine whether this is a 1 tail (a) or 2 tail (a/2) test. Determine the a risk for the test (typically .05). Determine the appropriate test statistic.

Determine the critical value from the appropriate test statistic table.

Gather the data.

Use the data to calculate the actual test statistic.

Compare the calculated value with the critical value.

If the calculated value is larger than the critical value, reject the nullhypothesis with confidence of 1-a (ie there is little probability (p


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Many test statistics use a and others use a/2 and often it is confusing to

tell which one to use. The answer is straightforward when you considerwhat the test is trying to accomplish.

If there are two bounds (upper and lower), the a probability must be splitbetween each bound. This results in a test statistic of a/2.

If there is only one direction which is under consideration, the error(probability) is concentrated in that direction. This results in a teststatistic of a.

CriticalValue

RareOccurrence

CommonOccurrence

a Probability1-a Probability

CriticalValue

CriticalValue

CommonOccurrence Rare

OccurrenceRare

Occurrence

a/2 Probability1-a Probabilitya/2 Probability

Test Fails in either direction = a/2Test Fails in one direction = a

a a/2Examples

Ho: 1

>2

Ho: s1


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t

n n

n s n s

n n

=+ - + -

+ -1 212

2]2

1 2

1 1 1 1

2

[( ) ( )| |tdCALC X XR R=-

+2 1 2

1 2

ZX X

n n

CALC =-

+

1 2

12

1

212

2

s s

n n1 2 20= n n1 2 30+ >n n1 2 30+

2 Sample Tau 2 Sample Z2 Sample t

(DF: n1+n2-2)

Use these formulas to calculate the actual statistic for comparisonwith the critical (table) statistic. Note that the only major determinatehere is the sample sizes. It should make sense to utilize the simpler

tests (2 Sample Tau) wherever possible unless you have a statisticalsoftware package available or enjoy the challenge.

X X-1 2

Hypothesis Testing Example (2 Sample Tau)

Several changes were made to the sales organization The weekly


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Ho: 1 = 2

Statistic Summary: n1 = n2 = 5

Significance: a/2 = .025 (2 tail) taucrit = .613 (From the table for a = .025 & n=5)

Calculation: R1=337, R2= 577

X1=2868, X2=2896

tauCALC=2(2868-2896)/(337+577)=|.06|

Test: Ho: tauCALC< tauCRIT Ho: .06


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Ho: 1 = 2

Statistic Summary: n1 = n2 = 5

DF=n1 + n2 - 2 = 8

Significance: a/2 = .025 (2 tail) tcrit = 2.306 (From the table for a=.025 and 8 DF)

Calculation: s1=130, s2= 227

X1=2868, X2=2896

tCALC=(2868-2896)/.63*185=|.24|

Test: Ho: tCALC< tCRIT Ho: .24 < 2.306 = true? (yes, therefore we will fail to reject the null hypothesis).

Conclusion: Fail to reject the null hypothesis (ie. The data does not support the conclusion that there is a significantdifference

orders were tracked both before and after the changes. Determine if the sampleshave equal means with 95% confidence.

Receipts 1 Receipts 23067 32002730 2777

2840 26232913 30442789 2834

t

n n n s n sn n

=+ - + -+ -1 2 1

2

2]

2

1 21 1 1 12[( ) ( )

X X-1 2


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Sample Variance vs Sample Variance (s2)Coming up with the calculated statistic...


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Use these formulas to calculate the actual statistic for comparisonwith the critical (table) statistic. Note that the only major determinateagain here is the sample size. Here again, it should make sense toutilize the simpler test (Range Test) wherever possible unless youhave a statistical software package available.

F sscalc

=2

2 =FRRCALC

MAX n

MIN n

,

,

1

2

Range Test F Test(DF1: n1-1, DF2: n2-1)

n1 < 10, n2 < 10 n1 > 30, n2 > 30

MAX

MIN


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HYPOTHESIS TESTING, PERCENT DEFECTIVE

n > 30 n > 30


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Use these formulas to calculate the actual statistic for comparisonwith the critical (table) statistic. Note that in both cases the individualsamples should be greater than 30.

( )Z

p p

p p

n

CALC =-

-

1 0

0 01Z

p p

np n p

n n

n p n p

n n n n

CALC =-

+

- +

+

1 2

1 1 2 2

1 2

1 1 2 2

1 2 1 21

1 1_ _

Compare to target (p0) Compare two populations (p1 & p2)

n1 > 30, n2 > 30

How about a manufacturing example?

We have a process which we have determined has a critical

8/3/2019 2555

2555571 six sigma and minitab 13

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