SIX SIGMA Green Belt Training

SIX SIGMA GREEN BELT
TRAINING
Shailja Chaudhry
National Institute of Technology, Kurukshetra

SIX SIGMA OVERVIEW AND
EVOLUTION

What is Six Sigma
A customer focused business
improvement process
Driven by teamwork, consensus &
logical reasoning
Structured methodology – DMAIC
Focuses on making the process
robust & reduce variations
Applies to any Process

What is Six Sigma
 Six sigma is a highly disciplined and
quantitative strategic business
improvement approach that seeks to
increase both customer satisfaction
and an organization’s financial health.
 Six Sigma helps a company focus on
developing and delivering near-
perfect products (durable goods or
services), to improve customer
satisfaction and the bottom line.

What Six Sigma is NOT
Six Sigma is NOT
A Quality Program
Cure for World
Hunger
Only for Technical
People
Just about
Statistics
Used when
solution is known
Used for
Firefighting

Six-Sigma – A note from
Originator of Six Sigma
 “Six Sigma is not an improvement
program. It is instead a business
philosophy that employs a step by
step approach to reducing variation,
increasing quality, customer
satisfaction, and in time, market
share”

Overview of Six Sigma
CULTURAL
CHANGE
TRANSFORM THE
ORGANIZATION
GROWTH
REDUCE COSTS
PAIN, URGENCY,
SURVIVAL
SIX SIGMA AS A
PHILOSOPHY
SIX SIGMA AS A
PROCESS
SIX SIGMA AS
A
STATISTICAL
TOOL

What is Six Sigma?
 Sigma is a measurement that indicates how
a process is performing
 Six sigma stands for Six Standard
Deviations (Sigma is the Greek letter used
to represent standard deviation in statistics)
from mean. Six Sigma methodology
provides the techniques and tools to
improve the capability and reduce the
defects in any process.
 Six Sigma is structured application of tools
and techniques applied on project basis to
achieve sustained strategic results.

What is Six Sigma
A Vision of a Six Sigma Company
Organizational
Issue
• Problem
Resolution
• Behavior
• Decision Making
• Process
Adjustment
• Supplier
Relationship
• Planning
• Design
• Employee Training
• Chain-of-command
• Direction
• Manpower
Traditional
approach
• Fixing (symptoms)
• Reactive
• Experience-based
• Tweaking
• Cost (piece price)
• Short-term
• Performance
• If Time Permits
• Hierarchy
• Seat-of-pants
• Cost
Six Sigma
Approach
• Preventing
(causes)
• Data-based
• Controlling
• Capability
• Long-term
• Producibility
• Mandated
• Empowered Teams
• Benchmarking and
metrics
• Asset
Traditional
approach
• Fixing (symptoms)
• Reactive
• Experience-based
• Tweaking
• Cost (piece price)
• Short-term
• Performance
• If Time Permits
• Hierarchy
• Seat-of-pants
• Cost
Six Sigma
Approach
• Preventing
(causes)
• Data-based
• Controlling
• Capability
• Long-term
• Producibility
• Mandated
• Empowered Teams
• Benchmarking and
metrics
• Asset

Character of 6s
Traditional Quality / Six Sigma Quality Method
ISSUE TRADITIONAL
APPROACH
SIX SIGMA
APPROACH
Index
Data
Target
Range
Method
Action
• % (Defect Rate)
• Discrete Data
• Satisfaction for
Mfg. Process
• Spec Outliner
• Experience + Job
• Bottom Up
• σ
• Discrete +
Continuous Data
• Customer
Satisfaction
• Variation
improvement
• Experience + Job
+ Statistical Ability
• Top Down

Aligning The Focus
Six Sigma Journey Started
(Traditional)
1000
Unassigned
Projects Six Sigma
Project
Strategic
DirectionTactical
Direction
Individual
Work
Group
(Lets do it) (Future)

What is Six Sigma Definition
2
Sigma
3
Sigma
4
Sigma
5
Sigma
6
Sigma
2σ
3σ
4σ
5σ
6σ
Sigma Level Defects/ Million Opportunities% Yield
308,537
66,807
6,210
233
3.4
69.1
93.3
99.4
99.98
99.9997

Six Sigma : The Statistical
Way
LSL USL LSL USL
LSL USL
Process of Target Excessive Variation
Reduce Variation % Center Process
Customers feel the
variation more than
the mean
Center
Proces
s
Reduc
e
Spread
Target
Target
Target

Six Sigma – Practical
Meaning
99% Good (3.8 Sigma) 99.99966% Good (6 Sigma)
• 20,000 lost articles of
mail per hour
• Unsafe drinking water
for almost 15 minutes
each day
• 5,000 incorrect surgical
operations per week
• Two short or long
landings at most major
airports each day
• 200,00 wrong drug
prescriptions each year
• No electricity for almost
seven hours each month
• Seven articles lost per
hour
• One unsafe minute every
seven months
• 1.7 incorrect operations
per week
• One short or long
landing every five years
• 68 wrong prescriptions
per year
• One hour without
electricity every 34 years

Philosophy of Six Sigma
 Know What’s Important to the
Customer (CTQ)
 Reduce Defects (DPMO)
 Center Around Target (Mean)
 Reduce Variation (σ)

Harvesting the fruit of Six
Sigma

History of Six Sigma
 Quality tools like SPC, Cost of Quality,
Control Charts, Process capability etc. are
known to industry for long time, much before
birth of Six Sigma.
 Quality Tools and Quality system
implementation was not in conjunction with
overall business Goals.
 Traditional Quality Tools have limitations to
orient the efforts on Quality Improvements to
the Organizational direction basically due to
approach.
 Motorola was the first Company to initiate the
Six Sigma breakthrough Strategy.

A Little Bit Of History
 Six Sigma was developed by Bill Smith, QM at
Motorola
 It’s implementation began at Motorola n 1987
 It allowed Motorola to win the first Baldrige Award
in 1988
 Motorola recorded more than $16 Billion savings
as a result of Six Sigma
 Several of the major companies in the world
have adopted Six Sigma since then….
Texas Instruments, Asea Brown Boveri,
AlliedSignal, General Electric, Bombardier, Nokia
Mobile Phones, Lockheed Martin, Sony, Polaroid,
Dupont, American Express, Ford Motor,…..
The Six Sigma Breakthrough Strategy has become
a Competitive Tool

Motorola Case Study
In early 1980’ Motorola was facing a
serious competitive challenge from
Japanese Companies.
Motorola was losing the market share
and customer confidence.
Motorola had not done any major
changes to their products.
The competitors from Japan were
offering much better product at much
lower price with no field failures.

Motorola Case Study….Continue
 When Motorola studies the competitors products,
it was revealed that the variation in key product
characteristics is very low.
 The competitors products were available at lower
price.
 The competitors products has very low warranty
failure rate.
 Motorola was not able to match the competitors
price mainly due to high cost of Poor quality
largely due to high reject rate, high rework /
repair rate, high inspection cost, high warranty
failure rate etc.
 THE TECHNICAL TEAM CONCLUDED THAT
THE COPETITORS ARE OFFERING BETTER
PRODUCT AT LOWER COST.

Motorola Case Study
Motorola requested to the competitors
from Japan to permit the Team from
Motorola to visit them fro Study.
Motorola sent the team of managers
to Japan to study the “Magic” of
Japanese companies.
What the team revealed?

Motorola Case Study
 What Motorola learning was as follows:
 Motorola was focusing too much on product
Quality i.e. Inspection, rework, repair etc.
 The internal defect rate was very high inside
Motorola.
 The reliability was slow since some of the
defects were passing on to the customer as
inspection lapses.
 A dissatisfied customer was shouting loudly
and was taking away min 10 potential
customers.
 As an effect of this, customers were lost to
the competitors.

Motorola Case Study
What was wrong?
 Japanese were concentrating on
◦ Customers
◦ Processes
◦ People
 Variation in product and process parameters was known and
controlled
 All people were well trained and highly motivated
 All activities and processes were highly standardized i.e. no
person dependence
 Defect free lines and robust processes
 Very less inspectors
 Yet, very low defect rate, internal rejection and customer
complaints
 VERY HIGH LEVEL OF CUSTOMER SATISFACTION

Motorola Case Study
WHAT WAS THE SECRET?
THE SECRET WAS CONTROL OVER VARIATION
Success factor:
Proactive Vs. Reactive Quality

The Impact Of Added
Inspection
3.4 ppm
100,000 ppm
6 ppm
If the likelihood of detecting the defect is
70% and we have 10 consecutive inspectors
with this level of capability, we would expect
about 6 escaping defects out of every
1,000,000 products produced
You can save yourself by producing quality not by

Motorola Case Study
In order to address these issues,
Motorola devised the Six Sigma
methodology.
Dr. Mikel Harry and Mr. Bill Smith were
pioneers in Developing and
implementing the Six Sigma
methodology at Motorola.
With implementation of Six Sigma,
Motorola could achieve:
4σ level in one and half year time
5σ level in following year

Six Sigma Progress
1985 1987 1992 1995 2002
Johnson & Johnson,
Ford, Nissan,
HoneywellGeneral Electric
Allied Signal
Motorola
Dr Mikel J Harry
wrote a Paper
relating early
failures to quality

What can it do?
Motorola:
 5-Fold growth in Sales
 Profits climbing by 20% pa
 Cumulative savings of $14 billion over 11
years
General Electric
 $2 billion savings in just 3 years
 The no. 1company in the USA
Bechtel Corporation:
 $200 million savings with investment of $30
million
It is high time, that Indian Companies also start
implementing Six Sigma for making
breakthrough improvements and to remain

Attempting to Define Quality
Experts’ definitions of quality fall into two
categories:
 Level one quality is a simple matter of producing
products or delivering services whose
measurable characteristics satisfy a fixed set of
specifications that are usually numerically
defined.
 Independent of any of their measurable
characteristics, level two quantity products and
services are simply those that satisfy customer
expectations for their use or consumption.
In short, level one quality means get it in the
specs,
and level two means satisfy the customer.

Quality Gap
Quality Gap
Understanding of
Needs
Customer
Perception of
Delivery
Customer
Expectations
Design of Products
Capability to Deliver
Design
Actual Delivery
Design Gap
Perception Gap
Operations Gap
Process Gap
Understanding the
Gap

Nine Dimensions of QUALITY
According to modern management
concepts, quality has nine dimensions:
1) Performance: main characteristics of
the product/service
2) Aesthetics: appearance, feel, smell,
taste
3) Special features: extra characteristics

4) Conformance: how well the
product/service
conforms to customer’s
expectations
5) Safety: risk of injury
6) Reliability: consistency of
performance

7) Durability: useful life of the
product/service
8) Perceived Quality: indirect evaluation
of quality (e.g.
reputation)
9) Service after Sale: handling of
Customer
complaints and
checking customer
satisfaction.

Evolution of Quality
Historically Contemporary
Reactive Quality
Quality Checks (QC) -
Taking the defectives
out of what is
produced
Proactive Quality
“Create process that
will produce less or
no defects”

Old Concept Of Quality
Past concepts of quality focused on
“conformance to standards”. This definition
assumed that as long as the company
produced quality products and services,
their performance standard was correct
regardless of how those standards were
met. Moreover, setting of standards and
measurement of performance was mainly
confined to the production areas and the
commercial and other service functions
were managed through command control.

Value Enrichment
The term ‘Value Enrichment’ for the
company means that they must strive to
produce highest quality products at the
lowest possible costs to be competitive
in the global markets.
For customers, the term ‘Value
Enrichment’ means that they have the
right to purchase high quality
products/services at the lowest cost.

Concept Of Value
Value to Customers
Value =
𝐖𝐡𝐚𝐭 𝐘𝐨𝐮 𝐑𝐞𝐜𝐞𝐢𝐯𝐞
𝐖𝐡𝐚𝐭 𝐘𝐨𝐮 𝐏𝐚𝐲
Price
+
Inconvenienc
e
Real
+
Perceived

Definitions
VALUE:
THOSE ACTIVITIES THAT CONVERT
MATERIALS OR IDEAS INTO GOODS OR
SERVICES THAT GENERATE CASH

Definition
ANYTHING THAT IS
NOT VALUE IS
WASTE

Six Sigma and Cost Of
Quality
Six Sigma has a very significant impact on
the cost of quality. As the Sigma level
moves up, the cost of quality comes down
and vice versa. Traditionally recorded
quality cost generally account for only 4 to
5 percent of sales which mainly comprise
of scrap, re-work and warranty.
There are additional costs of quality which
are hidden and do not appear in the
account books of the company, as they are
intangible and difficult to measure.

Visible And Hidden Costs
Visible
Costs
Hidden
Costs
• Scrap
• Rework
• Warranty Costs
• Conversion
efficiency of
materials
• Inadequate
resources
utilization
• Excessive use of
materials
• Cost of re-design
and re—
inspection
• Cost of resolving
customer
problems
• Lost customers /
Goodwill
• High Inventory

Cost OF Quality At Various
Levels Of Sigma
6 3.4 <10%
5 233 10-15%
4 6210 15-20%
3 66807 20-30%
2 308537 30-40%
1 6,90000 >40%
Sigma
Defect Rate
(PPM)
Cost Of
Quality
Competitive Level
World Class
Industry
Average
Non
Competitive

What is The Cost Of Quality?
 Cost of Quality: the cost of ensuring
that the job is done right + the cost of
not doing the job right.
Cost of Conformance + Cost of Non-
Conformance(Prevention and Appraisal) (Internal/External Defects)

Cost Of Quality
Prevention Costs
• Quality Planning
• Process Evaluation /
Improvement
• Quality Improvement Meetings
• Quality Training
External Failure Costs
• Complaint Handling
• Rework / Correction
• Re-Inspection
Internal Failure Costs
• Re-Inspection
• Internal Reject
• Loss of Business
Appraisal Costs
• Source Inspection
• In / End-Process Inspection
• Calibration
• Specialist Cost
Direct Costs
Prevention Costs
• Quality Planning
• Process Evaluation /
Improvement
• Quality Improvement Meetings
• Quality Training
External Failure Costs
• Complaint Handling
• Re-Inspection
Internal Failure Costs
• Re-Inspection
• Internal Reject
• Loss of Business
Appraisal Costs
• Source Inspection
• In / End-Process Inspection
• Calibration
• Specialist Cost

Fundamental Steps
There are 5 fundamental Steps involved
in applying the breakthrough strategy
for achieving Six Sigma. These steps
are :-
 Define
 Measure
 Analyze
 Improve
 Control

Define Phase
This phase defines the project. It
identifies critical customer requirements
and links them to business needs. It
also defines a project charter and the
business processes to be undertaken
for Six Sigma.

Define
Define D CM A I
Define Activities
Identify Project, Champion and Project Owner
Determine Customer Requirements and CTQs
Define Problem, Objective, Goals and Benefits
Define Stakeholders/Resource Analysis
Map the Process
Develop Project Plan
Define Quality Tools
Project Charter and Plan
Effort/Impact Analysis
Process Mapping
Tree Diagram
VOC
Kano Model
Pareto Analysis

Measurement Phase
This phase involves selecting product
characteristic, mapping respective
process, making necessary
measurements and recording the
results of the process. This is
essentially a data collection phase.

Measure – Operational Definition
Measure M CD A I
Measure Activities
Determine operational Definitions
Establish Performance Standards
Develop Data Collection and Sampling Plan
Validate the Measurements
Measurement System Analysis
Determine Process Capability and Baseline
Measure Quality Tools
Measurement Systems Analysis
Check Sheet
Process Capability
Process FMEA

Analysis Phase
In this phase an action plan is created
to close the “gap” between how things
currently work and how the organization
would like them to work in order to meet
the goals for a particular product or
service. This phase also requires
organizations to estimate their short
term and long term capabilities.

Analyze
Analyze A CMD I
Analyze Activities
Benchmark the Process or Product
Analysis of the Process Map
Brainstorm for likely causes
Establish Causal Relationships Using Data
Determine Root Cause(s) Using Data
Analyze Quality Tools
Cause and Effect or Event Diagram
Graphical Analysis
Statistical Analysis of Data
Hypothesis Testing
Correlation Regression
DOE

Improvement Phase
 This phase involves improving
processes/product performance
characteristics for achieving desired
results and goals. This phase involves
application of scientific tools and
techniques for making tangible
improvements in profitability and
customer satisfaction.

Improve
Improve I CMD A
Improve Activities
Develop Solution Alternatives
Assess Risks and Benefits of Solution Alternatives
Implement error-proofing solutions
Validate Solution using a Pilot
Implement Solution
Determine Solution Effectiveness using Data
Improve Quality Tools
Brainstorming
FMEA
Risk Assessment
Poka Yoke

Control Phase
This Phase requires the process
conditions to be properly documented
and monitored through statistical
process control methods. After a
“setting in” period, the process
capability should be reassessed.
Depending upon the results of such a
follow-up analysis, it may be sometimes
necessary to revisit one or more of the
preceding phases.

Control – Develop Standards
Control CIMD A
Control Activities
Determine Needed Controls (measurement, design, etc.)
Implement and Validate Controls
Develop Transfer Plan
Realize Benefits of Implementing Solution
Institutional Changes
Close Project and Communicate Results
Control Quality Tools
Statistical Process Control
Process Map and FMEA
Control Plans
5S
Control Charts

Why Project Selection is
Important?
 High leverage projects lead to largest
Savings
 Large returns are expected by
management to justify the investment
in time and effort
 Developing a Six Sigma culture
depend upon successful projects
having significant business impact

How To Focus Projects
Process Cost Savings Focus
Project Quality focus
Product focus (Six Sigma Design)
Problem Focus (Least Desirable Use)

Project Selection
Align with company objectives and
business plan (Annual Operating Plan)
– Voice of Customer/CT’s Inputs
– Quality (CTQ)/Cost (CTC)/ Delivery
(CTD)
– PPM / COPQ / RTY / Cycle Time
Consistent with principles of Six Sigma
– Eliminate process defects
Concentrate on “Common”
issues/opportunities …not “fir-fighting”
Large enough to justify the investment

Project Desirability
• Effort Required:- includes time required
of team members and expenditure of
money.
• Probability of Success:- An assessment
that takes into account various risk factors:
+ Time – uncertainty of the completion
date
+ Effort – uncertainty of the investment
required
+ Implementation – uncertainty of
roadblocks

Project Desirability Matrix
Hi
Med
Low
HiMedLow
Hi
Med
Low
IMPACT
EFFORT

Additional Project Considerations
 Projects must serve as a learning
experience for Green Belts to use the six
Sigma tools
 Projects scope should not be too large or
take too long to implement
 Projects scope should be manageable
and take at least 255 of the potential
Green Belt’s time.
 Pareto Chart may be used to Scope the
Project
 Desirable to have a measurable variable
for the primary project output/metrics

Additional Project Considerations
• Projects must serve as a learning
experience for Green Belts to use the six
Sigma tools
• Projects scope should not be too large or
take too long to implement
• Projects scope should be manageable and
take at least 255 of the potential Green
Belt’s time.
• Pareto Chart may be used to Scope the
Project
• Desirable to have a measurable variable for
the primary project output/metricsDO NOT try to Solve World Hunger

Strategy At Various Levels
Almost every Organization can be
divided into 3 basic levels:-
1. Business level
2. Operations level
3. Process level.
It is extremely important that Six Sigma
is understood and integrated at every
level.

Strategies At Various Levels
 Executives at the business level can use
Six Sigma for improving market share,
increasing profitability and organizations
long term viability.
 Managers at operations level can use
Six Sigma to improve yield and reduce
the labor and material cost.
 At the process level engineers can use
Six Sigma to reduce defects and
variation and improve process capability
leading to better customer satisfaction.

Factors To Control in
Improvement Project
 Resources
Team availability
The right tools
 Schedule
Be realistic
Be aggressive
Get buy-in
 Scope of Work
Watch for scope creep
Stay focused
Anticipate and mitigate risk
Control any two areas, the third floats in
response

Meetings – Make Them
Effective
 Defined goal for meeting
 Notice and agenda
 Decision makers prepare and participate
 Action Items
 Records
 Balance Sheet
– Focused on process, not topic
– What helped us get to our goal
– What could have been better
– Take appropriate action

Skills Needed
 People
Leadership behaviors
Communication
 Process
Time Management
Schedule Coordination
Problem Solving
Risk analysis and mitigation
Tactical Planning
 Technical
Six Sigma / Lean Tools
Business Knowledge

Voice of Customer –
CTP and CTQ

Establishing Customer Focus
• Customer – Anyone internal or
external to the organization who
comes in contact with the product
or output of work
• Quality – performance to the
standard expected by the Customer

Variation is the Enemy in
Achieving Customer Satisfaction
Variation
•Uncertainty
•Unknown
•Disbelief
•Risk
•Defect Rate

What is Variation
Variation is any deviation from the
expected outcome.

Something more on Variation
 Any process has variation
 There are two kinds of variation
Common cause variation
Special cause variation
 Variation is measured in terms of sigma
or standard deviation.

Variation and Standard
Deviation
If a good deal of variation exists in a process activity,
that activity will have a very large standard
deviation.
As a result, the distribution will be very wide and flat.
Less Variation More Variation

Types of Variation
 Special Cause: something different happening
at a certain time or place
Common Cause: always
present to some degree in the
process
We tamper with the system if we treat all variations as if it were special
cause

Dealing with Variation
 Eliminate special cause variation by
recognizing it and dealing with it
outside of the process
 Reduce common cause variation by
improving the process

Whom would you Prefer?
Operator - 1 Operator - 2

Critical To Quality (CTQ)
are the key measurable characteristics of
a product or process whose
performance standards or specification
limits must be met in order to satisfy the
customer.
They align improvement or design efforts
with customer requirements.

1. To put it in layman’s terms, CTQs are
what the customer experts of a
product...
2. ...the spoken needs of the customer.
3. The customer may often express this
in plain English, but it is up to us to
convert them to measurable terms
using tools such as QFD, DFMEA,
etc.

1. List customer needs.
2. Identify the major drivers for these
needs (major means those which will
ensure that the need is addressed).
3. Break each driver into greater detail.
4. Stop the breakdown of each level
when you have reached sufficient
information that enables you to
measure whether you meet the
customer need or not.

Example – CTQ Tree
Ease of Operation
Ease of Maintenance
Ease of
Operation
and
Maintenance
Operator Training Time
(hrs.)
Setup Time (minutes)
Operation Accuracy
(errors/1000 ops)
Mean Time to Restore
(MTTR)
# Special Tools
Required
Maintenance Training Time
(hrs.)
Need CTQsDrivers
SpecificGeneral
Hard to Measure Easy to Measure

Importance of Project Charter
A project charter is a written document and
works as an agreement between
management and the team about what is
expected.
The charter:
 Clarifies what is expected of the team.
 Keeps the team focused.
 Keeps the team aligned with
organizational priorities.
 Transfers the project from the
champion(s) to the project team.

Team Charter
 Problem Statement
– Currently we carry out reblows to the extent
of about 11-15% resulting in lower
converter life, lower productivity of
converter and increased Ferro-alloy and
oxygen consumption.
 Scope
– All batches and all converters in SMS 1.
 Project Goal and Measures
– Reblows should be less than 7.5% and 9%.
 Expected Business Results
– We hope to save Rs. Xxxxx lakhs per year
due to this reduction in reblows.

Team Charter
 Team Members
–Supervisor, two operators, technical
services, quality control
 Support Required
–Allow for weekly team meetings
–Team budget for quick wins
 Schedule
–Measure (7wks), Analyze (4wks),
Improve (6wks), Check (2wks), Control
(1wk), Standardise/Close (1wk)

Usual elements of a Project
Charter
 Project Description – Business Case
 Scope – Process/Product
 Goals and Measures (Key Indicators)
 Expected Business Results
 Team Members
 Support Required
 Expected Customer Benefits
 Schedule

Measurement Objective
The Measure phase aims to set a baseline
in terms of process performance
through the development of
clear and meaningful measurement
systems

The Measurement Process
TOOLS AND TECHNIQUS OF MEASURE
Develop
Process
Measures
Collect
Process
Data
Check
Data
Quality
Understand
Process
Behavior
Baseline
Process
Capability
and
Potential
How do you
measure
the
problem?
When and
where does
the data
come from?
How does
the process
currently
behave?
What is the
current
performance
of the process
with respect to
the customer
Does the data
represent
what you
think it does
 Statistics
 Operational
Definitions
 Data Worlds
 Process
Capability
 Cp, Cpk
 DPMO
 Distributions
 First pass
yield
 Short/long
term variation
 MSA
 Gage R&R
 Data
Collection
Methods
 Data
Collection
Plans
 Sampling

Statistical and Data World
If the data
is
AttributeCountContinuous
Relevant
statistical
model is …
Binomial
Distribution
Defects per
Unit (DPU)
Always
Poisson – if
process is
in control
Poisson
Distribution
When does
the
statistical
model
apply
Common
statistics
are…
Always
Binomial –
if process
is in control
Percentage
(Proportion)
Average (mean),
Standard
Deviation (sigma)
Not always –
validity of
normality needs
to be checked
Normal
Distribution

Statistics
 The science of:
–Collecting,
–Describing
–Analyzing
–Interpreting data...
And Making Decisions

What are Statistics?
 Descriptive Statistics
– Summarize and describe a set of data
– Mean, median, range, standard deviation,
variance, ....
 Analytical Statistical (or Statistics)
– Techniques that help us make decisions in
the face of uncertainty
– Use concepts of descriptive statistics as a
base
– Hypothesis testing, means comparisons,
variance comparisons, proportions
comparisons, ...

Sample Versus Population
 Using a small amount of data (Sample)...
to make assumptions (inferences)...
on a large amount of data (population).
 Population: the total collection of
observations or measurements that are if
interest.
 Sample: A subset of observations and
measurements taken form the population.
 Why do we use samples?
Time
Cost
Destructive testing (need product left to sell !!)
Other?

Measures of Central
Tendency
 What is the Median value of
Distribution?
– Median
 What value represents the
distribution?
– Mode
 What value represents the entire
distribution?
– Mean (x̄ )
 What is the best measures of central
tendency?

Data Distributions
 Mean: Arithmetic average of a set of
values
– Reflects the influence of all values
– Strongly influenced of all values
 Median: Reflects the 50% rank – the
center number after a set of numbers
has been sorted from low to high.
– Does not include all values in calculation
– Is “robust” to extreme scores
 Mode: The value or item occurring most
frequently in a series of observations or
statistical data.

Variable Data Location -
MeanMonth # of Units
Jan-2006 233
Feb-2006 281
Mar-2006 266
Apr-2006 237
May-2006 260
Jun-2006 250
Jul-2006 237
Aug-2006 275
Sep-2006 218
Oct-2006 279
Nov-2006 227
Dec-2006 246
Jan-2007 258
Feb-2007 272
Mar-2007 229
Apr-2007 240
May-2007 287
Jun-2007 260
Jul-2007 251
Aug-2007 288
Sep-2007 256
Oct-2007 219
Nov-2007 260
Dec-2007 249
n=24 = 𝟔𝟎𝟕𝟖
We have data on the monthly demand history of
one of our key product lines. Let’s calculate the
statistics for location.
 Mean (𝑿)
 Add all of the monthly numbers
 Divide by the number of months in the
sample.
 N=24, = 𝟔𝟎𝟖𝟕
𝑿 =
𝟔𝟎𝟖𝟕
𝟐𝟒
= 𝟐𝟓𝟑. 𝟐𝟓
Our average monthly shipment is 253 units
𝝁 =
𝒊=𝟏
𝑵
𝑿𝒊
𝑵
𝒙 =
𝒊=𝟏
𝒏
𝑿𝒊
𝒏
Populatio
n
Sample

MedianMonth # of Units
Jan-1999 233
Feb-1999 281
Mar-1999 266
Apr-1999 237
May-1999 260
Jun-1999 250
Jul-1999 237
Aug-1999 275
Sep-1999 218
Oct-1999 279
Nov-1999 227
Dec-1999 246
Jan-2000 258
Feb-2000 272
Mar-2000 229
Apr-2000 240
May-2000 287
Jun-2000 260
Jul-2000 251
Aug-2000 288
Sep-2000 256
Oct-2000 219
Nov-2000 260
Dec-2000 249
Month # of Units
Sep-1999 218
Oct-2000 219
Nov-1999 227
Mar-2000 229
Jan-1999 233
Jul-1999 237
Apr-1999 237
Apr-2000 240
Dec-1999 246
Dec-2000 249
Jun-1999 250
Jul-2000 251
Sep-2000 256
Jan-2000 258
May-1999 260
Jun-2000 260
Nov-2000 260
Mar-1999 266
Feb-2000 272
Aug-1999 275
Oct-1999 279
Feb-1999 281
May-2000 287
Aug-2000 288
Statistics for
location ~
Median (x)
• Sort the data from
lowest to highest
• If there is an even
number of
observations, the
median is the
average of the two
middle values
(𝟐𝟓𝟏 + 𝟐𝟓𝟔)
𝟐
= 𝟐𝟓𝟑. 𝟓

MedianMonth # of Units
Jan-1999 233
Feb-1999 281
Mar-1999 266
Apr-1999 237
May-1999 260
Jun-1999 250
Jul-1999 237
Aug-1999 275
Sep-1999 218
Oct-1999 279
Nov-1999 227
Dec-1999 246
Jan-2000 258
Feb-2000 272
Mar-2000 229
Apr-2000 240
May-2000 287
Jun-2000 260
Jul-2000 251
Aug-2000 288
Sep-2000 256
Oct-2000 219
Nov-2000 260
Dec-2000 249
Month # of Units
Sep-1999 218
Oct-2000 219
Nov-1999 227
Mar-2000 229
Jan-1999 233
Jul-1999 237
Apr-1999 237
Apr-2000 240
Dec-1999 246
Dec-2000 249
Jun-1999 250
Jul-2000 251
Sep-2000 256
Jan-2000 258
May-1999 260
Jun-2000 260
Nov-2000 260
Mar-1999 266
Feb-2000 272
Aug-1999 275
Oct-1999 279
Feb-1999 281
May-2000 287
Aug-2000 288
Statistics for
location ~
 Mode
• The most
frequently
occurring
value is the
mode
260 is the mode

Mode
 Notes on mean
–A measure of central tendency
–Limitations:
 Reflects the influence of all values
 Strongly influenced by extreme values
 Median (the centre number after sorting high to low) is
robust to extreme values.

Variable Data Description –
Range, Standard DeviationMonth # of Units
Jan-2006 233
Feb-2006 281
Mar-2006 266
Apr-2006 237
May-2006 260
Jun-2006 250
Jul-2006 237
Aug-2006 275
Sep-2006 218
Oct-2006 279
Nov-2006 227
Dec-2006 246
Jan-2007 258
Feb-2007 272
Mar-2007 229
Apr-2007 240
May-2007 287
Jun-2007 260
Jul-2007 251
Aug-2007 288
Sep-2007 256
Oct-2007 219
Nov-2007 260
Dec-2007 249
• Let’s use this same data to
calculate the statistics for
dispersion
 These statistics are
Range and Standard
Deviation

Example – commuting time
Commute time (mins)
19.5 22.4 20.7 18.8 18.2
20.0 19.6 19.8 21.0 19.8
20.7 21.9 22.0 22.6 19.4
22.8 18.1 17.5 21.3 19.1
18.4 19.8 21.0 18.5 19.2
19.2 19.4 19.3 24.8 21.2
21.2 18.3 18.2 17.4 19.9
21.0 18.9 16.4 17.6 19.5
19.2 23.9 20.6 21.9 18.7
19.5 20.1 17.1 22.1 19.2
19.6 20.3 20.8 20.7 22.4
19.9 21.1 20.4 16.7 19.1
18.3 22.4 27.1 17.6 18.8
22.5 19.9 21.8 20.4 17.7
21.3 17.8 18.7 15.8 18.9
21.7 20.1 19.6 18.4 21.7
18.7 18.8 20.5 18.6 20.9
22.0 15.8 19.4 20.2 18.7
23.6 21.0 19.9 20.1 18.3
21.9 19.7 21.1 19.9 22.9
• Collect over a hundred
occurrences.
• Tabulate in chronological
order.
• Does the data show variation?
• Can you make out anything
with this arrangement of data?
• Let us try and make some
sense of this data…

Measure of variation – Standard
Deviation and Range
Category 1
15 2018 22 24 25
19.50 19.75 20.00 20.25 20.50
..
Summary for Commute time Anderson – Darling Normality Test
A-Squared
P-Value
0.42
0.312
What are the relative merits and demerits of standard deviation over range?
Mean
St.Dev
Variance
Skewness
Kurtosis
N
Minimum
1st Quartile
Median
3rd Quartile
Maximum
95% Confidence Interval for Mean
19.632
20.006
1.884
3.550
0.54470
1.30256
100
15.754
18.714
19.819
21.186
27.054
20.380
19.448 20.263
1.654 2.189Mean
Median
One measure
of variation
(std. dev)
Another
Measure of
variation
(Range)
Outlier
*

Variable Data Dispersion –
Standard Deviation
“s” or “standard deviation”
 What does it mean?
–Standard deviation is a measure of
dispersion (or how our data is spread
out).
–Range will tell us the difference between
the highest and lowest values in a data
set, but nothing about how the data are
distributed.
–We need deviation to statistically describe
the distribution of values.

Standard Deviation
How we calculate it…
 A measure of how far each point deviates from the
mean
 We square each distance so that all the numbers
are positive
 The sum of the squares, divided by the sample size,
is equal to the variance
 The square root of the variance is the standard
deviation
– Variance can be added; standard deviations
cannot
𝜎 = 𝑖=1
𝑛
(𝑥𝑖 − 𝜇)2
𝑁
s = 𝑖=1
𝑛
(𝑥𝑖 − 𝑥)2
𝑛 − 1Population Sample

Standard Deviation CalculationMonth # of Units
Jan-2006 233
Feb-2006 281
Mar-2006 266
Apr-2006 237
May-2006 260
Jun-2006 250
Jul-2006 237
Aug-2006 275
Sep-2006 218
Oct-2006 279
Nov-2006 227
Dec-2006 246
Jan-2007 258
Feb-2007 272
Mar-2007 229
Apr-2007 240
May-2007 287
Jun-2007 260
Jul-2007 251
Aug-2007 288
Sep-2007 256
Oct-2007 219
Nov-2007 260
Dec-2007 249
-20.25
27.75
12.75
-16.25
6.75
-3.25
-16.25
21.75
-35.25
25.75
-26.25
-7.25
4.75
18.75
-24.25
-13.25
33.75
6.75
-2.25
34.75
2.75
-34.25
6.75
-4.25
-20.25
27.75
12.75
-16.25
6.75
-3.25
-16.25
21.75
-35.25
25.75
-26.25
-7.25
4.75
18.75
-24.25
-13.25
33.75
6.75
-2.25
34.75
2.75
-34.25
6.75
-4.25
𝑿𝒊 − 𝑿 (𝑿𝒊−𝑿) 𝟐
𝑺 = 𝒊=𝟏
𝒏
(𝒙𝒊 − 𝒙) 𝟐
𝒏 − 𝟏
𝑿 = 𝟐𝟓𝟑. 𝟐𝟓
Calculate
the Mean
Count the
Samples
n = 24
Square each
subtraction
result
Subtract the
mean from
each value
Sum the Squares
Calculate the Denominator
Complete the Calculation
𝒊=𝟏
𝒏
(𝒙𝒊 − 𝒙) 𝟐= 𝟗. 𝟖𝟑𝟏
𝒏 − 𝟏 = 𝟐𝟒 − 𝟏 = 𝟐𝟑
𝒔 =
𝟗. 𝟖𝟑𝟏
𝟐𝟑
= 𝟐𝟎. 𝟕

Standard Deviation
𝜎 = 𝑖=1
𝑛
(𝑥𝑖 − 𝜇)2
𝑁
 Standard deviation of a population
– If your data is from a population versus a
sample from a population, use this formula to
calculate standard deviation
– The difference is the denominator
“N” versus “n-1”

Fundamental Topic
The Normal Curve
◦ Processes have natural variation
◦ Many processes behave “normally”
◦ Characterized by Bell Shaped Curve
– Mean near peak
– Curve is symmetric
◦ Mean
◦ Standard Deviation
Histogram of Diameter, with Normal Curve
Diameter
Frequency

Measures of Variability
•The Range is the distance between the extreme
values of data set. (Highest – Lowest)
•The Variance(S ) is the Average Squared
Deviation of each data point from the Mean.
•The Standard Deviation (s) is the Square Root of
the Variance.
•The range is more sensitive to outliners than the
variance.
•The most common and useful measure of
variation is the Standard Deviation.

Sample of Statistics versus
Population Parameters
EstimateStatistics Parameters
µ = Population Mean
s = Sample Standard
Deviation
X = Sample Mean
σ = Population
Standard Deviation

Statistical Calculation
(Sample)
𝑋 =
𝑖=1
𝑛
𝑋𝑖
𝑛
𝑆2
=
𝑖=1
𝑛
(𝑋𝑖−𝑋)2
𝑛 − 1
𝜎 = 𝑅/𝑑2
s = 𝑖=1
𝑛
(𝑋𝑖 − 𝑋)2
𝑛 − 1
Standard Deviation
Standard Deviation
VarianceMean
n
2
3
4
6
𝒅 𝟐
1.128
1.693
2.059
2.326

Statistical Calculation
(Population)
𝜇 ≈ 𝑋𝝈 𝟐
=
𝒊=𝟏
𝑵
(𝑿𝒊 − 𝝁) 𝟐
𝑵
𝝈 = 𝒊=𝟏
𝑵
(𝑿𝒊 − 𝝁) 𝟐
𝑵
Standard Deviation
VarianceMean

Normal Distribution
Description of a NORMAL
DISTRIBUTION
LOCATION:
•The Central Tendency
•It is usually expressed as the
AVERAGE
SPREAD:
•The dispersion
•It is usually expressed as
standard deviation (Sigma)
LOCATION
SPREAD

Properties of Normal Distribution
•Normal Distribution is Symmetric
–Has equal number of points on both
sides
–Mean Median and Mode Coincide
•Normal Distribution is Infinite
–The chance of finding a point anywhere
on the plus and minus side (around the
mean) is not absolutely Zero.

Properties Of Normal Distribution
Normal Curve & Probability Areas
-3𝝈 -2𝝈 -1𝝈 0 1𝝈 2𝝈 3𝝈
68%
95%
99.73%

Let’s Summarize…
We need data study, predict and improve
the processes.
Data may be Variable or Attribute.
To understand a data distribution, we need
to know its Center, Spread and Shape.
Normal Distribution is the most common
but not the only shape.

Standard Deviation -
Graphically
0
1
2
3
4
5
Monthly Demand in Units
Frequency
Month # of Units
Jan-1999 233
Feb-1999 281
Mar-1999 266
Apr-1999 237
May-1999 260
Jun-1999 250
Jul-1999 237
Aug-1999 275
Sep-1999 218
Oct-1999 279
Nov-1999 227
Dec-1999 246
Jan-2000 258
Feb-2000 272
Mar-2000 229
Apr-2000 240
May-2000 287
Jun-2000 260
Jul-2000 251
Aug-2000 288
Sep-2000 256
Oct-2000 219
Nov-2000 260
Dec-2000 249
Let’s take our demand data and develop
a histogram
1. Set up the scale and limits per subdivision
2. Plot the count of values that fall within each
subdivision on the scale

Graphically
0
1
2
3
4
5
Frequency
If my data is normal… 𝑿 = 𝟐𝟓𝟑. 𝟐𝟓
1𝝈 = 𝟐𝟎. 𝟕 1𝝈 = 𝟐𝟎. 𝟕 1𝝈 = 𝟐𝟎. 𝟕 1𝝈 = 𝟐𝟎. 𝟕 1𝝈 = 𝟐𝟎. 𝟕 1𝝈 = 𝟐𝟎. 𝟕

Graphically
0
1
2
3
4
5
Frequency
If my data is normal…
±3𝝈 = 𝟗𝟗. 𝟕% 𝒐𝒇 𝒅𝒂𝒕𝒂
±2𝝈 = 𝟗𝟓. 𝟒% 𝒐𝒇 𝒅𝒂𝒕𝒂
±1𝝈 = 𝟔𝟖. 𝟑% 𝒐𝒇 𝒅𝒂𝒕𝒂

Standard Deviation – Simple
ApplicationFrequency
800
600
400
200
I have a process with mean of 43 and a standard deviation of
3
1200
1400
1000
37
43
42
41
40
39
38
48
47
46
45
44
49
35
36
51
50
68.3% of the data lies
between what points?
99.7% of the area lies

Standard Deviation – Simple
ApplicationFrequency
800
600
400
200
I have a process with mean of 43 and a standard deviation of
3
1200
1400
1000
37
43
42
41
40
39
38
48
47
46
45
44
49
35
36
51
50
between 40 and 45
𝑋 ± 1𝜎 = 43 ± 3
between 38 and 47
𝑋 ± 2𝜎 = 43 ± 6
99.7% of the area lies
between 36 and 49
𝑋 ± 3𝜎 = 43 ± 9
𝑿 = 𝟒𝟑
±3𝝈
±2𝝈
±1𝝈

Standard Deviation – Class
Exercise
 What is the probability that a random
sample taken from this process…
 Will have a value between 40 and 45?
68.3%
95.4%
99.7%

Probability theory And
Probability Distribution

Probability
What is the role of Probability in
Statistics?
 Any conclusion we reach on a
population, based on what we know
about a sample, is subject to
uncertainty.
 This uncertainty is calculated and
described using probability theory
 Every output (response) from a
process adds up to 100% of the

Probability Measure
 Every event (=set of outcomes) is assigned a
probability measure.
 The probability of every set is between 0 and 1,
inclusive.
 The probability of the whole set outcomes is 1.
 If A and B are two event with no common
outcomes, then the probability of their union is
the sum of their probabilities.

Probability Measure
 Probability of an event A = P (A)
 P (A) =
𝑪𝒉𝒂𝒏𝒄𝒆𝒔 𝒐𝒇 𝒇𝒂𝒗𝒐𝒓𝒊𝒏𝒈 𝒆𝒗𝒆𝒏𝒕
𝑻𝒐𝒕𝒂𝒍 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒆𝒗𝒆𝒏𝒕𝒔
Cards
 Events: a red card (1/2); a jack (1/13)
 Chances of calling correctly on toss of
a coin is ½ i.e. 0.5

Probability
Building an Understanding
 We’ll start with a pair of dice
 Our customer will only accept combinations
that equal 3,4,5,6,7,8,9,10 and 11.
 What is the probability of meeting his
requirement?

Probability
The customer defines a response of 2 or 12 as a defect
Die 1 Roll
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 1
0
5 6 7 8 9 1
0
11
6 7 8 9 1
0
11 1
2
Die2Roll
Calculate all possible
responses from the
combinations of inputs
How many total combinations exist?
How many times is my response a 2?
What is the probability of a response of 2?
How many times is my response a 12?
What is the probability of a response of
12?
What is the probability of a defect? (2 or 12)
𝟏𝑹𝒆𝒔𝒑𝒐𝒏𝒔𝒆 𝒊𝒏 𝟑𝟔 =
𝟏
𝟑𝟔
= 𝟎. 𝟎𝟐𝟕𝟖 = 𝟐. 𝟕𝟖%
𝟔 𝒅𝒊𝒆 𝟏 × 𝟔 𝒅𝒊𝒆 𝟐 = 𝟑𝟔 𝑻𝒐𝒕𝒂𝒍 𝑪𝒐𝒎𝒃𝒊𝒏𝒂𝒕𝒊𝒐𝒏𝒔
𝟏𝑹𝒆𝒔𝒑𝒐𝒏𝒔𝒆 𝒊𝒏 𝟑𝟔 =
𝟏
𝟑𝟔
= 𝟎. 𝟎𝟐𝟕𝟖 = 𝟐. 𝟕𝟖%
𝟎. 𝟎𝟐𝟕𝟖 + 𝟎. 𝟎𝟐𝟕𝟖 = 𝟎. 𝟎𝟓𝟓𝟔 = 𝟓. 𝟓𝟔%

Probability
Die 1 Roll
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 1
0
5 6 7 8 9 1
0
11
6 7 8 9 1
0
11 1
2
Die2Roll
• Another example
 What is the probability of rolling
a 7 using a fair pair of dice?
Die 1 Die 2 Probability
1 6 0.0278
2 5 0.0278
3 4 0.0278
4 3 0.0278
5 2 0.0278
6 1 0.0278
Total 0.1668
The probability of each roll
is included in each block
16.68% Probability

Probability
Value
(Response)
Frequency Probability
2 1 0.0278
3 2 0.0556
4 3 0.0833
5 4 0.1111
6 5 0.1389
7 6 0.1667
8 5 0.1389
9 4 0.1111
10 3 0.0833
11 2 0.0556
12 1 0.0278
Total 1.0000
Probability of any given value on Die
1 𝑭𝒐𝒓 𝒓𝒐𝒍𝒍𝒊𝒏𝒈 𝒕𝒉𝒆 𝒅𝒊𝒄𝒆, 𝒘𝒉𝒂𝒕 𝒊𝒔 𝒕𝒉𝒆
𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏?
𝟏
𝟔
= 𝟎. 𝟏𝟔𝟔𝟔 = 𝟎. 𝟏𝟔𝟔𝟕
Probability of any given value on Die
2 𝟏
𝟔
= 𝟎. 𝟏𝟔𝟔𝟔 = 𝟎. 𝟏𝟔𝟔𝟕
𝟏
𝟔
×
𝟏
𝟔
=
𝟏
𝟑𝟔
= 𝟎. 𝟎𝟐𝟕𝟕 = 𝟎. 𝟎𝟐𝟕𝟖
Probability of any given combination

Probability
0
2
4
6
8
10
12
14
16
18
2
3
4
5
6
7
8
9
10
11
12
This represents the
response of our system
(In Probability)
Response (Dice Total)
Probability
0.0278 0.0278
0.1389
0.1111
0.0833
0.0556
0.1667
0.0556
0.0833
0.1111
0.1389

Probability
 Our customer will only accept combinations that equal
3,4,5,6,7,8,9,10,11
 We have a 99.44% probability of
meeting the customers specification
 The curve of this distribution becomes
it’s Probability Density Function
0
5
10
15
20 2
3
4
5
6
7
LSL USL
Probability
Value
(Response)
Frequenc
y
Probabilit
y
2 1 0.0278
3 2 0.0556
4 3 0.0833
5 4 0.1111
6 5 0.1389
7 6 0.1667
8 5 0.1389
9 4 0.1111
10 3 0.0833
11 2 0.0556
12 1 0.0278

Probability
 Our customer will only accept combinations that equal
3,4,5,6,7,8,9,10,11
 We have a 99.44% probability of
meeting the customers specification
 The curve of this distribution becomes
it’s Probability Density Function
0
5
10
15
20
LSL USL
Probability
Value
(Response)
Frequenc
y
Probabilit
y
2 1 0.0278
3 2 0.0556
4 3 0.0833
5 4 0.1111
6 5 0.1389
7 6 0.1667
8 5 0.1389
9 4 0.1111
10 3 0.0833
11 2 0.0556
12 1 0.0278

Probability Theory
What is a Probability Density Function?
A Mathematical Function
 It models the probability density reflected in a histogram
With more observations
 Class intervals become narrower and more numerous
 The histogram of the variable takes on the appearance of
a smooth curve
The total area under the curve must equal 1.
The probability that a random variable will assume
a value between any two points is equal in value to
the area under the random variable’s probability
density function between these two points.
h
What does this mean to us?

Probability Theory
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Response Intervals
1
2
3
4
5
6
7
8
9
10
11
Frequency
• This histogram has 24 points distributed over 12 intervals

Probability Theory
Response Intervals
Frequency
400
300
200
100
• As the number of data increase, the intervals get
smaller
When we do this, the curve outlining the data gets smoother

Probability Theory
What do we know about Probability Distribution?
 The area under the curve always equals 1
 We can determine the probability that a value of a
random variable will fall between 2 points on the
curve by calculating the area under the curve
between the two points
Why would we
want to do this?
How do we do
this?

Using Probability Distribution
The Standard Normal Distribution
 Let’s take a look at the most important
PD…the standard normal distribution
 We can transform each point on our normal
curve into a standard normal curve value
using the Z transform
𝒁 𝒑𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒁 𝒔𝒂𝒎𝒑𝒍𝒆

• Standard Normal Curve Characteristics
𝑿 = 𝟎
𝑿 = 𝟐𝟓𝟑
𝟏𝝈 = 𝟏. 𝟎 𝟏𝝈 = 𝟏. 𝟎 𝟏𝝈 = 𝟏. 𝟎 𝟏𝝈 = 𝟏. 𝟎 𝟏𝝈 = 𝟏. 𝟎 𝟏𝝈 = 𝟏. 𝟎
𝟏𝝈 = 𝟐𝟏 𝟏𝝈 = 𝟐𝟏 𝟏𝝈 = 𝟐𝟏 𝟏𝝈 = 𝟐𝟏 𝟏𝝈 = 𝟐𝟏 𝟏𝝈 = 𝟐𝟏
It has a standard
deviation of 1.0
𝒁 𝒔𝒂𝒎𝒑𝒍𝒆 =
𝑿 − 𝑿
𝑺
It has a mean of 0.0
The area under the curve
equals 1
The curve is symmetrical
After the Z
Transform
The
Original
Distribution

The “How”
Find the points on the Standard Normal
Distribution that correspond to your values
Determine the area under the standard
normal curve that is between the points you
have found
If our data is normal, we can use the
Standard Normal Distribution
This saves us from having to do the
calculation for each specific situation!

A “Why” Example:
 The unit sales of Product A follows a
normal distribution and has a monthly
average of 253 units with a standard
deviation of 21 units
= 253
S = 21
What is the probability
that next months sales
will be greater than 300
units?
𝑿

 What is the probability that next month’s
unit sales will be greater than 300?
1. Find the point on the Standard Normal
Distribution that corresponds to 300
=25 S = 21
𝒁 𝒔𝒂𝒎𝒑𝒍𝒆 =
𝑋 − 𝑋
𝑆
=
300 − 253
21
= 2.24
𝑿
This is telling us that 300 is 2.24 standard deviations from the
mean

2) Determine the area under the
standard normal curve that is to the
right of 2.24
– How?
– Use the Table of the Standard Normal
Distribution
2.24

StandardNormalTable
Z was
2.24

 This table shows the area between 0 (the mean of a standard
normal table) and Z
 Because the curve is symmetric…
The area of each ½ is 0.500
The area to the right of a positive value is 0.500 minus the
area between 0 and the Z value
 For Z = 2.24 (the equivalent of 300)
Locate the row labeled .04
The area is 0.4875
 Subtract this area from 0.500
0.500 – 0.4875 = 0.0125
I have a 1.25% probability that my unit sales next month
will be greater than 300 units
2.24

Normal Distribution
If you know your average value ( ) and
your standard deviation (s) then for a
given specification limit, it is possible to
predict rejections (if any), that will occur
even if you keep your process in control.
Example:
= 2.85, s = 0.02 (The dimensions
relate to a punched part).
Lat us find the percentage rejection if the
specified value is 2.85±0.04 i.e. the part
is acceptable between 2.81-2.89

Normal Distribution
Applicable in real life:
Acceptable
RangeRejections Rejections
2.81
2.85
2.89

Normal Distribution
Let A, B and C represent the areas under the curve
for the following conditions:
A – rejections for undersize
B – acceptable range
C – rejections for oversize
Total Area = A+B+C
Total Rejections = A+C
BA C
2.85
2.892.81

Normal Distribution
We will introduce a concept called Z which
we can use with a one-sided
distribution to
determine the area
under A, B and C and
thus the percentage
rejections and acceptable
components.
BA C
2.85
2.892.81

Normal Distribution
 The area from Normal table
corresponding to 2 is 0.02275
 Hence Rejection for Over size (Area
C) = 2.275%
 Similarly one can find the rejection for
undersize

Discrete Probability
Distributions

Binomial Distribution
When applicable:
When the variable is in terms of attribute data
and in binary alternatives such as good or bad,
defective or non-defective, success or failure etc.
Conditions:
 The experiment consists of ‘n’ identical trials
 There are only two possible outcomes on each
trial. We denote as Success(S) and Failure(F).
 The probability of ‘S’ remains the same from trial
to trial and is denoted by ‘p’ and the probability of
‘F’ is ‘q’.
 p+q = 1
 The trials are independent

For a random experiment of sample size n where
there are two categories of events, the probability of
success of the condition x in one category (where
there is n-x in the other category) is
𝑃(𝑋 = 𝑥) =
𝑛
𝑥
𝑝 𝑥
(𝑞) 𝑛−𝑥
, 𝑥 = 0,1,2, , 𝑛
Where (𝒒 = 𝟏 − 𝒑) is the probability that the vent
will not occur.
Where
𝑛
𝑥
=
𝑛!
𝑥! 𝑛−𝑥 !

Consider now that the probability of having the
number “2” appear exactly three times in seven
rolls of a six die is
𝑷 𝑿 = 𝟑 =
𝒏
𝒙
𝒑 𝒙(𝟏 − 𝒑) 𝒏−𝒙
= 𝟑𝟓 𝟎. 𝟏𝟔𝟕 𝟑 𝟏 − 𝟎. 𝟏𝟔𝟕 𝟕−𝟑 = 𝟎. 𝟎𝟕𝟖𝟒

Poisson Distribution
When applicable:
 No. of accidents in a specified period of time
 No. of errors per 100 invoices
 No. of telephone calls in a specified period of time
 No. of surface defects in a casting
 No. of faults of insulation in a specified length of cable
 No. of visual defects in a bolt of cloth
 No. of spare parts required over a specified period of
time
 The no. of absenteeism in a specified no. of time
 The number of death claims in a hospital per day
 The number of breakdowns of a computer per month
 The PPM of Toxicant found in water or air emission from
a manufacturing plant

Two Properties of a Poisson Experiment
1) The Probability of an occurrence is he
same for any two intervals of equal length.
2) The occurrence or nonoccurrence in any
interval is independent of the occurrence or
nonoccurrence in any other interval.

Conditions:
 The experimental consists of counting
the number of times a particular event
occurs during a given unit of time or in a
given area or volume or weight or
distance etc.
 The probability that an event occurs in a
given unit of time is same for all the
units.
 The no. of events that occur in one unit
of time is independent of the number that
occur in other units.
 The mean no. of events in each unit will
be denoted by .

The Poisson Random Variable ‘X’ is the number of
events that occur in specified period of time.
𝑃 𝑋 = 𝑥 =
𝑒−𝝺
𝞴 𝑥
𝑥!
𝑥 = 0,1,2,3 …
A company observed that over several years they had a mean
manufacturing line shutdown rate of 0.10 per day. Assuming a
Poisson distribution, determine the probability of two
shutdowns occurring on the same day.
For the Poisson distribution, 𝝺 = 𝟎. 𝟏𝟎 occurrence/day and 𝐱 =
𝟐 results in the probability
𝑃 𝑋 = 2 =
𝑒−𝝺 𝞴 𝑥
𝑥!
=
𝑒−0.10.12
2!
= 0.004524

Suppose the number of breakdowns of machines
in a day follows Poisson Distribution with an
average number of breakdowns is 3.
Find the probability that there will be no
breakdowns tomorrow.
𝞴 = 3
𝐏(𝐗 = 𝟎) =
𝒆−𝟑
𝟑 𝟎
𝟎!
= 𝒆−𝟑
= 𝟎. 𝟎𝟒𝟕𝟗𝟕

Patients arrive at the emergency
room of Mercy Hospital at the
average rate of 6 per hour on
weekend evenings.
What is the probability of 4
arrivals in 30 minutes on a
weekend evening?
Example: Mercy Hospital

Process Accuracy And
Precision
 We have curves that describe our
process
 Some questions we may ask…
Is my process accurate?
Is my process precise?

Precision
 Accuracy describes
centering
 Is my process mean
at my target mean?
LSL USL
Target

Precision
LSL Target USL
•Precision describes
spread
•How does the spread
of my process compare
to the customer’s
specification limits?

Capability
In Statistic Terms…
LSL USLLSL USL
LSL USL LSL USL
Mean is not centered in Specification Mean is centered in Specification
SmallStandard
Deviation
LargeStandard
Deviation

SPC
PROCESS
The combination of people, equipment,
materials, methods, measurement and
environment that produce output – a
given product or service.
Process is transformation of given
inputs into outputs

SPC
VARIATION
The inevitable differences among
individual outputs of a process.
The sources of variation can be
grouped into two major classes,
Common Causes & Special Causes

SPC
COMMON CAUSE
A source of variation that affects all the
individual values of the process output
being studied
This is the source of the inherent
process variation.

SPC
Common Causes:
1. Plenty in Numbers
2. Results in less Variation
3. Part of the Process
4. Results in constant Variation
5. Predictable
6. Management Controllable
7. Statistics shall apply

SPC
Examples of Common Causes,
MAN
MACHINE
MATERIAL
Differences in Competency (setting,
operating & inspection) of Employees
working in shifts.
Difference in Quality of Product when
Production of same Part is being
carried out as per plan. UPS provided
for Electricity Supply
Difference in Mechanical & Chemical
Properties in 2 different lots of Material
of same grade received from suppliers
(Raw Material Manufacturers)

SPC
SPECIAL CAUSE:
A source of variation that affects only
some of the output of the process; it is
often intermittent and unpredictable. A
special cause is some times called
assignable cause. It is signaled by one
or more points beyond the control limits
or a non-random pattern of points within
the control limits.

SPC
Special Causes:
1. Few in numbers
2. Results in large variation
3. Visitors to the process
4. Variation due to external factors
5. Fluctuating Variation
6. Unpredictable
7. Controllable by Operating personnel
8. Statistics shall not apply
Recognize and deal with special causes outside the (Six Sigma)
process
Implement Corrective and Preventive Action (CAPC)

SPC
Examples of Special Causes,
MAN
MACHINE
MATERIAL
METHOD
MEASUREMENT
Untrained Employee working on the Machine
Production of Product on Conventional Lathe
machine where Product Run out requirement
is 2 microns. Major & frequent breakdowns of
Machine. Frequent Power Failures.
Use of different grade of raw material
Setting of process Parameters which are not
proven.
Tool breakage
Use of Micrometer having range of 0-25 mm
to check O.D. of 25 mm ± 0.1 mm.

Types of Control Charts
VARIABLE
𝑿, R
𝑿, s
𝑿, mR
CUSUM
ATTRIBUTE
p
np
c
u

Control Charts
Overview
The first step for control charting is to
identify the CTQ’s of the process which
is required to be brought under control
Types of Control Charts
Depends on the nature of the variable
needed to control:
 Variable Control Charts
 Attribute Control Charts

Variable Control Chart
CONTROL CHARTS

Xbar – Rbar
When to use:
When studying the behavior of a single measurable
characteristic produced in relatively high volumes.
How:
By plotting sample averages (X-bar) and ranges (R) on separate
charts. This allows for independent monitoring of the process
average and the variation about that average.
Conditions:
 Constant sample size.
 One characteristic per chart.
 Should have no less than 20 samples before calculating
control limits.

Xbar – Rbar
1. Most common type of control chart
for analyzing continuous variables.
2. The xbar part of the chart notes the
variation between the averages of
consecutive sub-groups of data
points.
3. The R part of the chart notes the
changes of variation within each of
the consecutive sub-groups.

RATIONAL SUBGROUP CONCEPT
 Subgroups or samples should be selected so
that if assignable causes are present, the chance
for differences between subgroups will be
maximized, while the chance for differences due
to these assignable causes within a subgroup will
be minimized.
 Time order is frequently a good basis for forming
subgroups because it allows to detect assignable
causes that occur over time.
 Two general approaches for constructing rational
subgroups:
◦ Construction units of production
◦ Random sample of all process output over the
sampling interval

Control Chart
 Reviewing plots & Analysis of trends:
Ensure that all points of both X and R charts
within control limits.
If any point touching to any of the control
limits, review process related remark
corresponding to particular sub-group.
This is assignable cause.
Study particular trends if any
◦ Case study:
◦ Consider process of side member sub-
assembly where critical dimensional
characteristics i.e. concentricity of mounting
holes is controlled.

Control Chart
TRENDS ANALYSIS IN SPC CHARTS
ALL POINTS WITHIN CONTROL LIMIT
S.NO Trend Type Meaning Precautions for
better process
control
1. All points within
control limits with
zigzag pattern
Process under control,
variation due to random
causes.
Zigzag pattern changing
with each point over
judgment
Let process continue. Try
to make it a natural
process
2. 7 more consecutive
points on one side
of center line
Process Centre shifted
towards one of the
specification limit
Do changes to bring
process to Centre
3. Cyclic trends Assignable cause
happening periodically
Study assignable cause
and reason. Study to
prevent
4. Continuous
inclination towards
one of the control
limits
Assignable cause for
process drift. If not
prevented, product may
go out of control
Study assignable cause,
set process to prevent
drifting

Control Chart
TRENDS ANALYSIS IN SPC CHARTS
ALL POINTS WITHIN CONTROL LIMIT
S.NO Trend Type Meaning Precautions for
better process
control
1. All points suddenly
going out of control
limits
Assignable cause present,
study specific process
event associated with
period of specific point
Study probable causes for
assignable cause taking
place try to resolve the
same
2. Any point going out
of control limits
with definite trend
Process going out of
control due to assignable
cause
Study the trend type &
establish controls to
prevent the assignable
cause occurring

Typical Out-Of-Control
Patterns
 Point outside control limits
 Sudden shift in process average
 Cycles
 Trends
 Hugging the center line
 Hugging the control limits
 Instability

Control Charts
PURPOSE OF CONDUCTING SPC
STUDIES:
 To study and analyze process variation
 To find out trends in processes
 To identify random & sporadic causes
 To manufacture products of consistent
quality
 To prevent wastage of material

Process Capability For
Continuous Data

Capability vs Stability
 Capability has a meaning only when a
process is stable.
 If a process is out of control, first we need to
stabilize the process.
 Improvement in the inherent variation can be
made only when the process is stable.
 Control Charts are used to study stability.
 The first job of Six Sigma practitioner is to
identify and remove Special Causes of
Variation.
 Once the process is made predictable, the
next job is to identify the causes of inherent
variation and remove them.

Calculating Capability
𝐶𝑎𝑝𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠
𝑇𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒
𝑃𝑟𝑜𝑐𝑒𝑠𝑠 𝑉𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛
LSL USL𝑿 ± 𝟏𝝈
𝑿 ± 𝟑𝝈
𝑿 ± 𝟐𝝈
𝑪 𝒑 =
𝑼𝑺𝑳 − 𝑳𝑺𝑳
𝟔𝝈
𝑪 𝒑 =
𝑻
𝟔𝝈

𝑴𝒂𝒓𝒈𝒊𝒏𝒂𝒍 𝑪𝒂𝒑𝒂𝒃𝒊𝒍𝒊𝒕𝒚
LSL USL𝑿 ± 𝟏𝝈
𝑿 ± 𝟑𝝈
𝑿 ± 𝟐𝝈
𝑪 𝒑 =
𝟔𝝈
𝟔𝝈
𝑪 𝒑 =
𝑻
𝟔𝝈
𝑪 𝒑 = 𝟏

LSL USL
𝑿 ± 𝟔𝝈
𝑪 𝒑 =
𝟏𝟐𝝈
𝟔𝝈
𝑪 𝒑 =
𝑻
𝟔𝝈
𝑪 𝒑 = 𝟐
• Six Sigma Capability
𝑿 ± 𝟑𝝈
𝟑𝝈 𝟑𝝈

0 2 4 6 8 10 12 14 16 18 20
𝑪 𝒑𝒌 = 𝑴𝒊𝒏
𝑿 − 𝑳𝑺𝑳
𝟑𝝈
,
𝑼𝑺𝑳 − 𝑿
𝟑𝝈
𝑪 𝒑𝑳 =
𝟑𝝈
𝑪 𝒑𝑼 =
𝟑𝝈
𝑿 − 𝑳𝑺𝑳 𝑼𝑺𝑳 − 𝑿
• Calculate 𝑪 𝒑 from Upper and Lower side

Calculating Performance
0 2 4 6 8 10 12 14 16 18 20
𝑷 𝑷𝒌 = 𝑴𝒊𝒏
𝟑𝝈
,
𝟑𝝈
𝑷 𝑷𝑳 =
𝟑𝝈
𝑷 𝑷𝑼 =
𝟑𝝈
𝑿 − 𝑳𝑺𝑳 𝑼𝑺𝑳 − 𝑿
• Calculate 𝑪 𝒑 from Upper and Lower side
𝑷 𝑷 =
𝟔𝝈

Calculating Performance
𝑪 𝑷 =
𝟔𝝈
𝑪 𝒑𝒌 = 𝑴𝒊𝒏
𝟑𝝈
,
𝟑𝝈
• If the formulae are same, what is the difference?
• The difference is in Sigma Calculation!
• Sigma in Capability covers Short Term Variation.
• Sigma in performance covers Long term Variation.
• How is the Data Collection Different?

Process Capability Ratios
ContinuousImprovement
LSL USL
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑 < 𝟏. 𝟎
LSL USL
IncreasedNumberofDefects
Process
Capability
Real
Capability
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑 = 𝟐. 𝟎
𝑪 𝒑𝒌 = 𝟐. 𝟎
𝑪 𝒑𝒌 < 𝟐. 𝟎
𝑪 𝒑𝒌 = 𝟏. 𝟎
𝑪 𝒑𝒌 = 𝟎. 𝟎
𝑪 𝒑𝒌 < 𝟎. 𝟎
𝑪 𝒑𝒌 < −𝟏. 𝟎
Understanding 𝑪 𝑷 and 𝑪 𝑷𝑲
𝑪 𝑷 only works for a process that is centered on the target
𝑪 𝑷𝑲 is a better measure for tracking performance

Capability Indices
𝑪 𝒑 =
𝟔𝑺
𝑪 𝒑𝒍 =
𝟑𝑺
𝑪 𝒑𝒖 =
𝟑𝑺
𝑪 𝒑𝒌 = 𝐦𝐢𝐧(𝑪 𝒑𝒖 , 𝑪 𝒑𝒍)
𝑪 𝒑 =
𝟎. 𝟑𝟏𝟕 − 𝟎. 𝟑𝟎𝟕
𝟔 × 𝟎. 𝟎𝟎𝟏𝟐
=
𝟎. 𝟎𝟏
. 𝟎𝟎𝟕𝟐
= 𝟏. 𝟑𝟖𝟗
𝑪 𝒑𝒖 =
𝟎. 𝟑𝟏𝟕 − 𝟎. 𝟑𝟏𝟓𝟓
𝟑 × 𝟎. 𝟎𝟎𝟏𝟐
=
𝟎. 𝟎𝟎𝟏𝟓
. 𝟎𝟎𝟑𝟔
= 𝟎. 𝟒𝟏𝟕
𝑪 𝒑𝒍 =
𝟎. 𝟑𝟏𝟓𝟓 − 𝟎. 𝟑𝟎𝟕
𝟑 × 𝟎. 𝟎𝟎𝟏𝟐
=
𝟎. 𝟎𝟎𝟖𝟓
. 𝟎𝟎𝟑𝟔
= 𝟐. 𝟑𝟔𝟏
𝑪 𝒑𝒌 = 𝐦𝐢 𝐧 𝟎. 𝟒𝟏𝟕, 𝟐. 𝟑𝟔𝟏 = 𝟎. 𝟒𝟏𝟕
Exampl
e
𝑿 𝒎𝒆𝒂𝒏 = 𝟎. 𝟑𝟏𝟓𝟓
𝒔 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 = 𝟎. 𝟎𝟎𝟏𝟐
𝑳𝑺𝑳 𝒍𝒐𝒘𝒆𝒓 𝒔𝒑𝒆𝒄 𝒍𝒊𝒎𝒊𝒕 = 𝟎. 𝟑𝟎𝟕
𝑼𝑺𝑳 𝒖𝒑𝒑𝒆𝒓 𝒔𝒑𝒆𝒄 𝒍𝒊𝒎𝒊𝒕 = 𝟎. 𝟑𝟏𝟕
0.306 0.308 0.310 0.312 0.314 0.316 0.318 0.320
LSL USL

Capability Indices
𝑪 𝒑 =
𝟔𝑺
𝑪 𝒑𝒍 =
𝟑𝑺
𝑪 𝒑𝒖 =
𝟑𝑺
𝑪 𝒑𝒌 = 𝐦𝐢𝐧(𝑪 𝒑𝒖 , 𝑪 𝒑𝒍)
𝑪 𝒑 =
𝟎. 𝟑𝟐𝟐 − 𝟎. 𝟑𝟎𝟐
𝟔 × 𝟎. 𝟎𝟎𝟏𝟐
=
𝟎. 𝟎𝟐
. 𝟎𝟎𝟕𝟐
= 𝟐. 𝟕𝟕𝟖
𝑪 𝒑𝒖 =
𝟎. 𝟑𝟐𝟐 − 𝟎. 𝟑𝟏𝟒𝟑
𝟑 × 𝟎. 𝟎𝟎𝟏𝟐
=
𝟎. 𝟎𝟎𝟕𝟕
. 𝟎𝟎𝟑𝟔
= 𝟐. 𝟏𝟑𝟗
𝑪 𝒑𝒍 =
𝟎. 𝟑𝟏𝟒𝟑 − 𝟎. 𝟑𝟎𝟐
𝟑 × 𝟎. 𝟎𝟎𝟏𝟐
=
𝟎. 𝟎𝟏𝟐𝟑
. 𝟎𝟎𝟑𝟔
= 𝟑. 𝟒𝟏𝟕
𝑪 𝒑𝒌 = 𝐦𝐢 𝐧 𝟐. 𝟑𝟏𝟗, 𝟑. 𝟒𝟏𝟕 = 𝟐. 𝟑𝟏𝟗
Exampl
e
𝑿 𝒎𝒆𝒂𝒏 = 𝟎. 𝟑𝟏𝟒𝟑
𝒔 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 = 𝟎. 𝟎𝟎𝟏𝟐
𝑳𝑺𝑳 𝒍𝒐𝒘𝒆𝒓 𝒔𝒑𝒆𝒄 𝒍𝒊𝒎𝒊𝒕 = 𝟎. 𝟑𝟎𝟐
𝑼𝑺𝑳 𝒖𝒑𝒑𝒆𝒓 𝒔𝒑𝒆𝒄 𝒍𝒊𝒎𝒊𝒕 = 𝟎. 𝟑𝟐𝟐
0.300 0.305 0.310 0.315 0.320
LSL USL

Let’s Summarize
 A process cannot be improved till it is
Stabilized.
 Capability data should be utilized for
stable processes
 Subgroups should contain consecutive
data, not random data.
 Performance calculations should be
done based on large amount of data
representing Long Term Variation.

Process Capability For
Attribute Data

Discrete Data Capability
A discrete defect is an attribute, which can
be counted.
Such as:
Scratches, Spots, Dent Marks, Cracks etc.
 In these cases ½ does not make sense.
 A defect is non conformance to the
standards.
 A defective unit can have more than one
defect.
 A sample of 100, may have 2 defectives
but 5 defects.

Defect Opportunities:
 Defect opportunities are various types of
defects, that may occur.
 These creates dissatisfaction to the
customers.
 This is different than defects that occur.
 Example : 12 type of defects that can occur
on painted part.
However, on a part produced, we may observe
0 to up to 12 defects.
Thus a part may be defect free or may have1to
12 defects.

Example:
A sample of 100 nos have been taken.
Following are the results of inspection:
No of Defectives – 3
No of defects – 10
No of Opportunities - 12

Example:
The capability can be calculated as
follows:
No of units = U =100
Defects = D =10
No of Opportunities = O = 12
Total defect opportunities = UxO =
100x12 =1200
DPO = Defects per opportunity =
10/1200 =1/2 = 0.0083

Example:
Defect per million opportunities (DPMO)
=DPO x 1,000,000
=0.0083 x 1,000,000
=8300 DPMO
From the tables, the corresponding
sigma level is 3.9.

The same formula also can be
expressed as
DPMO =
𝑵𝑶. 𝑶𝑭 𝑫𝑬𝑭𝑬𝑪𝑻𝑺×𝟏𝟎 𝟔
𝑵𝑶 𝑶𝒇 𝑼𝑵𝑰𝑻𝑺×𝑶𝑷𝑷./𝑼𝑵𝑰𝑻

Discrete Data Capability –
Example of DPMO
Suppose we observe 200 letters delivered
incorrectly to the wrong addresses in a
small city during a single day when a
total of 200,000 letters were delivered.
What is the DPMO in this situation?
DPMO =
𝟐𝟎𝟎 × 𝟏𝟎 𝟔
𝟐𝟎𝟎,𝟎𝟎𝟎 ×𝟏
= 𝟏, 𝟎𝟎𝟎
So, for every only million letters delivered this city’s postal
managers can expect to have 1,000 letters incorrectly sent to
the wrong address.
What is the Six Sigma Level for this
Process?

DPMO Example
 IRS tax form advice
 Survey of responses indicates
predicted error rate
 If 40% then:
DPO = 0.40
DPMO = 0.40 defects/opportunity *
1,000,000 opportunities/million
opportunities
400,000 DPMO = 1.75 Sigma

DPMO Example
Example of Rolled throughput yield
 If there are five processes with following yields:
 Rolled throughput yield for this process is =
0.9 × 0.99 × 0.95 × 0.96 × 1 = 0.7279 = 0.73 = 73%
Process No. Yield in %
1 90
2 99
3 95
4 96
5 100

DPMO - Exercise
 You have 100 documents
You take a sample of 10 documents
There are 10 opportunities for defect on each
document.
5 defects were found.
What is DPMO
Attendance Policy
June 23, 2000
Crane Operational
Excellence Program
All Operational
Excellence Leaders should
be aware.

Complexity and Capability
Payroll and Labor Tracking Process
Does complexity have an important impact on
process capability and quality?
There are many opportunities for defects…
Step 1
97.4%
Read
and
record
daily
start and
stop time
𝒀 𝑹𝑻
Output
79.1%
Step 6
99.9%
Create
payroll
checks
Step 5
95.5%
Transfer
hour
totals to
payroll
generatio
n system
Step 4
91.8%
Total
weekly
work
hours
and job
accounts
. Submit
time card
Step 3
98.0%
Total
daily
work
hours
Step 2
94.6%
Read
and
record
daily
start and
stop time
Rolled Throughput Yield Example
=

Payroll and Labor Tracking Process
Our goal, reduce the total number of opportunities and
increase the capability of remaining opportunities
Step 1
97.4%
Output
79.1%
Step 6
99.9%
Step 5
95.5%
Step 4
91.8%
Step 3
98.0%
Step 2
94.6%
=
𝒀 𝑹𝑻
Output
98.9%
Step 3
99.9%
Print payroll
checks from
computer
generated
database
Step 2
99.4%
Scan
employee
badge and
job card for
labor start
and stop
time
Step 1
99.6%
Scan
employee
badge for
start and
stop time
=

Notice any Difference?
Step 1
93.32%
Output
81.26%
Step 3
93.32%
Step 2
93.32%
Step 2
99.999997%
=
=
Output
79.1%
Step 2
99.999997%
Step 2
99.999997%x x
xx
A Three Sigma Process
A Six Sigma Process

Sigma Levels
SIGMA Defect per Million Opportunities
(DPMO)
1
690,000
2
308,537
3
66,807
4 6,210
5 233
6 3.4

Introduction To
Hypothesis Testing

Hypothesis Testing Concept
 Hypothesis testing is one of the most
scientific ways of decision making.
 It works very much like a court case.
 We have a suspect, we have to take
decision whether He / She is innocent or
guilty.
 Suppose there is person charged with
murder, and both sides (defense and
prosecution) do not have any evidence,
what would be decision?
 Innocent unless proven guilty?
 Guilty unless proven Innocent?
Null Hypothesis

Null Hypothesis
 Null hypothesis is represented by Ho
 It is statement of Innocence.
 It is something that has to be assumed
if you cannot prove otherwise.
 It is statement of No Change or No
Difference.

Null Hypothesis – A Court
Case
 Just Like a court case, we first assume the
accused (X) is innocent and then try to prove
it otherwise based on evidence (Data).
 If evidence (Data) does not show sufficient
difference, we cannot reject the
innocence(Ho)
 But if Evidence (Data) is strong enough, we
reject the Innocence (Ho) and pronounce the
suspect Guilty (Ha).
 The statement that will be considered valid if
null hypothesis is rejected is called Alternate
Hypothesis (Ha)

Null hypothesis – A Concept
 Hypothesis testing is a philosophy that
real life situations.
 You cannot prove two things equal.
 You cannot prove two things different by
proving only one difference
 If you cannot prove 2 things different,
you have to assume that they are equal.
 But if you cannot prove them Different,
are they really Equal?
 What is the RISK involved?

In Truth, the Defendant is:
Correct Decision
Innocent individual goes
Free
Incorrect Decision
Guilty Individual Goes Free
Incorrect Decision
Innocent Individual Is
Disciplined
Correct Decision
Guilty Individual Is
Disciplined
𝑯 𝑨: Guilty𝑯 𝒐: Innocent
Verdict
Innocent
Guilty

Correct Decision Incorrect Decision
Type II Error Probability = 𝛽
Incorrect Decision
Type I Error Probability = 𝛼
Correct Decision
𝑯 𝑨is True𝑯 𝒐is True
𝑯 𝒐is True
𝑯 𝑨is True
Decision
True, But Unknown State of the World

Hypothesis testing Justice System
 State the Opposing Conjectures, Ho and HA.
 Determine the amount of evidence required,
n, and the risk of committing a “type error”,
 What sort of evaluation of the evidence is
required and what is the justification for this?
(type of test)
 What are the conditions which proclaim guilt
and those which proclaim innocence/
(Decision Rule)
 Gather & Evaluate the evidence.
 What is the verdict? (Ho or HA?)
 Determine “Zone of Belief” : Confidence
Interval.
 What is appropriate justice? – Conclusions

Hypothesis Testing
1. Null Hypothesis (Ho) – statement of no change or
difference. The statement is assumed true until
sufficient evidence is presented to reject it.
2. Alternate Hypothesis (Ha) – statement of change or
difference. This statement is considered true if Ho is
rejected.
3. True I Error – the error in rejecting Ho when it is in
true fact, there is no difference.
4. Alpha Risk – the maximum risk or probability of
making a Type I Error. This Probability is always
greater then zero, and is usually established at 5%.
The researcher makes the decision to the greatest
level of risk that is acceptable for a rejection of Ho.
Also known as significant level.
5. Type II Error – The error in failing to reject Ho when it
in fact false, or saying there is no difference when
there really is a differerence.

6) Beta Risk – The risk probability or
making a Type II Error, or overlooking
an effective treatment or solution to the
problem.
7) Significant Difference – The term
where a difference is too large to be
reasonably attributed to chance.

𝛼 𝑎𝑛𝑑 𝛽 Risks
 𝛼 Risk is also called producer’s risk.
 𝛽 Risk is also called consumer’s risk.
 Can we commit both type I and type II
error at the same time?
 As it necessary that we will have both 𝛼
and 𝛽risks?
 Are𝛼 and 𝛽 risks equal?
 Is 𝛼 and 𝛽 = 1?
 Is there any relationship between 𝛼 and
𝛽?
 Which risk is more important?

𝛼 𝑎𝑛𝑑 𝛽 Risks
 An 𝛼 Risk of 5% is generally
accepted.
 An 𝛽 Risk of 10% is generally
Accepted.
 Since Ha cannot be proved, our
attempt is to try and reject it.
 What risk do we get in trying to reject
the Ho.
 Minitab represents 𝛼 risk by p-panel!

Steps in Hypothesis Testing
 Define Ho
 Define Ha
 Select Appropriate Test.
 Decide Significance Level (𝛼 and 𝛽)
 Decide Sample Size
 Collect Data
 Conduct Test
 Interpret!

Define Ho/Ha For following
Cases
To find if a distribution is normal or not.
 Ho =>
 Ha =?
To find if the defects from three machines
are same or different
 Ho =>
 Ha =>
To find if 2 groups of students from
different streams have differing IQ
 Ho =>
 Ha =>

Basic Concepts
Statistical Error

Statistical Error Definitions
 Null Hypothesis Ho:
 “Status quo”
 “Nothing is different”
 Equality
 We fail to reject Ho based
on statistical evidence
 Alternate Hypothesis Ha:
 “Something is different”
 Statement about the
population that requires
strong evidence to prove
 If we reject Ho, we in
practice accept Ha.
 Alpha Risk (𝛼)
 Also called type I Error
 Hypothesis the null
hypothesis when it is fact
true.
 Beta Risk (𝛽)
 Also called a Type II Error
 Accepting the null
Hypothesis when it is in fact
false.

Statistical Error
Typical 𝛼 𝑎𝑛𝑑 𝛽Risks
 Typically, the 𝛼 level is set at 0.05 and
the 𝛽 level is set at 0.10
 They can be set at any level
depending on what you want to know
The risk is also called the “p-value”
1-𝛼 = confidence that an observed outcome in
the sample is “real”
We typically look for a p-value of 0.05 because:
 1-0.05 = 0.95 (or 95% confidence)

The Central Limit Theorem
Normally
 Why are distributions normal?
When all factors are random
Some measurements are actually averages
over time of “micro-measurements”
In other words, what we see as a
measurement is
actually an average
The Central Limit Theorem explains why a
distribution of averages tends to be normal

Confidence
Sample statistics estimate the mean or standard deviation of a
population
The “True” population mean and standard are unknown
Confidence limits, levels, and intervals are used to determine the
population statistics
For means…
We use t distribution to
calculate limits, levels, and
intervals
For Standard Deviations…
We use the 𝑐2
distribution to
calculate limits, levels, and
intervals

Definition
Confidence Level:
 The level of risk we are willing to take
 How sure we want to be that the population mean or standard deviation
falls between the confidence level is typical
 95% confidence level is typical.
 95% chance that the population mean or standard deviation falls between
the limits.
 5% chance (alpha risk) that the population mean or standard deviation
isn’t contained within the calculated limits.

Risk (𝛼/2) Risk (𝛼/2) Risk (𝛼)

Definition
Confidence Limit
Upper and Lower limits that bracket the “true”
mean or standard deviation of a population
 Calculation from the sample data and the appropriate
test statistic.
 Test statistic is dependent on the risk we accept that
our results will be wrong.

Definition
Confidence Interval
The interval defined by the upper and
lower confidence limits.
A range of values based on
 Sample mean or sample standard deviation
 Sample size
 Confidence level
 Appropriate test statistic
Contains
 Population mean or
 Population standard deviation

Basic Concepts
Confidence Limits For
Means

Confidence Limit Formulas
Means
Lower Confidence Limit
Upper Confidence Limit
𝑋 - 𝑡(
𝛼
2
,𝑛−1)
𝑠
√𝑛
𝑋 + 𝑡(
𝛼
2
,𝑛−1)
𝑠
√𝑛
𝑋 = 𝑚𝑒𝑎𝑛
𝑠 = 𝑆𝑎𝑚𝑝𝑙𝑒 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑛 = 𝑆𝑎𝑚𝑝𝑙𝑒 𝑆𝑖𝑧𝑒
𝑡 = 𝑣𝑎𝑙𝑢𝑒 𝑓𝑜𝑟𝑚 𝑡ℎ𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡′
𝑠 𝑡 − 𝑡𝑎𝑏𝑙𝑒 𝑓𝑜𝑟
𝛼
2
𝑎𝑛𝑑 𝑛 − 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
(𝑓𝑜𝑟 𝑎 95% 𝑐𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙,
𝛼
2
= 0.025)
𝛼 = 𝐴𝑙𝑝ℎ𝑎 𝑅𝑖𝑠𝑘

Means
Confidence Interval
Lower Confidence Limit ≤ Mean ≤ Upper Confidence Limit
≤ 𝝁 ≤𝑋 - 𝑡(
𝛼
2
,𝑛−1)
(
𝑠
𝑛
) 𝑋 + 𝑡(
𝛼
2
,𝑛−1)
(
𝑠
√𝑛
)

Confidence Limit - Example
The tensioning device (rubber band) used on the Silobuster has come under scrutiny
 Two sets of tensioners are measured and descriptive statistics are run. What is
the 95% confidence interval for the variation?
≤ 𝝁 ≤
Set 1
Mean: 0.250”
Standard Deviation:
0.005”
Sample Size: 25
Set 2
Mean: 0.250”
Standard Deviation:
0.005”
Sample Size: 100
We are 95% confident that the
interval 0.2479 to 0.2521 brackets
the true process standard
deviation (0.0042 width)
𝑿 - 𝒕(
𝜶
𝟐
,𝒏−𝟏)(
𝒔
𝒏
) 𝑿 + 𝒕(
𝜶
𝟐
,𝒏−𝟏) (
𝒔
√𝒏
)
. 𝟐𝟓𝟎 ± 𝟐. 𝟎𝟔𝟑𝟗
. 𝟎𝟎𝟓
𝟐𝟓
= . 𝟐𝟒𝟕𝟗 𝒕𝒐 . 𝟐𝟓𝟐𝟏 . 𝟐𝟓𝟎 ± 𝟏. 𝟗𝟖𝟒𝟐
. 𝟎𝟎𝟓
𝟏𝟎𝟎
= . 𝟐𝟒𝟗𝟎 𝒕𝒐 . 𝟐𝟓𝟏𝟎

Basic Concepts
Confidence Limits For
Standard Deviation

Variation
Confidence Interval
- Lower Confidence Limit ≤ Standard Deviation ≤ Upper Confidence Limit
𝒔
𝒏 − 𝟏
𝒙 𝟐 𝒍𝒐𝒘𝒆𝒓 𝒗𝒂𝒍𝒖𝒆
≤ 𝝈
≤ 𝒔
𝒏 − 𝟏
𝒙 𝟐 𝒖𝒑𝒑𝒆𝒓 𝒗𝒂𝒍𝒖𝒆 Population
Standard Deviation

Variation
The tensioning device (rubber band) used on the Silobuster has come under scrutiny
 Two sets of tensioners are measured and descriptive statistics are run. What is
the 95% confidence interval for the variation?
𝒔
𝒏 − 𝟏
𝒙 𝟐 𝒍𝒐𝒘𝒆𝒓 𝒗𝒂𝒍𝒖𝒆
≤ 𝝈
≤ 𝒔
𝒏 − 𝟏
𝒙 𝟐 𝒖𝒑𝒑𝒆𝒓 𝒗𝒂𝒍𝒖𝒆
𝟎. 𝟎𝟎𝟓
𝟐𝟒
𝟑𝟗.𝟑𝟔
= 0.0039 and𝟎. 𝟎𝟎𝟓
𝟐𝟒
𝟏𝟐.𝟒𝟎
= .0070 𝟎. 𝟎𝟎𝟓
𝟗𝟗
𝟏𝟐𝟖𝟒𝟐
= 0.0044 and 𝟎. 𝟎𝟎𝟓
𝟗𝟗
𝟕𝟑.𝟑𝟔
= .0058
Set 1
Mean: 0.250”
Standard Deviation:
0.005”
Sample Size: 25
Set 2
Mean: 0.250”
Standard Deviation:
0.005”
Sample Size: 100

TEST OF HYPOTHESIS -
roadmap
You want to compare the averages/ medians of samples
of data to decide if they are statistically different
Are samples normally distributed
Compare median values
instead if average
How many samples do you
wan to compare
Kruskall Wallis
Test
For samples that do not
have any outliners
One-way ANOVA
For comparing averages of
three or more samples
against one another
1 Sample t-test
Comparing av. of
one sample against
target
Paired t-test
For comparing averages
of two samples that
contain data that is linked
in pairs
Two Sample t-test
For comparing averages
of two samples against
each other
Mood’s Median
Test
For samples that have
some outliners
Transform Data
Yes
2
1
or
3 or more
No
No
or

EXERCISE
 Represent the following data in
graphical form:
Temperature
100
100
120
120
Response
275
285
270
325
Pressure
250
300
250
300

EXERCISE - continued
a) Determine what parameter settings
yield the largest response.
b) Determine what parameter settings
of pressure would be bets if it were
important to reduce the variability of
the responses that results from
frequent temperature variations
between two extremes.

EXERCISE - continued
0
100
200
300
400
500
600
700
100 120
Pressure = 300
Pressure = 250
Response
Temperature

Design Of Experiments
Design of Experiments (DOE) is a valuable
tool to optimize product and process
designs, to accelerate the development
cycle, to reduce development costs, to
improve the transition of products from
research and development to
manufacturing and to effectively trouble
shoot manufacturing problems. Today,
Design of Experiments is viewed as a
quality technology to achieve product
excellence at lowest possible overall
cost.

Design of Experiments
General Comments
 Keep your experiments simple
 Don’t try to answer all the questions in one study
 Use 2 level designs to start
 Try potential business results to the project
 The best time to design an experiment is after the previous
one is finished
 Always verify results in a follow-up study (
verification)
 Be ready for changes
 A final report is a must to share the knowledge
 Avoid DoE infatuation…do your homework first!
 Measure & Analyze to reduce potential variables
 Use Graphical Analysis
 Use the basic tools of Operational Execllence

Design of Experiments
Be Proactive
 DOE is a proactive tool
 If DOE output is inconclusive:
You may be working with the wrong variables
Your measurement system may not be capable
The range between high and low levels may be sufficient
 There is no such thing as a failed experiment
Something is always learned
New data prompts us to ask new questions and generates
follow-up studies
 Remember to keep an open mind
Let the data/output guide your conclusions
Debunk or validate tribal knowledge
Don’t let yourself be “confused by the facts.”

Types of Experiments
Traditional
Approach
Six
Sigma
Approach
Very
Informal
Very
Formal
• Trial and Error Methods
 Introduce a change and see what happens
• Running Special Lots or Batches
 Produced under controlled conditions
• Pilot Runs
 Set up to produce a desired effect.
• One-Factor-at –a-Time Experiments
 Vary one factor and keep all other factors
constant
• Planned Comparisons of Two to Four Factors
 Study separate effects and interactions
• Experiment With 5 to 20 Factors
 Screening Studies
• Comprehensive Experimental Plan With Many
Phases
 Modeling, multiple factor levels,
optimization
Very
Informal
Very
Formal
• Trial and Error Methods
 Introduce a change and see what happens
• Running Special Lots or Batches
 Produced under controlled conditions
• Pilot Runs
 Set up to produce a desired effect.
• One-Factor-at –a-Time Experiments
 Vary one factor and keep all other factors
constant
• Planned Comparisons of Two to Four Factors
 Study separate effects and interactions
• Experiment With 5 to 20 Factors
 Screening Studies
• Comprehensive Experimental Plan With Many
Phases
 Modeling, multiple factor levels,
optimization

Barriers to Successful DoE’s
 Problem or objective unclear
 Results of the experiments unclear
 Be present during the DoE
 Identify and record unexpected noise or other variables
 Measurement Error
 Lack of Management Support
 Lack of Experimental Discipline
 Don’t use a DoE as the first pass to identify key X’s
 Manage the constants and the noise
 Process map, C&E, Constant or Noise or Experimental
 Unstable process prior to running DoE
 Process map, C&E, Constant or Noise or Experimental,
Manage the C’s and N’s to reduce extraneous variation

Objective
 Establish the objective for the
experiment
It should be stated in such a way to provide
guidance to those involved in designing the
experiment.

Planning the Experiment
 Team in involvement
 Maximize prior knowledge
 Pursue measurable objectives
 Plan the execution of all phases
 Rigorous sample size determination
 Allocate sufficient resources for data
collection and analysis.

The following are some of the objectives of
experimentation in an industry:
 Improving efficiency or yield
 Finding optimum process settings
 Locating sources of variables
 Correlating process variables with
product characteristics
 Comparing different processes,
machines, materials etc.
 Designing new processes and products.

Various Terms Used In
Experimentation
 Factor:
One of the controlled or uncontrolled variables whose
influence on the response is being studied. May ne variable
or classification data.
 Level:
The values or the factor being studied usually high(+) and
low(-)
 Treatment Combination:
An experiment run using a set of the specific levels or each
input variable
 Response Variable:
The variable that is being studied. “Y’ factor in the study.
Measured output variable.
 Interaction:
The combined effect of two or more factors that is observed
which is in addition to the main effect of each factor
individually.

Various Terms Used In
Experimentation
 Confounding:
One or more effects that can not unambiguously be
attributed to a single factor or interaction.
 Main effect:
Change in the average response observed during a
change from one level to another for a single factor.
 Replication:
Replication of the entire experiment. Treatment
combinations are not repeated consequently.
 Test run:
A single combination of factors that yields one or more
observation of the response.
 Treatment:
A single level assigned to a single factor during an
experiment.

Trial And Error
 Perhaps the most well known and used
methodology.
 The objective is to provide a quick fix to a
specific problem.
 The quick fix occurs by randomly and no-
randomly making changes to process
parameters.
 Often changing two or more parameters at the
same time.
 The result often is a “Band-Aid” fix as the
symptoms of the problem are removed, but the
cause of the problem goes undetected.
 In trial and error experimentation, knowledge is
not expanded but hindered.
 Implement multiple expensive fixes are not
necessary.

One-Factor-At-A-Time (OFAT)
The old dogma in experimentation is to hold
everything constant and vary only one-factor-
at-a-time.
◦ Assumes any changes in the response would be due
only to the manipulated factor.
But are they?
◦ Is it reasonable to assume that one can hold all
variables constant while manipulating one?
Experience tells us this is virtually impossible.
Imagine there area large number of possible
factors affecting the response variable:
◦ How long would OFAT take to identify critical factors
and where they should be run for best results?
◦ How much confidence would you have that the
knowledge gained would apply in the real world?

OFAT
Although OFAT may simplify the
analysis of results, the experiment
efficiency given up is significant:
◦ Don’t know the effects of changing one
factor while other factors are changing (a
reality).
◦ Unnecessary experiments may be run.
◦ Time to find casual factors (factors that
affect the response) is significant.

Classification Of Factors
1. Experimental Factors are those which we
really experiment with by varying them at
various levels.
2. Control Factors are those which are kept
at a constant (controlled) level throughout
experimentation.
3. Error or Noise factors are those which can
neither be changed at our will nor can be
fixed at one particular level. Effect of these
factors causes the error component in the
experiment and as such these factors are
termed as error or noise factors.

Experimental Design
Visualization of The 21
Design (2 Levels
– 1 factor)
This is often the method used today for
process optimization. It is the “only one
factor at a time” concept
High
Factor 1
Low

Experimental Design
Design (2 Levels
– 2 factor)
The most basic of true designs. There
are 4 runs.
High
Factor 1
HighFactor 2Low
Low

Experimental Design
Design (2 Levels
– 3 factor)
A little more complicated design but still
very practical. There are only 8 runs.
High
Factor 1
HighFactor 2Low
Low
High
Factor 3
Low

Experimental Design
Factor 4Low High
Design (2
Levels – 4 factor)

Experimental Design
Factor 4Low High
Factor 4
High
Factor 5
Low
Visualization of The 𝟐 𝟓
Design (2 Levels – 5 factors)
Here is where it’s time to stop drawing but it represents the complexity
associated with a 5 factor design.

Experimental Design
Three Factorial Design, without interaction
𝑿 𝟏 𝑿 𝟐 𝑿 𝟑 Y
- - - 𝒀 𝟏
- - + 𝒀 𝟐
- + - 𝒀 𝟑
- + + 𝒀 𝟒
+ - - 𝒀 𝟓
+ - + 𝒀 𝟔
+ + - 𝒀 𝟕
+ + + 𝒀 𝟖

Experimental Design
Three Factorial Design, with interaction
𝑿 𝟏 𝑿 𝟐 𝑿 𝟑 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 𝑿 𝟏 Y
- - - + + + - 𝒀 𝟏
- - + + - - + 𝒀 𝟐
- + - - + - + 𝒀 𝟑
- + + - - + - 𝒀 𝟒
+ - - - + + + 𝒀 𝟓
+ - + - + - - 𝒀 𝟔
+ + - + - - - 𝒀 𝟕
+ + + + + + + 𝒀 𝟖

Design Of Experiments (DOE)
0
10
20
30
40
50
60
70
CSR A CSR B
CALL DURATION 2
CALL DURATION 1
Change or address
(a) Only Interaction, no main effect
Cancel Amount

DOE: Continued
0
10
20
30
40
50
60
70
CSR A CSR B
CALL DUARTION
1
CALL DURATION
2
Change or address
Cancel Amount
(b) Only Main Effect, no Interaction

0
10
20
30
40
50
60
70
CSR A CSR B
CALL DURATION 2
CALL DURATION 1
Cancel Amount
Change or address
(c) Main Effect with Slight Interaction

Example : Full Factorial – Cookie
DOE
 Company HillsBerry produces premium
cookies.
 The company needs to increase throughput
in the bake process by 20% in order to keep
up with demand.
 Market research indicates customers require
the cookies have a taste index rating greater
than 45.
 They are interested in understanding the
effects of two factors, bake time and bake
temperature:
1) Bake Time (A)
2) Bake Temperature (B)
 The current bake process is 10 minutes at

Full Factorial – Cookie DoE
 Objective:
Reduce cook time by 20% while maintaining customer
satisfaction for taste.
 Response Variable:
Taste score (rated by multiple tasters)
Historical data indicates a score above 45 will delight the
customer
 Input Variables:
Bake Time (A).
Bake Time (B).
 What questions should be asked?
 Construct a 𝟐 𝟐 factorial design.
 How should the experiment be run?

Full Factorial – Cookie DoE
DOE Design and Results
Questions?
Which Factors are important?
How should the important factors be set?
Bake Time
(min)
A
6
10
6
10
Taste
41
50
47
35
Bake Temp
(Fº)
B
375
375
450
450

Full Factorial
 Sort by taste and look for trends in the factors
 Practically speaking, what observations can be made about the
data if:
 A change of 10 is required?
 A change of 50 is required?
 What preliminary actions would you recommend right now?
Bake Time
(min)
A
10 (+)
6 (-)
6 (+)
10 (-)
Taste
50
47
41
35
Bake Temp
(Fº)
B
375 (-)
450 (+)
375 (-)
450 (+)

Full Factorial
To begin analyzing the importance of each factor, code
the levels (+) and (-).
Using the (+) and (-) designation will be helpful in
understanding how designs are generated and allow simple
analysis of the DOE.
Bake Time
(min)
A
10 (+)
6 (-)
6 (+)
10 (-)
Taste
50
47
41
35
Bake Temp
(Fº)
B
375 (-)
450 (+)
375 (-)
450 (+)

Analyzing the Full Factorial
To calculate the importance of a factor, called the
“effect”, calculate difference between the average
results at each level of the variable:
Effect = (Av. at + level) – (Av. at – level) for each factor
For this problem, 𝑨 (+) - 𝑨(-) :
𝑨(+) = (50+35)/2 = 42.5
𝑨 (-) = (41+47)/2 = 44.0
Therefore the main effect of A = 42.5 – 44 = -1.5
The main effect of B (temp) is 𝑩(+) - 𝑩(-)
𝑩 (+) = (47+35)/2 = 41
𝑩 (-) = (41+50)/2 = 45.5
The main effect of B = 41- 45.5 = -4.5
Bake
Time
A
Bake
Time
B AB Taste
6 (-) 375 (-) + 41
10 (+) 375 (-) - 50
6 (-) 450 (+) - 47
10 (+) 450 (+) + 35
Notice how the
effects compare
to the graphs

Interactions
Interaction: 2 factors (input variables) interact if one
factor’s effect on the response is dependent upon the
level of other factor.
For Example: Time and temperature in Baking
Cookies
Bake
Time
A
Bake
Time
B AB Taste
6 (-) 375 (-) + 41
10 (+) 375 (-) - 50
6 (-) 450 (+) - 47
10 (+) 450 (+) + 35
Strong interactions are
indicated by lines
crossing at nearly 90˚
(perpendicular). As the
lines become more
parallel, the interactions
are weaker.
B (375)
B (375)
B (450)
B (450)
30
35
40
45
50
55
6 10
B (450)
B (375)
R(taste)
Too Raw
Very Good
Too Done
Interaction Plot
A (Time)
Good
Data Set:

Interactions
The (+) & (-) coding for the interaction can be created
by multiplying columns A and B:
Now we can calculate the importance (effect) of the
interaction of time and temperature on taste for
baking cookies:
To determine the interaction effect use the +,- column
to group the data just like before.
In this case: AB = (41+35)/2 – (50+47)/2 = -10.5
A x B = AB
1 - x - = +
2 + x - = -
3 - x + = -
4 + x + = +
Bake
Time
A
Bake
Time
B AB Taste
6 (-) 375 (-) + 41
10 (+) 375 (-) - 50
6 (-) 450 (+) - 47
10 (+) 450 (+) + 35
Notice how the
effects compare
to the graphs

Analysis Summary
Now that we have calculated the effect for time, temperature and the
time/temperature interaction we can compare the relative importance of
each:
In this experiment, the Time x Temperature Interaction has the largest
effect and is therefore the most important.
Practical Significance:
Controlling Time or Temperature alone will not ensure good tasting
cookies. Time and Temperature should be analyzed and set together to
ensure the best tasting cookies.
Factor Effect
Time 1.5
Temperature 4.5
Time x
Temperature
10.5
B (375)
B (375)
B (450)
B (450)
30
35
40
45
50
55
6 10
B (450)
B (375)
R(taste)
Too Raw
Very Good
Too Done
Interaction Plot
A (Time)
Good

Analysis Summary
Y = Grand average + ((effect A)/2) *A+ ((effect B)/2)*B + ((effect AB)/2)*AB
Taste = 43.24 – ((1.5)/2)*A – ((4.5)/2)*B – ((10.5)/2)*AB
What is the optimum setting of time and temperature that will yield a 20%
improvement in throughput and maintain the taste index above 45?
 Time (factor A) needs to be 8 minutes and Y (taste) needs to be > 45
 Plug into predictive equation, and solve for temperature
 Remember the equation is based on -1 to +1 coded data
Factor Effect
Time (A) -1.5
Temperature (B) - 4.5
Time x
Temperature
-10.5
B (375)
B (375)
B (450)
B (450)
30
35
40
45
50
55
6 10
B (450)
B (375)
R(taste)
Too Raw
Very Good
Too Done
Interaction Plot
A (Time)
Good

Creating The Predictive
Equation
 Solve for time reduction at 80% of 10 minutes = 8
minutes
 Actual time = (hi + lo)/2 + ((hi – lo)/2)*Coded value(A)
 Actual time = (10+ 6)/2 + ((10 – 6)/2)*Coded value(A)
 8 = 8 + 2*(0) coded value
 The coded value for 8 minutes is 0
 Y = 45
 A = 0
 Use the predictive equation and solve for B
(temperature)
 -45 = 43.24 – ((1.5)/2)*((4.5)/2)*B – ((10.5)/2)*(0)*B
 Solve for B
 Coded value for B = -.78
Converting back to actual temperature
Actual temp = (hi + lo)/2 + ((hi – lo)/2)*Coded Value(B)
Actual temp = (450 + 375)/2 + ((450 – 375)/2)*(-.78)
Actual temp = 383 F
8 minutes at 383F will achieve the desired throughput and delight the
customers
Determine the
optimal
settings for
the bake
process

Example 2: 23
 3 factors at 2 levels (Time-A, Temperature-B &
Pressure-C)
 8 runs minimum (2x2x2)
 3 factor columns:
 3 factor columns:
# A B C AB AC BC ABC
1 - - - + +
2 + - - - -
3 - + - - +
4 + + - + -
5 - - + + -
6 + - + - +
7 - + + - -
8 + + + + +
• Complete the
interaction
columns
• How would you
calculate the
importance
(effect) of each
factor?

Full Factorial
The limitations of a full factorial
experiment are not in theory but bin
practically. The resources (i.e. time, cost)
necessary to run full factorial designs can
be significant.
Full factorials can be used when
investigating a small number of variables
(2-4), but are not recommended when
investigating a larger number of factors (5
or more)
The number of runs needed increases
exponentially with the number of factors.
Therefore, the time and costs involved in
running the experiment become
prohibitive.
This does not mean that full factorials are
not useful, but that they should be used at
the proper time (see experiment
Factor
s
Level
s
Symbol Experiments
2 2 𝟐 𝟐 4
3 2 𝟐 𝟑 8
7 2 𝟐 𝟕 128
15 2 𝟐 𝟏𝟓 32,768
2 3 𝟑 𝟐 9
3 3 𝟑 𝟑 27
7 3 𝟑 𝟕 2,187

DOE
1) Define goals/objectives:
 Specific
 Measurable
2) Determine response
variables(s):
 Investigate measurement system
 Consider multiple responses
3) Select factors:
 Basic logical problem solving
techniques (Process map, FMEA,
C/E, ranking etc.)
 Consider the strategy
4) Identify factor levels:
 Realistic (would you want to run at
the selected levels?)
 Move the response variable
5. Choose appropriate design
and assign factors:
 Increase resolution if possible by
properly assigning factors
 Consider where interactive effects
will show up
6. Plan experiment:
 Sampling, replication, repetition…
 Data collection methods
7. Perform Experiment:
 Note observations during the
experiment
8. Analyze results: Six Sigma
rules of Analysis
 Practical
 Graphical
 Analytical
9. Make decisions:
 Goals met?
 Plan next experiment?
 Management presentation?

Summary
 Learning requires two necessary elements: a
significant event and a perceptive observer.
 Learning is accelerated by creating significant
events and observing them. Designed
experiments are most effective.
 Learning is an iterative process.
 We make hypothesis based on our current knowledge
 We test these hypothesis by collecting data and information
 We make deductions based on information collected, and
form new hypothesis.
 Repeat collecting data until we are confident we have
learned what we need to
 Factorial designs help
 They are efficient tests
 Information from factorials may be combined to get better
results faster

Benchmarking
Benchmarking is Quality by
Comparison for achieving better
standards. In the global movement
today, the competition is improving at a
faster rate, and the only way to improve
your relative quality and move upwards
to find out and implement the best
industry practices.

Benchmarking
Benchmarking :-
Stages of Improvement
World Class
Recognized as the best
Benchmarked by others
Efficient
Meets all internal requirements for cost margins,
asset utilization, cycle-time and measures of
excellence
Effective
Satisfies all customer
requirements
Incapable
Is inefficient, ineffective and at the risk of failing.
Needs major redesign
Best in Class
Exceeds customer expectations,
outperforms all competitors and has clear
competitive edge

Benchmarking Methodology
Benchmarking can be of various types –
1. Competitive benchmarking
2. Product benchmarking
3. Process benchmarking
4. Best practices benchmarking
Whatever be the category chosen by the
organization, the benchmarking methodology
remains the same in each case.

A. Identify Processes To
Benchmark
 Select Processes to Benchmark
 Measure current process capability
and define goal.
 Understand detailed process which
needs improvement.

B. Select Organization To
Benchmark
 Select organizations which perform
your process.
 Compile a list of world class process
parameters.

C. Compile the required
Information
 Develop a detailed questionnaire to
obtain detailed information.
 Obtain the desired information. The
information can be obtained from
various sources viz: internet, trade
journals, professional associates,
industry publications, industry experts,
libraries etc.

D. Analyze Gaps
 Too much information is as bad as too
little.
 Gather and analyze only the
information you need to make a direct
comparison of performance.
 Compare like with it.
 Identify the performance gaps and
develop an action plan to close the
gaps.
 Also highlight and quantify the
consequences of not closing the gap.

E. Develop An Action Plan
 Review observations of the gap
analysis.
 Set new performance standards.
 Develop an action plan for meeting the
new performance standards, identify
process owners with their
responsibilities and move to
improvement phase.

Mistake Proofing
Mistake proofing is a scientific
technique for improvement of operating
systems including materials, machines
and methods with an aim of preventing
problems due to human error. The term
“error” means a sporadic deviation from
standard procedures resulting from loss
of memory, perception or motion.

Defect Vs Errors
It is important to understand that
defects and error s are not the same
thing. A defect is the result of an error,
or an error is the cause of defects as
explained below.
ERROR DEFECT
Cause Result

Prevention of Defects
Machine or
human error
Zero
defec
t
Take
action
feedback
Detect
error
Cause
Stop errors from
turning into defects
End ResultIntermediate Result

Type of Errors
 Error in memory of PLAN: Error of
forgetting the sequence/ contents
operations required or restricted in
standard procedures.
 Error in memory of EXECUTION : Errors
of forgetting the sequence/ contents of
operations having been finished.
 Error in PERCEPTION of type : Error of
selecting the wrong object in type of
quantity.

Types of Error
 Error in MOVEMENT : Error in
misunderstanding/ misjudging the shape,
position, direction or other characteristics
of the objects.
 Error in motion of HOLDING : Error in
failing to hold objects.
 Error in motion of CHANGING : Errors of
failing to change the shape, position,
direction or other characteristics of object
according to the standard.

Are Errors Unavoidable?
Traditional view: Errors are inevitable
 People are only human
 There is variation in everything
 Lack of standard operating procedures result
in each person having their own way to do
things
 Inspection is necessary
Six Sigma View: Errors can be eliminated
 Not all errors can be eliminated, but many
can and others can be reduced
 The more errors we can eliminate, the better
our quality
 The need for inspection can be reduced or
eliminated

Basic Functions of Poka Yoke
DEFECTS
WARNING
FLOW
CHART
SHUTDOWN
WARNING
CONTROL
SHUTDOWN
OCCURRED
ABOUT TO
OCCUR
STATE
Normal functions stopped
when defect predicted
FUNCTION
Even intentional errors
are impossible
Signals that abnormality or
errors are about to occur
Normal functions stopped
when defect detected
Defective items cannot
pass on to next stage
Signals that defects
have occurred

Principles of Poka Yoke
Respect the intelligence of workers.
Take over repetitive tasks or actions
that depend on constantly being alert
(vigilance) or memory.
Free a worker’s time and mind to
pursue more creative and value-
adding activities.
It is not acceptable to produce even a
small number of defects or defective
products.
The objective is zero defects.

Ten Types of Human Errors
1. Forgetfulness (not concentrating)
2. Errors due to misunderstanding (jump to
conclusions)
3. Errors in identification (view incorrectly…too far
away)
4. Errors made by untrained workers
5. Willful errors (ignore rules)
6. Inadvertent errors (distraction, fatigue)
7. Errors due to slowness (delay in judgment)
8. Errors due to lack of standards (written & visual)
9. Surprise errors (machine not capable,
malfunctions)
10. Intentional errors (sabotage – least common)
Use Mistake Proofing to Eliminate these

Human Error Provoking
Conditions
1. Adjustments
2. Tooling/ tooling change
3. Dimensionality/ specification/ critical condition
4. Many parts/ mixed parts
5. Multiple Steps
6. Infrequent Production
7. Lack of, or ineffective standards
8. Symmetry
9. Asymmetry
10. Rapid Repetition
11. High volume/ extremely high volume
12. Environment conditions
a. Material/process handling
b. Housekeeping
c. Foreign matter
d. Poor lighting

Three levels of Poka Yoke
1. Catching errors before they create
defects,
2. Catching errors during the process of
creating defects,
3. Catching errors that have created
defects and keeping the defects from
further in the process.

Poka Yoke – How to use it?
Step by step process in applying Poka-Yoke
 Identify the operation or process – based on a
Pareto.
 Analyze the 5-whys and understand the ways a
process can fail.
 Decide the right poka-yoke approach, such as
using a shut out type (preventing an error being
made), or an attention type (highlighting that an
error has been made) poka-yoke.
 Take a more comprehensive approach instead of
merely thinking of poka-yoke as limit switches, or
automatic shutoffs. A poka-yoke can be electrical,
mechanical, procedural, visual, human or any other
from that prevents incorrect execution of a process
step.

Where Poka Yoke works well
 Where manual operations and thereby worker
vigilance is needed.
 Where mis-positioning can occur.
 Where adjustment is required.
 Where teams need common-sense tools and not
another buzz-word.
 Where SPC is difficult to apply or apparently
ineffective.
 Where training cost and employee turnover are
high.
 Where customers make mistakes and blame the
service provider.
 Where special causes can reoccur.
 Where external failure costs dramatically exceed
failure costs.

Poka Yoke advantages
 No formal training programs required.
 Eliminates many inspection
operations.
 Relieves operators from repetitive
tasks.
 Promotes creativity and value adding
activities.
 Results in defect-free work.
 Provides immediate action when
problems arise.

Human Error Provoking
Situations
 Inadequate written standards
 Too many parts
 Mix up
 Too many steps
 Specifications or critical conditions
 Too many adjustments
 Tooling change
 Frequent change
 Environment factors

Principles Of Mistake Proofing
The principles of mistake proofing can
be categorized into groups.
A. Prevention of occurrence
B. Minimization of effects

A. Prevention Of Occurrence
Methods under this principle aim to
prevent the occurrence of human errors
from all stages of operations and make
corrections unnecessary. This can be
done through the following 3 methods :-
 Elimination
 Replacement
 Facilitation

Elimination
Elimination method aims to remove the
system properties which generate
operations/restrictions susceptible to
human errors so as to make them
unnecessary. Consider the error of an
operator touching a high temperature pipe
and getting burnt. One method of
preventing this error is to make the pipe
safe by covering it with insulating material.
This improvement removes the property of
the pipe which generates the restriction.

Replacement
The replacement method aims to use
more reliable methods than operators
and make it unnecessary for operators
to perform such functions as memory,
perception or motion. The use of
specially partitioned parts boxes holding
specified number of parts to prevent the
error of supplying incorrect number of
parts is an example of foolproof method
based on replacement.

Facilitation
The principles of elimination and
replacement described above make it
unnecessary for operators to perform
operations or note restrictions. In
contrast to these, the purpose of
facilitation is to make such functions as
memory, perception and motion
required in operations easy for the
operators, and thus reduce the
occurrence rate of human errors.

B. Minimization Of Effects*
Methods under this principle aim to minimize the
effects of human error and focus on processes
where the error develop into serious problems of
quality, safety or efficiency. This can be divided
into two categories :-
1. Detection and
2. Mitigation
*This Principle is also known as Stop-in-time.

Detection
The concept of detection is that even if a human
error occurs, the deviations from standard states
caused by it can be detected and correlated in the
succeeding operation steps. The methods of
correcting the detected deviations are classified
into two types :-
1. Operators are informed or find by themselves
the deviations and take necessary corrective
actions.
2. The deviations are automatically corrected
without operators.
There are many methods of informing the operator.
The first method is to bring the operator’s notice by
alarms or the remained parts and the second
method is by making the successive operations

Mitigation
The mitigation method aims to make
operations redundant or incorporate shock
absorbing/protecting materials so as to
mitigate the effects of human errors in their
development process.
Example: consider the error of burning out
the motor of a production machine by
forgetting to switch off the power supply.
The mitigation countermeasure is to install
a fuse which cuts off the power supply
when the temperature reaches certain
point.

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