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7QC Tools
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7 QC TOOLS7 QC TOOLS
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Types of 7QC tools
Check Sheet
Pareto Diagram
Cause & Effect diagram
Stratification
Scatter Diagram
Graph &Control charts
Histogram
7 QC Tools
Types
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Check sheetCheck sheet
Check sheet
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What is a check sheet?
Why is a check sheet necessary?
Check sheet
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Check sheets are forms used for
standardizing
checking results of work
verifying and collecting data
Check sheet
Check sheet
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CheckSheet
MeasuredData
CountedData
PrimaryData
Point ScaleData
OrderedData
Indiscrete value such as height,weight, length, time & temp., Etc.
Discrete value such as no. Ofrecording errors, no. ofItem sold
& Rejections etc.
YES / NO or / X - Type
1st, 2nd Order Very Good, Good, No Good
- Type
1 Point, 2 Point etc.
Types of Check Sheet
Check sheet
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Below items can be added , as necessary
1. The purpose of the checks
2. The items being checked
3. The methods of the checks
4. The dates and times of the checks
5. The person to perform the checks
6. The results
Check points for check sheets preparation
Check sheet
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Defect check sheet
Check sheet
Example of check sheet
Month ,day
Component
1
2
34
5
6
7
8
9
10
4/1 2 3 4
No. ofNo. ofdefectsdefects
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You have to cut down your
house expenditure by 20% / month
How will you do it ?
Now lets do an exercise!
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BRAIN STORMING !
Brain storming
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BRAIN STORMING
Brain storming can be conducted in two ways
1. Structured
Every person in a group must give an idea as their turnarises.
Forces even shy people to participate.
Creates a certain amount of pressure to contribute.
Brain storming
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2. Unstructured
Group members simply give ideas as they come to mind. Creates more relaxed atmosphere
Risks domination.
Thumb rule : 5 15 minutes works well
Brain storming
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BRAIN STORMING
Brain storming is a technique to obtain creative ideas froma group of persons in a shortest possible time on an effect.
Brain storming plays an important role to build a cause and
effect diagram
To identify the problem - to identify the causes
To find solution - to prevent problem
Brain storming
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BRAIN STORMING SESSION
Let all the members speak freely and give ideas
Encourage wild ideas
Quantity rather than Quality ideas
Suspend judgment on Good orBad
Ride on anothers ideas
Never criticize other persons opinions
Never prohibit a person from speaking
See the problem from different angles/facets
Write down all the viewpoints
List the cause/ideas
Think of the countermeasures to eliminate the causes
Leader/facilitator need to guide the members in generating ideas
Whenever necessary non members can also be involved
Brain storming
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WHY BRAIN STORMING?
TO IDENTIFY THE PROBLEM
TO IDENTIFY CRITICAL CAUSES
TO FIND THE SOLUTION
TO PREVENT THE PROBLEM
Brain storming
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Cause & EffectCause & Effectdiagramdiagram
Cause & effect diagram
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To identify and systematically list the different causes
that can be attributed to a problem (or an effect)
To identify the reasons why a process goes out of
control
To decide which causes to investigate for process
improvement.
Why Cause & Effect ?
Cause & effect diagram
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What is Effect ?
EFFECT = A Result or an outcome
EFFECT is What happens
Cause & effect diagram
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What is cause ?
CAUSE = Reason orFactor contributing to the EFFECT
CAUSE is WHY it happens
Cause & effect diagram
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The analysis of
why? for what?
is cause and effect
diagram
Cause & effect diagram
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In 1953, Kaoru Ishikawa, Professor of theUniversity of Tokyo, used the Cause & effectdiagram for the first time.A cause & effect diagram is also called a fishbone diagram since it looks like the skeleton
of a fish.
Cause & effect diagram
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Cause & effect diagram
The EFFECT or PROBLEM is stated on the right side of
the diagram and the majorINFLUENCES or CAUSES arelisted to the left.
Main problem
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Cause & effect diagram
2
There are two kinds of cause & effect diagrams:
1. Cause & effect diagrams for identifying causes.
2. Cause & effect diagrams for systematically listing causes
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Cause & effect diagram for identifying the causes
Cause & effect diagram
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More emphasis should be given to this type of cause &effect diagram.
This is because this type of diagram is a great tool to list
ALL the causes.No effort should be made while preparing this diagram toclassify the causes. (For example under the headings ofman, machine, material etc).
Cause & effect diagram
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Special effort should be made to identify as many causes aspossible
The ideal method to identify as many causes as possiblewould be by brainstorming with the team members.
Cause & effect diagram
The classification of the causes should be taken up duringthe preparation of the second cause & effect diagram whichis specifically meant for classification of the causes.
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Remember
All the ideas/causes suggested by
members during the brainstormingsession should be noted howeverinsignificant they may initially appear.Brainstorming is a session speciallymeant for free flow of ideas.
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Effect
Causes
1
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Cause & effect diagram
DimensionalVariation
Poor operator skill
Machinevibration
Wrong inspection method
Wronginspectioninstrument
Improperclamping onjig or fixture
Operatorfatigue
Wrong setting of job on locator
Insufficienttraining
Too muchtightening of job
Improper material storage
No inboundinspection ofraw material Raw material
dimension tooclose to finaldimension
Improper/warped shape of raw material
Wrong spindle speed
Wrongfeed
Example
1
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Cause & effect diagram
Cause & effect diagram for systematically listing causes
The following steps can be followed to make a successfulCause & effect diagram for systematically listing causes
Step 1
List all the causes that have been suggested by teammembers as a part of brain storming.
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Cause & effect diagram
Step 3
Assign an importance to each factor, & mark the particularlyimportant factors which seem to have a significant effect.
Step 4Draw the diagram & continually look for improvement.
Step 2
Sort out the relations among the causes & connect the subcauses to the main causes. The main causes should thenbe connected to the effect.
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Cause & effect diagram
MACHINEMAN
MATERI
AL
METHOD
Fatigue
Dimensional
VariationLocation
JIGS & FIXTURES
STABILITY
INSPECTION
Method
Attentiveness
SPIRIT
Experience
HEALTH
SKILL
Storage
QUALITY
Inspection
ShapeFORM
Dimension
Feed
WORKING
Spindle speed
Illness Training
ConcentrationInstrument
Clamping
Imbalance
Vibration
SETTING
Degree of tightening
Placement on locator
2
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A combination of Pareto diagram & cause and effectdiagram is an ideal way to arrive at the main problem & itscauses.
Take the biggest problem from the pareto diagram & put iton the right side in the cause & effect diagram.
Derive the causes for the same.
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Whydefects?
D
People
1st Why
Plant
PoliciesProcedure
5th Why
C
3rd Why
4th Why
Thorough investigation
of causes
2nd Why
Cause & effect diagram
ASK WHY? 5 TIMES
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Cause Verification
4M Cause Specification Investigation Analysis
Man No focused trainingFunctionwise
trainingGeneric
No method to measure
operator's skills
Skill matrix for
each workmenNo skill matrix
No OJTPractical training
at genbaNo OJT
Workmen not trained in
specific jobs
Need based
training
Common
module given
MaterialModule content is
academic oriented
Content should bespecific need
based
Theory based
Method
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Pareto diagramPareto diagram
Pareto
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Do you remember this? (14th March 2001 - Eden gardens )
Pareto
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Do you remember this? (14th March 2001 - Eden gardens )
Pareto
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Now lets look at the second innings score board:
IndiaSS Das hit wicket b Gillespie 39
S Ramesh c ME Waugh b Warne 30
VVS Laxman c Ponting b McGrath 281
SR Tendulkarc Gilchrist b Gillespie 10SC Ganguly c Gilchrist b McGrath 48
R Dravid run out 180
N R Mongia b McGrath 4
Zaheer Khan not out 23
Harbhajan Singh not out 8
Total 657
Pareto
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When we observe which batsmen got the maximum runswe can see that Laxman & Dravid got 461 out of the 657runs. That is about 70% of the runs.
22% of the 9 batsmen who batted got 70% of the runs!
Lets now observe the Australian 2nd innings score card
Pareto
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Australia
M Hayden lbw b Tendulkar 67M Slater c Ganguly b Harbhajan 43
J Langer c Ramesh b Harbhajan 28
M Waugh lbw b Raju 0
S Waugh c Sub B Harbhajan 24R Ponting c Das b Harbhajan 0
A Gilchrist lbw b Tendulkar 0
J Gillespie c Das b Harbhajan 6
S Warne lbw b Tendulkar 0M Kasprowicz not out 13
G McGrath lbw b Harbhajan 12
Total 212
Pareto
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Who got the wickets?
O M R W
Zaheer Khan 8 4 30 0
V Prasad 3 1 7 0
Harbhajan Singh 30.3 8 73 6
V Raju 15 3 58 1
S Tendulkar 11 3 31 3
S Ganguly 1 0 2 0
Pareto
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When we observe which bowlers got the maximum wickets we
can see that Harbhajan & Tendulkar got 9 out of the 11wickets. That is about 80% of the wickets.
30% of the 6 bowlers who bowled got 80% of the wickets! Thisillustrates the Pareto principle
Paret
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Vilfredo Pareto was an Italian engineer in the 19th Centurywho studied the number of people in various income classes &declared
20% of the people own 80% of the countrys wealth;80% of the people own 20% of the countrys wealth
Pareto
Pareto
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Pareto Principle
Pareto principle holds good to the present day in variousapplications
A fewcauses lead to manydefects;
manycauses lead to fewdefects.
The few causes that lead to many defects are the vital few.
The many causes that lead to few defects are the trivial many.
Pareto
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Get to the biggest problems first
Solve the vital few
Pareto
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1. Collect data
2. Arrange data in the descending order
3. Calculate the relative % for individual data
4. Calculate the cumulative % for individual data
5. Draw a graph with scales on both axis
6. Draw bar chart based on data
7. Using cumulative % data, draw cumulative curve
8. Identify the VITALFEW (thumb rule > 70%)
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2015 12
10 8 6 5 4 4 2 2 1
33
55.5
65.5
73
79
8488
9193.5
95.597.5 98.5
99.5 100
0
25
50
75
100
125
150
175
200
Ma
teria
ls
Marke
ting
Plan
t
Ma
intenance
Finance
Serv
ice
Pro
duc
tion
Eng
ineering
Personne
l
Informa
tion
Sys
tems
Stores
Researc
h&
Deve
lopmen
t
Others
Qua
lity
Fac
tory
pro
duc
tion
Manu
fac
turing
Plann
ing
Dept
Innos
0
25
50
75
100
In%
Pareto
Creating a Pareto Diagram
Steps
7QC Tools Pareto
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Data collection through check sheetPeriod : Week No. 45 To 50
No. Of External Phone CallsSl.No Department No. Of cells regd.
1 Production Engineering 10
2 Quality 2
3 Service 12
4 Marketing 455 Plant Maintenance 20
6 Factory production 2
7 Manufacturing Planning 1
8 Stores 5
9 Personnel 810 Materials 66
11 Finance 15
12 Research & Development 4
13 Information & Systems 6
14 Others 4
STEP 1 6645
2015 12
10 8 6 5 4 4 2 2 1
33
55.5
65.5
73
79
8488
9193.5
95.597.5 98.5
99.5 100
0
25
50
75
100
125
150
175
200
Ma
teria
ls
Marke
ting
Plan
t
Ma
intenance
Finance
Serv
ice
Pro
duc
tion
Eng
ineering
Personne
l
Informa
tion
Sys
tems
Stores
Researc
h&
Deve
lopmen
t
Others
Qua
lity
Fac
tory
pro
duc
tion
Manu
fac
turing
Plann
ing
Dept
Innos
0
25
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75
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In%
Pareto
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Sl.No Department No. Of cells reqd.
1 Materials 66
2 Marketing 45
3 Plant Maintenance 20
4 Finance 15
5 Service 12
6 Production Engineering 10
7 Personnel 8
8 Information Systems 6
9 Stores 5
10 Research & Development 4
11 Others 4
12 Quality 2
13 Factory production 2
14 Manufacturing Planning 1
200
66
45
2015 12
10 8 6 5 4 4 2 2 1
33
55.5
65.5
73
79
8488
9193.5
95.597.5 98.5
99.5 100
0
25
50
75
100
125
150
175
200
Ma
teria
ls
Marke
ting
Plan
t
Ma
intenance
Finance
Serv
ice
Pro
duc
tion
Eng
ineering
Personne
l
Informa
tion
Sys
tems
Stores
Researc
h&
Deve
lopmen
t
Others
Qua
lity
Fac
tory
pro
duc
tion
Manu
fac
turing
Plann
ing
Dept
Innos
0
25
50
75
100
In%
Pareto
STEP 2 Arrange data in the descending orderNo. Of External Phone Calls
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Sl.No Department Nos. Relative %
1 Materials 66 33.0
2 Marketing 45 22.5
3 Plant Maintenance 20 10.0
4 Finance 15 7.5
5 Service 12 6.0
6 Production Engineering 10 5.0
7 Personnel 8 4.0
8 Information Systems 6 3.0
9 Stores 5 2.5
10 Research & Development 4 2.0
11 Others 4 2.0
12 Quality 2 1.0
13 Factory production 2 1.0
14 Manufacturing Planning 1 0.5
200
66
45
2015 12
10 8 6 5 4 4 2 2 1
33
55.5
65.5
73
79
8488
9193.5
95.597.5 98.5
99.5 100
0
25
50
75
100
125
150
175
200
Ma
teria
ls
Marke
ting
Plan
t
Ma
intenance
Finance
Serv
ice
Pro
duc
tion
Eng
ineering
Personne
l
Informa
tion
Sys
tems
Stores
Researc
h&
Deve
lopmen
t
Others
Qua
lity
Fac
tory
pro
duc
tion
Manu
fac
turing
Plann
ing
Dept
Innos
0
25
50
75
100
In%
Pareto
STEP 3 Calculate the relative % for individualNo. Of External Phone Calls
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Sl.No Department Nos. Relative % Cumulative %
1 Materials 66 33.0 33.0
2 Marketing 45 22.5 55.5
3 Plant Maintenance 20 10.0 65.5
4 Finance 15 7.5 73.0
5 Service 12 6.0 79.0
6 Production Engineering 10 5.0 84.0
7 Personnel 8 4.0 88.0
8 Information Systems 6 3.0 91.0
9 Stores 5 2.5 93.5
10 Research & Development 4 2.0 95.5
11 Others 4 2.0 97.5
12 Quality 2 1.0 98.5
13 Factory production 2 1.0 99.5
14 Manufacturing Planning 1 0.5 100.0
200 100
66
45
2015 12
10 8 6 5 4 4 2 2 1
33
55.5
65.5
73
79
8488
9193.5
95.597.5 98.5
99.5 100
0
25
50
75
100
125
150
175
200
Ma
teria
ls
Marke
ting
Plan
t
Ma
intenance
Finance
Serv
ice
Pro
duc
tion
Eng
ineering
Personne
l
Informa
tion
Sys
tems
Stores
Researc
h&
Deve
lopmen
t
Others
Qua
lity
Fac
tory
pro
duc
tion
Manu
fac
turing
Plann
ing
Dept
Innos
0
25
50
75
100
In%
Pareto
STEP 4 Calculate the cumulative % for individualData No. Of External Phone Calls
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Sl.No Department Nos. Relative % Cumulative %
1 Materials 66 33.0 33.0
2 Marketing 45 22.5 55.5
3 Plant Maintenance 20 10.0 65.5
4 Finance 15 7.5 73.0
5 Service 12 6.0 79.0
6 Production Engineering 10 5.0 84.0
7 Personnel 8 4.0 88.0
8 Information Systems 6 3.0 91.0
9 Stores 5 2.5 93.5
10 Research & Development 4 2.0 95.5
11 Others 4 2.0 97.5
12 Quality 2 1.0 98.5
13 Factory production 2 1.0 99.5
14 Manufacturing Planning 1 0.5 100.0
200 100
TRIVIAL
MANY
VITAL FEW
Pareto
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Sl.No Department Nos. Relative % Cumulative %
1 Materials 66 33.0 33.0
2 Marketing 45 22.5 55.5
3 Plant Maintenance 20 10.0 65.5
4 Finance 15 7.5 73.0
5 Others 60 27 100
200 100
Pareto
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Pareto
66
45
2015
54
0
30
60
90
120
150
180
Materials
Marketing
PlantMaintenance
Finance
others
Dept
Inn
os
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66
45
2015
5433
55.5
65.5
73
100
0
30
60
90
120
150
180
Materials
Marketing
PlantMaintenance
Finance
others
Dept
Inn
os
0
25
50
75
100
Cumulative%
Pareto
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Pareto
66
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2015
5433
55.5
65.5
73
100
0
30
60
90
120
150
180
Materials
Marketing
Plant
Maintenance
Finance
others
Dept
Innos
0
25
50
75
100
Cum
ulative%
Vital
Few
70 %
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To Clearly prioritise the magnitude of the problem.
To identify the vital few and trivial many problems.
To find 80/20 rule which states that 80% of the
problems are created by 20% of the causes.
Pareto
Why pareto ?
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1. The most important problem
2. The rate of each problem to the whole
3. The degree of improvement action
4. The comparison of improvement level
5. Before & after remedial action taken
Pareto
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10 8 6 5 4 4 2 2 1
33
55.5
65.5
73
79
8488
9193.5
95.597.5 98.5
99.5 100
0
25
50
75
100
125
150
175
200
Ma
teria
ls
Marke
ting
Plan
t
Ma
intenance
Finance
Serv
ice
Pro
duc
tion
Eng
ineering
Personne
l
Informa
tion
Sys
tems
Stores
Researc
h&
Deve
lopmen
t
Others
Qua
lity
Fac
tory
pro
duc
tion
Manu
fac
turing
Plann
ing
Dept
Innos
0
25
50
75
100
In%Pareto diagram is used to find out
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You have to cut down your
house expenditure by 20% / month
How will you do it ?
Pareto
Same problem, but different approach
ParetParet
oo
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Make a check list of all the expenses
in your home & the amount youspend on these expenses
Pareto
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Sl.No Expense Amount
1 House Rent 3000
2 Electricity Bill 5003 WaterBill 280
4 Cable TV Bill 210
5 News paper bill 120
6 Milkman 300
7 Maid servant 1508 Groceries 2000
9 Entertainment & Lifestyle 1500
10 Travel 200
11 Educational 1500
12 Hospital 20013 Insurance Premium 500
14 Loan repayment 0
15 Clothes 200
16 Petrol 1300
17 Others 300
Pareto
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Arrange these expenses & amounts
in an order, with the highestexpense being the first & lowest
expense being the last
Pareto
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Q
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Sl.No Expense Amount
1 House rent 3000
2 Groceries 2000
3 Entertainment 1500
4 Educational 1500
5 Petrol 1300
6 Electricity bill 500
7 Insurance premium 500
8 Milkman 300
9 Others 300
10 Water bill 280
11 Cable TV 210
12 Travel 200
13 Clothes 200
14 Hospital 200
15 Maid servant 150
16 News paper 120
17 Loan repayment 0
12260
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Q
Q7T/PPT- 65
Calculate the percentage contribution of each of theseexpenses.
Percentage can be calculated by the formula
Individual expense
Total expenseX 100
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Sl.No Department Nos. Relative %
1 House rent 3000 24.47
2 Groceries 2000 16.32
3 Entertainment 1500 12.23
4 Educational 1500 12.23
5 Petrol 1300 10.6
6 Electricity bill 500 4.08
7 Insurance premium 500 4.088 Milkman 300 2.45
9 Others 300 2.45
10 Water bill 280 2.28
11 Cable TV 210 1.72
12 Travel 200 1.63
13 Clothes 200 1.63
14 Hospital 200 1.63
15 Maid servant 150 1.22
16 Newspaper 120 0.98
17 Loan repayment 0 0
12260 100
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Sl.No Department Nos. Relative % Cumulative %
1 House rent 3000 24.47 24.47
2 Groceries 2000 16.32 40.793 Entertainment 1500 12.23 53.02
4 Educational 1500 12.23 65.25
5 Petrol 1300 10.6 75.85
6 Electricity bill 500 4.08 79.93
7 Insurance premium 500 4.08 84.01
8 Milkman 300 2.45 86.46
9 Others 300 2.45 88.91
10 Water bill 280 2.28 91.19
11 Cable TV 210 1.72 92.91
12 Travel 200 1.63 94.54
13 Clothes 200 1.63 96.1714 Hospital 200 1.63 97.8
15 Maid servant 150 1.22 99.02
16 Newspaper 120 0.98 100
17 Loan repayment 0 0 100.0
12260 100 100
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Sl.No Department Nos. Relative % Cumulative %
1 House rent 3000 24.47 24.47
2 Groceries 2000 16.32 40.793 Entertainment 1500 12.23 53.02
4 Educational 1500 12.23 65.25
5 Petrol 1300 10.6 75.85
6 Electricity bill 500 4.08 79.93
7 Insurance premium 500 4.08 84.01
8 Milkman 300 2.45 86.46
9 Others 300 2.45 88.91
10 Water bill 280 2.28 91.19
11 Cable TV 210 1.72 92.91
12 Travel 200 1.63 94.54
13 Clothes 200 1.63 96.1714 Hospital 200 1.63 97.8
15 Maid servant 150 1.22 99.02
16 Newspaper 120 0.98 100
17 Loan repayment 0 0 100.0
12260 100 100
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0
2000
4000
6000
8000
10000
12000
HouseR
ent
Grocerie
s
Ente
rtainment
Educ
atio
nal
Petro
l
Others
Expenses
Amount
0
10
20
30
40
50
60
70
80
90
100
Cumulative%
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Histogram
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In quality control, we try to discover facts by collecting data &then take necessary action based on those facts.
The data is not collected as an end in itself, but as a meansof finding out the facts behind the data.
Data
FACTS
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Histogram shows a bar chart of accumulated data and
provides the easiest way to evaluate the distribution of
data .
The sizes of the vertical bars reflects the number of data
that fall into these spaces.
What is histogram ?
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How to make a histogram?Let us make a histogram using an example.
Example:
To investigate the distribution of thediameters of steel shafts produced inthe grinding process, the diametersof 90 shafts are measured asshown in the table.
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Sample
Number
1 - 10 2.51 2.517 2.522 2.522 2.51 2.511 2.519 2.532 2.543 2.525
11 - 20 2.527 2.536 2.506 2.541 2.512 2.515 2.521 2.536 2.529 2.524
21 - 30 2.529 2.523 2.523 2.523 2.519 2.528 2.543 2.538 2.518 2.53431 - 40 2.52 2.514 2.512 2.534 2.526 2.53 2.532 2.526 2.523 2.52
41 - 50 2.535 2.523 2.526 2.525 2.532 2.522 2.502 2.53 2.522 2.514
51 - 60 2.533 2.51 2.542 2.524 2.53 2.521 2.522 2.535 2.54 2.528
61 - 70 2.525 2.515 2.52 2.519 2.526 2.527 2.522 2.542 2.54 2.528
71 - 80 2.531 2.545 2.524 2.522 2.52 2.519 2.519 2.529 2.522 2.513
81 - 90 2.518 2.527 2.511 2.519 2.531 2.527 2.529 2.528 2.519 2.521
Results of Measurement
Diameter after grinding
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Step 1Calculate the range (R)
Obtain the largest & smallest of observed values & calculate R.
R=
(the largest observed value) (the smallest observed value)
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R = 2.545 2.502 = 0.043
Step 1
Calculate the range (R)Obtain the largest & smallest of observed values & calculate R.
R = (the largest observed value) (the smallest observed value)
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Step 2
Determine the class interval & interval breadth.The class interval is calculated by the formula
Class interval = n where n is total number of observations
Here, n=
90Therefore, n = 9.48.
Rounding to nearest integer,
ClassI
nterval=
9.Interval breadth = =
= 0.005
R
n
0.043
9
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Step 3
Prepare a frequency table form
Prepare a form as shown below on which class, mid point,frequency marks, frequency etc can be recorded
Class Midpoint Frequency marks(tally)
Frequency
Total
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Step 4
Determine the class boundariesInclude so that they include the smallest & the largest ofvalues.
1. Determine the lower boundary of the first class & add
the interval breadth.Therefore, 2.5005 + interval breadth
2.5005 + 0.005 = 2.5005
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Therefore first class boundary
2.005 2.5055
The second class boundary
2.5055 2.5105
Since the class interval is 9, the last class boundary will bethe 9th. Note that this has to contain the largest recordedvalue. Therefore,
9th class boundary
2.5405 2.5455
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Step 5Calculate the mid point of the class
Sum of the upper & lower boundaries ofeach class
2
Mid point of each class =
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Step 6
Obtain the frequenciesRead the observed values one by one & record the frequencies
Falling in each class using tally marks, in groups of five asfollows:
Frequency Frequency notation
1 /
2 //
3 ///4 ////
5 ////
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Class Midpoint Frequency marks
(tally)
Frequency
1
2
34
5
6
78
9
2.5005 2.5055
2.5055 2.5105
2.5105 2.51552.5155 2.5205
2.5205 2.5255
2.5255 2.5305
2.5305 2.53552.5355 2.5405
2.5405 2.5455
2.503
2.508
2.5132.518
2.523
2.528
2.5332.538
2.543
/
////
//// //////// //// ////
//// //// //// //// //
//// //// //// ////
//// ////////
//// /
1
4
914
22
19
105
6
Total 90
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How to draw a histogram
Step 1Mark the horizontal axis with a scale. The scale need not be on the base ofthe class interval. A unit of measurement of data can be used. In thecurrent example we can take 0.01mm of diameter= 10mm on thehistogram scale. Leave a space equal to the class interval on the
horizontal axis on each side of the scale.
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Step 2Mark the left hand vertical axis with a frequency scale.
Step 3
Draw the bar chart as per the data in the frequency table.
Step 4
Draw a line on the histogram to represent the mean, & also draw a
line representing the specification limit, if any.
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0
5
10
15
20
25
2.50 2.51 2.52 2.53 2.54 2.55
X= 2.5247
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The characteristics of the frequency distribution areshown more clearly when results are plotted in form of
block diagram
The horizontal axis is divided into segmentscorresponding to ranges of the group
On each segment a rectangle is constructed whose
height is proportional to the frequency in the group Higher bar represents more data values
Lower bar represents less data values
Overview of Histogram
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To analyze processes and discover items to be improved
To research process capability
To verify effects of an improvement To tell relative frequency of occurrence.
To easily see the distribution of the data.
To see if there is variation in the data. To make future predictions based on the data.
Application
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The shape of the distribution gives a more elusive conceptthan mean or standard deviation
From the distribution we can deduce the peak value offrequency and symmetry of the data range
(i) Normal distribution
Normal distribution is commonly used
type.Here the values are symmetric
about the center
Types of distribution
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(ii) Positively skewed
Values are more concentrated in one
side nearer to origin of x line.
Here most of the values lies in thelower part of the values of histogram
(iii) Negatively skewed
Values are more concentrated in oneside far from the origin
Values lies in the higher part of thevalues of histogram
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StratificationStratification
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Stratification is the act of fine tuning the data in order to
make sure of the significance of the assured factors, to the
grass root level.
Stratification
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Problem : More No. of Accidents
Let us stratify the the data regarding the accidents
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STATISTICSSTATISTICS
REPORTABLE ACCIDENTSREPORTABLE ACCIDENTS :: 0808
NONNON--REPORTABLE ACCIDENTSREPORTABLE ACCIDENTS :: 3333
NEAR MISS INCIDENTSNEAR MISS INCIDENTS :: 2121
LOST TIME INJURIESLOST TIME INJURIES :: 4141
MANDAYS LOSTMANDAYS LOST :: 187187
Rep.acct.Rep.acct. Operator not reporting back to dutyOperator not reporting back to duty
for more than 48hrsfor more than 48hrs
NonNon--reportable acct.reportable acct. Operator disablement extendingOperator disablement extending
beyond the day of shift but lessbeyond the day of shift but less
than 48 hrsthan 48 hrs
Lost time injuryLost time injury Reportable + NonReportable + Non--reportablereportable
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ANALYSISANALYSIS REPORTABLE ACCIDENTREPORTABLE ACCIDENT
Total no of reportable accident :Total no of reportable accident : 88
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ACCORDING TO CATEGORYACCORDING TO CATEGORY
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ACCORDING TO CATEGORYACCORDING TO CATEGORY
Temp.workman
(3)
38%
Regular
Employee (4)
49%
Contract
Labour (1)
13%
Total no.of Reportable accidents :Total no.of Reportable accidents : 88
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ACCORDING TO PHENOMENONACCORDING TO PHENOMENON
Total no.of Reportable accidents :Total no.of Reportable accidents : 88
0
0
0
0
0
0
0
2
6
0 1 2 3 4 5 6 7
wrong assembly
Contact with chemical
Fall from Two wheeler
Fall from Height
others
Hit by objects/Fallen objects
Hit against object
Wrong handling of material handlingequipment
Adjusting/Cleaning/Loading/Unloadingwhile M/C running
No of Accidents
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Total no.of Reportable accidents :Total no.of Reportable accidents : 88
ACCORDING TO BODY PARTS INJUREDACCORDING TO BODY PARTS INJURED
Finger (5)
62%Hand (1)
13%
Leg (2)
25%
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Plant-1 (0)
0%
Plant-2 (3)
37%
Plant-3 (2)
24%
R & D (1)
13%
Sp. Wh (1)
13%
Others (1)
13%
ACCORDING TO PLANTACCORDING TO PLANT
Total no.of Reportable accidents :Total no.of Reportable accidents : 88
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ACCORDING TO SHIFTACCORDING TO SHIFT
4
2 2
0 00
1
2
3
4
5
I II III GEN OT
No.o
facc
iden
ts
Total no.of Reportable accidents :Total no.of Reportable accidents : 88
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ACCORDING TO FAULTACCORDING TO FAULT
3
2
3
0
1
2
3
4
OPERATORS FAULT SUPERVISORY FAULT SYSTEM AND
ENVIRONMENT FAULT
No.of
acc
iden
ts
Total no.of Reportable accidents :Total no.of Reportable accidents : 88
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The data has been stratified
1. According to employee category
2. According to phenomenon
3. According to body parts injured
4. According to plant
5. According to shift
6. According to fault
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Stratification over a sufficient number of units
often gives rise topatterns.
Location of these patterns often contains much
information about the causes of defects
GOAL Generation ofINFORMATION
through DATA ANALYSIS
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Scatter diagramScatter diagram
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In actual practice, it is often essential to study the relation ofTWO corresponding variables.
For example,
* Relation between the dimension of a machined part onthe cutting speed of a lathe
* Relation between insert life and cutting speed
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To study the relation ofTWO variables we can use a Scatter
diagram.
KEY QUESTION How does change in one variable affect
the outcome of second variable
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The two variables we will deal with are:
a) A quality characteristic & a factor affecting it,
b) Two related quality characteristics, or
c) Two factors relating to a single quality characteristic.
Lets consider the steps in making a scatter diagram
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Example
A manufacturer of plastic tanks who made them using theblow moulding process encountered problems with defectivetanks that had thin tank walls. It was suspected that thevariation in air pressure, which varied from day to day, was thecause of the defective thin walls. The table shows data on
blowing pressure & percent defective. Let us draw a scatterdiagram using this data according to the steps givenpreviously.
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Step 1
Collect paired data (x,y) between which you want to studythe relations & arrange the data in a table. It is desirableto have at least 30 pairs of data.
7QC Tools
Data of blowing air pressure & percent defectivef l ti t k
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Date Air pressure Percent
(kgf/cm2) Defective
Oct-01 8.6 0.8892 8.9 0.8843 8.8 0.8744 8.8 0.8915 8.4 0.8746 8.7 0.8867 9.2 0.9118 8.6 0.9129 9.2 0.89510 8.7 0.89611 8.4 0.89412 8.2 0.86413 9.2 0.92214 8.7 0.90915 9.4 0.90516 8.7 0.89217 8.5 0.87718 9.2 0.88519 8.5 0.86620 8.3 0.89621 8.7 0.89622 9.3 0.92823 8.9 0.88624 8.9 0.90825 8.3 0.88126 8.7 0.88227 8.9 0.90428 8.7 0.91229 9.1 0.925
30 8.7 0.872
of plastic tank
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Step 1
As seen in the table, we have 30 pairs of data.
Step 2
In this example, let blowing air pressure be indicated by X(horizontal axis), & percent defective by Y(vertical axis).
Then,
The maximum value ofX:Xmax= 9.4 (kgf/cm2)
The minimum value ofX:Xmin = 8.2 (kgf/cm2)
The maximum value ofY: Ymax= 0.928 (%)
The minimum value ofY: Ymin = 0.864 (%)
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We mark off
the horizontal axis in 0.5(kgf/cm2) intervals, from 8.0 to9.5 (kgf/cm2) and
the vertical axis in0.01(%) intervals, from 0.85 to 0.93(%)
Step 3
Plot the data.
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Step 4
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Step 4
Enter the time interval of the sample obtained (oct.1 oct 30) number ofsamples (n = 30), horizontal axis (blowing air pressure [kgf/cm2]), verticalaxis (percent defective [%]), and title of diagram (scatter diagram ofblowing air pressure & percent defective).
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
8 8.5 9 9.5
(Oct 1 Oct 30)
n=30
Blowing air pressure
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How to read scatter diagrams
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0
5
10
15
2025
30
35
0 5 10 15 20
Series1
0
50
100
150
200250
300
350
0 100 200 300 400
Series1
Positive correlation Negative correlation
How to read scatter diagrams
You can grasp the correlation between pairs of data just by
looking at the shape of a scatter diagram. 5 examples aregiven below
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0
10
20
30
40
0 5 10 15 20
Series1
0
100
200
300
400
500
0 100 200 300 400
Series1
Positive correlation may be present Negative correlation may be present
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0100
200
300
400
500
600
700
0 100 200 300 400
Series1
No correlation
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If a relation is shown between two variables,
does itNECESSARILY mean that the two
variables ARE related?
GOAL Generation ofINFORMATION through DATAANALYSIS