7 quality control tool / qc
TRANSCRIPT
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By:Maruti Center for Excellence
(MACE)On 20th June’2008
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ULTIMATE GOAL OF AN
ORGANIZATION
- Making Profits- Survival &- Growth
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SURVIVAL AND GROWTH:
HOW?- Products & Services must be
preferable to Customers overCompetitors.
- Organization must continually meetNeeds & Expectations of Customersi.e. Voice of Customers also called
QUALITY
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WHAT IS A PROBLEM?
A “problem” is the gap between the presentSituation And the ideal situation or objective
C o n t r o l
c h
a r a c t e r i s t i c
Good
Ideal situation or objective
Gap Problem (ideal situation or objective) -=(present level)
Present Level
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PROBLEM DEFINED …
Deviation from the expectation- In maintaining the status quo.
- In improving the status quo
An opportunity for improvement
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The Problem-Solving Process
Expose problem
Experience,intuition, nerve,
inspiration
Implementcountermeasures
Expose problem
Analyze causes
Implementcountermeasures
• Grasp problem• Set target• Identify gap betweenexisting situation and target
• Investigate cause
• Plan countermeasures• Implement countermeasure• Institutionalize
1. The conventional problemsolving approach
2. The QC problem-solvingapproach
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TOOLS ARE USED TO IDENTIFY, ANALYSEAND RESOLVE PROBLEMS
TOOLS ARE SIMPLE, VERY POWERFUL
AND HELP TO IDENTIFY THE CAUSES FOR
WORK RELATED PROBLEMS AND TO FIND
SOLUTIONS FOR THE SAME IN A
SYSTEMATIC MANNER.
WHAT ARE SEVEN QC TOOLS :
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1 Cause and Effect Diagram 2 Graphs / Flowcharts
3 Pareto analysis 4 Checksheets
5 Control charts
6 Scatter diagram
7 Histogram
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PARETO DIAGRAM
• Prioritisation Tool – It tells where to concentrate first.
• Vilfredo Pareto (1848-1923) Italianeconomist – 20% of the population has 80% of the
wealth
• Juran used the term “vital few, trivialmany”. He noted that 20% of the quality
problems caused 80% of the dollar loss.
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PRINCIPLE OF PARETO
Isolate vital few from trivial many.
80/20 principle. 80% improvement can be achieved by
working on 20% of the causes.
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WHAT IS PARETO DIAGRAM :
It is a column graph. Differentiating major
factors contributing to problem from other
factors which have less contribution. Thus it
helps fixing priority to take first. OR
A technique to segregate vital few from
trivial many
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DATA TALLY SHEET
DEFECT TALLY Total
A //// //// //// ....... //// 150
B //// //// //// …… //// 60
C //// //// …. //// 45D //// //// …. //// 30
E //// //// 9
F //// / 6TOTAL No.of defects
300
Total No. of defectives : //// //// //// ……. //// 150
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What is a defective : A unit that contains at least one defect.
What is a defect :
An output of a process that does not meeta defined specification.
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DATA SHEET FOR PARETO DIAGRAM
S.No.
Defect No. ofdefects
Cumulativetotal
Percentageof overall
total
Cumulativepercentage
1) A 150 150 50 50
2) B 60 210 20 70
3) C 45 255 15 85
4) D 30 285 10 95
5) E 9 294 3 98
6) F 6 300 2 100
Total 300 - 100 -
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Pareto diagram
6930
4560
150
100989585
70
50
0
50100
150
200
A B C D E F
Type of defect
N o .
o f d e f e c t s
0
20
40
60
80
100
C u m m u
l a t i v e
%
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DATA TALLY SHEET
DEFECT TALLY TotalFlow mark //// //// //// ....... //// 200Sink Mark //// //// //// …… //// 100Warpage //// //// …. //// 50
Silver streak //// //// …. //// 25Crack //// //// //// 15Flash //// //// 10
TOTAL 400
Total No. of defectives : //// //// //// ……. //// 250
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DATA SHEET FOR PARETO DIAGRAM
S.No.
Defect No. ofdefects
Cumulativetotal
Percentageof overall
total
Cumulative
percentage
1) Flow
mark200 200 50 50
2) Sinkmark
100 300 25 75
3) Warpage
50 350 12.5 87.5
4) Silverstreak 25 375 6.25 93.755) Crack 15 390 3.75 97.5
6) Flash 10 400 2.5 100
Total 400 - 100 -
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Pareto diagram
200
100
5025 15 10
10097.593.75
87.575
50
050
100
150200250
F l o w m a r k
S i n k m
a r k
W a r p a
g e
S i l v e r s t r e a
k C r a
c k F l a s h
Type of defect
02040
6080100
C u m m u
l a t i v e %
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• Shows the relationship between a problemand its possible causes.
• Developed by Kaoru Ishikawa (1953)
• Also known as … – Fishbone diagram – Ishikawa diagram
CAUSE AND EFFECTDIAGRAM
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WHAT IS CAUSE AND EFFECT DIAGRAM :
A systematic arrangement of all possible causes
which give rise to the effect are made. The
causes are first divided into major sources (4Ms)
i.e., MAN, MACHINE, METHOD & MATERIAL.
Then each source is divided into sub-sourcesand
so on. It helps to find out the root cause of the
roblem.
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METHODOLOGY : How to reach to rootcause of the problem
Step 1 : Make fish bone diagram and write all possiblecauses in 4 M’s after brainstorming session.
Step 2 : Find out suspect or potential causes from thepossible causes. Highlight it.
Step 3 : Validate all the suspect or potential causes andwrite the judgement. All the NG judgements are the
main causes of the problem.
Step 4 : Do the why why analysis of the main causes andfind out the root causes of the problem.
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METHODOLOGY : Step 1 : Possible
causes
Step 2 : Suspect orPotential causes
Step 3 : Maincauses
Step 4 : Root causes
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Cause & Effect diagram - Major and subsidiary causes
Material Methods Environment
Men MeasurementMachine
Quality
Assemblies
Components
Suppliers
Consumables
Procedures
Policies
Accounting
Noise level
Humidity
Temperature
Lighting
Training
Experience
Skill
Attitude
Variability
Tooling
Fixtures
Technology
Instruments
Gauging
Counting
Tests
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CAUSE & EFFECT DIAGRAM
POORWASHINGQUALITY
TRAINING
WORK LOAD
CAR SHAMPOORUBBING CLOTH
MATERIALMETHOD
MACHINE MAN
JOB SKILL
PUMP PRESSURE
PROCESSTIME
NOZZLE DIAEXPERIENCE
A P
DC
WATER
PROCESS SEQUENCE
POLISH QUALITYBRUSHES
VACUUM CLEANER
LAYOUT
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VALIDATION OF POTENTIAL CAUSES
S.No.
Potential Cause Validation Conclusion
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SAMPLE SIZE FOR VALIDATION
Formulae : np = 5 or n= 5 / p
Where n= number of sample sizep= no. of defects coming out of
100Example : Suppose 1% defect is there,
then sample size will ben= 5*100 / 1 = 500
5 WHY FOR ROOT CAUSE
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5 WHY FOR ROOT CAUSEASK “WHY” FIVE TIMES.
When a problem has just been discovered, think about it by asking the
question ―why?‖ five times. A haphazard idea of the cause cannot becounted on. Thinking it over repeatedly will discover the root cause.
For example , imagine that a bracket has broken from a pipe. 1. Why did the bracket broken?( It was not welded properly)
2. Why wasn’t it welded properly?( The bracket wasn’t set in the proper position when it was welded)
3. Why wasn’t the bracket placed in the proper position for welding?
( The welding jig was loose)
4. Why was the welding jig loose?( The welding jig is worn)
5. Why is the welding worn?( The material specification of the jig was not proper)
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WHAT IS A SCATTER DIAGRAM :
A graphical technique to show the dependency oftwo
variables.
It is used to study the variation of two
corresponding variables. For example, what
extent surface finish of a machined part be
varied by the change in speed of a lathe.
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SCATTER DIAGRAM - Exampleof Positive Correlation
0
5
10
15
0 5 10 15
X - Axis
Y -
A x i s
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SCATTER DIAGRAM - Example
of Negative Correlation
0
5
10
15
0 5 10 15
X - Axis
Y -
A x i s
Scatter Diagram
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Scatter Diagram
variation of strength with varying air
pressure
1
1.2
1.4
1.6
1.8
3.5 4 4.5 5 5.5 6
Air pressure
S t r e n g
t h
02-Mar 03-Mar 04-Mar 05-Mar 06-Mar
EXAMPLE OF SCATTER DIAGRAM
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Two variables :
a) Speed of the car in kms / Hr.b) Petrol consumption in kms / Lt.
Speed ofthe car inkms / Hr.
Petrolconsumption in kms /Lt.
30 15, 15.5
35 16, 16.5
40 17, 17.5
45 18, 18.550 19, 19.5
55 20, 20.5
60 22, 22.5
65 21, 21.5
70 20, 20.4
75 19, 19.6
80 18, 18.6
85 18, 18.5
90 17, 17.4
95 16, 16.2
100 16, 16.1
CAUSE
E F F E C T
Here
Cause : Speed of the car in kms / Hr.
Effect : Petrol consumption in kms / Lt.
EXAMPLE OF SCATTER DIAGRAM :
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16
16.1
16
16.2
17
17.418
18.5
18
18.6
19
19.6
20
20.421
21.5
22
22.5
20
20.5
19
19.5
18
18.5
17
17.5
16
16.5
15
15.5
1012
14
16
18
20
22
24
20 30 40 50 60 70 80 90 100 110
SPEED IN KMS / HR.
F
U E L A V E R A G E
I N K M S / L T
EXAMPLE OF SCATTER DIAGRAM :
Positive and negative co-relation
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OTHER EXAMPLES OF SCATTER DIAGRAM :
In machining process :1) Cause : Speed of the machine and
effect : surface finish of thecomponent.
2) Cause : Feed of the machine andeffect : surface finish of the component
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WHAT ARE CONTROL CHARTS :
Control charts serve to detectabnormal trends with the help of linegraphs.
Control charts differ from standardline graphs as they have control limitlines at the center, top and bottomlevels.
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WHAT IS X CHART
X chart shows the centering of theprocess,i.e. it shows the variation in theaverage of samples. It is the most
commonly used variable chart.
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WHAT IS R CHART
R chart shows the uniformity orconsistency of the process i.e. it
shows the variation in the range ofsamples.
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Diameter of Shaft: 23.75 + 0.1 mm
No. of samples : 6
The diameter of shafts are as given below :
1 st day 2 nd day 3 rd day 4 th day 5 th day 6 th day 7 th day 8 th day
23.77 23.80 23.77 23.79 23.75 23.78 23.76 23.76
23.80 23.78 23.78 23.76 23.78 23.76 23.78 23.79
23.78 23.76 23.77 23.79 23.78 23.73 23.75 23.77
23.73 23.70 23.77 23.74 23.77 23.76 23.76 23.72
23.76 23.81 23.80 23.82 23.76 23.74 23.81 23.78
23.75 23.77 23.74 23.76 23.79 23.78 23.80 23.78
Diameter of Shaft: 23.75 + 0.1 mm
No. of samples per day : 6
The diameter of shafts are as given below :
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Construct the X and R chart:
Average diameter for the first day
X1 = X 1+X 2+X 3+X 4+X 5+X 6
6
= 23.77+23.80+23.78+23.73+23.76+23.75
6= 23.765
Similarly, the averages for each day are calculated:
X1 X2 X3 X4 X5 X6 X7 X8
23.765 23.77 23.7716
23.7767
23.7717
23.7583
23.7767
23.7667
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Now X = X = 190.1567 = 23.7696
N 8
Ranges :
M
R1 R 2 R3 R 4 R5 R6 R7 R 8
.07 .11 .06 .08 .04 .05 .06 .07
R = R = 0.0675
N
For X chart :
UCL X =X + A 2R
=23.7696 + 0.48 x 0.0675
(A2= 0.48 for subgroup of from table )
M
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= 23.7696 + 0.0324 = 23.802LCLX =X - A2R
= 23.7696 – 0.0324 = 23.7322For R chart :
UCL R = D 4R
= 2 x 0.0675 = 0.1350LCR R = D 3R
= 0 (D 3 = 0 for subgroup of 6 or less)
Process capability : Sigma = R / d2 = 0.0675 / 2.534 = 0.0266
( for subgroup of 6, d2 = 2.534
Xmax = (USL) upper specification limit, Xmin= (LSL) lower
specification limit
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• Xmax – Xmin = 0.2 mm from data• Process capability Cp = (USL – LSL)/ 6
sigma• Cp= 0.2 / 6* 0.0266 = 0.2 / 0.15982 =
1.25
• Cpk = (USL – X ) / 3 sigma or (X – LSL) / 3 sigma• Cpk = (23.85 – 23.7696)/ 3*0.0266=1.0• Cpk = (23.7696 – 23.65)/ 3*0.0266 =
1.49• Cpk = 1.0 or 1.49 ( 1.0 is minimum )
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X CHART
UCL —23.802
x—23.7696
LCL—23.7322
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R chart
UCL – 0.1350 R1 R2 R3 R4 R5 R6 R7 R8
R – 0.0675
LCL – 0
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No. I II III IV 1 1.38 1.42 1.42 1.42 1.38 1.39 1.42 1.41
3 1.41 1.41 1.39 1.384 1.36 1.45 1.41 1.395 1.36 1.42 1.46 1.376 1.38 1.45 1.39 1.407 1.40 1.40 1.41 1.398 1.40 1.41 1.38 1.399 1.40 1.40 1.40 1.39
10 1.41 1.41 1.39 1.38
11 1.41 1.40 1.39 1.4212 1.44 1.34 1.36 1.3813 1.37 1.43 1.41 1.3814 1.38 1.44 1.42 1.4015 1.39 1.46 1.40 1.4016 1.37 1.47 1.40 1.4017 1.38 1.42 1.45 1.3718 1.39 1.39 1.39 1.4519 1.38 1.44 1.46 1.3720 1.32 1.40 1.41 1.40
DIMENSION : 1.4 + / - 0.3
Draw X bar and R chart
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No. I II III IV X 1 1.38 1.42 1.42 1.4 1.4052 1.38 1.39 1.42 1.41 1.4
3 1.41 1.41 1.39 1.38 1.39754 1.36 1.45 1.41 1.39 1.40255 1.36 1.42 1.46 1.37 1.40256 1.38 1.45 1.39 1.40 1.4057 1.40 1.40 1.41 1.39 1.48 1.40 1.41 1.38 1.39 1.3959 1.40 1.40 1.40 1.39 1.397510 1.41 1.41 1.39 1.38 1.397511 1.41 1.40 1.39 1.42 1.40512 1.44 1.34 1.36 1.38 1.3813 1.37 1.43 1.41 1.38 1.397514 1.38 1.44 1.42 1.40 1.4115 1.39 1.46 1.40 1.40 1.4125
16 1.37 1.47 1.40 1.40 1.4117 1.38 1.42 1.45 1.37 1.40518 1.39 1.39 1.39 1.45 1.40519 1.38 1.44 1.46 1.37 1.412520 1.32 1.40 1.41 1.40 1.3825
N l l t th R
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Now we calculate the Range:-
Range:- Highest value of sample – Smallest value of sampleFor example:-
R1 1.42 – 1.38
0.04
Similarly we calculate for all the samples.
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No. I II III IV X R 1 1.38 1.42 1.42 1.4 1.405 0.042 1.38 1.39 1.42 1.41 1.4 0.04
3 1.41 1.41 1.39 1.38 1.3975 0.034 1.36 1.45 1.41 1.39 1.4025 0.095 1.36 1.42 1.46 1.37 1.4025 0.16 1.38 1.45 1.39 1.40 1.405 0.077 1.40 1.40 1.41 1.39 1.4 0.028 1.40 1.41 1.38 1.39 1.395 0.039 1.40 1.40 1.40 1.39 1.3975 0.0110 1.41 1.41 1.39 1.38 1.3975 0.0311 1.41 1.40 1.39 1.42 1.405 0.0312 1.44 1.34 1.36 1.38 1.38 0.0113 1.37 1.43 1.41 1.38 1.3975 0.0614 1.38 1.44 1.42 1.40 1.41 0.0615 1.39 1.46 1.40 1.40 1.4125 0.07
16 1.37 1.47 1.40 1.40 1.41 0.117 1.38 1.42 1.45 1.37 1.405 0,0818 1.39 1.39 1.39 1.45 1.405 0.0619 1.38 1.44 1.46 1.37 1.4125 0.0920 1.32 1.40 1.41 1.40 1.3825 0.09
M
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Now Calculate X XN
X1 + X2 + X3+ X4 + X5 + X6……………………..X19 +X20 20
1.405 + 1.4 + 1.3975 + 1.4025 + 1.4025 + 1.4……………………….1.4125 + 1.3825 20
1.4011
Now Calculate R RN
R1+ R2+R3+R4………….R19+R20 20
0.04 + 0.04 + 0.03 + 0.09……………………………..0.06 + 0.09 + 0.09 20
0.06
M
M
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Calculation For Average (X-bar) Chart
Upper Control Limit(UCL) = X +A 2 R
= 1.4011+0.73*0.06= 1.4449= 1.44
Lower Control Limit (LCL) =X - A 2 R
= 1.4011 - 0.73 *0.06= 1.3573= 1.36
For future control it is advise to set theprocess at target value i.e., (USL + LSL)/2.
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LCL = D 3 X R= 0 * 0.06= 0
UCL = D 4 X R= 2.28 * 0.06
= 0.1368= 0.14
Calculation for Range(R) Chart
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THE X-bar & R CHARTS 1.44
1.40
1.36
0.14
0.06
0
THE X-bar CHART
THE R CHART
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1.44
1.36X-bar CHART
R CHART
1.40
0.14
0.06
0
The Control Chart
TWO KINDS OF VARIATIONS
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WO DS O V O S
1) Variation due to chance causes
2) Variation due to assignable causes.1) Variation due to chance causes:Variations due to chance causes are inevitable in
any process or product. They are difficult to trace anddifficult to control even under best conditions ofproduction. Since these variations may be due tosome inherent characterstic of the process ormachine which functions at random. For example, alittle play between nut and screw at random maylead to back-lash error and may cause a change indimension of a machined part.
2) Variation due to assignable causes:
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2) Variation due to assignable causes:These variations possess greater magnitude
as compared to those due to chancecauses and can be easily traced ordetected. The variations due to assignablecauses may be because of the followingfactors:
a) Differences among machines.b) Differences among workersc) Differences among materialsd) Change in working conditions
d
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How to read CONTROL CHARTS :
Whether a process is in the controlled state or not is
judged by the following criteria from the controlchart.
Process not in control :
a) Point out of control limit.b) Seven points on one side of the average (run)c) Seven points in a row continously increasing
and decreasing (trend).d) Points very close to control limits and close to
average.e) When the curve repeatedly shows an up and
down trend for the same interval (periodicity).
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• A diagram that graphically depicts the
variability in a population.
WHAT IS HISTOGRAM :
WHAT IS HISTOGRAM :
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WHAT IS HISTOGRAM :
The frequency data obtained frommeasurements display a peak around acertain value. The variation of qualitycharacterstics is called distribution. The
figure that illustrates frequency in the forma pole is referred to as a Histogram.
POPULATION AND SAMPLE
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POPULATION AND SAMPLE
• The entire set of items is called thePopulat ion.
• The small number of items taken from thepopulation to make a judgment of thepopulation is called a Sample .
• The numbers of samples taken to make this judgment is called Sam p le si ze.
SAMPLE OF
SIZE THREEPOPULATION
Histogram steps
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Histogram – steps
1.Obtain a set of 50 ~ 100 observations as shownbelow:SampleNumber
Results of Measurement
1-10 2.510 2.517 2.522 2.522 2.510 2.511 2.519 2.532 2.543 2.525
11-20 2.527
2.536 2.506 2.541 2.512 2.521 2.521 2.536 2.529 2.524
21-30 2.529
2.523 2.523 2.523 2.519 2.538 2.543 2.538 2.518 2.534
31-40 2.520
2.514 2.512 2.534 2.526 2.532 2.532 2.526 2.523 2.520
41-50 2.535
2.523 2.526 2.525 2.532 2.530 2.502 2.530 2.522 2.514
51-60 2.533
2.510 2.542 2.524 2.530 2.535 2.522 2.535 2.540 2.528
61-70 2.52 2.515 2.520 2.519 2.526 2.542 2.522 2.542 2.540 2.528
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Histogram – steps 2. Obtain the maximum value and minimum value:SampleNumber
Results of Measurement Maximumvalue of theline
Minimum value of theline
1-10 2.510 2.517
2.522
2.522
2.510
2.511
2.519
2.532
2.543
2.525
2.543 2.510
11-20 2.527 2.536
2.506
2.541
2.512
2.521
2.521
2.536
2.529
2.524
2.541 2.506
21-30 2.529 2.523
2.523
2.523
2.519
2.538
2.543
2.538
2.518
2.534
2.543 2.518
31-40 2.520 2.514
2.512
2.534
2.526
2.532
2.532
2.526
2.523
2.520
2.534 2.512
41-50 2.535 2.523
2.526
2.525
2.532
2.530
2.502
2.530
2.522
2.514
2.545 2.502
51-60 2.533 2.510
2.542
2.524
2.530
2.535
2.522
2.535
2.540
2.528
2.542 2.510
61-70 2.525 2.515
2.520
2.519
2.526
2.542
2.522
2.542
2.540
2.528
2.542 2.515
71-80 2.531 2.545
2.524
2.522
2.520
2.519
2.519
2.529
2.522
2.513
2.545 2.513
81-90 2.518 2.527
2.511
2.519
2.531
2.527
2.529
2.528
2.519
2.521
2.531 2.511
The largest value2.545
The smallest value 2.502
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Histogram- Steps3. Determine the number of classes: Let these be between 5-12,
Say it is “K”. Generally to decide the number of classes, dividethe range by 1,2 or 5 (their fractions or multiples).
In this case, the range is 0.043 and if it divided by 0.002 or 0.005or 0.010, we will get
- 0.043 / 0.002 = 21.5 ~ 22- 0.043 / 0.005 = 8.6 ~ 9- 0.043 / 0.010 = 4.3 ~ 4Therefore, number of interval of classes be taken as 9.
4. Determine class width rounded off to a convenient figure. Sothat it covers maximum and minimum value both
C = Max- Min = 0.043 / 9 ~ 0.005K
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Histogram- Steps
5. Calculate the class boundaries so that itcovers minimum and maximum value both.In the case, let the boundary be 2.5055.
Note: First class boundary should containssmallest value and boundary value falls onhalf of the unit of measurement .
6. Calculate the mid point of first class by sumof upper and lower boundaries of first classi.e. (2.5005 + 2.5055)/2 = 2.503. Mid point ofsecond class shall be (2.5055+2.5105)/2=2.508 and so on.
Histogram – Steps
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g p7. Make a frequency table as given below:
Class Mid-Point of class x Frequency
1 2.5005-2.5055 2.503 1
2 2.5055-2.5105 2.508 4
3 2.5105-2.5155 2.513 9
4 2.5155-2.5205 2.518 14
5 2.5205-2.5255 2.523 22
6 2.5255-2.5305 2.528 19
7 2.5305-2.5355 2.533 10
8 2.5355-2.5405 2.538 5
9 2.5405-2.5455 2.543 6
Total ……… 90
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Histogram-Steps
8. Mark the horizontal axis with the classboundary values.
9. Mark the vertical axis with a frequencyscale.
10. Erect the rectangles over the classinterval having area proportion to thefrequencies.
11 Draw a line on the Histogram to representMean, number of data points and standarddeviation.
Types of Histogram
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Types of Histogram
0
5
10
15
20
25
N=90Mean=2.5247
S.D=0.00906
2.51 2.52 2.53 2.54
2.5247
Histogram of shaft Dia
Shaft Dia
F r e q u e n c y
Normal Distribution
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ử 3o99.7%
ử 2o95.4%
ử o68.3%
ử -3o ử -2o ử -o ử ử+o ử+2o ử+3o
Normal Distribution
USLLSL
Out of
Spec.
Out of
Spec.
o
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Histogram for grade wisedistribution in a class
10
23
35
25
15
5
C- C B B+ A A +
Grade
N o .
o f s
t u d
e n
t s
WHAT IS CHECK SHEET :
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WHAT IS CHECK SHEET :A check sheet is a paper form on which items
to bechecked have been printed so that data can
be
collected easily and concisely. Its mainpurpose is
twofold.• To make data gathering easy• To arrange data automatically so that they
can be used easily later on.
EXAMPLE 1 OF CHECK SHEET
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Shown a check sheet used in final inspectionprocess of a certain molded plastic product. Atthe end of the day we can immediately calculatethe total number and types of defects that haveoccurred.
EXAMPLE -1 OF CHECK SHEET
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Product: Date
Manufacturing stage: final insp. SectionType of defect: sink mark, silver Inspector’s name Streak,flow mark, crack, flash Lot no:.Total no. inspected: 1525
Remarks: all items inspected Order no:.
Type Check Subtotal
Sink markSilver streak
Flow markCrackFlash
//// //// //// // //// //// /
//// //// //// //// //// / /// ////
1711
2635
Total defects 62
Total Rejects //// //// //// //// //// //// //// //// // 42
Defective Item Check Sheet
CHECK SHEET
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It is necessary to decide clearly how to record thedefects. When these are found in a product. We
shouldgive proper instructions to the staff regarding the
formatin which the defects are to be gathered. In this
case, 42out of 1525 components were found defective.
Howeverthe total nos of defects was 62 because two or
more
defects were found on the same piece.
EXAMPLE -2
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Check Sheet
Product : Date :
Manufacturing stage : Final Inspection Section :
Type of defect :scratch, dents, door gap uneven Inspector's name :
Total no. inspected : 1500
Remarks : all itemsinspected
Type of defect Tally Mark Frequency
Scratch IIII IIII IIII 15
Dents IIII IIII 10 Door gap uneven IIII 4
Others III 3
Total No. Of Defects 32
Total No. Of Defectives IIII IIII IIII IIII 20
h k h
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Check Sheet
Purpose of data recording : Variation in the Quality Characteristics of a proces
• Applicable for Histogram, Run chart / Control Charts
Difference between Groups/ Batches/ Machines• Applicable for Pareto Diagram / Bar-chart.
Relationship between two characteristics• Applicable for Scatter Diagram
Flow Chart / Graphs
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Flow Chart / Graphs
Flow Chart :
A Tool that graphically represents the stepsof a process
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Flow Chart
Different icons/symbols to indicate thedifferent types of actions in the process.
Start / End :Process Activities :Decision points :Movement :Storage :
Fl Ch t f li
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Flow Chart of annealing process:
Annealing Process :
Rodreceipt
Start
Storage
Testing
Inspection
Storage
Base & BatchPreparation
Heating
EndPurging
FurnaceLoading
Parameter
setting
Soaking
Controlledcooling
Unloading
Controlledcooling
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GRAPH : It is one of the meansof data stratification.
• Bar Graph
• Line Graph• Pie Graph• Radar Graph
Purpose of Graph: A picture is worth morethan thousand words
BAR GRAPH : A graph to compare the
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18.8
12.8
21.6
8.6
14.8 13.69.8
0
5
10
15
20
25
R E A R B O D Y
C T R P L L R
F R O N T
P A N E L
R O O F
F R
. D O O R
R R
. D O O R
B . D
O O R
g p pdifference in numeric quantity.
% O
F D E N T S
DENT ANALYSIS
Pie Chart) :
A graph for the proportion of the different classifications
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AREA OMNI W/B PERIOD 05/01/2001 NO.OF.VEHICLE 150
Gun TouchDents 25%
5
20% Spot Dent
4
Others 10%
2
HandlingDents 45%
9
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SYSTEM AUDIT REPORT
A dit Points % Achie ed
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Audit Points % Achieved
1. Production preparation 40%
2. Initial supply control 60%
3. Initial control- Changed parts 40%
4. Preventive measures for defects 70%
5. Education and training 80%
6. Quality audit 80%
7. Supplier control 70%
8. Control of drawings and engg. Changes 100%
9. Inspection standard & PCS 100%
10. Operation standard 100%
11. Observance of operation standard 100%
12. Role of Manager/ Supervisor 100%
13. Quality improvement of process 0%
14. Control of Manufacturing machine/ jig 50%
15. Control of Inspection equipment and Jig 40%
16. Statistical method 100%
17. Prevention of missing process/ wrong assembling 75%
18. Control of non conforming product 100%
19. Storage of product 100%
20. First-in , First-out 100%
21. History management of A parts 100%
22. Implementation of inspection 100%
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Radar Graph
7. Supplier control, 70.0%
8. Control of drawings and engg. Changes,
100.0%
6. Quality audit, 80.0%
9. Inspection standard & PCS, 100.0%
3. Initial control- Changed parts, 40.0%
4. Preventiv e measures for defects, 70.0%
5. Education and training, 80.0%
2. Initial supply control, 60.0%
10. Operation standard, 100.0%
11. Observ ance of operation standard, 100.0%12. Role of Manager/ Supervisor, 100.0%
21. History m anagement of A parts, 100.0%
22. Implementation of inspection, 100.0%1. Production preparation, 40%
20. First-in , First-out, 100.0%
19. Storage of product, 100.0%
18. Control of non conforming product, 100.0%
17. Prevention of missing process/ w rong
assembling, 75.0%
14. Control of Manufacturing machine/ jig,
50.0%
13. Quality improvement of process, 0.0%
15. Control of Inspection equipment and Jig,40.0%
16. Statistical method, 100.0%
0%
10%
20%
30%
40%50%
60%
70%
80%
90%
100%
Series1
Radar Graph : To see the total sharp image as a compositegraph.