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1
Technical Note 8
Process Capability and
Statistical Quality Control
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2
Process Variation
Process Capability
Process Control Procedures
Variable data
Attribute data
Acceptance Sampling
Operating Characteristic Curve
OBJECTIVES
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3
Basic Forms of VariationAssignable variation
is caused byfactors that can beclearly identified
and possiblymanaged
Common variationisinherent in theproduction process
Example: A poorly trained
employee that creates
variation in finished
product output.
Example: A moldingprocess that always leaves
burrs or flaws on a
molded item.
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4Taguchis View of Variation
Incremental
Cost of
Variability
High
Zero
Lower
Spec
Target
Spec
Upper
Spec
Traditional View
Incremental
Cost of
Variability
High
Zero
Lower
Spec
Target
Spec
Upper
Spec
Taguchis View
Exhibits
TN8.1 &
TN8.2
Traditional view is that quality within the LS and US is good
and that the cost of quality outside this range is constant, where
Taguchi views costs as increasing as variability increases, so seek
to achieve zero defects and that will truly minimize quality costs.
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5Process CapabilityProcess limits
Specification limits
How do the limits relate to one another?
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6
Process Capability Index, Cpk
3
X-UTLor
3
LTLXmin=Cpk
Shifts in Process Mean
Capability Index shows
how well parts being
produced fit into design
limit specifications.
As a production process
produces items small
shifts in equipment or
systems can cause
differences inproduction
performance from
differing samples.
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A simple ratio:
Specification Width_________________________________________________________
Actual Process Width
Generally, the bigger the better.
Process Capability A Standard
Measure of How Good a Process Is.
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Process Capability
This is a one-sided Capability Index
Concentration on the side which is closest to
the specification - closest to being bad
3
;3
XUTLLTLXMinCpk
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9The Cereal Box Example
We are the maker of this cereal. Consumer reportshas just published an article that shows that we
frequently have less than 15 ounces of cereal in a
box.
Lets assume that the government says that wemust be within 5 percent of the weight advertised
on the box.
Upper Tolerance Limit = 16 + .05(16) = 16.8 ounces
Lower Tolerance Limit = 16 .05(16) = 15.2 ounces
We go out and buy 1,000 boxes of cereal and find
that they weight an average of 15.875 ounces with a
standard deviation of .529 ounces.
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Cereal Box Process Capability
Specification orTolerance Limits Upper Spec = 16.8 oz
Lower Spec = 15.2 oz
Observed Weight Mean = 15.875 oz
Std Dev = .529 oz
3
;3
XUTLLTLXMinCpk
)529(.3
875.158.16;
)529(.3
2.15875.15MinCpk
5829.;4253.MinCpk
4253.pkC
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What does a Cpk of .4253 mean?
An index that shows how well the units
being produced fit within the specification
limits.This is a process that will produce a
relatively high number of defects.
Many companies look for a Cpk of 1.3 orbetter 6-Sigma company wants 2.0!
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Types of Statistical Sampling
Attribute (Go or no-go information) Defectives refers to the acceptability of product
across a range of characteristics.
Defects refers to the number of defects per unit
which may be higher than the number ofdefectives.
p-chart application
Variable (Continuous) Usually measured by the mean and the
standard deviation.
X-bar and R chart applications
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Statistical
Process
Control(SPC) Charts
UCL
LCL
Samples
over time
1 2 3 4 5 6
UCL
LCL
Samples
over time
1 2 3 4 5 6
UCL
LCL
Samples
over time
1 2 3 4 5 6
Normal Behavior
Possible problem, investigate
Possible problem, investigate
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14Control Limits are based on theNormal Curve
x
0 1 2 3-3 -2 -1z
m
Standarddeviation
units or z
units.
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15Control LimitsWe establish the Upper Control Limits (UCL) and
the Lower Control Limits (LCL) with plus or minus
3 standard deviations from some x-bar or mean
value. Based on this we can expect 99.7% of our
sample observations to fall within these limits.
xLCL UCL
99.7%
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Example of Constructing a p-Chart:Required Data
1 100 4
2 100 2
3 100 5
4 100 3
5 100 6
6 100 4
7 100 3
8 100 7
9 100 1
10 100 2
11 100 3
12 100 2
13 100 2
14 100 8
15 100 3
Sample
No.
No. of
Samples
Number of
defects found
in each sample
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Statistical Process Control Formulas:Attribute Measurements (p-Chart)p =
T o t al N u m b e r o f D e fe c tiv e s
T o ta l N u m b e r o f O b s e rv a tio n s
n
s
)p-(1p=p
p
p
z-p=LCL
z+p=UCL
s
s
Given:
Compute control limits:
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1. Calculate the
sample proportions,
p (these are what
can be plotted on thep-chart) for eachsample
Sample n Defectives p1 100 4 0.04
2 100 2 0.02
3 100 5 0.05
4 100 3 0.03
5 100 6 0.066 100 4 0.04
7 100 3 0.03
8 100 7 0.07
9 100 1 0.01
10 100 2 0.02
11 100 3 0.03
12 100 2 0.02
13 100 2 0.02
14 100 8 0.08
15 100 3 0.03
Example of Constructing a p-chart: Step 1
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2. Calculate the average of the sample proportions
0.036=
1500
55=p
3. Calculate the standard deviation of the
sample proportion
.0188=100
.036)-.036(1=
)p-(1p=p
n
s
Example of Constructing a p-chart: Steps 2&3
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4. Calculate the control limits
3(.0188).036
UCL = 0.0924LCL = -0.0204 (or 0)
p
p
z-p=LCL
z+p=UCL
s
s
Example of Constructing a p-chart: Step 4
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Example of Constructing a p-Chart: Step 5
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Observation
p
UCL
LCL
5. Plot the individual sample proportions, the averageof the proportions, and the control limits
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Example of x-bar and R Charts:Required DataSample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5
1 10.68 10.689 10.776 10.798 10.714
2 10.79 10.86 10.601 10.746 10.779
3 10.78 10.667 10.838 10.785 10.723
4 10.59 10.727 10.812 10.775 10.73
5 10.69 10.708 10.79 10.758 10.6716 10.75 10.714 10.738 10.719 10.606
7 10.79 10.713 10.689 10.877 10.603
8 10.74 10.779 10.11 10.737 10.75
9 10.77 10.773 10.641 10.644 10.725
10 10.72 10.671 10.708 10.85 10.71211 10.79 10.821 10.764 10.658 10.708
12 10.62 10.802 10.818 10.872 10.727
13 10.66 10.822 10.893 10.544 10.75
14 10.81 10.749 10.859 10.801 10.701
15 10.66 10.681 10.644 10.747 10.728
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Example of x-bar and R charts: Step 1.Calculate sample means, sample ranges,mean of means, and mean of ranges.Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Avg Range
1 10.68 10.689 10.776 10.798 10.714 10.732 0.116
2 10.79 10.86 10.601 10.746 10.779 10.755 0.259
3 10.78 10.667 10.838 10.785 10.723 10.759 0.171
4 10.59 10.727 10.812 10.775 10.73 10.727 0.221
5 10.69 10.708 10.79 10.758 10.671 10.724 0.1196 10.75 10.714 10.738 10.719 10.606 10.705 0.143
7 10.79 10.713 10.689 10.877 10.603 10.735 0.274
8 10.74 10.779 10.11 10.737 10.75 10.624 0.669
9 10.77 10.773 10.641 10.644 10.725 10.710 0.132
10 10.72 10.671 10.708 10.85 10.712 10.732 0.179
11 10.79 10.821 10.764 10.658 10.708 10.748 0.16312 10.62 10.802 10.818 10.872 10.727 10.768 0.250
13 10.66 10.822 10.893 10.544 10.75 10.733 0.349
14 10.81 10.749 10.859 10.801 10.701 10.783 0.158
15 10.66 10.681 10.644 10.747 10.728 10.692 0.103
Averages 10.728 0.220400
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Example of x-bar and R charts: Step 2. DetermineControl Limit Formulas and Necessary TabledValuesx Chart Control Limits
UCL = x + A R
LCL = x - A R
2
2
R Chart Control Limits
UCL = D R
LCL = D R
4
3
From Exhibit TN8.7
n A2 D3 D4
2 1.88 0 3.27
3 1.02 0 2.57
4 0.73 0 2.28
5 0.58 0 2.11
6 0.48 0 2.00
7 0.42 0.08 1.92
8 0.37 0.14 1.869 0.34 0.18 1.82
10 0.31 0.22 1.78
11 0.29 0.26 1.74
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25Example of x-bar and R charts: Steps 3&4.Calculate x-bar Chart and Plot Values
10.601
10.856
=).58(0.2204-10.728RA-x=LCL
=).58(0.2204-10.728RA+x=UCL
2
2
10.550
10.600
10.650
10.700
10.750
10.800
10.850
10.900
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Sample
Means
UCL
LCL
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Example of x-bar and R charts: Steps 5&6.Calculate R-chart and Plot Values
00.46504
)2204.0)(0(RD=LCL)2204.0)(11.2(RD=UCL
3
4
0 . 0 0 0
0 . 1 0 0
0 . 2 0 0
0 . 3 0 0
0 . 4 0 0
0 . 5 0 0
0 . 6 0 0
0 . 7 0 0
0 . 8 0 0
1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5
S a m p l e
RUCL
LCL
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Basic Forms of Statistical Samplingfor Quality Control
Acceptance Sampling is sampling
to accept or reject the immediate lot
ofproduct at hand
Statistical Process Control is
sampling to determine if the
process is within acceptable limits
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Acceptance Sampling
Purposes Determine quality level
Ensure quality is within predetermined level
Advantages
Economy Less handling damage
Fewer inspectors
Upgrading of the inspection job
Applicability to destructive testing Entire lot rejection (motivation for
improvement)
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Acceptance Sampling (Continued)Disadvantages
Risks of accepting bad lots and
rejecting good lots Added planning and documentation
Sample provides less information
than 100-percent inspection
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Acceptance Sampling:Single Sampling PlanA simple goal
Determine (1) how many units, n,to sample from a lot, and (2) themaximum number of defective
items, c, that can be found in thesample before the lot is rejected
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Risk Acceptable Quality Level (AQL)
Max. acceptable percentage of defectives
defined by producer
The a (Producers risk)
The probability of rejecting a good lot Lot Tolerance Percent Defective (LTPD)
Percentage of defectives that defines
consumers rejection point The (Consumers risk)
The probability of accepting a bad lot
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Operating Characteristic Curve
n = 99
c = 4
AQL LTPD
00.1
0.2
0.3
0.4
0.50.6
0.7
0.8
0.9
1
1 2 3 4 5 6 7 8 9 10 11 12
Percent defective
Probability
ofac
ceptance
=.10(consumers risk)
a = .05 (producers risk)
The OCC brings the concepts of producers risk, consumers
risk, sample size, and maximum defects allowed together
The shape
or slope of
the curve is
dependent
on a
particular
combination
of the four
parameters
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Example: Acceptance Sampling ProblemZypercom, a manufacturer of video interfaces,purchases printed wiring boards from an outside
vender, Procard. Procard has set an acceptable
quality level of 1% and accepts a 5% risk of rejecting
lots at or below this level. Zypercom considers lotswith 3% defectives to be unacceptable and will assume
a 10% risk of accepting a defective lot.
Develop a sampling plan for Zypercom and determinea rule to be followed by the receiving inspection
personnel.
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Example:Step 1. What is given and what is not?
In this problem, AQL is given to be 0.01 and LTPD
is given to be 0.03. We are also given an alpha of
0.05 and a beta of 0.10.
What you need to determine is your sampling
plan is c and n.
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Example: Step 2. Determine c
First divide LTPD by AQL.
LTPD
AQL
=.03
.01
= 3
Then find the value for c by selecting the value in the
TN7.10 n(AQL)column that is equal to or just greater than
the ratio above.
Exhibit TN 8.10
c LTPD/AQL n AQL c LTPD/AQL n AQL
0 44.890 0.052 5 3.549 2.6131 10.946 0.355 6 3.206 3.286
2 6.509 0.818 7 2.957 3.981
3 4.890 1.366 8 2.768 4.695
4 4.057 1.970 9 2.618 5.426
So, c = 6.
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Example: Step 3. Determine Sample Size
c = 6, from Table
n (AQL) = 3.286, from Table
AQL = .01, given in problem
Sampling Plan:Take a random sample of 329 units from a lot.
Reject the lot ifmore than 6 units are defective.
Now given the information below, compute the sample
size in units to generate your sampling plan
n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)
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Computers in Quality Control
Records about quality testing and results limit
a firms exposure in the event of a product
liability suit.
Recall programs require that manufacturers
Know the lot number of the parts that are
responsible for the potential defects
Have an information storage system that can tie
the lot numbers of the suspected parts to the finalproduct model numbers
Have an information system that can track the
model numbers of final products to customers
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Computers in Quality Control
With automation, inspection and testing canbe so inexpensive and quick that
companies may be able to increase sample
sizes and the frequency of samples, thusattaining more precision in both control
charts and acceptance plans
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Question BowlA methodology that is used to show how well parts
being produced fit into a range specified by
design limits is which of the following?
a. Capability index
b. Producers riskc. Consumers risk
d. AQL
e. None of the above
Answer: a. Capability index
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Question BowlOn a quality control chart if one of the values plotted falls outside
a boundary it should signal to the production manager to dowhich of the following?
a. System is out of control, should be stopped and fixed
b. System is out of control, but can still be operated without any
concern
c. System is only out of control if the number of observations
falling outside the boundary exceeds statistical expectations
d. System is OK as is
e. None of the aboveAnswer: c. System is only out of
control if the number of observationsfalling outside the boundary exceeds
statistical expectations
(We expect with Six Sigma that 3 out of 1,000 observations will fall outside
the boundaries normally and those deviations should not lead managers
to conclude the system is out of control.)
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Question BowlYou want to prepare a p chart and you
observe 200 samples with 10 in each,
and find 5 defective units. What is the
resulting fraction defective?
a. 25b. 2.5
c. 0.0025
d. 0.00025e. Can not be computed on data above
Answer: c. 0.0025 (5/(2000x10)=0.0025)
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Question BowlYou want to prepare an x-bar chart. If the number of
observations in a subgroup is 10, what is the
appropriate factor used in the computation of
the UCL and LCL?
a. 1.88
b. 0.31
c. 0.22
d. 1.78
e. None of the above
Answer: b. 0.31 (from Exhibit TN8.7)
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Question BowlYou want to prepare an R chart. If the number of
observations in a subgroup is 5, what is the
appropriate factor used in the computation
of the LCL?
a. 0b. 0.88
c. 1.88
d. 2.11
e. None of the above
Answer: a. 0 (from Exhibit TN8.7)
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Question BowlYou want to prepare an R chart. If the
number of observations in a subgroup
is 3, what is the appropriate factor
used in the computation of the UCL?
a. 0.87b. 1.00
c. 1.88
d. 2.11
e. None of the above
Answer: e. None of the above (from Exhibit
TN8.7 the correct value is 2.57)
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Question BowlThe maximum number of defectives that can be
found in a sample before the lot is rejected is
denoted in acceptance sampling as which of
the following?
a. Alphab. Beta
c. AQL
d. c
e. None of the above
Answer: d. c
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