Transcript Document

Slides Prepared by
JOHN S. LOUCKS
St. Edward’s University
© 2002 South-Western College Publishing/Thomson Learning
Slide 1
Chapter 20
Statistical Methods for Quality Control

Statistical Process Control
Acceptance Sampling
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UCL
CL
LCL
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Slide 2
Quality Terminology

Quality is “the totality of features and characteristics
of a product or service that bears on its ability to
satisfy given needs.”
Slide 3
Quality Terminology



Quality assurance refers to the entire system of
policies, procedures, and guidelines established by
an organization to achieve and maintain quality.
The objective of quality engineering is to include
quality in the design of products and processes and
to identify potential quality problems prior to
production.
Quality control consists of making a series of
inspections and measurements to determine whether
quality standards are being met.
Slide 4
Statistical Process Control (SPC)



The goal of SPC is to determine whether the process
can be continued or whether it should be adjusted to
achieve a desired quality level.
If the variation in the quality of the production
output is due to assignable causes (operator error,
worn-out tooling, bad raw material, . . . ) the process
should be adjusted or corrected as soon as possible.
If the variation in output is due to common causes
(variation in materials, humidity, temperature, . . . )
which the manager cannot control, the process does
not need to be adjusted.
Slide 5
SPC Hypotheses




SPC procedures are based on hypothesis-testing
methodology.
The null hypothesis H0 is formulated in terms of the
production process being in control.
The alternative hypothesis Ha is formulated in terms
of the process being out of control.
As with other hypothesis-testing procedures, both a
Type I error (adjusting an in-control process) and a
Type II error (allowing an out-of-control process to
continue) are possible.
Slide 6
Decisions and State of the Process

Type I and Type II Errors
State of Production Process
Decision
H0 True
In Control
Ha True
Out of Control
Continue
Process
Correct
Decision
Type II Error
Adjust
Process
Type I Error
Adjust in-control
process
Allow out-of-control
process to continue
Correct
Decision
Slide 7
Control Charts


SPC uses graphical displays known as control charts
to monitor a production process.
Control charts provide a basis for deciding whether
the variation in the output is due to common causes
(in control) or assignable causes (out of control).
Slide 8
Control Charts



Two important lines on a control chart are the upper
control limit (UCL) and lower control limit (LCL).
These lines are chosen so that when the process is in
control, there will be a high probability that the
sample finding will be between the two lines.
Values outside of the control limits provide strong
evidence that the process is out of control.
Slide 9
Types of Control Charts





An x chart is used if the quality of the output is
measured in terms of a variable such as length,
weight, temperature, and so on.
x represents the mean value found in a sample of the
output.
An R chart is used to monitor the range of the
measurements in the sample.
A p chart is used to monitor the proportion defective
in the sample.
An np chart is used to monitor the number of
defective items in the sample.
Slide 10
x Chart Structure
x
UCL
Center Line
Process Mean
When in Control
LCL
Time
Slide 11
Control Limits for an x Chart

Process Mean and Standard Deviation Known
UCL =   3 x
LCL =   3 x
Slide 12
Example: Granite Rock Co.

Control Limits for an x Chart: Process Mean
and Standard Deviation Known
The weight of bags of cement filled by Granite’s
packaging process is normally distributed with a
mean of 50 pounds and a standard deviation of 1.5
pounds.
What should be the control limits for samples of
9 bags?
Slide 13
Example: Granite Rock Co.

Control Limits for an x Chart: Process Mean
and Standard Deviation Known
 = 50,  = 1.5, n = 9
x 
n
 1.5
9
 0.5
UCL = 50 + 3(.5) = 51.5
LCL = 50 - 3(.5) = 48.5
Slide 14
Control Limits for an x Chart

Process Mean and Standard Deviation Unknown
UCL = x  A2 R
LCL = x  A2 R
where:
=
x_ = overall sample mean
R = average range
A2 = a constant that depends on n; taken from
“Factors for Control Charts” table
Slide 15
Factors for x and R Control Charts

Factors Table (Partial)
n
5
6
7
8
9
10
.
.
d2
2.326
2.534
2.704
2.847
2.970
3.078
.
.
A2
0.577
0.483
0.419
0.373
0.337
0.308
.
.
d3
0.864
0.848
0.833
0.820
0.808
0.797
.
.
D3
0
0
0.076
0.136
0.184
0.223
.
.
D4
2.114
2.004
1.924
1.864
1.816
1.777
.
.
Slide 16
Control Limits for an R Chart
_
UCL = RD
_ 4
LCL = RD3
where:
_
R = average range
D3, D4 = constants that depend on n; found
in “Factors for Control Charts”
table
Slide 17
Factors for x and R Control Charts

Factors Table (Partial)
n
5
6
7
8
9
10
.
.
d2
2.326
2.534
2.704
2.847
2.970
3.078
.
.
A2
0.577
0.483
0.419
0.373
0.337
0.308
.
.
d3
0.864
0.848
0.833
0.820
0.808
0.797
.
.
D3
0
0
0.076
0.136
0.184
0.223
.
.
D4
2.114
2.004
1.924
1.864
1.816
1.777
.
.
Slide 18
Example: Granite Rock Co.

Control Limits for x and R Charts: Process Mean
and Standard Deviation Unknown
Suppose Granite does not know the true mean
and standard deviation for its bag filling process. It
wants to develop x and R charts based on twenty
samples of 5 bags each.
The twenty samples resulted in an overall
sample mean of 50.01 pounds and an average range
of .322 pounds.
Slide 19
Example: Granite Rock Co.

Control Limits for R Chart: Process Mean
and Standard Deviation Unknown
_
=
x = 50.01, R = .322, n = 5
_
UCL = RD
= .322(2.114) = .681
_ 4
LCL = RD3 = .322(0)
= 0
Slide 20
Example: Granite Rock Co.
R Chart
A
D
R BChart for C
Granite Rock
Co. E
F
0.80
0.70
Sam ple Range R

UCL
0.60
0.50
0.40
0.30
0.20
0.10
LCL
0.00
0
5
10
Sam ple Num ber
15
20
Slide 21
Example: Granite Rock Co.

Control Limits for x Chart: Process Mean
and Standard Deviation Unknown
=
x = 50.01, R = .322, n = 5
UCL = x= + A2R = 50.01 + .577(.322) = 50.196
=
LCL = x - A2R = 50.01 - .577(.322) = 49.824
Slide 22
Example: Granite Rock Co.
x Chart
x Chart for Granite Rock Co.
50.3
UCL
50.2
Sample
Mean

50.1
50.0
49.9
49.8
LCL
49.7
0
5
10
Sample Number
15
20
Slide 23
Control Limits for a p Chart
UCL = p  3 p
LCL = p  3 p
where:
p 
p( 1  p )
n
assuming:
np > 5
n(1-p) > 5
Note: If computed LCL is negative, set LCL = 0
Slide 24
Example: Norwest Bank
Every check cashed or deposited at Norwest
Bank must be encoded with the amount of the check
before it can begin the Federal Reserve clearing
process. The accuracy of the check encoding process
is of upmost importance. If there is any discrepancy
between the amount a check is made out for and the
encoded amount, the check is defective.
Slide 25
Example: Norwest Bank
Twenty samples, each consisting of 250 checks,
were selected and examined when the encoding
process was known to be operating correctly. The
number of defective checks found in the samples
follow.
4
2
1
8
5
5
3
3
2
6
7
4
4
2
5
5
2
3
3
6
Slide 26
Example: Norwest Bank

Control Limits for a p Chart
Suppose Norwest does not know the proportion
of defective checks, p, for the encoding process when
it is in control.
We will treat the data (20 samples) collected as
one large sample and compute the average number
of defective checks for all the data. That value can
then be used to estimate p.
Slide 27
Example: Norwest Bank

Control Limits for a p Chart
Estimated p = 80/((20)(250)) = 80/5000 = .016
p(1  p)
.016(1  .016)
.015744
p 


 .007936
n
250
250
UCL = p  3 p  .016  3(.007936)  .039808
LCL = p  3 p  .016  3(.007936)  -.007808  0
Slide 28
Example: Norwest Bank
p Chart
p Chart for Norwest Bank
0.045
0.040
Sample Proportion p

UCL
0.035
0.030
0.025
0.020
0.015
0.010
0.005
LCL
0.000
0
5
10
Sample Number
15
20
Slide 29
Control Limits for an np Chart
UCL = np  3 np( 1  p)
LCL = np  3 np(1  p)
assuming:
np > 5
n(1-p) > 5
Note: If computed LCL is negative, set LCL = 0
Slide 30
Interpretation of Control Charts



The location and pattern of points in a control chart
enable us to determine, with a small probability of
error, whether a process is in statistical control.
A primary indication that a process may be out of
control is a data point outside the control limits.
Certain patterns of points within the control limits
can be warning signals of quality problems:
• Large number of points on one side of center line.
• Six or seven points in a row that indicate either an
increasing or decreasing trend.
• . . . and other patterns.
Slide 31
Acceptance Sampling


Acceptance sampling is a statistical method that
enables us to base the accept-reject decision on the
inspection of a sample of items from the lot.
Acceptance sampling has advantages over 100%
inspection including: less expensive, less product
damage, fewer people involved, . . . and more.
Slide 32
Acceptance Sampling Procedure
Lot received
Sample selected
Sampled items
inspected for quality
Quality is
satisfactory
Results compared with
specified quality characteristics
Quality is not
satisfactory
Accept the lot
Reject the lot
Send to production
or customer
Decide on disposition
of the lot
Slide 33
Acceptance Sampling


Acceptance sampling is based on hypothesis-testing
methodology.
The hypothesis are:
H0: Good-quality lot
Ha: Poor-quality lot
Slide 34
The Outcomes of Acceptance Sampling

Type I and Type II Errors
State of the Lot
Decision
H0 True
Good-Quality Lot
Ha True
Poor-Quality Lot
Accept H0
Accept the Lot
Correct
Decision
Type II Error
Consumer’s Risk
Reject H0
Reject the Lot
Type I Error
Producer’s Risk
Correct
Decision
Slide 35
Probability of Accepting a Lot

Binomial Probability Function for Acceptance
Sampling
f ( x) 
n!
p x ( 1  p)(n x)
x !(n  x)!
where:
n = sample size
p = proportion of defective items in lot
x = number of defective items in sample
f(x) = probability of x defective items in sample
Slide 36
Example: Acceptance Sampling
An inspector takes a sample of 20 items from a lot.
Her policy is to accept a lot if no more than 2 defective
items are found in the sample.
Assuming that 5 percent of a lot is defective, what is
the probability that she will accept a lot? Reject a lot?
n = 20, c = 2, and p = .05
P(Accept Lot) = f(0) + f(1) + f(2)
= .3585 + .3774 + .1887
= .9246
P(Reject Lot) = 1 - .9246
= .0754
Slide 37
Example: Acceptance Sampling

n
20
Using the Tables of Binomial Probabilities
x
0
1
2
3
4
5
6
7
8
9
.05
.3585
.3774
.1887
.0596
.0133
.0022
.0003
.0000
.0000
.0000
.10
.1216
.2702
.2852
.1901
.0898
.0319
.0089
.0020
.0004
.0001
.15
.0388
.1368
.2293
.2428
.1821
.1028
.0454
.0160
.0046
.0011
.20
.0115
.0576
.1369
.2054
.2182
.1746
.1091
.0545
.0222
.0074
p
.25
.0032
.0211
.0669
.1339
.1897
.2023
.1686
.1124
.0609
.0271
.30
.0008
.0068
.0278
.0716
.1304
.1789
.1916
.1643
.1144
.0654
.35
.0002
.0020
.0100
.0323
.0738
.1272
.1712
.1844
.1614
.1158
.40
.0000
.0005
.0031
.0123
.0350
.0746
.1244
.1659
.1797
.1597
.45
.0000
.0001
.0008
.0040
.0139
.0365
.0746
.1221
.1623
.1771
.50
.0000
.0000
.0002
.0011
.0046
.0148
.0370
.0739
.1201
.1602
Slide 38
Selecting an Acceptance Sampling Plan


In formulating a plan, managers must specify two
values for the fraction defective in the lot.
• a = the probability that a lot with p0 defectives will
be rejected.
• b = the probability that a lot with p1 defectives will
be accepted.
Then, the values of n and c are selected that result in
an acceptance sampling plan that comes closest to
meeting both the a and b requirements specified.
Slide 39
Probability of Accepting the Lot
Operating Characteristic Curve
1.00
.90
a
.80
n = 15, c = 0
.70
p0 = .03, p1 = .15
.60
a = .3667, b = .0874
.50
.40
(1 - a)
.30
.20
p0
p1
.10
b
0
5
10
15
20
Percent Defective in the Lot
25
Slide 40
Multiple Sampling Plans



A multiple sampling plan uses two or more stages of
sampling.
At each stage the decision possibilities are:
• stop sampling and accept the lot,
• stop sampling and reject the lot, or
• continue sampling.
Multiple sampling plans often result in a smaller
total sample size than single-sample plans with the
same Type I error and Type II error probabilities.
Slide 41
A Two-Stage Acceptance Sampling Plan
Inspect n1 items
Find x1 defective items in this sample
x1 < c1 ?
Reject
the lot
Yes
Yes
No
Accept
the lot
x1 > c2 ?
No
Inspect n2 additional items
Find x2 defective items in this sample
No
x1 + x2 < c3 ?
Yes
Slide 42
End of Chapter 20
Slide 43