ken black qa ch18

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Business Statistics, 5th ed.

by Ken Black

Chapter 18

Statistical Quality Control

Discrete Distributions

PowerPoint presentations prepared by Lloyd Jaisingh, Morehead State University



•

Understand the concepts of quality, qualitycontrol, and total quality management.

• Understand the importance of statisticalquality control in total quality management.

• Learn about process analysis and someprocess analysis tools.

• Learn how to construct x-bar charts, R charts, p charts, and c charts.

• Understand the theory and application of acceptance sampling.

Learning Objectives



Quality

• Quality is when a product delivers what is stipulated for in its specifications

• Crosby: “quality is conformance torequirements”

• Feigenbaum: “quality is a customerdetermination”



Garvin’s Five Dimensions of Quality

• Transcendent quality: “innate excellence”

• Product quality: quality is measurable

• User quality: quality is determined by the consumer

• Manufacturing quality: quality is measured by the manufacturer's ability to target the

product specifications with little variability• Value Quality: did the consumer get his or

her money’s worth?



Quality Control• Quality control is the collection of strategies,

techniques, and actions taken by an organization to assure themselves that they are producing a quality product.

• After-process quality control involves inspecting the attributes of a finished product to determine whether the

product is acceptable, is in need of rework, or is to be rejected and scrapped. – reporting of the number of defects per time period – screening defective products from consumers

• In-process quality control techniques measure product

attributes at various intervals throughout the manufacturing process in an effort to pinpoint problem areas.



Deming’s Fourteen Points (Total Quality

Management, TQM)

1. Create constancy of purpose for improvement of productand service.2. Adopt a new philosophy.3. Cease dependence on mass inspection.4. End the practice of awarding business on price tag alone.5. Improve constantly and forever the system of production

and service.6. Institute training.7. Institute leadership.8 Drive out fear.9. Break down barriers between staff areas.

10. Eliminate slogans.11. Eliminate numerical quotas.12. Remove barriers to pride of workmanship.13. Institute a vigorous program of education and retraining.14. Take action to accomplish the transformation.



Important Quality Concepts

• Benchmarking – examine and emulate the best practices and techniques used in

the industry – a positive, proactive process to make changes that will effect

superior performance• Just-In-Time Inventory Systems

– necessary parts for production arrive “just in time” –

reduced holding costs, personnel, and space needed for inventory – no extra raw materials or inventory of parts for production arestored

• Reengineering – complete redesign of the core business process in a company

• Six sigma – Total quality approach that measures the capacity of a process to

perform defect -free work•

Team Building: – employee groups take on managerial responsibilities – quality circle



Process Analysis•

A process is a series of actions, changes or functionsthat bring about a result.

• Flowcharts - schematic representation of all the activities and interactions that occur in a process

• Pareto Analysis -quantitative tallying of the number and

types of defects that occur with a product• Pareto Chart - ranked vertical bar chart with most

frequently occurring on the left

• Fishbone Diagram - display of potential cause-and-effect relationships

• Control Charts - graphical method for evaluatingwhether a process is or is not in a “state of statistical control”



Flowchart Symbols

Input/Output Symbol

Processing Symbol

Decision Symbol

Start/Stop Symbol

Flow line Symbol



0

10

20

30

4050

60

70

80

90100

Poor

Wiring

Short in

Coil

Defective

Plug

Other

F r e q u e

n c y

0%

10%

20%

30%

40%50%

60%

70%

80%

90%100%

Pareto Charts



Cause-and-Effect Diagram

Raw MaterialsEquipment

WorkersMethodology

Poor

Wiring

Wiring Scheme

Pland Layout

Maintenance

Tools

Out-of-AdjustmentOut-of-Date

Vendor

Transportation

Inventory

Training

Attitude

Absenteeism



MINITAB Pareto Chart

C o u n t

P e r c e n t

Defects

Count

15.4 7.7 3.8

Cum % 42.3 73.1 88.5 96.2 100.0

22 16 8 4 2

Percent 42.3 30.8

O t h e

r

W r o n g P

r e s c r i

p t i o n

W r i t t

e n

M i s r

e a d H

a n d w

r i t i n g

I m p r o p

e r l y S

t o r e d

W r o n g

M e d i c i n

e A d m i n i

s t e r e d

50

40

30

20

10

0

100

80

60

40

20

0

Pareto Chart of Medication Errors in a Hospital



Types of Control Charts

• Control charts for measurements

– charts

– R charts• Control charts for compliance items

– P charts – c charts

X



Control Chart

Monitor process location (center)1. Decide on the quality to be measured.2. Determine a sample size.3. Gather 20 to 30 samples.4. Compute the sample average for each sample.

5. Compute the sample range for each sample.6. Determine the average sample mean for all samples.7. Determine the average sample range (or sample

standard deviation) for all samples.8 Using the size of the samples, determine the value of A2

or A3.9. Compute the UCL and the LCL

X



Control Chart: Formulas

X and Charts

X X

k

UCL X R

LCL X R

R R

k

LCL R

UCL R

A A

D

D

R

2

2

3

4

X and S Charts

X X

k

UCL X R

LCL X R

SS

k

UCL R

LCL R

A A

B

B

3

3

4

3

X



Data for Demonstration Problem

18.1: Samples 1 - 10

1 2 3 4 5 6 7 8 9 10

5.13 4.96 5.21 5.02 5.12 4.98 4.99 4.96 4.96 5.03

4.92 4.98 4.87 5.09 5.08 5.02 5.00 5.01 5.00 4.99

5.01 4.95 5.02 4.99 5.09 4.97 5.00 5.02 4.91 4.964.88 4.96 5.08 5.02 5.13 4.99 5.02 5.05 4.87 5.14

5.05 5.01 5.12 5.03 5.06 4.98 5.01 5.04 4.96 5.11

4.97 4.89 5.04 5.01 5.13 4.99 5.01 5.02 5.01 5.04

4.9933 4.9583 5.0567 5.0267 5.1017 4.9883 5.0050 5.0167 4.9517 5.0450

0.25 0.12 0.34 0.10 0.07 0.05 0.03 0.09 0.14 0.18

XR



Data for Demonstration Problem

18.1: Samples 11 - 20

11 12 13 14 15 16 17 18 19 20

4.91 4.97 5.09 4.96 4.99 5.01 5.05 4.96 4.90 5.04

4.93 4.91 4.96 4.99 4.97 5.04 4.97 4.93 4.85 5.03

5.04 5.02 5.05 4.82 5.01 5.09 5.04 4.97 5.02 4.97

5.00 4.93 5.12 5.03 4.98 5.07 5.03 5.01 5.01 4.99

4.90 4.95 5.06 5.00 4.96 5.12 5.09 4.98 4.88 5.05

4.82 4.96 5.01 4.96 5.02 5.13 5.01 4.92 4.86 5.06

4.9333 4.9567 5.0483 4.9600 4.9883 5.0767 5.0317 4.9617 4.9200 5.0233

0.22 0.11 0.16 0.21 0.06 0.12 0.12 0.09 0.17 0.09XR



Demonstration Problem 18.1:

Control Chart Computations

X and Charts

X X

k

UCL X R

LCL X R

R R

k LCL R

UCL R

A A

D

D

R

4 9933 4 9583 50566 50233

20500215

500215 0 483 0136 506784

5 00215 0 483 0136 4 93646

0 25 012 0 34 0 09

20

0136

0 0136 0

2 004 0136 0 2725

2

2

3

4

. . . ..

. . . .

. . . .

. . . ..

.

. . .



Sigma level: 3

2019

1817

1615

1413

1211

109

87

65

43

21

Bearing Diameter

UCL = 5.0679

Average = 5.0022

LCL = 4.9364

Control Chart: Bearing Diameter

Mean

5.10963

5.05590

5.00217

4.94844

4.89471


Control ChartX



R Chart

Monitor process variation1. Decide on the quality to be measured.

2. Determine a sample size.

3. Gather 20 to 30 samples.

4. Compute the sample range for each sample.5. Determine the average sample mean for all

samples.

6. Using the size of the samples, determine the

values of D3 and D4.7. Compute the UCL and the LCL



R Chart Formulas

R Charts

R

R

k

LCL R

UCL R

D

D

3

4




R Control Chart

Control Chart: Bearing Diameter

Sigma level: 3

2019

1817

1615

1413

1211

109

87

65

43

21

Range

.4

.3

.2

.1

0.0

Bearing DiameterUCL = .2725

Average = .1360

LCL = .0000



P Charts

Monitor proportion in noncompliance

1. Decide on the quality to be measured.

2. Determine a sample size.

3. Gather 20 to 30 samples.4. Compute the sample proportion for each

sample.

5. Determine the average sample proportion for

all samples.6. Compute the UCL and the LCL



P Chart Formulas

:

:

:

pn

the number of noncomplying items in the sample

the number of items in the sample

=

the sample proportion

the number of samples

UCL=P+3

P Q

n

LCL P 3P Q

n

non

non

n

n

where

n

P pk

where p

k

where Q P1




Twenty Samples of Bond Paper

Sample n

Number Out

of

Compliance Sample n

Number Out

of

Compliance

1 50 4 11 50 2

2 50 3 12 50 63 50 1 13 50 0

4 50 0 14 50 2

5 50 5 15 50 1

6 50 2 16 50 6

7 50 3 17 50 28 50 1 18 50 3

9 50 4 19 50 1

10 50 2 20 50 5




Preliminary Calculations

Sample n nnon Sample n nnon

1 50 4 0.08 11 50 2 0.042 50 3 0.06 12 50 6 0.123 50 1 0.02 13 50 0 0.004 50 0 0.00 14 50 2 0.045 50 5 0.10 15 50 1 0.026 50 2 0.04 16 50 6 0.12

7 50 3 0.06 17 50 2 0.048 50 1 0.02 18 50 3 0.069 50 4 0.08 19 50 1 0.02

10 50 2 0.04 20 50 5 0.10

pp




Centerline, UCL, and LCL Computations

P p

k

UCL PP Q

n

LCL P P Qn

LCL

. . . ..

.. .

.

. . . .

08 06 02 10

20053

3 053 3053 947

50148

3 053 3 053 94750

042

0




MINITAB P Control Chart

Sample

P r o p o r t i o n

2018161412108642

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

_ P=0.053

UCL=0.1480

LCL=0

P Chart of Out of Compliance



c Charts

Monitor number of nonconformances per item1. Decide on nonconformances to be evaluated.

2. Determine the number of items to be studied (atleast 25).

3. Gather items.4. Determine the value of c for each item by

summing the number of nonconformances in theitem.

5. Determine the average number of nonconformances per item.

6. Determine the UCL and the LCL



c Chart Formulas

cc LCL

ccUCL

where

ic

cccc i

3-

3+=

itemperitiesnonconformof number

itemsof number=i :

ci

321




Number of Nonconformities in Oil Gauges

ItemNumber

Number of Nonconformities

ItemNumber

Number of Nonconformities

1 2 14 2

2 0 15 1

3 3 16 4

4 1 17 05 2 18 2

6 5 19 3

7 3 20 2

8 2 21 1

9 0 22 310 0 23 2

11 4 24 0

12 3 25 3

13 2




c Chart Calculations

ci

UCL c c

LCL c c

LCL

c c c ci

1 2 3 2 0 3 3

252 0

3 2 0 3 2 0 62

3 2 0 3 2 0 2 2

0

.

. . .

. . .

= +

-



Demonstration Problem 18.4: c Chart

01234

567

0 5 10 15 20 25

Item Number

c

UCL = 6.2

LCL = 0

c = 2.0




MINITAB c Chart

Sample

S a m p l e C o u n t

24222018161412108642

7

6

5

4

3

2

1

0

_ C=2

UCL=6.243

LCL=0

C Chart of Number of Nonconformances



Interpreting Control Charts•

Points are above UCL and/or below LCL• Eight or more consecutive points fall above or

below the centerline. Ten out of 11 points fallabove or below the centerline. Twelve out of 14points fall above or below the centerline.

•

A trend of 6 or more consecutive points(increasing or decreasing) is present• Two out of 3 consecutive values are in the outer

one-third.• Four out 5 consecutive values are in the outer

two-thirds.• The centerline shifts from chart to chart.



Interpreting Control Charts:

Points above UCL and/or below LCL

UCL

LCL

Centerline



Interpreting Control Charts: 8 Consecutive

Points on One Side of the Centerline

UCL

LCL

Centerline




7 Consecutive Increasing Points

UCL

LCL

Centerline




4 out of 5 Consecutive Points in Outer 2/3

UCL

LCL

Centerline



Acceptance Sampling

• Acceptance sampling is the inspection of asample from a lot of goods to determine if the lot will be accepted or rejected. –

N = the lot size – n = the sample size

• Single Sample Plan

• Double-Sample Plan

• Multiple-Sample Plan



Rules for Sampling Plans

Accept lot if

Reject lot if

x c

x c

First sample:Accept if

Reject if

Take second sample if

Second sample:Accept if

Reject if

1

1

1

1

1

xx

c

x

x

1

1

1 1

2 2

2 2

c

r

x r

x c

x c

Single Sample

Plan

Double Sample

Plan



Producer’s and Consumer’s Risk

State of Nature

Null True Null False

Actions

Fail toReject Null

Type II errorCorrectDecision --

Consumer’s Risk

Reject Null Type Ierror --Producer’s Risk

CorrectDecision

H0: the lot is of acceptable quality



Bicycle Manufacturer Example•

N = 3,000 (Braces arrive at the manufacturer’s plant in lots of 3,000.)

• n = 15 (The bicycle manufacturer randomly selects a sample of 15 bracesto inspect.)

• X is the number of nonconforming braces in thesample of 15.

•

A 2% nonconformance rate is acceptable to theconsumer (the bicycle manufacturer).• If the lot contains 60 nonconforming braces, what is

the probability that the consumer will reject the lot(producer’s risk)?

•

If the lot contains 360 nonconforming braces, what isthe probability that the consumer will not reject thelot (consumer’s risk)?



Bicycle Manufacturer Example:

Sampling Plan

n

c

x

x

15

1

1

1

=

Accept lot if

Reject lot if





Analysis for 2% Nonconforming Braces

0

0 15 1 14

02

0 1 150

151

1 9647 0353

02 98 02 98

p

P x P x

.

.

. .

. . . .

Probability of accepting

Probability of rejecting

9647

Producer’s Risk




Analysis for 12% Nonconforming

Braces

1

0 15 1 14

12

0 115

0

15

14476

1 4476 5524

12 88 12 88

p

P x P x

.

.

. .

. . . .

Probability of accepting

Probability of rejecting

Consumer’s

Risk




OC Curve for n = 15 and c = 1

0.00

0.20

0.40

0.60

0.80

1.00

0% 10% 20% 30% 40%

Percent nonconforming

Probability

of

acceptance

2%

.9647 } .0353 Producer’s Risk

.4476 Consumer’s Risk

12%

.4476

i f




OC Curve for n = 15 and c = 0

0.00

0.20

0.40

0.60

0.80

1.00

0% 10% 20% 30% 40%

Percent nonconforming

Probability

of

acceptance

2%

.74

12%

.21



Demonstration Problem 18.5



Demonstration

Problem 18.5

n = 20

c = 2

p P(Accept) 0.00001 =BINOMDIST(B$2,B$1,A5,TRUE)

0.01 =BINOMDIST(B$2,B$1,A6,TRUE) 0.02 =BINOMDIST(B$2,B$1,A7,TRUE) 0.03 =BINOMDIST(B$2,B$1,A8,TRUE) 0.04 =BINOMDIST(B$2,B$1,A9,TRUE) 0.05 =BINOMDIST(B$2,B$1,A10,TRUE) 0.06 =BINOMDIST(B$2,B$1,A11,TRUE)

0.07 =BINOMDIST(B$2,B$1,A12,TRUE)

0.08 =BINOMDIST(B$2,B$1,A13,TRUE) 0.09 =BINOMDIST(B$2,B$1,A14,TRUE) 0.1 =BINOMDIST(B$2,B$1,A15,TRUE) 0.11 =BINOMDIST(B$2,B$1,A16,TRUE) 0.12 =BINOMDIST(B$2,B$1,A17,TRUE) 0.13 =BINOMDIST(B$2,B$1,A18,TRUE) 0.14 =BINOMDIST(B$2,B$1,A19,TRUE) 0.15 =BINOMDIST(B$2,B$1,A20,TRUE)

0.16 =BINOMDIST(B$2,B$1,A21,TRUE) 0.17 =BINOMDIST(B$2,B$1,A22,TRUE) 0.18 =BINOMDIST(B$2,B$1,A23,TRUE) 0.19 =BINOMDIST(B$2,B$1,A24,TRUE) 0.2 =BINOMDIST(B$2,B$1,A25,TRUE)



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ken black qa ch18

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