business statistics, 4e, by ken black. © 2003 john wiley & sons. 18-1 business statistics, 4e...

Business Statistics, 4e, by Ken Black. © 2003 John Wiley & Sons. 18-1

Business Statistics, 4eby Ken Black

Chapter 18

Statistical Quality Control

Discrete Distributions


• Understand the concepts of quality, quality control, and total quality management.

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

• Learn about process analysis and some process analysis tools, including Pareto charts, fishbone diagrams, and control chars.

• Learn how to construct charts, R charts, P charts, and c charts.

• Understand the theory and application of acceptance sampling.

Learning Objectives

X


Quality

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

• Crosby: “quality is conformance to requirements”

• Feigenbaum: “quality is a customer determination”


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 Points1.Create constancy of purpose fro improvement of product and 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.8Drive 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

• 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 functions

that 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 evaluating

whether 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


Pareto ChartPareto Chart

0

10

20

30

40

50

60

70

80

90

100

PoorWiring

Short inCoil

DefectivePlug

Other

Fre

qu

ency

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%


Cause-and-Effect Diagram

Raw MaterialsEquipment

WorkersMethodology

PoorWiring

Wiring Scheme

Pland Layout

MaintenanceTools

Out-of-AdjustmentOut-of-Date

Vendor

Transportation

Inventory

Training

Attitude

Absenteeism


Control Chart

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Sample Number

Sample Mean

UCL

LCL

Centerline

XX


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.8Using the size of the samples, determine the value of A2 or A3.9.Compute the UCL and the LCL

XX


Control Chart: Formulas

X and Charts

XX

k

UCL X R

LCL X R

RR

k

LCL R

UCL R

AA

DD

R

2

2

3

4

X and S Charts

XX

k

UCL X R

LCL X R

SS

k

UCL R

LCL R

AA

BB

3

3

4

3

XX


Data for Demonstration Problem 18.1: Samples 1 - 10

1 2 3 4 5 6 7 8 9 105.13 4.96 5.21 5.02 5.12 4.98 4.99 4.96 4.96 5.034.92 4.98 4.87 5.09 5.08 5.02 5.00 5.01 5.00 4.995.01 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 5.145.05 5.01 5.12 5.03 5.06 4.98 5.01 5.04 4.96 5.114.97 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 5.04500.25 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 19 204.91 4.97 5.09 4.96 4.99 5.01 5.05 4.96 4.90 5.044.93 4.91 4.96 4.99 4.97 5.04 4.97 4.93 4.85 5.035.04 5.02 5.05 4.82 5.01 5.09 5.04 4.97 5.02 4.975.00 4.93 5.12 5.03 4.98 5.07 5.03 5.01 5.01 4.994.90 4.95 5.06 5.00 4.96 5.12 5.09 4.98 4.88 5.054.82 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 5.02330.22 0.11 0.16 0.21 0.06 0.12 0.12 0.09 0.17 0.09

XR


Demonstration Problem 18.1:Control Chart Computations

X and Charts

XX

k

UCL X R

LCL X R

RR

k

LCL R

UCL R

AA

DD

R

4 9933 4 9583 5 0566 5 0233

205 00215

5 00215 0 483 0136 5 06784

5 00215 0 483 0136 4 93646

0 25 012 0 34 0 09

200136

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

Demonstration Problem 18.1: Control ChartXX


R ChartMonitor process Monitor process variationvariation

1.1.Decide on the quality to be measured.Decide on the quality to be measured.2.2.Determine a sample size.Determine a sample size.3.3.Gather 20 to 30 samples.Gather 20 to 30 samples.4.4.Compute the sample range for each sample.Compute the sample range for each sample.5.5.Determine the average sample mean for all Determine the average sample mean for all samples.samples.6.6.Using the size of the samples, determine the Using the size of the samples, determine the values of Dvalues of D33 and D and D44..7.7.Compute the UCL and the LCLCompute the UCL and the LCL

Monitor process Monitor process variationvariation1.1.Decide on the quality to be measured.Decide on the quality to be measured.2.2.Determine a sample size.Determine a sample size.3.3.Gather 20 to 30 samples.Gather 20 to 30 samples.4.4.Compute the sample range for each sample.Compute the sample range for each sample.5.5.Determine the average sample mean for all Determine the average sample mean for all samples.samples.6.6.Using the size of the samples, determine the Using the size of the samples, determine the values of Dvalues of D33 and D and D44..7.7.Compute the UCL and the LCLCompute the UCL and the LCL


R Chart Formulas

R Charts

RR

k

LCL R

UCL R

DD

3

4


Demonstration Problem 18.2: 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 Diameter

UCL = .2725

Average = .1360

LCL = .0000


P ChartsMonitor Monitor proportion in noncomplianceproportion in noncompliance

1.1.Decide on the quality to be measured.Decide on the quality to be measured.2.2.Determine a sample size.Determine a sample size.3.3.Gather 20 to 30 samples.Gather 20 to 30 samples.4.4.Compute the sample proportion for each Compute the sample proportion for each sample.sample.5.5.Determine the average sample proportion Determine the average sample proportion for all samples.for all samples.6.6.Compute the UCL and the LCLCompute 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 + 3P Q

n

LCL P 3P Q

n

non

non

n

n

where

n

Pp

kwhere p

k

where Q P1


Demonstration Problem 18.3: Twenty Samples of Bond Paper

Sample n

Number Out of

Compliance Sample n

Number Out of

Compliance1 50 4 11 50 22 50 3 12 50 63 50 1 13 50 04 50 0 14 50 25 50 5 15 50 16 50 2 16 50 67 50 3 17 50 28 50 1 18 50 39 50 4 19 50 1

10 50 2 20 50 5


Demonstration Problem 18.3: 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 0.127 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


Demonstration Problem 17.3:Centerline, UCL, and LCL Computations

Pp

k

UCL PP Q

n

LCL PP Q

nLCL

. . . ..

.. .

.

.. .

.

08 06 02 10

20053

3 053 3053 947

50148

3 053 3053 947

50042

0


Demonstration Problem 17.3:P Control Chart

0.000.020.040.060.080.100.120.140.16

0 5 10 15 20

Sample Number

P = .053

UCL = .148

LCL = 0

p


c ChartsMonitor number of nonconformances per item

1.Decide on nonconformances to be evaluated.2.Determine the number of items to be studied (at least 25).3.Gather items.4.Determine the value of c for each item by summing the number of nonconformances in the item.5.Determine the average number of nonconformances per item.6.Determine the UCL and the LCL


c Chart Formulas

ci

where

UCL c c

LCL c c

c c c ci

1 2 3

3

3

: i = number of items

number of nonconformities per item

= +

-

ic


Demonstration Problem 18.4:Number of Nonconformities in Oil Gauges

Item Number

Number of Nonconformities

Item Number

Number of Nonconformities

1 2 14 22 0 15 13 3 16 44 1 17 05 2 18 26 5 19 37 3 20 28 2 21 19 0 22 3

10 0 23 211 4 24 012 3 25 313 2


Demonstration Problem 18.4: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 6 2

3 2 0 3 2 0 2 2

0

.

. . .

. . .

= +

-


Demonstration Problem 18.4: c Chart

01234567

0 5 10 15 20 25Item Number

c

UCL = 6.2

LCL = 0

c = 2.0


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


Interpreting Control Charts:7 Consecutive Increasing Points

UCL

LCL

Centerline


Interpreting Control Charts:2 out of 3 Consecutive Points in Outer 1/3

UCL

LCL

Centerline


Interpreting Control Charts:4 out of 5 Consecutive Points in Outer 2/3

UCL

LCL

Centerline


Acceptance Sampling

• Acceptance sampling is the inspection of a sample 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

cxx

1

1

1 1

2 2

2 2

cr

x rx cx c

Single SamplePlan

Double SamplePlan


Producer’s and Consumer’s Risk

State of Nature

Null True Null False

Actions

Fail toReject Null

Type II errorCorrectDecision --

Consumer’sRisk

Reject NullType Ierror --Producer’sRisk

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 braces

to inspect.)• X is the number of nonconforming braces in the

sample of 15.• A 2% nonconformance rate is acceptable to the

consumer (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 is the probability that the consumer will not reject the lot (consumer’s risk)?

Business Statistics, 4e, by Ken Black. © 2003 John Wiley & Sons.

Bicycle Manufacturer Example:Sampling Plan

n

c

x

x

15

1

1

1

=

Accept lot if

Reject lot if

n

c

x

x

15

1

1

1

=

Accept lot if

Reject lot if


Bicycle Manufacturer Example:Analysis for 2% Nonconforming Braces

0

0 15 1 14

02

0 115

0

15

1

1 9647 0353

02 98 02 98

p

P x P x

.

.

. .

. . . .

Probability of accepting

Probability of rejecting

9647

Producer’s RiskProducer’s Risk


Bicycle Manufacturer Example: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’sRisk


Bicycle Manufacturer Example: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


Bicycle Manufacturer Example: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



OC Curven = 20; c = 2

0.00.10.20.30.40.50.60.70.80.91.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0Fraction Defectives



n = 20c = 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)

business statistics, 4e, by ken black. © 2003 john wiley & sons. 18-1 business statistics, 4e...

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