list of topics in lecture 1 ioe 466 statistical quality …. shi, the university of michigan,...

1J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

Statistical Quality ControlIOE 466

Statistical Quality Control(MW 12:00 – 1:30pm, Chrysler Center media classroom #165)

Instructor: Prof. Jianjun Shi1784 IOE

Department of Industrial and Operations EngineeringThe University of Michigan

[email protected], 734-763-5321(O), 734-764-3451(Fax)


List of Topics in Lecture 1

• What are the course grading rules?• What do you expect to learn from this course?• Introduction of SPC

– Why is quality control important?– Introduce quality engineering terminology– Evolution of SPC methods

• What are the statistical methods for quality improvement?• Examples: How to use SPC in real applications?


Background Overview• INSTRUCTORS

- Background- Availability

• GSIs- Justin Wayne Kile <[email protected] Yu-Li Huang <[email protected]>

• CPD Student Support- Hongbin Jia <[email protected]>

• TEXT- Author- Prerequisites- Other References- Coursepack

• COURSE- Attendance Policy- Computers


Introduce Yourself to Others(2 minutes)

• Name• Department• Undergraduate or graduate students?• Other Background and more?


Prerequisites: Review Questions

• Do you know what is Normal distribution?• Do you know the difference between mean and median?• Do you know how to calculate the variance of a sample?• Do you know the meaning of p-value in hypothesis

testing?• If X follows normal distribution with mean 2 and

standard deviation 3, do you know how to use table to get the probability of X<0?

• Do you know what is type I and type II errors in hypothesis testing?

• Do you know what is partial derivative?• Can you calculate the integral and derivative of x2?• Do you know how to calculate the inverse of a 2-by-2

matrix and the product of any two matrices by hand?

How many “Yes” do you get?


More Questions

• Are you familiar with hypogeometric distribution, Binomial distribution and Poisson distribution?

• Do you know how to estimate and test the difference in variances of two normal distribution?

• Are you familiar with OC-curves?• Do you know when should use S chart and when use R

chart to monitor process variability?• Do you know what’s the advantage of CUSUM and EWMA

chart to Shewhart control chart?• Do you know PCR and PCRk?• Do you know p-chart?


Grading Policy

• Homework 30% • Exam 1 35%• Exam 2 35%

- Homework should be handed in during class on the due date;- No late homework is acceptable;


IOE 466Statistical Quality Control

• Fundamentals of Engineering Statistics• Statistical Methods in Quality Improvements• Statistical Process Control• Introduction to Advanced Quality Control

Topics


Objectives

• Introduce statistical tools and concepts that are useful for product/process quality improvements

• Demonstrate the procedures of implementation of the quality engineering tools in various applications

• This is NOT a course on mathematical statistics


PRODUCTIONPROCESS

• Leave Alone• Adjust• Stop

• Accept• Rework• Scrap

Analysis ofVariations

(Samples)

"Feedback" "Feedforward"

OURSE VERVIEWC O

"Charting" "Process Capability Analysis"

Measurement


Contents

• Statistical Methods: Modeling & Inferences Ch.2 - Ch3• Statistical Process Control:

– Philosophy Ch.4– Control Charts Ch.5,Ch.6 & Ch8

• Process Capability Analysis Ch.7• Advanced Topics Ch 9, 10, 11

– SPC for Short Run – Multivariate Quality Control– SPC with Correlated Data– Frontier of the Current Research Ideas

• Acceptance Sampling Ch14


WHY IS QUALITY IMPORTANT? --- Quality, Productivity and Cost

1. Consumer awareness and quality / performance sensitive.

2. Product liability laws.3. Costs of labor, energy, and materials.4. Competition is doing it.5. Quality, Productivity and Cost are

complementary !


Quality Engineering Terminology— “Quality” Definitions

• Quality means fitness for use• Quality is evaluated by the variability,

which is inversely proportional to the variability

• Quality improvement is the reduction of variability in processes and products


How to describe a product not meet “Quality” requirements?

• Nonconformity—A departure of a quality characteristic from its intended level or state that occurs with a severity sufficient to cause an associated product or service not to meet a specificationrequirement.

• Nonconforming unit—A unit of product or service containing at least one nonconformity .

• Defect—A departure of a quality characteristic from its intended level or state that occurs with a severity sufficient to cause an associated product or service not to satisfy intended normal, or reasonably foreseeable usage requirements.

• Defective (Defective Unit)—A unit of product or service containing at least one defect, or having several imperfections that in combination cause the unit not to satisfy intended normal, or reasonably foreseeable, usage requirements.


TWO COMPONENTS OF QUALITY

Manufacturing Industries Service Industries

Product features

Performance AccuracyReliability TimelinessDurability CompletenessEase of use Friendliness and courtesyServiceability Anticipating customer needsAesthetics Knowledge of serverAvailability of options Esthetics

and expandability ReputationReputation

Freedom from deficiencies

Product free of defects Service free of errorsand errors at delivery, during original and futureduring use, and during service transactionsservicing Sales, billing, and other

business processes freeof errors


Quality Engineering and Process Characteristics

• Quality Engineering:– A set of operational, managerial and engineering activity to ensure the

quality characteristics at a nominal level • Attributes/Variables depend on the measurements of the

quality characteristics– attributes: discrete data

• to judge each product as either conforming or non-conforming, or to count the number of nonconformities appearing on a unit of product

– variables:continuous measurement of quality characteristics

• Specifications: desired measurements for the quality characteristics

– upper/lower specification limits (USL/LSL)


WHO’s Responsible for Quality?

1. Product planning, marketing, and sales. 2. Development engineering.3. Manufacturing engineering. 4. Purchasing. 5. Manufacturing management.6. Manufacturing employees.7. Inspection and test. 8. Packaging and shipping. 9. Customer service.


It’s NOT My Job!!!

• This is a story about four people named EVERYBODY, SOMEBODY, ANYBODY, and NOBODY.

• There was an important job to be done, and EVERYBODY was sure that SOMEBODY would do it. ANYBODY could have done it but NOBODY did it. SOMEBODY got angry because it was EVERYBODY'S job. EVERYBODY thought ANYBODY could do it, but NOBODY realized that EVERYBODY wouldn’t do it. It ended up that EVERYBODY blamed SOMEBODY when NOBODY did what ANYBODY could have done.


What are the statistical methods for quality improvement?


A Shewhart Control Chart

201000-3

-2

-1

0

1

2

3

Time

LCL

Average

UCL

mm

Dr. Shewhart first proposed usage of control chart in 1924, which is the start of “statistical process control”


Statistical Quality Control Methods

• Three major quality control methods are – SPC, – DOE (design of experiments), and – acceptance sampling.

• This course will cover two of them: SPC and acceptance sampling. DOE is covered in IOE 465.

• A simple example to illustrate the three methods: A typing example.


Typing Example

• Consider improvement of typing quality, which is measured by typing accuracy and clearness.

• Acceptance sampling—several sample pages are inspected from every “lot” (for example, every 100 typed pages). If the selected sample pages have satisfactory quality, the whole “lot” is accepted. Otherwise, the whole lot is rejected and rework should be done.

• Statistical process control (SPC)—Every hour one page is selected and its quality is measured. Plot the measurements from each hour on a control chart. If a shift of quality is detected, the root cause of this shift (such as typist tiredness, lack of ink) is identified and fixed.

• Design of experiment (DOE)—Conduct experiments with combinations of different typists, typewriters, papers, working schedules. The best combination of these factors are selected toachieve optimal typing quality.


AcceptanceSampling

Perc

ent o

f App

licat

ion

Time

DesignedExperiments

ProcessControl

0

100

Phases in Quality Engineering Methods


Variation Reduction Approaches


Statistical Process Control Procedures

Process

DecisionFormulate action Data analysis

EvaluationFaulty discovery

Diagnosis

Take actionImplementation

Data collectionObservation


Quality Improvement Example

Problem: Cowl side reinforcement panel I/O variation

2L (y=2.91) 5L (y=3.16)

SPC ExampleSPC Example


Data Analysis: spikes with a cycle of 26

Process knowledge: welding robot changing tip dressing after 26 welding


200

180

160

140

120

10080604020

00.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.004/10/92 Sample of 160

AFTER CORRECTION

BEFORE CORRECTION

Time

Y

[mm

]

SPC Example (Cont’d)SPC Example (Cont’d)


Root causes and action:

Welding robot #3 after tip dressing, after 26 panels, skip two welding spots

Welding spots

Missed welding spots

Cowlside Reinforcement panel

Evaluations

Sensors6 SIGMA 6 SIGMA

BEFORE AFTER

2 L_Y 2.91 2.02 3.16 1.44

1.0 1.4 2.0 0.9

2 L_Z5 L_Y5 L_Z


SPC Example (Cont’d)SPC Example (Cont’d)


Total Quality Management (TQM)

• Statistical techniques must be implemented within a management system that is quality driven.

• One of the managerial frameworks used is total quality management.• TQM is a strategy for implementing and managing quality improvement

activities on an organization-wide basis.• TQM emphasizes on continuous improvements.


Total Quality Person

Companies today are turning to total quality management to improve their capabilities. To adapt to TQM, management styles have had to change to new form of employee-employer relationships. But what about the individuals involved in this transformation? Are they TQM people?


Personal Leadership

1. I take on responsibility for my actions and don’t rely on others to plan my future.

Rarely Sometimes Always

1 2 3 4 5 6 7 8 9 10

2. I enjoy the people and things in my environment.


1 2 3 4 5 6 7 8 9 10


Planning

3. Every day I take time to plan my daily activities around that which is important to me.


1 2 3 4 5 6 7 8 9 10

4. I have a good sense of how personal values, strengths, and weaknesses align with what I am doing.


1 2 3 4 5 6 7 8 9 10


Improvement

5. I constantly strive to measure whether I am meeting my personal goals.


1 2 3 4 5 6 7 8 9 10

6. I celebrate my successes and improvements.


1 2 3 4 5 6 7 8 9 10


Interpret Your Score12-17: Grade F. You might want to adopt some of these individual total quality strategies to get your life back on track.

18-25: Grade D. You might want to analyze your daily living patterns and goals in life. You do not demonstrate and individual total quality philosophy.

26-31: Grade C. You demonstrate some patterns of a total quality person but need to more consistent on daily basis.

32-45: Grade B. You have a good individual foundation in total quality principles and could serve as a role model for others.

46-60: Grade A. You are a great total quality role model, with a solid set of principles in leadership, planning, and continuous improvement.


Six-Sigma Program

• First developed by Motorola in the late 1980s.• To reduce the process variability so that the

specification limits are six standard deviations from the mean. Then there will only be about 2 parts per billion defective.

• Four phases of six sigma project: Measure Analyze Improve Control

• SPC is a major tool of Six-Sigma• An introduction to six-sigma is posted on course

web site.


Chapter 2: Probability Review- Fundamentals of Engineering Statistics

• Describing Variation– Frequency Distribution &

Histogram– Numerical Summary of Data– Probability Distribution

• Important Distributions• Some Useful Approximations


Need for Statistics

• Some variation is inevitable in manufacturing processes.

• Variation reduction is one of the major objectives in quality control

• Variation needs to be described, modeled, and analyzed

How to do it?


Describing VariationDescribing VariationMethod 1: Frequency Distribution & Histogram


An Example:Forged Piston Rings for Engines (Text Book P43-46)

• Variable Data (Table 2-2, P44): – the inside diameter of forged piston rings(mm)– 125 observations, 25 samples of 5 observations each.


Frequency Table & Frequency Histogram

• To construct a frequency table1. Find the range of the data

– start the lower limit for the first bin just slightly below the smallest data value

– b0<min(x), bm=max(x), (m: # of bins)

2. Divide this range into a suitable number of equal intervals– m=4 ~ 20, or (N is the total number of observations)

3. Count the frequency of each interval– if bi-1< x ≤ bi,

N


Frequency Distribution for Piston-Ring Diameter Table 2-3 (P45)

• Data range b0=73.965, bN=74.030– Min[x(i,j)]=73.967(i=14, j=2); max[x(i,j)]=74.030 i=1, j=1

• # of Bin m=13, Interval=(74.030-73.965)/13=0.005

• count for each bin: bi-1< x ≤ bi,


Fig. 2-4 (P44)Histogram for Piston-ring Diameter Data

- A graphical display of the frequency table


Interpretation based on the Frequency Histogram

Visual Display of Three Properties of Sample Data• Shape:

– roughly symmetric and unimodal• The center tendency or location

– the points tend to cluster near 74mm.• Scatter or spread range

– variability is relatively high (min=73.967; max=74.030)


Describing VariationDescribing VariationMethod 2: Numerical Summary of Data

• Center Tendency: sample average

• Scatter: sample variance or sample standard deviation

• Shape: skewness and kurtosis– skewness: measure the lack of symmetry of the distribution

symmetry; mean>median; mean<median– kurtosis: indicates the heaviness of the tails of the data distribution

larger has a heavier tail

n

)xx(M;

MMˆ;

)M(Mˆ

n

1i

ji

j22

422/3

2

31

∑=

−==β=β

n

xx

n

ii∑

== 1

;1

)(ˆ 1

2

22

−

−==σ

∑=

n

xxS

n

ii

0ˆ1 =β 0ˆ

1 <β0ˆ1 >β

2β


Describing VariationDescribing VariationMethod 3: Probability Distribution

• A Probability distribution is a mathematical model that relates the value of the variable with the probability of occurrence of thatvalue in the population.

• Two types of distributions: – Continuous: if the variable being measured is expressed on a

continuous scale– discrete :if the parameter being measured can only take on

certain values, e.g.. 1,2,3,4,..

p(xi)

p(x1)p(x2)

p(x5)p(x4)

p(x6)

p(x3)

p(x7)

x1 x2 x5x4 x6x3 x7x

f(x)

a bx

∫+∞

∞−

= 1dx)x(f ∑∞

=

=1i

i 1)x(p


Theoretical & Sampling Distribution

• If we consider each interval as one unit, then each rectangle in the histogram will have an area equal to its relative frequency. The total area of the rectangles will be unity

• If we could increase samples and make the intervals much smallerand still maintain enough data for each interval. When the intervals become narrower, the histogram will appear smoother.

• Extending this concept to the extreme case where the histogram becomes a smooth curve. This smooth curve is called a “theoretical probability distribution” or “theoretical distribution”.

Increase samples

Smoother

∫+∞

∞−

= 1dx)x(f


Review of Probability Distribution Calculation

Continuous Distribution Discrete Distribution

Probability ∫=≤≤b

a

dx)x(f)bxa{P )x(p)x(P ii =

Distribution mean ∫

+∞

∞−

=µ dxxxf )( ∑

∞

=

=µ1

)(i

ii xpx

Distribution variance ∫

+∞

∞−

µ−=σ= dxxfxxV )()()( 22 ∑∞

=

µ−=σ=1

22 )()()(i

ii xpxxV

Sample mean

Sample variance

n

xx

n

1ii∑

==

1n

)xx(Sˆ

n

1i

2i

22

−

−==

∑=σ


Important Distributions

1. Discrete Probability Distribution• Hypergeometric distribution • Binomial distribution• Poisson Distribution

2. Continuous Probability Distribution • Normal distribution • Chi-Square distribution• Student t distribution


Hypergeometric Distribution• Suppose that there is a FINITE population consisting of N items. Some

number , say D (D≤N), of these items fall into a class of interest. A random sample of n items is selected from the population without replacement, and the number of items in the sample that fall into the class of interest, say x, is observed. Then x is a Hypergeometric random variable with the probability distribution:

• Used as a model when selecting a random sample of n items without replacement from a lot of N items of which D are noncomforming or defective

• Excel function: HYPGEOMDIST(x,n,D,N)

⎟⎠⎞

⎜⎝⎛

−−

⎟⎠⎞

⎜⎝⎛ −=σ

1NnN

ND1

NnD2

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

nN

xnDN

xD

)x(p x=0, 1,…,min(n,D)

NnD

=µ

)!(!!

baba

ba

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛


Example 1: Special-purpose circuit boards are produced in lots of size N = 20. The boards are accepted in a sample of n =3 if all are conforming. The entire sample is drawn from the lotat one time and tested. If the lot contains D=3 nonconforming boards, what is the probability of acceptance?


Example 2: (Textbook Problem 2-28) A lot of size N = 30 contains five nonconforming units. What is the probability that a sample of five units selected at random contains exactly one nonconforming units? What is the probability that it contains one or more nonconformances?


Binomial Distribution

p(x) = ⎝⎜⎛

⎠⎟⎞n

x px (1 – p)n – x x = 0,1,2,...,n 0 ≤p≤ 1

E(x) = np V(x) = np(1 – p) [Note: V(x) < E(x)]

Bernoulli trials: A sequence of n independent trials, where the outcome of each trial is either a “success” or a “failure”

Binomial Distribution: If the probability of a success on any trial is a constant, p, the number of "success" x in n Bernoulli trials has the Binomial distribution

Assumption: (1) Constant probability of success p; (2) Two mutually exclusive outcomes; (3) All trials statistically independent; (4) Number of trials n is known and constantApplication: used as a model when sampling from an infinitely large population. The constant p represents the fraction of defective or nonconforming items in the population

Excel Function: BINOMDIST(x,n,p,false) (True:accumulative probability)


• is the ratio of the observed number of defective or nonconforming items in a sample x to the sample size n

• the probability distribution of is obtained from the binomial

Estimation of Binomial Distribution Parameter

nxp =

p

xnx]na[

0x)p1(p

xn

}nax{P}anx{P}ap{P −

=

−⎟⎟⎠

⎞⎜⎜⎝

⎛=≤=≤=≤ ∑

p

n)p1(p2

p−

=σpp =µ


Example 1: Sixty percent of pulleys are produced using Lathe #1, 40% are produced using Lathe #2. What is the probability that exactly three out of a random sample of four production parts will come from Lathe #1 ?

Ans:


Example 2: (Textbook problem 2-23)A production process operates with 2% nonconforming output. Every hour a sample of 50 units of product is taken, and the number of nonconforming units counted. If one or more nonconforming units are found, the process is stopped and the quality control technician must search for the cause of nonconforming production. Evaluate this decision rule.


Example 3: A firm claims that 99% of their products meet specifications. To support this claim, an inspector draws a random sample of 20 items and ships the lot if the entire sample is in conformance. Find the probability of committing both of the following errors: (1) Refusing to ship a lot even though 99% of the items are in conformance.(False alarm) (2) Shipping a lot even though only 95% of the items are conforming.(Miss detection)


Example 4: (Textbook problem 2-25) A random sample of 100 units is drawn from a production process every half hour. The fraction ofnonconforming product manufactured is 0.03. What is the probability that if the fraction nonconforming really is 0.03?04.0p ≤=


Poisson Distribution

The number of random events occurring during a specified time period:

p(x) = e–λλx

x! x = 0, 1, 2,. . .

E(x) = λ V(x) = λ β1 = 1

λ β2 = 3 + 1

λ

Uses:a. number of "defects" per unitb. number of "defects" per unit of areac. number of random occurrences per unit of timed. approximate the binomial distribution with λ= np when np Š 5 and p Š .1 or �

.9

Assumptions: 1. The average occurrence rate (λ) is known and constant.2. Occurences are equally likely to occur during any time interval.3. Occurences are statistically independent.

λ=β

11

Excel Function: POISSON(x,λ, false) (True:accumulative probability)

PoissonBinomialthenttanconsnp,0p;nif

→=→∞→


Example 1: Arrivals of parts at a repair station are Poisson distributed, with a mean rate of 1.2 per day. What is the probability of no repairs in the next day?


Example 2: What is the probability that today the number of parts requiring repair will exceed the average by more than one standard deviation?


Example 3: Glass bottles are formed by pouring molten glass into a mold. The molten glass is prepared in a furnace lined with firebrick. As the firebrick wears, small pieces of brick are mixed into the molten glass and finally appear as defects (called "stones") in the bottle. If we can assume that stones occur randomly at the rate of 0.00001 per bottle, what is the probability that a bottle selected at random will contain at least one such defect?


Example 4: The billing department of a major credit card company attempts to control errors (clerical, keypunch, etc.) on customers' bills. Suppose that errors occur according to a Poisson distribution with parameter λ = 0.01. What is the probability that a customer's bill selected at random will contain one error?


Normal Distributionf(x) =

1

2πσ2 e–(x–µ)2/2σ2

– ∞ ≤ x ≤ ∞

E(x) = µ V(x) = σ2

β1 = 0 β2 = 3

f(x)

xµ

σ2

Pr(µ−σ≤x≤µ+σ)=68.26%Pr(µ−2σ≤x≤µ+2σ)=95.46%Pr(µ−3σ≤x≤µ+3σ)=99.73%)(}zPr{}axPr{

σµ−

Φ=σ

µ−≤=≤

aa

)1,0(N~z;),(N~x 2σµ

If x1, x2 are independently normally distributed variables, then y=x1+x2also follows the normal distribution, i.e. y~N(µ1+µ2,σ1

2+ σ22)

The Center Limit Theorem: if x1, x2, …, xn are independent random variables, with mean µi and variance σi

2, and if y=x1+x2+…+xn, then the distribution approaches the N(0,1) distribution as n approaches infinite.

Excel Function: NORMDIST(x,µ,σ,true)

∑∑==

σµ−n

1i

2i

n

1ii /)y(


Example 1: The tensile strength of a metal part is normally distributed with mean 40 LB. and standard deviation 8 LB. If 50,000 parts are produced, approximately how many would fail to meet a minimum specification limit of 34-LB tensile strength? Approximately how many would have a tensile strength in excess of 48 LB?


Example 2: Three shafts are made and assembled in a linkage. The length of each shaft, in centimeters, is distributed as follows:

Shaft 1: N ~ (75, 0.09) Shaft 2: N ~ (60, 0.16) Shaft 3: N ~ (25, 0.25)

(a) What is the distribution of the linkage? (b) What is the probability that the linkage will be longer than 160.5 cm?


Example 3: A quality characteristic of a product is normally distributed with mean µ and standard deviation one. Specifications on the characteristic are 6 < x < 8. A unit that falls within specifications on this quality characteristic results in a profit of C0. However, if x < 6, the profit is –C1, while if x > 8, the profit is –C2. Find the value of µ that maximizes the expected profit.


Chi–Squared Distribution (with degrees of freedom ν)

E(x) = ν V(x) = 2ν β1=8/ν, β2=3+12//ν,

• The Chi-squared Distribution is associated with squared normalrandom variables.

• The most popular use of this distribution is for testing hypothesesabout variances of samples from normal distributions.

2/y1)2/n(2/n

2

ey)2/n(2

1)y(f −−

Γ= Γ(

ν2 ) = (

ν2 – 1) (

ν2 – 2)... 3 • 2 • 1 for ν even

= (ν2 – 1) (

ν2 – 2)...

52 •

32 •

π2 for ν odd

y>0

2n

22

21 xxxy +++=

Y follows If x1, x2, …, xn are normally and independently distributed random variables

21n−χ


Student t Distribution (with degrees of freedom ν)

f(x ) = 1π ν

Γ ⎣⎢

⎡⎦⎥⎤ν + 1

2

Γ ⎣⎢⎡

⎦⎥⎤ν

2

⎝⎜⎛

⎠⎟⎞

1 + x 2

ν

– (ν + 1 )2

E (x ) = 0 V (x ) = ν

ν – 2 β 1 = 0

β 2 = 3 + 6

n – 4 fo r n > 4

N o te : A s n → ∞ th e d is tr ib u tio n o f x (d is tr ib u ted a s a S tu d en t tran d o m v a riab le ) ap p ro ach e s th a t o f a s tan d a rd n o rm a l ran d o mv aria b le .

Γ (ν2 ) = (

ν2 – 1 ) (

ν2 – 2 )... 3 • 2 • 1 fo r ν e v en

= (ν2 – 1 ) (ν

2 – 2 )... 52 •

32 • π

2 fo r ν o d d

Application: If x and y are independent standard normal and chi-square random variable respectively, then is distributed as t with k degrees of freedom. k/y

xt =


INTERRELATIONSHIPS BETWEEN DISTRIBUTIONSHypergeometric, Binomial, Poisson, Normal

N: population sizen:sample size

Sampling without replacementin finite population

The sum of a sequence of n Bernoulli trials in infinite population with probability of success p

Number of defects per unit

p=D/N, n

λ=np constant

Poissonif λ ≥15

µ= λ, σ2= λ

If np>10 and p ≥0.5µ=np, σ2=np(1-p)

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−

Φ−⎟⎟⎠

⎞⎜⎜⎝

⎛

−−+

Φ≈=)1(

5.0)1(

5.0)Pr(pnpnpa

pnpnpaax

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−

Φ−⎟⎟⎠

⎞⎜⎜⎝

⎛

−−+

Φ≈≤≤)1(

5.0)1(

5.0)Pr(pnpnpa

pnpnpbbxa

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−µ

Φ−⎟⎟⎠

⎞⎜⎜⎝

⎛

−−ν

Φ≈ν≤≤µnpp

pnpp

pp/)1(/)1(

)ˆPr(

Normal

Binomialif larger n, smaller p <0.1

Hypergeometricfinite population

if n/N≤0.1


Example 1: (Textbook Problem 2-27) An electronic component for a laser range-finder is produced in lots of size N = 25. An acceptance testing procedure is used by the purchaser to protect against lots that contain too many nonconforming components. The procedure consists of selecting five components at random from the lot (without replacement) and testing them. If none of the components is nonconforming, the lot is accepted.

a. If the lot contains three nonconforming components, what is the probability of lot acceptance?

b. Calculate the desired probability in (a) using the binomial approximation. Is this approximation satisfactory'? Why or why not?

c. Suppose the lot size was N=150. Would the binomial approximation be satisfactory in this case?

d. Suppose that the purchaser will reject the lot with the decision rule of finding one or more nonconforming components in a sample of size n, and wants the lot to be rejected with probability at least O.95 if the lot contains five or more nonconforming components. How large should the sample size n be?


Example 2: A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly located on those pages. Find the probability that a random selection of 50 pages will contain no errors. Find the probability that 50randomly selected pages will contain at least two errors.


Example 3: A sample of 100 units is selected from a production process that is 2% nonconforming. What is the probability that will exceed the true fraction nonconforming by k standard deviations, where k = 1, 2, and 3?

p


Interrelations Normal, Chi-Squared, Student t, F

1. N(0,1)

χ2(ν)/ν = t(ν)

2. χ2(ν) / ν = F(ν,∞)

3. F(α,ν1,ν2) = 1F(1-α,ν2,ν1)

4. χ2(α,ν) = ν F(α,ν,∞)

5. t(α/2, ν) = F(α,1,ν) 1/2

6. t(∞) = N(0,1)


Chapter 3 Inference About Process Quality- Statistics Review

• Motivation• Estimation

– point estimation– interval estimation

• Hypothesis Testing– Definition– Testing on means

• known and Unknown variance

– Testing on Variance


The need of “Statistical Inference”

• The parameters of a probability distribution are unknown.– Estimation of Process Parameters

• The parameters of a process can be time varying, how do we identify a process change?– Hypothesis Testing


Random Samples

• Random Sample: – Sampling from an infinite population or finite

population with replacement: A sample is selected so that the observations are independently and identically distributed.

– Sampling n samples from a finite population of N items without replacement if each of the possible samples has an equal probability of being chosen

⎟⎟⎠

⎞⎜⎜⎝

⎛nN


Terminology Definition

• Estimate: a particular numerical value of an estimator, computed from sample data.

– Point estimator: a statistic that produces a single numerical value as the estimate of the unknown parameter

– Interval estimator: a random interval (or called confidence interval) in which the true value of the parameter falls with some level of probability.

• Statistic: – any function of the sample data that does not contain unknown

parameters.

• Sampling distribution: – The probability distribution of a statistic.


METHODS FOR ESTIMATION1. METHOD OF MOMENTS (MOM):

Principle:if E{|x|r}<∝, then sample rth moment converges with probability 1 to the population rth moment when sample size is larger enough.

Analysis procedures:• If p.d.f has k unknown parameters, equating the first k

population moments to the first k sample moments.• Solve k parameters from these simultaneous equationsProperty:• Simple to generate but may not have desired

properties


METHOD OF MOMENTS (MOM): Example

Example: f(x) = λ e –λx , x ≥ 0 , λ > 0

Population Mean (First Moment): E(x) = 1λ

Sample Mean: x−

Estimate: λ = 1

x−

Exponential:

Poisson: E(x)=λ, thus, x=λ


METHODS FOR ESTIMATION2. Method of Maximum Likelihood Estimation(MLE)

a.) Given x1, x2, ..., xn from f(x) define

L = ∏i=1

nf(xi) or L* = ln(L)

b.) Maximize L or L* usually by setting dL*

d(parameter of interest) = 0 and

c.) Solve system of simultaneous equations.

Usually preferred to MOM since the MLE's are

1. Consistent2. Asymptotically Normal3. Asymptotically Efficient4. May not be unbiased.


METHOD OF MLE: Example

Exponential: Suppose f(x) = λ e –λx , x ≥ 0 , λ > 0

L = ∏i

n λ e –λxi = λn e –λΣxi

L* = ln(L) = n ln λ – λ ΣxidLdλ

* =

nλ – Σxi = 0

Thus the estimate λ = 1

x− (same as MOM) .


Assessment of Estimation

A. PROPERTIES

1. UNBIASED: An estimate θ∧

of parameter θ is unbiased if

E( θ∧

) = θ

2. CONSISTENT: An estimate θ∧

of parameter θ is consistent if

E( θ∧

– θ )2 → 0 as n → ∞

3. EFFICIENT: θ∧

1 is more efficient than θ∧

2 if

E( θ∧

1 – θ1 )2 < E( θ∧

2 – θ2 )2

it has a minimum variance


METHODS FOR ESTIMATION3. Interval Estimation

• Estimate the interval between two statistics that include the true value of the parameter with some probability– Example: Pr{ L≤ µ ≤ U}=1-α– The interval L≤ µ ≤ U is called a 100(1- α)% confidence interval (C.I.)

for the unknown mean µ– two side C.I. (L is lower confidence limit, U is upper confidence limit)– single side C.I.:

• lower side L≤ µ , Pr{ L≤ µ }=1-α• upper side µ ≤ U, Pr{ µ ≤ U}=1-α

• Analysis procedures:– get the samples– compute the statistic– determine the statistic reference distribution– select confidence level– find the lower and/or upper confidence limits based on the reference

distribution

xµ UL

α/2α/2


If x is a random variable with unknown mean µ and known variance σ2, what is estimation interval for mean µ?

– Select a statistic– The approximate distribution of is regardless of the

distribution of x per the central limit theorem.– Given confidence level α, then

• 100(1-α)% two-side confidence interval on µ is:

• 100(1-α)% upper confidence interval on µ is:

• 100(1-α)% lower confidence interval on µ is:

Interval Estimation

∑=

=n

ii nxx

1/)(

)/,( 2 nN σµ

nZx

nZx σ

+≤µ≤σ

− αα 2/2/ 2/}Pr{ 2/ α=≥ αZzwhere

nZx σ

+≤µ α

µ≤σ

− α nZx

x


Example 1: The strength of a disposable plastic beverage container is being investigated. The strengths are normally distributed, with a known standard deviation of 15 psi. A sample of 20 plastic containers has a mean strength of 246 psi. Compute a 95% confidence interval for the process mean.


Example 2: A chemical process converts lead to gold. However, the production varies due to the powers of the alchemist. It is known that the process is normally distributed, with a standard deviation of 2.5 g. How many samples must be taken to be 90% certain that an estimate of the mean process is within 1.5 g of the true but unknown mean yield?


Interval Estimation of the Binomial Distribution Parameter with A Larger Sample

Size

• From the central limit theorem: p =x/n~ Normal (p, p(1-p) /n )

Example 1: (Textbook Problem 3-18) A random sample of 200printed circuit boards contains 18 defective or nonconforming units.Estimate the process fraction nonconforming. Construct a 90% two-sided confidence interval on the true fraction nonconforming in theproduction process.


Hypothesis Testing

• Statistical hypothesis:– a statement about the values of the parameters of a

probability distribution• Hypothesis testing:

– Making a hypothesis concerning what we believe to be true and then use sampled data to test it.

• Two Hypotheses (Two Competing Propositions)– Null Hypothesis H0: will be rejected if the sample data do not

support it.– Alternative Hypothesis H1: a hypothesis different from the null

hypothesis


Hypothesis Testing Procedures

1) State the null and alternative hypothesis, and define the test statistic.

2) Specify the significance level α.3) Find the distribution of the test statistic

and the rejection region of H0.4) Collect data and calculate the test

statistic.5) Compare the test statistic with the

rejection region.6) Assess the risk.


TESTS FOR COMPARING ONE POPULATION MEAN WITH A

STANDARDA ssum ing K now n P opu lation V ariance

x− – µσ / n ~ N (0 ,1 )

R ejec t H 0: µ ≠ µ0 if ⎪⎪⎪⎪

⎪⎪⎪⎪x− – µ0

σ / n > Z (α /2 )

R ejec t H 0: µ < µ0 if x− – µ0σ / n < -Z (α )

R ejec t H 0: µ > µ0 if x− – µ0σ / n > Z (α )

nZx

nZx σ

+≤µ≤σ

− αα 2/2/

00 :H µ=µ H1


Assuming Unknown Population Variance

x− – µs/ n ~ t(n-1)

Reject H0: µ ≠ µ0 if ⎪⎪⎪⎪

⎪⎪⎪⎪x− – µ0

s/ n > t(α/2, n-1)

Reject H0: µ < µ0 if x− – µ0s/ n <- t(α, n-1)

Reject H0: µ > µ0 if x− – µ0s/ n > t(α, n-1)

TESTS FOR COMPARING ONE POPULATION MEAN WITH A STANDARD

nstx

nstx 2/2/ αα +≤µ≤−

00 :H µ=µ H1


Example 3: The mean time it takes a crew to restart an aluminum rolling mill after a failure is of interest. The crew was observed over 25 occasions, and the results were = 26.42 minutes andvariance S2 =12.28 minutes. If repair time is normally distributed, find a 95% confidence interval on the true but unknown mean repair time.

x


Example 4: The life of a battery used in a cardiac pacemaker isassumed to be normally distributed. A random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lives are obtained.

Construct a 90% two-sided confidence interval on mean life in the accelerated test.

2 5 .5 h 2 6 .1 h

2 6 .8 2 3 .2

2 4 .2 2 8 .4

2 5 .0 2 7 .8

2 7 .3 2 5 .7


TESTS FOR COMPARING TWO POPULATION MEANS

nnxxnnxx2

22

1

21

2/2

_

1

_

212

22

1

21

2/2

_

1

_

-- σ+

σ+≤µ−µ≤

σ+

σ− αα ZZ

Assume Known Population Variances

)1,0(~-

nn

xx

2

22

1

21

2

_

1

_

Nσ

+σ

Reject 210 :H µ≠µ if 2/

2

22

1

21

2

_

1

_

xx -α>

σ+

σZ

nn

Reject 210 : µµ <H if α−<σ

+σ

Z

nn 2

22

1

21

2

_

1

_

xx -

Reject 210 : µµ >H if α>σ

+σ

Z

nn 2

22

1

21

2

_

1

_

xx -

210 :H µ=µ H1


Example 5: A bakery has a line making Binkies, a big-selling junkfood. Another line has just been installed, and the plant managerwants to know if the output of the new line is greater than that ofthe old line, as promised by the bakery equipment firm. 12 days ofdata are selected at random from line 1 and 10 days of data are

selected at random from line 2, with x– 1

= 1124.3 cases and

x– 2

= 1138.7. It is known that σ12= 52 and σ2

2 = 60. Test the

appropriate hypotheses at α = 0.05, given that the outputs arenormally distributed.


1. Assume Unknown Population Variances

a) Assume Homogeneity (22

21

2 σσσ == )

)2(~11

-21

21

2

_

1

_

xx −++

nnt

nnS p

where 2

)12( 22

)11( 21

21

2

−+

−+−=

nn

snsnS p

Reject 210 :H µ≠µ if )2,2/(11

-21

21

2

_

1

_

xx −+α>+

nnt

nnSp

Reject 210 : µµ <H if )2,(

11-

21

21

2

_

1

_

xx −+α−<+

nnt

nnS p

Reject 210 : µµ >H if )2,(

11-

21

21

2

_

1

_

xx −+α>+

nnt

nnS p


nnxxnnxx21

2,2/2

_

1

_

2121

2,2/2

_

1

_ 11-11-2121

++≤µ−µ≤+− −+α−+α pnnpnn StSt;

210 :H µ=µ H1


2. Assume Unknown Population Variances

b) Assume Heterogeneity (22

21 σσ ≠ )

)(~-

2

22

1

21

2

_

1

_

xx vt

ns

ns

+ where 2

1

)/

1

)/

)//

2

22

2

2

1

21

2

1

22

2

21

2

1

(((

−

++

+

+=

n

n

n

n

nnv

ssss

Reject 210 :H µ≠µ if ),2/(-

2

22

1

21

2

_

1

_

xx vt

ns

ns

α>

+

Reject 210 : µµ <H if ),(-

2

22

1

21

2

_

1

_

xx vt

ns

ns

α−<

+

Reject 210 : µµ >H if ),(-

2

22

1

21

2

_

1

_

xx vt

ns

ns

α>

+


210 :H µ=µ H1


Textbook problem: 3-11. Two quality-control technicians measured the surface finish of a metal part, obtaining the data shown below. Assume that the measurements are normally distributed. Technician 1 Technician 2

1.45 1.54 1.37 1.41 1.21 1.56 1.54 1.37 1.48 1.20 1.29 1.31 1.34 1.27

1.35 Assuming that the variances are equal, construct a 95% confidence interval on the mean difference in surface-finish measurements.


να,t


TESTS FOR COMPARING ONE "NORMAL" POPULATION VARIANCE WITH A STANDARD

2

2s)1n(σ− ~ χ2

(n – 1)

Reject H0: σ ≠ σo

if (n - 1) s2

σo2 > χ2

(α/2,n - 1) or (n - 1) s2

σo2 < χ2

(1 - α/2,n - 1)

Reject H0: σ < σo if

(n - 1) s2

σo2 < χ2

(1-α,n - 1)

Reject H0: σ > σo if

(n - 1) s2

σo2 > χ2

(α,n - 1)

2/}Pr{,)1()1( 21,2/

212

1,2/1

22

21,2/

2

α=χ≥χχ

−≤σ≤

χ−

−α−−α−−α

nnnn

SnSn

00 :H σ=σ H1

21n,2/2

22

1n,2/1S)1n(

−α−α− χ≤σ−

≤χ


Consider the data in Exercise 3-3. Construct a 90% two-sided confidence interval on the variance of battery life. Convert this into a corresponding confidence interval on the standard deviation of battery life.

2 5 .5 h 2 6 .1 h

2 6 .8 2 3 .2

2 4 .2 2 8 .4

2 5 .0 2 7 .8

2 7 .3 2 5 .7


TESTS FOR COMPARING TWO NORMAL POPULATION VARIANCES

1,122

22

21

21

21~

//

−−σσ

nnFSS With H0: σ1

2 = σ2

2

for H1: σ12

≠ σ22

Reject H0 if s1

2

s22 > F(α/2,n1–1,n2–1) or

s12

s22 < F(1–α/2,n1–1,n2–1)

for H1: σ12

< σ22

Reject H0 if s2

2

s12 > F(α,n2–1,n1–1)

for H1: σ12

> σ22

Reject H0 if s1

2

s22 > F(α,n1–1,n2–1)

µνανµα−−−α−−α− =≤σσ

≤ ,,2/,,2/11,1,2/22

21

22

21

1,1,2/122

21 /1,

1212FFF

SSF

SS

nnnn


(Textbook Problem 3-11 - AGAIN) Two quality-control technicians measured the surface finish of a metal part, obtaining the data shown below. Assume that the measurements are normally distributed. b. Construct a 95% confidence interval estimate of the ratio of the variances of technician measurement error. c. Construct a 95% confidence interval on the variance of measurement error for Technician 2.

Ans: x 1 = 1.3829, S1 = .11485, n1 = 7x 2 = 1.3763, S2 = .1249, n2 = 8

S p2 = 6 (.11485) 2 + 7 (.1249) 2

13 = .0145


The Use of P-Values in Hypothesis Testing

1. Traditional hypothesis testing:– Given α to determine whether the null hypothesis was rejected– Disadvantage:

• No information on how close to/far away from the rejection region• predefined α may not reflect different decision maker’s risk assessments

2. P-Value approach– P-Value: the smallest level of significance would lead to rejection of the

null hypothesis– if the predefined α>P= αmin, reject the null hypothesis

f(x)

xµ=0 Z0>0Z0<0

1-Φ(Z0)Φ(Z0)


Use of P Value for the Normal Distribution

H0: µ=µ0 , standard statistic Z0~N(0,1) – P=2[1-Φ|Z0|] for two- tailed test with H1: µ≠µ0

– P=1-Φ(Z0) for an upper-tailed test with H1: µ>µ0

– P=Φ(Z0) for an lower-tailed test with H1: µ<µ0

– e.g. Textbook Page 100, P=1-Φ(Z0)=0.00023, If α>P then rejected. If α=0.01 rejected; however, If α=0.00001, not rejected.

f(x)

xµ=0 Z0>0Z0<0

1-Φ(Z0)Φ(Z0)


%)95(.I.Csideone1)z(%)90(.I.Csizetwo2/1)z(

α−=Φα−=Φ


Testing on Binomial Parameters

• To test whether the parameter p of a binomial distribution equals a standard value p0

• The test is based on the normal approximation to the binomial distribution

• The null hypothesis is rejected if |z0|>Zα/2

01

00

::

ppHppH

≠=

⎪⎪⎩

⎪⎪⎨

⎧

>−−−

<−−+

=0

00

0

000

0

0

)1()5.0(

)1()5.0(

npxifpnpnpx

npxifpnpnpx

Z

npppxZ

/)1( 00

00 −

−=Or using the central limit theorem

211

210

::

ppHppH

≠=

21

2211

21

210

ˆˆˆ;)11)(ˆ1(ˆ

ˆˆnn

pnpnp

nnpp

ppZ++

=+−

−= 21 ppif =

2/0 Z|Z| α>H0 is rejected if

Example 3-7, p108 108J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

Test on Poisson Distribution

• A random sample of n observation is taken, say x1, x2, ..,xn. Each {xi} is Poisson distributed with parameter λ. Then the sum x= x1+ x2+...+xn is Poisson distributed with parameter nλ.Example 3-9 show how to use Poisson distribution to do hypothesis test directly

• If n is large, =x/n is approximately normal with mean λ and variance λ/n

• Test hypothesis H0: λ =λ0H1: λ ≠λ0

• The null hypothesis would be rejected if |Z0|>Zα/2.

n/xZ

0

00 λ

λ−=

x


Two Types of Hypothesis Test Errors

• Type I error ( producer’s risk):– α = P{type I error} = P{reject H0 |H0 is true}

=P{product is rejected| but product is good}

• Type II error (consumer’s risk):– β = P{type II error} = P{fail to reject H0 |H0 is false}

=P{product is not rejected|although product is bad}

• Power of the test:– Power = 1- β = P{reject H0 |H0 is false}


The Probability of Type II Error— Detection of a mean shift

• Type II error= β=Pr{H0 |H1 |}=Pr{within the control limits|has a mean shift}H0: µ = µ0

H1: µ = µ1≠ µ0 with known σ2 0if,01 >δδ+µ=µ

)nZ()nZ(

}H|n/Zxn/ZPr{

}H|HPr{

2/2/

12/02/0

10

σδ

−−Φ−σ

δ−Φ=

σ+µ≤≤σ−µ=

=β

αα

αα

OC curve see Fig. 3-7 P109• The larger the mean shift, the smaller the type II error• The larger the sample size, the smaller the type II error


Ch 4. Methods and Philosophy of SPC

• Chance Causes and Assignable Causes of Variations

• Statistical Basis of Control Charts• Implementation of SPC and Examples


MotivationHow to do?• Distinguish two process variations:

– Chance causes and assignable causes• Decide the status of a process

– in control– out of control

• Continuously improve quality


Chance and assignable causes of variationTextbook Fig. 4-1 P155


Chance Cause & Assignable Cause

• Chance causes/common causes/system faults/chronic problems– system problems/inherent problems (natural

variation/background noise)– “in statistical control”

• Assignable causes/special causes/local causes/sporadic problems– problems arise in somewhat unpredictable fashion (operator

error, material defects, machine failure)– “out of statistical control”

Textbook Fig. 4-1 P155


Purpose of Using Control charts- Improve Process and Reduce Process Variation

1. Most processes do not operate in a state of statistical control.

2. Consequently, the routine and attentive use of control charts will identify assignable causes. If these causes can be eliminated from the process, variability will be reduced and the process will be improved.

3. The control chart will only detect assignable causes. Management, operator, and engineering action will usually be necessary to eliminate the assignable cause.


Objectives of SPC

• To quickly detect the occurrence of assignable causes or process shifts so that investigations of the process and corrective actions may be undertaken before many nonconforming units are manufactured.

• Process Variation Reduction


Procedures of implementing SPC

– Monitoring the process and detecting process changes– Diagnosing the assignable causes– Providing corrective actions plans– Dealing with resistance to changes/actions– Instituting controls to hold the gains

Problem/VariationProcess Measurement/

ObservationData

Analysis Evaluation

Find RootCauses

Formulate Action

Chances Causes

Assignable CausesTake

Action


Continuous Improvement• "Continuous improvement" (called Kaizen by the Japanese) –

enduring efforts to act upon both chronic and sporadic problems and to make refinements to processes. – For sporadic problems, it means taking corrective action on

periodic problems; – For chronic problems, it means achieving better and better

levels of performance each year (move mean to target); – For process refinements, it means taking such action as

reducing variation around a target value.

Problem/VariationProcess Measurement/

ObservationData

Analysis Evaluation

Find RootCauses

Formulate Action

Chances Causes

Assignable CausesTake Action

Control mean close to the targetReduce variation


Example: Implementation of Continuous Quality Improvement

•Special cause: tool broken, operator injury•easy to fix•manufacturer problem

•Chronic problem: design problem, degradation•continuous improvement


Concept of Control Charts• Control Chart: is a graphical display of a quality characteristic that has

been measured or computed from a sample versus the sample number or time.

• Center Line – represents the average value of the quality characteristic corresponding to the in-control state (only chance causes are present.)

• Upper Control Limit (UCL), Lower Control Limit (LCL) – are chosen so that if the process is in control, nearly all of the sample points will fall between them.

201000-3

-2

-1

0

1

2

3

Time

LCL

Average

UCL

mm


The Basis of Control Charts

201000-3

-2

-1

0

1

2

3

Time

LCL

Average

UCL

mm

Distribution ofindividual

measurements x:

Distribution ofsample subgroupmean x-bar

n/1σ

n/2σ

n/3σ


General Model for a Control Chart

Let w be a sample statistic that measures some quality characteristics ofinterest, and suppose that the mean of w is µw and the standard deviationof w is σw. Then the center line, the upper control limit, and the lowercontrol limit become

UCL = µw + k σw

Center line = µw

LCL = µw - k σw

where k is the "distance" of the control limits from the center line,expressed in standard deviation units

3 sigma control limits:• Action limits: K=3 (p=0.0027)• Warning limits: K=2 (p=0.0455)

Probability limits (Western Europe):• Action limits: 0.001 limits (p=0.002)• Warning limits: 0.025 limits (p=0.050)


Control Charts vs. Hypothesis Test

• Control Charts• Control chart has UCL & LCL• The process is out of control if the data beyond the control limits

f(x)

xµ0 UCLLCL

α/2α/2

H0: µ = µ0H1: µ≠ µ0

if ⎪⎪⎪⎪

⎪⎪⎪⎪x− – µ0

σ/ n > Z(α/2)Reject H0

• Hypothesis Testing• Hypothesis testing has a rejection region• H0 is rejected if the data follow in the rejection region


Review of Two Types of Hypothesis Test Errors• Type I error ( producer’s risk):

– α = P{type I error} = P{reject H0 |H0 is true}=P{conclude out of control|although the process is truly in control}

• Type II error (consumer’s risk):– β = P{type II error} = P{fail to reject H0 |H0 is false}

=P{conclude in control|although the process is truly out of control}

• Power of the test:– Power = 1- β = P{reject H0 |H0 is false}

β

µ1

f(x)

xµ0 UCLLCL

α/2α/2


The Probability of Type II Error— Detection of a mean shift with a known σ

• Type II error= β=Pr{H0 |H1 |}=Pr{within the control limits|has a mean shift}H0: µ = µ0

H1: µ = µ1≠ µ0 with known σ2 0if,01 >δδ+µ=µ

)nZ()nZ(

}H|n/Zxn/ZPr{

}H|HPr{

2/2/

12/02/0

10

σδ

−−Φ−σ

δ−Φ=

σ+µ≤≤σ−µ=

=β

αα

αα

n/σδ


• The larger the mean shift, the smaller the type II error• The larger the sample size, the smaller the type II error

increased

OC curve with α=0.05 (P119, Fig. 3-9)

σδ= /||d


Two Types of Errors

– Type I error:– Concluding the process out of control when the process is

really in control– Type II error:

– Concluding the process in control when it is really out of control.

β

µ1

f(x)

xµ0 UCLLCL

α/2α/2


Summary of Type I and Type II Errors

You Conclude :"In Control"

Nature :"Out of Control"

"In Control"

"Out of Control"

Confidence 1–α

Producer Error, α

Power 1–β

Consumer Error, β


• ARL: The average number of points that must be plotted before a point indicates an out-of-control condition.

• Example: ARLin-control = 1/α= 1/0.0027 = 370. Even the process is in control, an out-of-control signal will be generated every 370 samples on the average.

Average Run Length (ARL)— In Control

The following table illustrates the possible sequences leading toan "out of control" signal:Run length Probability

1 α2 α (1– α)3 α (1– α)2

: : :k α (1– α)k–1

α=−

1ARL controlin


Average Run Length (ARL)—Out of Control

Remark: we want 1–β to be large. Thus, the "out of control"condition can be quickly detected.

• If the process is actually “out-of-control”, and the probability that the shift will be detected on

• the first sample is 1-β• the second sample is β(1-β)• the rth sample is βr-1(1-β)

•The expected number of samples taken before the shift is detected is

∑∞

=

−−− β−

=β−β=1r

1rcontrolofout 1

1)1(rARL


Example: Suppose that a control chart with 2-sigma limits is used to control a process. If the process remains in control, find the average run length until a false out-of-control signal is observed. Compare this with the in-control ARL for 3-sigma limits and discuss.


Sample Size and Sample Frequency- Operating-Characteristic (OC) Curve

• Strategies:– Small samples at short intervals (favorite in High volume or

more problem processes)– larger samples at longer intervals – Adaptive or variable sampling interval

• An OC curve shows the relationship between a process parameter (the mean for an X bar chart) and the probability of a type II error

• Average run length (average time to signal=ARL*sampling interval) is considered in design and then check the detection power.


Process Out of Control

Out of control process:• When one or more points fall beyond the control limit• Plotted points exhibit some nonrandom pattern of behavior

Description of nonrandom pattern• a nonrandom pattern with a longer run up or run down, or a run of

length 8 or more – Run up: a sequence of increasing observations– Run Down: a sequence of decreasing observations– Run: a sequence of observations of the same type – Run of length: the number of samples in a run

• a periodic pattern


Control Chart Patterns and Causes JUMPS IN PROCESS LEVEL

1. NEW SUPPLIER2. NEW WORKER3. NEW MACHINE4. NEW TECHNOLOGY5. CHANGE IN METHOD OR PROCESS6. CHANGE IN INSPECTION DEVICE OR METHOD

HIGH PROPORTION OF POINTS NEAR OUTER LIMITS

1. OVER CONTROL2. LARGE DIFF IN MATERIAL QUALITY, TEST METHOD3. CONTROL OF 2 OR MORE PROC. ON ONE CHART4. MIXTURES OF MATERIALS OF DIFFERENT QUALITY5. MULTIPLE CHARTERS6. IMPROPER SUBGROUPING

STRATIFICATION (LACK OF VARIABILITY)

1. INCORRECT CALCULATION OF CONTROL LIMITS2. SYSTEMATIC SAMPLING


Control Chart Patterns and Causes RECURRING CYCLES

1. TEMPERATURE AND OTHER CYCLIC ENVIRONMENTAL EFFECTS2. WORKER FATIGUE3. DIFFERENCES IN MEASURING DEVICES USED IN ORDER4. REGULAR ROTATION OF MACHINES OR OPERATORS5. SCHEDULED PREVENTIVE MAINTENANCE (R CHART)6. TOOL WEAR (R CHART)

TRENDS1. GRADUAL EQUIPMENT DETERIORATION2. WORKER FATIGUE3. ACCUMULATION OF WASTE PRODUCTS4. IMPROVEMENT OR DETERIORATION OF WORKER SKILL/EFFORT

(ESPECIALLY IN R CHART)5. DRIFT IN INCOMING MATERIALS QUALITY


SUMMARY OF OUT-OF-CONTROL CRITERIA— Western Electric Rules (Zone Rules for Control

Charts)

Enhance the sensitivity of control charts for detecting a small shift or other nonrandom patterns

1. One point plot outside 3σ limits.2. Two successive points plot outside 2σ limits3. Four consecutive points plot at a distance of 1σ or beyond

from the center line (one side)4 A run of length eight points

More other sensitizing rules for Shewhart control chart; Table 4-1, P176

The final type I error: the process is concluded out of control if any one of the rules is applied

∏=

α−−=αk

ii

1

)1(1 αi is the type I error of using one rule i alone


Rational Subgroups• Want each subgroup as homogeneous as possible• Want maximum opportunity for variation between groups.• Should be time ordered.• Should consist of items produced together for detection of a mean shift.


Implementing SPC- Magnificent SEVEN

1. Histogram2. Check Sheet3. Pareto Chart4. Cause and Effect Diagram5. Defect Concentration Diagram6. Scatter Diagram7. Control Chart


2. Check Sheet Example


3. Pareto Chart Example


4. Cause and Effect Diagram Example


5. Defect Concentration Diagram Example


6. Scatter Plot Example


CASE Example (Grant/Levenworth 1988) : Thread grinding for fitting used in aircraft hydraulic system.

Process• Inspection thread pitch diameter based on a given specifications

(37.5±12.5)• Total 20hrs with each hour 5 items

Question:• Is the process capable of producing such quality products?• Can the process or quality be further improved?


Sample Thread Pitch Diameter DataAircraft Fittings (Thread Pitch Diameter)

5 items sampled each hourValues in .0001 inches excess of 0.4000 in.

Sample Avg. R1 36 35 34 33 32 34.0 42 31 31 34 32 30 31.6 43 30 30 32 30 32 30.8 24 32 33 33 32 35 33.0 35 32 34 37 37 35 35.0 56 32 32 31 33 33 32.2 27 33 33 36 32 31 33.0 58 23 33 36 35 36 32.6 139 43 36 35 24 31 33.8 19

10 36 35 36 41 41 37.8 611 34 38 35 34 38 35.8 412 36 38 39 39 40 38.4 413 36 40 35 26 33 34.0 1414 36 35 37 34 33 35.0 415 30 37 33 34 35 33.8 716 28 31 33 33 33 31.6 517 33 30 34 33 35 33.0 518 27 28 29 27 30 28.2 319 35 36 29 27 32 31.8 920 33 35 35 39 36 35.6 6

33.6 6.2146J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

20

30

40

50

20

30

40

50

Thr

ead

Pitc

h

0 5 10 15 20 25

Sample Number

PITCH DIAMETER FOR AIRCRAFT FITTINGS

Upper Tolerance Limit

Lower Tolerance Limit

Nominal Target


MEAN 33.5500

LWL 31.1659

UWL 35.9341

LCL 29.9738

UCL 37.1262

27.5

30

32.5

35

37.5

40

Fitt

ing

Dim

ensi

on

0 5 10 15 20 25

Sample Number

Aircraft Fittings X-Bar Chart


MEAN 6.2000

LWL 1.5940

UWL 10.8060

UCL 13.1090

0

5

10

15

20

Fitt

ing

Dim

ensi

on

0 5 10 15 20 25

Sample Number

Aircraft Fittings Range Chart


Summary: What can a control chart do?

• Is a proven technique for improving productivity• Is effective in defect prevention• Prevent unnecessary process adjustment• Provide diagnostic information• Provide information about process capability


Ch 5 Control Charts for Variables

• Control Chart for and R• Control Chart for and S• Operating-Characteristic Function• Relationship of NTL, CL, and SL

XX


Need for Control of Both Mean and Variability• The number of nonconforming product is dependent on both

mean shift and larger variation (Textbook Fig. 5-1, P208)

Normal mean and variance

Larger mean and normal variance

normal mean and larger variance

• Mean is monitored by X bar chart• Variability is monitored by either S chart (standard deviation) or R chart (range)


Review of the Basic Model of Control Charts

Let w be a sample statistic that measures some quality characteristic of interest, andsuppose that the mean of w is µw and the standard deviation of w is σw. Then the centerline, the upper control limit, and the lower control limit become

UCL = µw + k σw

Center line = µw

LCL = µw - k σw

where k is the "distance" of the control limits from the center line, expressed in standarddeviation units


• Statistical Basis of the Charts– suppose {xij, i=1,…,m, j=1,…,n} are normally distributed with

xij,~N(µ,σ2), thus,

• X bar chart monitors between-sample variability (variability over time) and R chart measures within-sample variability (instantaneous variability at a given time)

• If µ and σ are known, X bar chart is

Control Chart for and R— Known µ,σ

X

))n/(,(N~X 2σµ

σ±µ⇒σ

±µ⇒σ±µ An

33 x

n3A =

σ+µ=µ=

σ−µ=

AULCCL

ALCL


• Range Ri=max(xij)-min(xij) for j=1,..n• If µ and σ are known, the statistical basis of R charts is as follows:

– Define the relative range W=R/σ. The parameters of the distribution of W are a function of the sample size n.

– Denote µW =E(W)=d2, σW =d3, • (d2 and d3, are given in Table VI of Textbook on Page 761)

– µR =d2σ, σR=d3 σ, which are obtained based on R=W σ– R chart control limits

Control Chart for and R— Known µ,σ (Cont’s)

X

σ±⇒σ±σ⇒σ±µ )d3d(d3d3 3232RR

σ=σ=

σ=

2

2

1

DULCdCL

DLCL

322

321

d3dDd3dD

+=

−=


• Need to estimate µ and σ

• X bar chart

Control Chart for and R— Unknown µ and σ

X

m

RR;

dRˆ;

mn

x

m

xXˆ

m

1ii

2

m

1i

n

1jij

m

1ii

x

∑∑∑∑== == ==σ===µ

RAxnd/R3x

nˆ

3xˆ3ˆ 22

xx ±⇒±⇒σ

±⇒σ±µ

RAxULCxCL

RAxLCL

2

2

+=

=

−=

nd3A

22 =


Control Chart for and R— Unknown µ and σ (cont’s)

X

233R

m

1ii

R dRdˆdˆ;

m

RRˆ =σ=σ==µ

∑=

• Need to estimate based on R=W σ, σW =d3,

• R chart

RR , σµ2d

Rˆ =σ

R)dd31(

dRd3Rˆ3ˆ

2

3

2

3RR ±⇒±⇒σ±µ

RDULCRCL

RDLCL

4

3

=

=

=

2

34

2

33

dd31D

dd31D

+=

−=


Procedures for Establishment of Control Limits— Unknown µ and σ

• If µ and σ are unknown, we need to estimate µ and σ based on the preliminary in-control data (normally m=20~25).

• The control limits established using the preliminary data are called trial control limits, which are used to check whether the preliminary data are in control.

• First check R or S chart to ensure all data in-control, and then check X bar chart (P213, Example 5-1)

Collect Preliminary Data

EstablishControl Limits

EstimateorX SR

Check Preliminary Data

Eliminate the Outliers due to Assignable Causes

Update Estimation

In-control FutureMonitoring

Out-of-control


Review of Type I and Type II Error

You Conclude :"In Control"

Nature :"Out of Control"

"In Control"

"Out of Control"

Confidence 1–α

Producer Error, α

Power 1–β

Consumer Error, β


Review of the ARL - “In Control”

The following table illustrates the possible sequences leading to an"out of control" signal:Run length Probability

1 α2 α (1– α)3 α (1– α)2

: : :k α (1– α)k–1

∑∞

=

−− α

=αα−=1r

1kcontrolin

1)1(kARL


Review of the ARL - “Out of Control”

• If the process is actually “out-of-control”, the probability that the shift will be detected on

• the first sample is 1-β• the second sample is β(1-β)• the rth sample is βr-1(1-β)

• The expected number of samples taken before the shift is detected is

∑∞

=

−−− β−

=β−β=1r

1rcontrolofout 1

1)1(rARL


OC Curve for x bar and R Chart• X bar chart

• The expected number of samples taken before the shift is detected

If sample interval is very small, the small shift may still be detected reasonably fast although perhaps not on the first sample following the shift

• If process is in control: ARL is the expected number of samples until a "false alarm” occurs

[ ] [ ]nkLnkL

}|LCLxPr{}|UCLxPr{}k|UCLxLCL{P

11

01

−−Φ−−Φ=

µ≤−µ≤=σ+µ=µ≤≤=β

;n/LLCL;n/LUCL,k 0001 σ−µ=σ+µ=σ+µ=µ

∑∞

=

−−− β−

=β−β=1r

1rcontrolofout 1

1)1(rARL

( ) ( )α

=αα−= ∑∞

=

−−

11kARL1i

1kcontrolin


OC Curve for X bar and R chart (Cont’s)

P234 Figure 5-15

P236 Figure 5-16


ARL for X bar

P237 Figure 5-17

P238 Figure 5-18

I=n*ARLARL


Estimation of the Process Capability

• Get process specification limits (USL, LSL)• Estimate σ based on (R chart) or (S chart)• Estimate the fraction of nonconforming products p (or p×106PPM)

• Process-Capability Ratio

PCR=1 means the process uses up 100% tolerance band with 0.27% (2700PPM) nonconforming units

• Percentage of the specification band that the process uses upP=(1/PCR)100%

2d/Rˆ =σ 4c/Sˆ =σ

)ˆ

xUSL(1)ˆ

xLSL(}USLxPr{}LSLxPr{pσ

−Φ−+

σ−

Φ=>+<=

;ˆ6LSLUSLPCR;

6LSLUSLPCR

σ−

=σ−

=


Differences among NTL, CL and SL and Impact on Process Capability

µ USL3σ 3σ

LSL UNTLLNTL

µ USL3σ 3σ

LSL UNTLLNTL

µ

USL3σ 3σLSL

UNTLLNTL

PCR>1, P<100%

PCR<1, P>100%

PCR=1, P=100%µ

USL

3 σ3 σ

LSL

UNTL

LNTL

Externallydetermined


Distribution of individualprocess measurement

UCL

LCL Distribution of x bar values

n3σ

Center lineon x bar

• There is no relationship betw een control lim its and specification lim its.• PCR is an index relating natural to lerance lim its to specification lim its.


Interpretation of X bar and R Chart

• First check the R chart and eliminate the assignable causes from R chart, and then check the X bar chart

• Check non-random pattern– Cyclic pattern due to temperature, regular rotation of

operators or machines, maintenance schedules, tool wear (Fig. 5-10, P230)

– Mixture pattern when the plotted points tend to fall near or slightly outside the control limits. Two overlapping distributions are resulted from too often process adjustment(Fig 5-11, p230).

– Shift in process level due to introduction of new workers, methods, materials, or inspection standard (Fig. 5-12, P231)

– Trend pattern due to gradual tool wear (Fig. 5-13, P231)– Stratification pattern for the points to cluster around the center

line due to incorrect calculation of Control limits or inappropriate reasonable sampling group (Fig. 5-14, P232)

167J. Shi, the University of Michigan, [email protected], 734-763-5321(O) 168J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

Application Conditions of X bar and R chart

• Underlying distribution of the quality characteristics is normal– X bar chart is more robust to nonnormality than R chart– samples of 4 or 5 are sufficient to ensure reasonable

robustness to the normality assumption for X bar chart• Calculation accuracy of Type I error is dependent on the

distribution• X bar chart (n=4, 5, 6) is not effective to detect a small mean shift

(less than 1.5 σ) on the first sample following the shift• R chart is insensitive to small or moderate shifts (σ1/σ0 <2.5) for

the sample size of n=4, 5, or 6. If n>10, a s chart should be used instead of a R chart


• If known µ and σ, X bar chart is the same as that in and R chart

• S chart– Denote µS =E(S)=c4 σ,

(c4 is a constant that depends on the sample size n, which,is given in Table VI of Textbook on Page 761)

– control limits for S chart

Control Chart for and S— Known µ and σ

X

σ−±⇒−σ±σ⇒σ±µ )c13c(c13c3 244

244SS

σ=σ=

σ=

6

4

5

BULCcCL

BLCL

2446

2445

c13cB

c13cB

−+=

−−=

X

24S c1−σ=σ )]2/1n[(

)2/n(1n

2c2/1

4 −ΓΓ

⎟⎠⎞

⎜⎝⎛

−=


• Need to estimate µ and σ.

• X bar chart

Control Chart for and S— Unknown µ and σ

X

;cS

cˆˆ

Sm1S;

1n

)xx(S;

mn

x

m

xXˆ

44

S

m

1ii

n

1j

2iij

i

m

1i

n

1jij

m

1ii

=µ

=σ

=−

−====µ ∑

∑∑∑∑=

== ==

(based on µS =E(S)=c4 σ)

SAxcS

n3x

nˆ

3xˆ3ˆ 34

x ±⇒±⇒σ

±⇒σ±µ

nc3A

43 =

SAxULCxCL

SAxLCL

3

3

+=

=

−=


• Need to estimate µs and σs.

• S chart

X

1n

)xx(S;S

m1Sˆ

n

1j

2iij

i

m

1iiS −

−===µ

∑∑ =

=

S)c

c131(c1

cS3Sˆ3ˆ

4

242

44

SS

−±⇒−±⇒σ±µ

24

44

24

43

c1c31B

c1c31B

−+=

−−=

SBULCSCL

SBLCL

4

3

=

=

=

Control Chart for and S— Unknown Standard µ,σ (Cont’s)

24

4

24S c1

cSc1ˆˆ −=−σ=σ ;

cSˆ

4

=σ

Please review Example 5-3, p243172J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

SBULCSCL

SBLCL

4

3

=

=

=

;cSˆ

4

=σ SAxULCxCL

SAxLCL

3

3

+=

=

−=

σ=σ=

σ=

6

4

5

BULCcCL

BLCL

RDULCRCL

RDLCL

4

3

=

=

=

RAxULCxCL

RAxLCL

2

2

+=

=

−=

σ=σ=

σ=

2

2

1

DULCdCL

DLCL

σ+µ=µ=

σ−µ=

AULCCL

ALCL

2dRˆ =σ

Xˆ =µ

Xˆ =µ

σµknown

X bar chart R chart S chartProcess Parameters

X bar & R chart

X bar & S chart

Summary of Control Charts


Textbook 5-43:A normally distributed quality characteristic is monitored through use of an X bar and an R chart. These charts have the following parameters (n = 4):

x bar chart: UCL = 626.0;Center line = 620.0LCL = 614.0

R chart: UCL = 18.795 Center line = 8.236 LCL = 0

Both charts exhibit control.(a) What is the estimated standard deviation of the process?(b) Suppose an S chart were to be substituted for the R chart. What would be the appropriate parameters of the S chart? (c) If specifications on the product were 610 ± 15, what would be your estimate of the process fraction nonconfornting?(d) What could be done to reduce this fraction nonconforming? (e) What is the probability of detecting a shift in the process mean to 610 on the first sample following the shift (sigma remains constant)? (f) What is the probability of detecting the shift in part (e) by at least the third sample after the shift occurs?


X bar and S Control Chart with Variable Sample Size

• Use a weighted average approach in calculating and

– A3, B3, and B4 will use the corresponding sample size of each subgroup (Please see Example 5-4, P245; Fig. 5-20, P248)

• Use an average sample size , or use the most often sample sizeif ni are not very different

x S

∑

∑

=

== m

1ii

m

1iii

n

xnx

2/1

m

1ii

m

1i

2ii

mn

S)1n(S

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

∑

∑

=

=

n


areChart S


Comparison of R Chart and S Chart

• R Chart– simple for hand calculation;– good for small sample size;– lose information between xmin and xmax;

– not used for variable sample size. • S Chart

– when the sample size is large (n>10);– Used for variable sample size ;– Computation complexity can be simplified by using a

computer.

Summary of x bar, R and S chart formula (Table 5-9, Table 5-10; P260)

n RelativeEfficiency

2 1.000

3 0.992

4 0.975

5 0.955

6 0.930

10 0.850

The relative efficiency of R to S2


Application Conditions of X bar and R chart

• Underlying distribution of the quality characteristics is normal– X bar chart is more robust to nonnormality than R chart– samples of 4 or 5 are sufficient to ensure reasonable

robustness to the normality assumption for X bar chart• Calculation accuracy of Type I error is dependent on the

distribution• X bar chart (n=4, 5, 6) is not effective to detect a small mean shift

(less than 1.5 σ) on the first sample following the shift• R chart is insensitive to small or moderate shifts (σ1/σ0 <2.5) for

the sample size of n=4, 5, or 6. If n>10, a s chart should be used instead of a R chart


Example #3: (Problem 5-16) Control charts for X and R are maintained for an important qualitycharacteristic. The sample size is n = 7; X and R are computed for each sample. After 35 samples, wehave found that

x i = 7805 and Ri =1200i=1

35

∑i =1

35

∑ (a) Set up X and R charts using these data. (b) Assuming that both charts exhibit control, estimate the process mean andstandard deviation. (c) If the quality characteristic is normally distributed and if the specifications are

220 ± 35, can the process meet the specifications? Estimate the fractionnonconforming.

(d) Assuming the variance to remain constant, state where the process mean should be located to minimizethe fraction nonconforming. What would be the value of the fraction nonconforming under theseconditions?


Applications of Variables Charts

• Ex. 5-7: Improving Suppliers' Processes• Ex. 5-8: Purchasing a Machine Tool• Ex. 5-9: Short Run Job Shop• Ex. 5-10: Nonmanufacturing Application• Ex. 5-11: Care in Selecting Rational Subgroups

This is a nice collection of “successful stories” describing howcharting can be used effectively to improve quality.


Ch6 Control Charts for Attributes

• p and np chart• c and u chart• Variable sample size• OC curve and ARL


Terminology

• Nonconforming: Defective • Conforming: Nondefective• Attributes: Quality characteristics of conforming or non

conforming• The fraction nonconforming: the ratio of the number of

nonconforming items in a population to the total number of itemsin that population.

samplesof#Totalgminnonconforof#

nDp i

i ==


Review of Binomial Distribution

Let x = # defects in a sample of size n where defects follow aBernoulli Process (two outcomes, p-constant, x - independent)

f (x) =nx

⎛ ⎝ ⎜ ⎞

⎠ px (1 − p)n − x

E(x) = np

V(x) = np(1 − p)


Sample Estimation of p

Let ˆ p = xn

E( ˆ p ) =E(x)

n=

npn

= p

V( ˆ p ) =1n2 V(x) =

p(1 − p)n


Review of the Basic Model of a Control Chart

Let w be a sample statistic that measures some quality characteristic of interest, andsuppose that the mean of w is µw and the standard deviation of w is σw. Then the centerline, the upper control limit, and the lower control limit become

UCL = µw + k σw

Center line = µw

LCL = µw - k σw

where k is the "distance" of the control limits from the center line, expressed in standarddeviation units


Fraction Nonconforming Control Chart (p-Chart)

For np large ⇒ ˆ p ~ N p,p(1 − p)

n⎛ ⎝

⎞ ⎠

UCL ˆ p = p + 3p(1 − p )

n Centerline = p

LCL ˆ p = p − 3 p(1 − p)n

Binomial (n>10, p close to 0.5) ⇒ normal

If LCLp <0, set LCLp=0

Remarks: When data points are plotted below LCL, they generally do not represent a real improvement. Actually, they are often caused by errors in the inspection rather than a process improvement


How to Establish a p-Chart?

• Is there an assignable cause for out-of-control points or a nonrandom pattern? If so, find the root causes and delete these points, and then update control limits.(Example 6-1, p. 288)

If p unknown, conduct a test and trial control limits with

p)p(Em

p

mn

Dp

m

1ii

m

1ii

=

==∑∑

==

pp =

m=20-25samples for constructing trial control limits


A P-Chart Example

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 10 20 30 40 50 60 70 80 90 100

Revised control chart after making adjustments

Out-of-control point


Example The following data give the number of nonconforming ROM chips in samplesof size 200. Construct a p chart for these data. Assume that any values beyond the controllimits have an assignable cause and revise the control limits as appropriate.

Sample Nonconforming Sample Nonconforming 1 19 12 182 7 13 173 11 14 214 29 15 165 24 16 166 24 17 237 15 18 148 25 19 49 11 20 21

10 10 21 2411 37 22 10


Example (Textbook problem 6-18) A fraction nonconforming control chart with center line0.10, UCLp = 0.19, and LCLp = 0.01 is used to control a process.

a. If 3-sigma limits are used, find the sample size for the control chart.

b. Use the Poisson approximation to the binomial to find the probability of type I error.

c. Use the Poisson approximation to the binomial to find the probability of type II error if theprocess fraction defective is actually p = 0.20.


Design of the Fraction Nonconforming Control Chart (p Chart)

• Three key parameters:– the sample size; the frequency of sampling; the width of the

control limits

• General Guidelines:– select n so that the probability of finding at least one

nonconforming unit per sample is at least r Pr{D≥1} ≥r– p small ---> n large

– Duncan approach: 50% chance of detecting a process shift, i.e. p1=p0+δ (δ>0). P{x>UCL| p1}=0.5

− δ small⇒ n large

– a positive lower control limit

)p1(pLn2

−⎟⎠⎞

⎜⎝⎛

δ=

round n to integer2L

pp1n −

>


Variable Sample Size• Variable width of control limits

corresponding to each sample size– not appropriate for nonrandom pattern check

• Constant width of control limits using average sample size– future sample size should not differ greatly

• Standardized Control Chart– can be used to check a nonrandom pattern– no reference to the actual process fraction defective

ip

ip n

)p1(p3pLCL;n

)p1(p3pUCLii

−−=

−+=

nsobservatioof#totaldefectsof#total

n

Dp m

1ii

m

1ii

==

∑

∑

=

=

m

nn

m

1ii∑

==

0CL;3LCL;3UCL;pp;

n)p1(p

ppZ

i

ii =−===

−−

=

MEAN 0.223

LWL

UWL

LCL

UCL

0

0.1

0.2

0.3

0.4

0.5

Def

ects

0 10 20 30

Row Numbers

P Chart Using Cricketgraph III

Remarks: Use Approaches 1 and 3 together (Example, Table 6-4, Figs. 6-6, 6-7, 6-8; P299-P301)


Table 6-4, P267I n(i) D(i) pi=D(i)/n(i) sigma=sqrt(pbar*(1-pbar)/n(i)) LCL (ni) UCL(ni) LCL (n bar) UCL(n bar) Zi LCL(stand)UCL(stand)1 100 12 0.120 0.029 0.007 0.184 0.007 0.185 0.81 -3 32 80 8 0.100 0.033 0.000 0.194 0.007 0.185 0.12 -3 33 80 6 0.075 0.033 0.000 0.194 0.007 0.185 -0.64 -3 34 100 9 0.090 0.029 0.007 0.184 0.007 0.185 -0.20 -3 35 110 10 0.091 0.028 0.011 0.180 0.007 0.185 -0.18 -3 36 110 12 0.109 0.028 0.011 0.180 0.007 0.185 0.47 -3 37 100 11 0.110 0.029 0.007 0.184 0.007 0.185 0.48 -3 38 100 16 0.160 0.029 0.007 0.184 0.007 0.185 2.17 -3 39 90 10 0.111 0.031 0.003 0.188 0.007 0.185 0.49 -3 310 90 6 0.067 0.031 0.003 0.188 0.007 0.185 -0.94 -3 311 110 20 0.182 0.028 0.011 0.180 0.007 0.185 3.06 -3 312 120 15 0.125 0.027 0.015 0.176 0.007 0.185 1.08 -3 313 120 9 0.075 0.027 0.015 0.176 0.007 0.185 -0.78 -3 314 120 8 0.067 0.027 0.015 0.176 0.007 0.185 -1.09 -3 315 110 6 0.055 0.028 0.011 0.180 0.007 0.185 -1.48 -3 316 80 8 0.100 0.033 0.000 0.194 0.007 0.185 0.12 -3 317 80 10 0.125 0.033 0.000 0.194 0.007 0.185 0.88 -3 318 80 7 0.088 0.033 0.000 0.194 0.007 0.185 -0.26 -3 319 90 5 0.056 0.031 0.003 0.188 0.007 0.185 -1.30 -3 320 100 8 0.080 0.029 0.007 0.184 0.007 0.185 -0.54 -3 321 100 5 0.050 0.029 0.007 0.184 0.007 0.185 -1.56 -3 322 100 8 0.080 0.029 0.007 0.184 0.007 0.185 -0.54 -3 323 100 10 0.100 0.029 0.007 0.184 0.007 0.185 0.14 -3 324 90 6 0.067 0.031 0.003 0.188 0.007 0.185 -0.94 -3 325 90 9 0.100 0.031 0.003 0.188 0.007 0.185 0.13 -3 3

sum 2450 234average 98 pbar= 0.096

p bar=total defective/total samples

Fig 6-6 control chart with variable sample size

0.0

0.1

0.2

0.3

1 4 7 10 13 16 19 22 25

sample index

p

0.0

0.1

0.1

0.2

0.2

0.3

1 4 7 10 13 16 19 22 25

p

standardized control chart

-4.0-3.0-2.0-1.00.01.02.03.04.0

1 4 7 10 13 16 19 22 25 28

index

zi


np Control Chart (The number of nonconforming items)

Rather than plotting the fraction nonconforming, we plot the number ofnonconforming items with an “np Chart”:

UCLX = np + 3 np(1–p) Center line = npLCLX = np – 3 np(1–p)

• This chart, too, can be "standardized", adjusted for unequal ni

, etc.

nsobservatioof#totaldefectsof#total

n

DP m

1ii

m

1ii

==

∑

∑

=

=pp =

Example 6-2, P298


Example: The number of transmission cases that required deburring in a 16-day sample of 100 each was as follows:

Day Number Day Number1 5 9 42 4 10 63 3 11 154 2 12 45 6 13 56 3 14 77 9 15 38 6 16 6

Prepare an np chart with trial control limits. Assume that any points plotting out of control have assignable causes, and continue to refine the control limits until no points plot out of control.


I Di UCL(trial) LCL(trial) UCL LCL1 5 12.34 0 11.32 0.002 4 12.34 0 11.32 0.003 3 12.34 0 11.32 0.004 2 12.34 0 11.32 0.005 6 12.34 0 11.32 0.006 3 12.34 0 11.32 0.007 9 12.34 0 11.32 0.008 6 12.34 0 11.32 0.009 4 12.34 0 11.32 0.00

10 6 12.34 0 11.32 0.0011 15 12.34 0 11.32 0.0012 4 12.34 0 11.32 0.0013 5 12.34 0 11.32 0.0014 7 12.34 0 11.32 0.0015 3 12.34 0 11.32 0.0016 6 12.34 0 11.32 0.00

sum 88 UCL(trial) LCL(trial)p bar=88/(100*16)= 0.055 12.34 -1.34 set to zeronp 5.500eliminate point 11pbar=(88-15)/(100*15) 0.048667 UCL LCLnp 4.866667 11.32 -1.59 set to zero

np chart with trial control limits

0.0

5.0

10.0

15.0

20.0

1 3 5 7 9 11 13 15

np chart after eliminate outliers

0.0

5.0

10.0

15.0

20.0

1 3 5 7 9 11 13 15

Di

UCL(trial)LCL(trial)UCL

LCL


np Chart Properties

• Advantage– np chart is a scaling of the vertical axis by the constant n,

provide the same information as p chart– np chart needs less calculation ( no need to calculate Di/ni)– often used when n is constant and p is small

• Limitation– not easy for interpretation when n is varied (UCL LCL and Ctr

line all vary)– only plot # of defects without considering sample size, hard to

take action


Example #8: (Textbook Exercise 6-15) A control chart is used to control the fraction nonconforming for a plasticpart manufactured in an injection molding process. Ten subgroups yield the following data:

Sample Number Sample Size No. Nonconforming1 100 102 100 153 100 314 100 185 100 266 100 127 100 258 100 159 100 810 100 8

a. Set up a control chart for the number nonconforming in samples of n = 100.b. For the chart established in part (a), what is the probability of detecting a shift in the process fractionnonconforming to 0.30 on the first sample after the shift has occurred?


OC Curve and ARL

• Type II error for the p chart (OC curve see Fig 6-11, P307)

• ARL0=ARLin-control=1/α• ARL1=ARLout-of-control=1/(1-β)

}|{}|{}|ˆ{}|ˆ{ 1111 pnLCLDPpnUCLDPpLCLpPpUCLpP pppp ≤−<=≤−<=β


Example (Textbook Exercise 6-20) Consider the control chart in Exercise 6-18. Find the average run length if the process fraction nonconforming shifts to 0.20.


• Why need it?– Control the total number of nonconformities in a sample or the

average number of nonconformities per unit• nonconformity/defect: a specification of the quality characteristic is not

satisfied result in a defect or nonconformity, e.g., – weld spots on a car– paint dent on a car body

• A unit may not be “nonconforming”, even though it has several nonconformities. So, nonconforming ≠defects or nonconformities

• Assumption: The occurrence of nonconformities in samples of constant size is well modeled by the Poisson distribution.

– The number of potential location for nonconformities is the infinitely large, and the probability of occurrence of a nonconformity at any location is small and constant

Control Charts for Nonconformities (Defects)- C and U Charts


Statistical Basis of Control Charts for Nonconformities (Defects)

- c and u ChartsLet X = total # of defects in a sample

Assume X ~ Poisson ( E(X) = C )For large C ⇒ X ~ N ( E(X) = C , Var(X) = C )

Further, suppose that a sample size of n inspection units(e.g. 100 yd2 , 144 microprocessors, 1100 employees)Note: n is not necessarily integer.Then, Y = average # of defects per unit in a sample = X / n

E ( Y ) = C / n = UVar ( Y ) = C / n2 = U / n

Note: C vs. U distinction is similar to X vs X–

common in "normal" analyses

• c chart: total number of defects in a sample • u chart: average number of defects per unit in a sample size of n inspection units

X=0,1,2,...


Control Charts for Nonconformities (Defects)- c Chart

• Control limits for the c chart with a known c

• If unknown c, c is estimated from preliminary samples of inspection units for constructing trial control limits

• The preliminary samples are examined by the control chart using the trial control limits for checking out-of-control points

Example 6-3, P310

c3cUCL

cCLc3cLCL

+=

=+=

If LCL<0, set LCL=0

samplesofnumbersamplesallindefectsof#total

m

ccc

m

1ii

===∑

=



Control Charts for Nonconformities Per Unit- u Chart

• c: total nonconformities in a sample of n inspection units (n is not necessary be integer)

• u: average # of nonconformities per inspection unit in a sample

• If unknown u, is estimated from preliminary samples of inspection units for constructing trial control limits

Example 6-4, P317

nu3uUCL

uCLnu3uLCL

+=

=

−=

m

uu;

ncu

m

1ii

i

ii

∑===

u


Variable Sample Size of Control Charts for Nonconformities

• If sample size varies, it is always to use a u chart rather than a c chart

• Approaches– Control limits varies with each sample size, but the center

line is constant

– Use a control limits based on an average sample size

– Use a standardized control chart (this is preferred option), with UCL=3, LCL=-3, Center line=0.

• This chart can be used for pattern recognition

uCL;nu3uUCL;

nu3uLCL

ii

=+=−=

m

nn

m

1ii∑

==

i

ii

nu

uuZ −=

Example 6-5, P319


OC Curve and ARL for c and u Charts

• Type II error for the c chart (OC curve see Fig 6-20, P291)

• ARL0=ARLin-control=1/α• ARL1=ARLout-of-control=1/(1-β)

}|{}|{}|{}|{

11

11

unLCLcPunUCLcPuLCLxPuUCLxP

uu

uu

≤−<=≤−<=β


Example (Textbook Exercise 6-45) Find the 3-sigma control limitsfor:a. A c chart with process average equal to four nonconformities.b. A u chart with c = 4 and n = 4.


Following are the number of nonconformities in 20 samples of 50 letter-quality printer cases.Develop the trial control limits for a c chart and if any values are out of control, assume that thecause is assignable. Modify the control limits accordingly.

Sample Nonconf. Sample Nonconf.1 19 11 372 21 12 163 14 13 44 23 14 285 13 15 176 21 16 297 15 17 258 24 18 159 20 19 1110 19 20 19


Day RollsProduced

Number ofImperfections

Day RollsProduced

Number ofImperfections

1 18 12 11 18 182 18 14 12 18 143 24 20 13 18 94 22 18 14 20 105 22 15 15 20 146 22 12 16 20 137 20 11 17 24 168 20 15 18 24 189 20 12 19 22 2010 20 10 20 21 17

Example (Textbook Problem 6-37 ) A paper mill uses a control chart to monitor the imperfection infinished rolls of paper. Production output is inspected for 20 days, and the resulting data are shown below.Use these data to set up a control chart for nonconformities per roll of paper. Does the process appear to bein statistical control? What center line and control limits would you recommend for controlling currentproduction?


Example (Textbook Problem 6-56) A control chart is to be established on a process producingrefrigerators. The inspection unit is one refrigerator, and a control chart for nonconformities is tobe used. As preliminary data, 16 nonconformities were counted in inspecting 30 refrigerators.

a. What are the 3-sigma control limits'?b. What is the α-risk for this control chart?c. What is the β-risk if the average number of defects is actually 2 (i.e., if c = 2.O)?d. Find the average run length if the average number of defects is actually 2.


Chapter 8 - CUSUM and EWMA Control Charts

• Why need them– CUSUM chart– EWMA chart – Weighted/Moving Average chart

• What are they?• How to set up the control limits?


Needs for CUSUM and EWMA Control Charts

Why? ... To expedite detection of a small mean shift in the process.

• Shewhart chart – takes a long time to detect a small mean shift (shift<1.5σ)

• only uses the information about process contained in the last plotted point and ignores any information given by the entire sequence of points

– is not suitable for the sample with a single observation

• Shewhart chart with other supplemental sensitizing rules– can increase detection sensitivity but reduce simplicity and ease

of interpretation of the Shewhart control chart


What is CUSUM Control Chart?

• The CUSUM chart is the “most powerful chart” for detecting a small mean shift in the process, which was first proposed by Page (1954).

• CUSUM chart: directly incorporates all the information in the sequence of sample values by plotting the cumulative sums (CUSUM) of deviations of the sample values from a target value

– : the average of the jth sample– µ0: the target for the process mean– Ci: the cumulative sum up to and including the ith sample– n≥1: cusum could be constructed for individual observations

∑=

µ−=i

1j0ji )x(C

jx


Interpretation of the CUSUM chart

• µ=µ0, Ci is a random walk with mean zero• µ>µ0, Ci is an upward drift trend• µ<µ0, Ci is a downward drift trend

Remark: a trend of Ci is an indication of the process mean shift.

)x(C)x()x()x(C 0i1i0i

1i

1j0j

i

1j0ji µ−+=µ−+⎥

⎦

⎤⎢⎣

⎡µ−=µ−= −

−

==∑∑


Example #1: Consider the following tension test data. This process has a mean increase at sample 20, do you see it?


The process mean in this case could be modeled as: µ1=µ + δσ

UCL

LCL

µµ + δσ

X

t


∑=

−=i

0jji )10X(C


How to Construct a CUSUM Control Chart?

• Monitor the mean of a process :– Tabular (algorithmic) cusum (preferable way)– V-mask form of cusum

• cusum can be constructed for both individual observations and for the averages of rational subgroups.

• For individual observation:

)x(C)x(Cxx 0i1i

i

1j0jiii µ−+=µ−=⇒= −

=∑


• Statistic / : one side upper/lower cusum– / : accumulate derivations from µo that are greater than K,

with both quantities reset to zero upon becoming negative• K: reference value (allowance or slack value)

– chosen about halfway between the target µo and the out-of-control value of the mean µ1 that we are interested in detecting quickly

• Decision Rules:– If either or exceeds the decision interval H (Generally H=5σ),

the process is considered to be out of control

Construct a CUSUM Control Chart–— Tabular CUSUM

0CC

]Cx)K(,0max[C

]C)K(x,0max[C

00

1ii0i

1i0ii

==

+−−µ=

++µ−=

−+

−−

−

+−

+

+C −C

2||

2K|| 0101

01µ−µ

=δσ

=⇒σ

µ−µ=δ⇒δσ±µ=µ

+C −C

+C −C

Example 8-1, P411


0CC

]Cx)K(,0max[C

]C)K(x,0max[C

00

1ii0i

1i0ii

==

+−−µ=

++µ−=

−+

−−

−

+−

+

5.022

||K 01 =δσ

=µ−µ

= ; H=5σ=5

HCorHCifcheckto ii >> −+


Procedures for Construction of CUSUM

• Select K and H:• Construct one side upper and lower cusum and represented in the two

separate columns of the table (Table 8-2, P412)– Calculate xi-(µ0+K) and µ0-K- xi

– Calculate the accumulative derivations and – Count the number of consecutive periods that the cusum or

have been nonzero, which are indicated by N+ and N- respectively

+C −C

22||K 01 δσ=

µ−µ= ; H=5σ

+C −C


Interpretation of CUSUM

• Find the data point iout at which or exceeds the decision interval H (H=5σ)

• If the out-of-control data corresponds to an assignable cause, then to determine the location of the last in-control data

iin=iout - N+out or iin=iout - N-

out

where N+ out and N-

out correspond to N+ and N- at data point iout

• Estimate the new process mean

• Plot a CUSUM status chart for visualization (Fig. 8-3, P413) – however, the other sensitizing rules cannot be safely used for the CUSUM chart

because and are not independent

⎪⎪⎩

⎪⎪⎨

⎧

>−−µ

>++µ=µ

−−

−

++

+

HCifNCK

HCifNCK

ˆ

iout

i0

iout

i0

+C −C

+C −C


Design of CUSUM Based on ARL

• The reference value of K and the decision interval H have an effect on ARL0 and ARL1

– k=0.5δ (K=kσ): to minimize the ARL1 value for fixed ARL0

– choose h (H=hσ): to obtain the desired in-control ARL0 performanceTables 8-3 & 8-4, P415-416

ARL0ARL1

Shewhart chart ARL1=43.96


ARL of the CUSUM Chart

• Siegmund’s approximation for ARL– one side ARL+ or ARL- for or

– total ARL

166.1hb;*;k*

;2

1b2eARLorARL

01

2

b2

+=σ

µ−µ=δ−δ=∆

∆−∆+

=∆−

−+ If ∆=0, ARL=b2

−+ +=ARL

1ARL

1ARL

1

δ*=0: ARL0 (one side) δ*<0: ARL-

δ*>0: ARL+

Example, P417

+C −C


Standardized CUSUM

Advantage of a standardized cusum:• does not depend on σ. So, many cusum charts can

now have the same values of k and h• naturally represent the process variability

;0CC

]Cyk,0max[C]Cky,0max[C

;xy

00

1iii

1iii

0ii

==

+−−=

+−=σ

µ−=

−+

−−

−

+−

+


Improvement of CUSUM

• Fast initial response (FIR): set– If a shift occurred at the beginning, it can detect the shift more quickly

to decrease ARL1

– If in control at the beginning, cusum will quickly drop to zero, little effect on the performance

• Example:µ0=100, K=3, H=12, 50%headstart value

0CC 00 ≠= −+

62/HCC 00 === −+

In-control data µ1=105


More Discussion on CUSUM

• Rational subgroup: the cusum often work best with n=1– if n>1, replace by , replace σ with

• One side cusum in each direction can be designed differently • CUSUM chart is not as effective as the Shewhart chart in detecting

large shifts– combined cusum-Shewhart procedure (Shewhart limits use

3.5σ) can improve the ability of detecting larger shifts, and has only slightly decreased ARL0 (Table 8-5, P418).

n/x σ=σix ix


CUSUM for Monitoring Process Variability

• Create a new standardized quantity (Hawkins, 1981,1993), which is sensitive to variance changes.

• In-control distribution of νi is approximately N(0,1)

• Selection of h and k and the interpretation of cusum are similar to the cusum for controlling the process mean

]Sk,0max[S

]Sk,0max[S

;xy);1,0(N~;349.0

822.0|y|

1iii

1iii

0iii

ii

−−

−

+−

+

+ν−−=

+−ν=σ

µ−=ν

−=ν


EXPONENTIALLY WEIGHTED MOVING AVERAGE (EWMA)

• To exponentially forget the past data, we want to attach more weight to the most recent data

• It is a weighted average: “a geometric series of weights”

0i

ji

1i

0j

ji Z)1(X)1(Z λ−+λ−λ= −

−

=∑

XZ;Z)1(XZ 001iii =µ=λ−+λ= −

0<λ≤1, Z0=µ0

Assume xt are independent random variables, with E(xt)=µ0, Var(xt)=σ2


Var (Zt ) = ⎝⎜⎛

⎠⎟⎞σ2

n ⎝⎜⎛

⎠⎟⎞λ

2–λ ⎝⎛ ⎠⎞1 – (1 – λ)2t

As t becomes large:

Var (Zt ) = ⎝⎜⎛

⎠⎟⎞σ2

n ⎝⎜⎛

⎠⎟⎞λ

2–λ

• Note: for λ=1 we have Shewhart Chart.

In general,

UCLZt = µ0 + L σ ])1(1[

n)2(t2λ−−

λ−λ

LCLZt = µ0 – L σ ])1(1[

n)2(t2λ−−

λ−λ

How to Construct a EWMA Chart?

CL=µ0

Note : Different from the textbook, here we use the sample averages (n>1) rather than the individual samples (n=1)

(from Eq. 8-21)

Assume xt are independent random variables, with E(xt)=µ0, Var(xt)=σ2


UCLZt = µ0+ L σ ])1(1[

n)2(t2λ−−

λ−λ

LCLZt = µ0 – L σ ])1(1[

n)2(t2λ−−

λ−λ

XZ;Z)1(XZ 001iii =µ=λ−+λ= −

n=1


Design of EWMA Control Chart

• Selection of parameters: L and λ– Affect the ARL performance (see Table 8-10, P431)– Smaller λ to detect smaller shift

• recommend to use 0.05≤λ≤0.25, especially, λ=0.05, 0.10, 0.20)

– Generally L=3, but for small λ≤0.1, L=2.6~2.8 • approximately ARL0=500, ARL1=10.3 for detection of 1σ shift

ARL0

ARL1

Shewhart chart ARL1=43.96

σ


Performance of EWMA Control Chart

• Compared to Shewhart chart and CUSUM chart, EWMA chart is– effective on detection of small mean shifts like CUSUM, – less effective on larger shift detection than the Shewhart chart,

but generally superior to the CUSUM chart (particularly if λ>0.1)

• EWMA is very insensitive to the normality assumption. So, it is an ideal control chart for individual observations.

Recommendation:To combine the Shewhart chart with the EWMA and use a wider control limits ( L=3.25 or 3.5) for the Shewhart chart


EWMA for Monitoring Process Variability

• The exponentially weighted mean square error (EWMS)

• For larger i and independent observation, it follows the chi-square distribution with (2- λ)/ λ degrees of freedom and

• Construct exponentially weighted root mean square (EWRMS) control chart to check data point in-control or not.

• To be insensitive to process mean change, it is suggested to replace µ with zi (a prediction of xi+1)

21i

2i

2i S)1()x(S −λ−+µ−λ=

22i )S(E σ=

2iS

νχ

σ=

νχ

σ=

α−ν

αν

2/1,0

2/,0

LCL

UCL

),(N~x 2i σµ

i1i zx =+


Extension of EWMA

• EWMA for time i is equal to the EWMA for time i-1 plus a fraction λof the forecast error of the mean shift at time i

• More general expression is:

i1iii1i1ii1i1iii ez)xx(z)zx(zz)1(xz λ+=−λ+=−λ+=λ−+λ= −−−−−

1ii zx −= iii xxe −=

i3

i

1jj2i11ii eeezz ∇λ+λ+λ+= ∑

=−

proportional Integral First differentialPredication at i+1

i1i zx =+

One time ahead predication

Remarks: Generally, zi is plotted at time period i+1 on the EWMA control chart


Moving Average Control Charts

• Different from EWMA, use an unweighted moving average Mt = ( X t + X t-1 + X t-2 +...+ X t-w+1) / w • This window of size, w, incorporates some of the memory of

the past data information by dropping the oldest data and adding the newest data


Construct Moving Average Control Charts

• The moving averatge can be written recursively as,

Mt = Mt-1 + ( X t – X t-w)/w

w = window size, n = # samples in X

Var (Mt ) = 1w2 • ∑

i=t-w+1

t Var( X i) =

σ2

nw

UCL Mt = µ + 3

σnw

LCL Mt = µ - 3

σnw

Remarks: The window size of w and the magnitude of the shift of interest are inversely related.

in30 ⋅

σ±µ

At the beginning, if i<w


in30 ⋅

σ±µ

UCL Mt = µ + 3

σnw

LCL Mt = µ - 3

σnw

n=1


Example 4 : Compare MA and X-bar Chart

Let w = 4, suppose process shifts from µ to µ + σn

Pr(detect) = Pr( X > µ0 +3 σn | µ = µ0 +

σn ) = Pr (z > 2) = .0227

• for Shewhart Chart (this value does not change from sample to sample)

Pr(detect) = Pr (Mt > UCL Mt | µ = µ0 +

σn )

• for Moving Average Chart (this value changes from sample to sample since,

E(Mt) = 1w {(µ + σn

)+ (w-1) µ} [for 1st sample]

= 1w {2(µ + σn )+ (w-2) µ} [2nd sample]

= 1w {w(µ + σn

)+ (w-w) µ} [wth sample]


Ch 9 Other Univariate Statistical Process Monitoring and Control Techniques


9-1 SPC for Short and Small Production Runs

I. Why Need Short/Small Run SPC?

• Job shops• Batch processes• Large, unique products (e.g., Hubble Space Telescope, Space Shuttles)• Start-up processes, new machines, new production systems, new products• Flexible manufacturing, Just-In-Time inventory control systems, and numerically

controlled processes with build-to-order production schedules• Teaching/learning processes• Many non-traditional applications of SPC where the prior knowledge of the

system/process is not available and one wants to monitor/control the process assoon as possible.


What do we mean by Short and Small Runs?

• Short Run: the time between different jobs/product types is short– It is an environment that has a large number of jobs per operator in a

production cycle (typically a week or month), with each job involving different products. (e.g., automobile maintenance shops)

• Small Run: the number of the same product type is small– It is a situation in which only a very few products of the same type are

to be produced (e.g., space shuttles).• An example of short and small run

– The computer numerical control (CNC) machines in an aerospace firm producing only a small number of guided missiles each month.

• Short runs need not be small runs– a can manufacturing line can produce more than 100,000 cans in an

hour or two.• Small runs are not necessarily short runs

– One wing of an aircraft may requires hundreds of machining hours.


X-Bar and R Chart for Short Production Run— Simplest Technique: Deviation from Nominal (DNOM)

• A particular process/product characteristic can be identified for our SPC purposes. If the following assumptions hold for that parameter, we can use deviation from nominal (DNOM) (xi=measurementi - targetproduct) as the process characteristic to construct control charts as we did before.– Process standard deviation across different product types is

approximately the same (otherwise, use a standardized control chart).– The nominal specification is the desired target value for the process.– More than twenty samples (after aggregation) have been collected– assumptions for a general X bar or R chart hold

• The assumption of a normal distribution holds• DNOM is IID for different nominal values• constant sample size constrained for the R chart

Note: It is customary to use a division line to separate different products on the control charts (see Fig. 8-1 P351, Data Table 8-1 P350)


Review of Previous Variable Charts

Charts Advantages Disadvantagesx and R charts Easy to

construct/implementEasy to interpret

Not sensitive to small shiftsNeed lots of dataNeed prior knowledge

Cusum charts Sensitive to small shifts Difficult to construct/implementDifficult to interpretNeed prior knowledge

EWMA charts Sensitive to small shiftsProvide “predictive” info.

Difficult to construct manuallyNeed prior knowledge


Standardized Variable Control Chartsfor Short Runs

• Let and Ti be the average range and nominal value of x for part i

• Plot Rs on a standardized R chart with control limits at LCL=D3and UCL=D4

• Plot on a standardized x-bar chart with control limits at LCL=-A2 and UCL=A2

• Use traditional sensitizing rules to detect out-of-control signals

iR

is R/RR =

sx

i

is

RTxx −

=

• Why need it?If process standard deviation across different product types isdifferent, the general DNOM chart is not appropriate

Analysis procedures:


Standardized Attribute Control Chartsfor Short Runs

ip

ipn

ic

iu

p

pn

c

u

n/)p1(p −

)p1(pn −

cn

n/u

n/)p1(pppZ i

i −−

=

)p1(pnpnpnZ i

i −−

=

cccZ i

i−

=

n/uuuZ i

i−

=

Attribute Target Value

StandardDeviation

Statistic to plot on the control chart


• Cusum and EWMA control charts with n=1• Modified DNOM (error reported as percentage of the reading) • Q chart (other transform)• Kalman filter

Some Other Methods


Some Useful References

1. Castillo, E.D. and Montgomery, D.C. (1994) Short-Run Statistical ProcessControl: Q-Chart Enhancements and Alternative Methods, Quality and ReliabilityEngineering International, 10(2) 87-97.

2. Farnum, N.R. (1992) Control Charts for Short Runs: Nonconstant Process andMeasurement Error. Journal of Quality Technology, 24(3) 138-144.

3. Foster, G. (1988) Implementing SPC in Low Volume Manufacturing. 1988 AnnualQuality Congress Transactions, Milwaukee, WI: ASQC. pp. 261-267.

4. Hillier, F.S. (1969) X and R -Chart Control Limits Based on a Small Number ofSubgroups, Journal of Quality Technology, 1(1) 17-26.

5. Holmes, D. (1990) SPC Techniques for Low-Volume Batch Processes. CeramicBulletin, 69(5) 818-821.

6. Proschan F. and Savage, I.R. (1960) Starting a Control Chart, Industrial QualityControl, Sep, 1960, 12-13.

7. Pyzdek, T. (1993) Process Control for Short and Small Runs. Quality Progress,26(4) 51-60.

8. Quesenberry, C.P. (1991) SPC Q Charts for Start-Up Processes and Short or LongRuns. Journal of Quality Technology, 23(3), 213-224.

9. Quesenberry, C.P. (1991) SPC Q Charts for A Binomial Parameter P: Short orLong Runs. Journal of Quality Technology, 23(3), 239-246.

10. Quesenberry, CP.. (1991) SPC Q Charts for A Poisson Parameter λ: Short orLong Runs. Journal of Quality Technology, 23(4), 296-303.

11. Wheeler, D.J. (1991) Short Run SPC. Knoxville, Tennessee: SPC Press, Inc.


Section 9-4 SPC With AutoCorrelated Process Data

Why need to study it?– Less defects rates and high quality require

100% inspection. This generally results in correlated inspection data.

– Recently the sensing techniques and computer capacity become more powerful, thus, 100% inspection becomes more important.

– Applications: • Autobody assembly process using OCMM (perceptron) • Painting process using interferometer (Autospec)• Semiconductor using optical/image measurement


Example• Sampling one out of every five products: individual x and moving R chart.

sequence # x mR

1 29.336 32.68 3.35

11 28.94 3.7416 27.63 1.3121 30.80 3.1726 33.22 2.4231 29.04 4.1836 30.29 1.2541 32.43 2.1446 28.12 4.3151 27.99 0.1356 28.78 0.7961 33.61 4.8366 25.12 8.4971 32.01 6.8976 24.28 7.7381 28.17 3.8986 16.56 11.6191 26.14 9.5896 31.68 5.54

Average 28.841 4.492


Based on Data to Construct Control Charts

• The traditional way of constructing the control charts and control limits:

(1) Individual x chart:UCL = 28.841 + 3(4.492)/1.128 = 40.788

CL = 28.841LCL = 28.841 - 3(4.492)/1.1.28 = 16.894

(2) Moving R chart:UCL = 4.492(3.267) = 14.675

CL = 4.492LCL = 0

16.89

28.84

40.79

1 11 21 31 41 51 61 71 81 91

individual x chartUCL

CL

LCL

0

2

4

6

8

10

12

14

16

6 16 26 36 46 56 66 76 86 96

moving R chart

UCL

CL

LCLWe could do lots of things with the data(estimate process mean, process variation,process stability, process capability, etc.)


Suppose, due to a breakthrough of the inspection technology, 100% inspection becomes feasible. Allthe parts are measured:

sequence # x mR

1 29.332 19.98 9.353 25.76 5.784 29 3.245 31.03 2.036 32.68 1.657 33.56 0.888 27.50 6.069 26.75 0.7510 30.55 3.8011 28.94 1.6112 28.50 0.44• • •• • •

95 31.95 0.4296 31.68 0.2797 29.10 2.5898 23.15 5.9599 26.74 3.59

100 32.44 5.70average 28.569 3.212


Comparison Results

• Compare the estimated process parameters from two data sets:The estimated process standard deviations are very different!

• Why are they different? Is this purely a coincidence? Which one should we use?

sampled data complete data

x 28.841 28.569

R 4.492 3.212R /d2 3.982 2.848

15

25

35

45

1 11 21 31 41 51 61 71 81 91

UCL

LCL

CL

Using sampled data to estimate the parameters

15

25

35

45

1 11 21 31 41 51 61 71 81 91

UCL

LCL

CL

Using complele data to estimate the parameters


Autocorrelation

• What is autocorrelation?– Some times called serial correlation– Autocorrelation describes the dependence of data over

time – Recall one of the basic assumptions for Shewhart

charts to work is data independence. This assumption would usually hold if the products are being sampled.

– It is important to recognize, however, that all manufacturing processes are driven by inertial elements, and when the interval between samples becomes small relative to these forces, the observations on the process will be correlated over time.


Mathematical Definition of Autocorrelation

• The correlation over a series of time-oriented observations is measured by the autocorrelation functions:

– where the covariance of observations is k times periods apart. We have assumed that the observations have constant variance.

• All autocorrelation functions are bounded by [-1, 1].

• The sample autocorrelation functions are:

( )( ) ,...2,1,0k;xV

x,xcov

t

kttk ==ρ − 10 =ρ

( )( )

( ),...2,1,0k;

xx

xxxxˆ

n

1t

2t

n

1ktktt

k =−

−−=ρ

∑

∑

=

+=−


How to determine if the data are autocorrelated?

There are several ways of checking autocorrelations.1) Graphical method: Scatter plot

– Just plot xt vs xt-1 and visually inspect if there is any correlation. In our example, we can have scatter plots for both the sampled and the complete data:

0

10

20

30

40

0 10 20 30 40

x(t)

x(t-5)

Scatter plot of the sampled data. No autocorrelation

0

10

20

30

40

0 10 20 30 40

x(t)

X(t-1)

Scatter Plot of the complete data. Autocorrelation exists.


How to determine if the data are autocorrelated?

2) Numerical method: Sample autocorrelation functions• Determine the sample autocorrelation functions for our sample data

using Excel's CORREL function. • In a cell of your choice, type:

– CORREL(array1,array2)– where array1 is the cell range of the data except the first entry, x(1)

and array2 is the cell range of the data except the last entry, x(100).


• The autocorrelation for the above example:(sampled data) = 0.018; (complete data) = 0.510

Yes! Autocorrelation does play an important role in this particular process.

• What about the order of the autocorrelation function, k?– We can compute the second and higher-order sample

autocorrelation functions: In our example, we can use the function, CORREL to compute by assigning x(1) ~ x(98) as array1 and x(3) ~ x(100) as array2.

– After getting a series of , we can plot them on a column chart:

1ρ 1ρ

( )( )

( )∑

∑

=

=−

−

−−=ρ n

1t

2t

n

3t2tt

2

xx

xxxxˆ

2ρ

-1.0

0.0

1.0

1 2 3 4 5

Sample autocorrelation functions, rho(k)

lag, k


How large an autocorrelation function is large?

• A simple rule of thumb: any absolute sample autocorrelations, , greater than are large

that is, they are statistically significant, with α= 0.05 roughly).

• In our example, n = 100, is the boundary.

kρ

n ns,observatio ofnumber 2

2.0n/2 =

-1.0

-0.20.00.2

1.0

1 2 3 4 5

Sample autocorrelation functions, rho(k)

lag, k


When should consider autocorrelation? • Dynamic systems:

– A dynamic system is one that experiences a memory effect in response to the external disturbances to the system.

– A dynamic system can be analyzed and described in a continuous time frame or a discrete time frame in terms of the transfer functions. If a discrete time frame is being used, the data being recorded with a small time interval usually have very strong autocorrelation.

• Static systems: – on the other hand, is one that does not remember its past behavior and

response to the external disturbances instantaneously. If the external disturbances are purely random variables (i.e., white noises, asmechanical engineers call them), the system output would be just a series of independent random numbers. No autocorrelation would be observed, the traditional control charts would work just well.

• Sampling intervals:– When the sampling intervals get larger and larger, the autocorrelation,

even if it exists, will gradually disappear in the data. We use system time constant to characterize the time a system takes to reach a new state after being “excited”. (e.g., the time you have to wait when taking your body temperature.) If, in an intuitive sense, the sampling interval is longer than the system time constant, one would not expect to detect autocorrelation in the observed data. This was exactly what happened in our example.


How to analyze data with autocorrelation?

(1) The statistician’s approach• Worst suggestion: Increase sampling intervals to avoid autocorrelations.

– Why not good? It is counter-evolutionary. Even though autocorrelation complicates our data analysis procedure, we should not just throw away the important information about the system dynamics and go back to the “good old days” in which everything was assumed IID normal.

• Expand the control limits to compensate for the underestimation of the process variance.

– If the sample first-order autocorrelation function is roughly between -0.5 and +0.5, Hunter (1988) suggests the following modified control limits:

nˆ)1n(21

nˆ

3xUCL 12 ρ−

+σ′

+=n

ˆ)1n(21nˆ

3xLCL 12 ρ−

+σ′

−=

where ˆ ′ σ =R d2

is the standard deviation estimated from the range chart, and n is

the sample size. If n = 1 (for the individual x chart) use n = 2 in the second squareroot.


In the example:

08.3951.0185.2357.28n

ˆ)1n(21nˆ

3xUCL 12

=+××+=ρ−

+σ′

+=

06.1851.0185.2357.28n

ˆ)1n(21nˆ

3xLCL 12

=+××−=ρ−

+σ′

−=

15.00

30.00

45.00

1 11 21 31 41 51 61 71 81 91

CL

UCL, sampled

UCL, complete

UCL, modified

LCL, complete

LCL, modified

LCL, sampled

Comparison of different control limits. 266J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

How to analyze data with autocorrelation?

2. Time Series methodology: • A useful tool for analyzing system dynamics and data

autocorrelations in the discrete time domain.– Time series methodology is extensively covered in ME563/IOE565.

• Basic ideas:– Once the autocorrelations are detected (utilizing previous

methods), we can use some types of computer coded algorithm to fit an ARMA (autoregressive moving average) model [ARMA (n, m)] for the data:

mtm2t21t1tntn2t21t1t ...x...xxx −−−−−− εθ−−εθ−εθ−ε+ξ=φ−−φ−φ−


More about ARMA(n,m) Model

For AR(1) models, the extreme value φ1 =1 means the process has a slowwondering mean. It is called a random walk. Stock market prices are anexcellent example.

ε t , so called disturbances or residuals, should be IID N(0, σε2 ) if the empirical

model fits the real system well enough.

mtm2t21t1tntn2t21t1t ...x...xxx −−−−−− εθ−−εθ−εθ−ε+ξ=φ−−φ−φ−

Where ξ is the process mean; xt is the output readings. εt is the estimated unmeasurable disturbances. This procedure, called system identification. φi’s and θi’s are the model parameters. Note that φi is not ρi.

In order to have accurate estimates of the parameters, it is suggested to have at least 200 data points.


How to analyze data with autocorrelation?— Using ARMA modeling

• Fit an appropriate ARMA model to the data and plot the residuals on the Shewhart charts.

– Sounds good. As argued above, should be IID normal, and therefore is a “perfect” candidate for Shewhart control charts in the sense that all the requirements for the Shewhart charts are met.

• Limitations:– does not have physical meanings except that it provides a check for

model validity. An out-of-control signal on a control chart tells us that the empirical model no longer fits the true system. Is it showing that the process is out-of-control? May be, may be not.

– It is sensitive to the accuracy of ARMA modeling• In our example, It is told that the appropriate model is (using computer

codes to do model fitting, you’d learn how to do it in IOE565)

The residuals are approximately N 0,3.202( ) (recall the raw data has standarddeviation ~ 3.50. The ARMA model has reduced the standard deviation by ~10%)

tttt axxx +−+= −− 21 492.0777.062.20ˆ


• In our example, It is told that the appropriate model is (using computer codes to do model fitting, you’d learn how to do it in IOE565)

ˆ x t = 20.62 + 0.777xt −1 − 0.492xt −2 +at

The residuals are approximately N 0,3.202( ) (recall the raw data has standarddeviation ~ 3.50. The ARMA model has reduced the standard deviation by ~10%)

-10.00

0.00

10.00

1 11 21 31 41 51 61 71 81 91

UCL

CL

LCL

Plot the residuals on the control chart:


Different Views of the Autocorrelation• Almost 99% of the statisticians would think of autocorrelations as the

“trouble-makers” in a sense that their existence makes the statistical inferences about the data very difficult and tricky. – The traditional control charts fail terribly when being applied to

autocorrelated data.• As engineers, however, we should think of autocorrelations as the

opportunity of process improvement.– When autocorrelations are present, it means that the processes are

somewhat predictable. Since the data is NOT totally random (as being required by the traditional SPC tools), the process behavior is following a specific pattern. This observation opens up a whole different viewpoint of controlling the process variability. We should consider taking a proactive approach to control the process.

• The traditional SPC methodology is a reactive approach. – The process is left alone as long as no out-of-control signal is

detected. The opportunity of process improvement (reducing process variability and/or bringing the process mean to target) appears only when the process is being diagnosed as out-of-control. Even when an alarm goes off, it is not guaranteed that the process will be improved. The signal might be a false alarm, or no assignable causes can be identified or removed.


Different Views of the Autocorrelation (Cont’s)• On the other hand, the Engineering Process Control (EPC) approach, takes

full advantages of system dynamics (i.e., process predictability) and proactively manipulates the process to minimize the process variability and maintain the process mean.– It is beyond the scope of this course to go further and discuss how to

perform engineering control. Basically, you have to identify one or more controllable process variables (the input), and adjust the input in order to compensate any possible disturbances that would drive the process away from target. A “closed-loop” control scheme can effectively reduce the process variability.

• You should know that when you are faced with autocorrelation, do not just come up with some fancy control charts that sound statistically correct , but do little in improving processes. Work together with the engineers and try to perform proactive controls if at all possible.

• A future trend: Combined the SPC (Statistical Process Control) and APC (Automatic Process/feedback Control).

Process

Disturbances,

Input Output,

ε t

x t

Controller


Ch. 10 Multivariate Quality Control

• Monitoring of Process Mean– Chi-square Control Chart– Hotelling T2 control chart

• sample size n>1• Sample size n=1

– Interpretation of out-of-control signals• Monitoring Process Variability

– test covariance matrix– test sample generalized variance


Need for Multivariate Control Charts• Simultaneous monitoring or control of two or more related

quality characteristics is necessary• Using individual control charts to monitor the independent p

variables separately can be very misleading– Decision rule 1: if one variable is out of control, then

system is concluded to be out-of-control• use x-bar chart for each variable, let Pr{Type I error}=α, then the Type

I error for the joint control of p independent variables could be very large for a large p

α’=1-(1-α)p

– Decision rule 2: if all variables are out of control, then the system is concluded to be out-of-control

• use x-bar chart for each variable, let Pr{Type II error}=β, then the Type II error for the joint control of p independent variables could be very large for a large p

β’=1-(1- β)p

• For dependent variables, there is no easy way to measure the distortion in the joint control rules


Chi-Square Control Chart— Known process mean & variance matrix

• Assumption: The joint probability distribution of the p variables is the p-variate normal distribution

• Sample statistic and control limits

• How to use control charts: – plot of each sample on the control chart and check

with the UCL – check any nonrandom patterns

2p,

1T20

UCL

)()(n

α

−

χ=

−−=χ µΣµ XX

20χ


Geometry Interpretation of Chi-square Control Chart

• For p=2, it forms a control ellipse

– If σ12=0, the principal axes parallel to the – If σ12=0 and σ1=σ2, the control ellipse becomes a circle

• a pair of sample average plotting inside of the ellipse indicates that the process is in control.

)]x)(x(2)x()x([n221112

222

21

211

222

1222

21

20 µ−µ−σ−µ−σ+µ−σ

σ−σσ=χ

)x,x( 21

21 x,x


Hotelling T2 Control Chart(sample size n>1)

• T2 statistic: )()(nT 1T2 XXSXX −−= −

∑∑

∑∑

∑∑

==

==

==

=−−−

=

=−−

=

==

m

1kjhkjhhkihkjk

n

1iijkjhk

m

1k

2jk

2j

2jk

n

1iijk

2jk

m

1kjkj

n

1iijkjk

Sm1S;)xx)(xx(

1n1S

Sm1S;)xx(

1n1S

;xm1x;x

n1x

ith observation (i=1,…,n)j, hth variable (quality characteristic) (j,h=1,…p, j≠h)kth sample (k=1,…,m)

Sample average Total average

mean

variance

covariance


Construct a Hotelling T2 Control Chart

• Two distinct phases of control chart usage– phase I: testing whether the process was in control

when the m preliminary subgroups were drawn and the sample statistics and S computed (m>20)

– Phase II: monitor the future observation

• if large number of preliminary samples are used, it is customary to use as the upper control limit inboth phases

0;1)1)(1(

1,, =+−−

−−= +−−α LCLF

pmmnnmpUCL pmmnp

0;1)1)(1(

1,, =+−−

−+= +−−α LCLF

pmmnnmpUCL pmmnp

2p,UCL αχ=

X

Reference: Alt, F. B. (1985), “Multivariate Quality Control,” Encyclopedia of Statistical

Science, Vol. 6, edited by N. L. Johnson and S. Kotz, John Wiley, New York, pp. 110-122.


Example:

)]xx)(xx(s2)xx(s)xx(s[sss

nT 2211122

2221

211

222

1222

21

2 −−−−+−−

=


Interpretation of Out-of-Control Signals

How to find which of the p variables is responsible for the out-of-control signal?– Simultaneous univariate control chart– Simultaneous confidence intervals– Control chart for p principle component– Decomposition of T2 control chart


Monitor Process Variability

• Repetitively test the significance of the hypothesis that the process covariance matrix is equal to a particular matrix of constants Σ

• Test on the sample generalized variance |Si|

• Remarks:– It lost the individual variance information.– Recommended to use univariate control charts for variability

in conjunction with the control chart for |S|

22/)1p(p,

iii1

ii

UCL

S)1n(A;)A(tr|)|/|Aln(||n)nln(pnpnW

+α

−

χ=

−=Σ+Σ−+−=

)b3b(||LCL

||bCL)b3b(||UCL

21

1

21

−Σ=

Σ=

+Σ=

∏ ∏ ∏

∏

= = =

=

⎥⎦

⎤⎢⎣

⎡−−+−−

−=

−−

=

p

1i

p

1j

p

1jp22

p

1ip1

)jn()2jn()in()1n(

1b

)in()1n(

1b

221 ||b|)S(|V|;|b|)S(|E Σ=Σ=

1b/|S||ˆ| =Σ


Ch7 Process Capability Analysis

• PCR, PCRk, PCRkm

• C.I. and Hypothesis testing of PCR• Gage Capability Study• Setting Specification limits


Ch7 Contents

7-1 Introduction7-2 Process Capability Analysis Using a Histogram7-3 Process Capability Ratio7-4 Process Capability Analysis Using a Control Chart7-5 Process Capability Analysis Using a Designed Experiments7-6 Gage and Measurement System Capability Analysis7-7 Setting Specification Limits7-8 Natural Tolerance Limits


Why Need to Study It?

1) Predicting ability to hold tolerances.2) Assist product developers/designers in selecting /designing processes.3) Assist in establishing an interval between sampling.4) Specifying requirements of new equipment.5) Selecting between vendors.6) Planning production sequences with inter-active effects of processes on tolerances.7) Reducing variability in manufacturing process.

The above general activity is called process capability analysis, involving:• product and process design• vendor sourcing• production or manufacturing planing• manufacturing process control


Process Capability Analysis and Techniques

• Three primary techniques– histograms or probability plot to know the distribution, and get the

parameter estimation • product characterization can only use this method

– PCR analysis using control charts to obtain estimation of the parameters (will be discussed in the lecture)

• better approach for understanding process capability by monitoring process variability

– design experiments to isolate and estimate the sources of process variability

• effective way for understanding the effect of process variables on the outputs to finally realize optimization of process design and control


Description of Process Capability

• Process capability refers to the uniformity of the process, representing by– natural or inherent variability at a specified time: “instantaneous” variability– variability over time

• Natural tolerance limit– It is customary to take the 6-sigma spread in the distribution of the product

quality characteristic as a measure of process capability

• PCR: process ability to manufacture products that meet specification

LNTL = µ – 3 σ UNTL = µ + 3 σFor normal distribution, we expect 99.73% of data between UNTL and LNTL.


Review of PCR

• Process-Capability Ratio

R chart S chart samples

PCR=1 means the process uses up 100% tolerance band with 0.27% (2700PPM) nonconforming units for a normal distribution

• Percentage of the specification band used up by the process

P=(1/PCR)100%

;ˆ6LSLUSLPCR;

6LSLUSLPCR

σ−

=σ−

=

242 Sor;c/Sˆor;d/Rˆ =σ=σ


Review of SL, CL and NTL

A s lo n g as U S L – L S L > U N T L – L N T Lw e a re ca p ab le o f p ro d u c in g a la rg ep ro p o rtio n o f g o o d item s b y c o n tro llin g(c en te rin g ) th e p ro c es s m e an .

µ USL3σ 3σLSL UNTLLNTL

µUSL3σ 3σLSL UNTLLNTL

µ

USL

3σ 3σ

LSL

UNTLLNTL

PCR>1, P<100%

PCR<1, P>100%

PCR=1, P=100%µ

USL

3σ3σ

LSL

UNTL

LNTL



Distribution of individualprocess measurement

UCL

LCL Distribution of x bar values

n3σ

Center lineon x bar

• There is no relationship betw een control lim its and specification lim its.• PCR is an index relating natural to lerance lim its to specification lim its.


More about Process Capability Ratio

• Two- sided specifications

• One-sided specification

• Estimation of PCR are obtained by replacing µ and σ with the estimates of and

σ−

=6

LSLUSLPCR

;3LSLPCR;

3USLPCR LU σ

−µ=

σµ−

=

µ σ


PCR vs Defective PPM NormallyNormally Distributed Process with the centeredcentered mean

PPM: defective or nonconforming parts per million

Table 7-3, P360


Recommended Minimum PCR

Two Spec. One Spec.

Existing Process 1.33 1.25New Process 1.50 1.45Critical Existing 1.50 1.45Critical New Process 1.67 1.60

Table 7-4, P361


ExampleSheet metal components for an autombile (Doors, Fenders, Hood, etc.) are placed in afixture to measure conformance to specifications. The fixture is 3mm larger than thenominal dimensions of the part. The gap between the part and the fixture is measured in"critical" positions. The data are reported as deviations from 3mm (i.e. deviations from thenominal).

A sample of 30 parts are mounted in the fixture and measured. Summary data (in mm) forone critical position follow:

USL: .75LSL: -.75

X = -.54σ = .08

UNTL = -.54 + 3 (.08) = -.30LNTL = -.54 – 3 (.08) = -.78

Note: USL – LSL > UNTL – LNTL

PCR^

= USL – LSL6σ

= .75 – (-.75)6(.08) = 3.125

Note: This process has the potential to have virtually 100% to specification

beyond LSL


PCRk (Off–Center Processes)

• To take process centering into acount, define

PCRk = min { PCRU, PCRL }

where PCRU = USL – µ

3σ ; PCRL = µ – LSL

3σ

• PCRk is just one side PCR for the specification limit closest to µ(use one side specification on Table 9-3 to get PPM).

• If the process is centered , then PCR = PCRk ; If not centered, then PCRk < PCR.• PCR measures potential capability, PCRk measures actual capability


Example: to calculate PCR and PCRk for the following two processes with LSL=35, USL=65Process A: µA=50, σA=5Process B: µB=57.5, σB=2.5


More about Process Capability Index

τ : square root of expected square deviation from target

PCRkm

= USL – LSL

6 τ

τ 2 = E[ (x – T) 2 ] = σ2 + (µ – T) 2 ; T = USL–LSL

2

• Note: Variance is composed of two components

Hence, PCRkm

= USL – LSL

6 σ2

+(µ–T)2 =

PCR

1 + ξ2

ξ = T – µ

σ

+


Discussion on PCRkm

• PCRkm range: 0≤ PCRkm ≤ PCR (however, PCRk could be negative)• When µ=T, PCRk=PCRkm

• A given value of PCRkm places a constraint on the |µ−T|– A necessary condition for PCRkm≥1 is:

• The estimation of PCRkm is

• Third generation process capability ratio: increase the sensitivity to departures of µ from T,

SxTV;

V1PCRPCR

2km−

=+

=

)LSLUSL(61|T| −<−µ

2k

2k

pkm1

PCR

T1

PCRPCRξ+

=

⎟⎠⎞

⎜⎝⎛

σ−µ

+

=PCR: the first generationPCRk: the second generation


Problem 7-6: A process is in control with

.2 and 75 == Sx

The process specifications are at 80±8. N=5

a. Estimate the potential capability.

b. Estimate the actual capability.

c. How much could process fallout be reduced by shifting the mean to the nominal dimension?

Assume that quality characteristic is normally distributed


Problem 9-7: Consider the two processes shown below:

Process A: X A = 100 S A = 3

Process B: X B = 105 S B =1

Specifications are at 100 ± 10.Calculate PCR, PCRk, and PCRkm and interpret these ratiosWhich process would you prefer to use?

n=5


Problem 9-8: Suppose that 20 of the parts manufacturedby the processes in example 2 were assembled so thattheir dimensions were additive; that is,x = x1 +x2 +.....x20

Specifications on x are 2000 ± 200. Would you prefer toproduce the parts using process A or process B? Why?Do the capability ratios computed in problem 9-7 provideany guidance for process selection?


Confidence Intervals for PCR

Based on σ−

=6

LSLUSLPCR , thus

11

1,2/2

4

1,2/12

4 −χ

≤≤−

χ −α−α−

ncPCRPCR

ncPCR nn


textbook, P368



• Why need it?• Total variability decomposition

Gage and Measurement System Capability

2gage

2product

2total σ+σ=σ

2ilityreproducib

2ityrepeatabil

2gage

2error_tmeasuremen σ+σ=σ=σ

inherent precision of gage different operators or conditions


Gage Capability Criteria

• Precision to tolerance ratio or P/T ration

• gage error as a percentage of the product variability

1.0LSLUSL

ˆ6TP gage <

−

σ=

%100ˆˆ

product

gage ×σ

σ


Example 7-7 P378

2gage

2product

2total σ+σ=σ

• X-bar chart represents variability between different product units• R chart represents the gage measurement variability:

2gage d

Rˆ =σ


Example 7-8 P380

)x,x,xmin(x)x,x,xmax(x

19.0693.132.0

|dRˆ32.028.2260.22xxR

02.1128.115.1

|dRˆ15.1)2.125.11(

31)RRR(

31R

321min

321max

3n2

xilityreproducibminmaxx

2n2ityrepeatabil321

=

=

===σ=−=−=

===σ=++=++=

=

=

(1) average of all ranges

(2) Difference among operators

(3) Each operator’s average308J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

Gage and Measurement System Capability• Variation Decomposition

2ilityreproducib

2ityrepeatabil

2gage

2tmeasuremen

2gage

2product

2total σ+σ=σ=σ⇒σ+σ=σ

Use R chart for estimationr: # of operatorsm: # of samplesn: # of repeated measurementsxkij :i: sample indexj: repeated measurement indexk: operator index

rmn

xx

1rmn

)xx(

r

1k

m

1i

n

1jkij

r

1k

m

1i

n

1jkij

2total

∑∑∑

∑∑∑

= = =

= = =

=

−

−=σ

2ityrepeatabil d

Rˆ =σ

r

RR

r

1kk∑

==

)x(min)x(maxRm

RR

kijjkijjki

m

1iki

k

−=

=∑

=

)x,x,xmin(x

)x,x,xmax(x

;xxR

r,21min

r,21max

minmaxX

…

…

=

=

−=

2

Xilityreproducib d

Rˆ =σ

mn

x

m

xx

m

1i

n

1jxij

m

1iki

k

∑∑∑= == ==


• Gage capability: precision-to-tolerance ratio (P/T ratio)– Generally, an adequate gage capability: P/T≤0.1

• gage variability-to-product variability ratio– independent of specification limits

Gage and Measurement System Capability (Cont’s)

LSLUSLˆ6

TP gage

−σ

=

%100ˆˆ

product

gage ×σσ

2totalσ

2ilityreproducib

2ityrepeatabil

σ

σ 2gageσ

2productσ2

ilityreproducib2

ityrepeatabil2gage σ+σ=σ

2gage

2total

2product σ−σ=σ


Setting Specification Limits on Discrete Components

N am e of lim it

Toler anc e S et by the engineer ing des ign f unct ion to de f ine the minimum and m axim um values allowa ble f or the pr oduct to w or k pr oper ly

S tat is tic al toler ance Calcula ted f rom pr oces s data to def ine the amount of var ia tiontha t the pr oce ss exhibits ; thes e limits w il l contain a spe cif ied pr opor tion of the total populat ion

P redic tion Calcula ted f rom pr oces s data to def ine the lim its w hich w il l contain a ll of k f utur e obs er vat ions

Conf ide nce Calcula ted f rom data to def ine an inter va l w ithin which a popula tion pa r ame ter l ies

Contr ol Calcula ted f rom pr oces s data to def ine the lim its of cha nce ( random ) var iation ar ound s ome c entr al value

(SPL)

(NTL)

(C. I)

(CL)

Meaning


Example

Suppose three components were manufactured to the specificationsindicated below:

Assembly

A B C

1.000 0.500 2.000± 0.001 ± 0.0005 ± 0.002

A “logical” specification on the assembly length would be 3.500 ± 0.0035.

Maximum Minimum1.0010 0.99900.5005 0.49952.0020 1.9980 3.5035 3.4965


Example (cont’s)

The approach of adding component tolerances is too conservative!

Suppose that about 1% if A component is below lower tolerancelimits. Likewise for B and C.

If assemblies are made at random and if the components aremanufactured independently, then the chance that an assembly willhave all three components simultaneously below the lowertolerance limit is

1100

x1

100x

1100

=1

1,000,000

Thus, setting component and assembly tolerances based on thesimple stacking of tolerances is conservative; failing to recognizelow “joint probabilities”.


Recall the variance of sums of independent events:

σ result = σ cause A2 + σ cause B

2 + σ cause C2 +

For the assembly example:

σ assembly = σA2 + σ B

2 + σ C2 +

Suppose for each component the tolerance range is equal to ± 3standard deviations (natural tolerance limits; process capability =1.0). The variance relationship may be rewritten as

Tassembly = TA2 + TB

2 + TC2

Thus the squares of tolerances are added to determine the square ofthe tolerance for the overall result.


AssumptionsThe formula is based on several assumptions:

• Each component dimension is independent and the components are assembled randomly.(usually met in practice).

• Each component dimension should be normally distributed. (robust to this assumption)

• The component has mean = nominal spec.

Bender (1975) has studied these assumptions for some complex assembly cases andconcluded based on a "combination of probability and experience" that a factor of 1.5should be included to account for the assumptions, i.e.,

Tresult = 1.5 TA2 + TB

2 + TC2 +

Variation Simulation Analysis (VSA) uses simulation to analyze tolerances with/withoutnormality and with/without independence; see Rowzee and Holmes (1986).


7-7.1 Linear Combinations

Now let’s generalize a bit. For linear, independent assemblies

y = a1x1 + a2 x2 + anxn

We recall,

E(y) = a1E(x1 ) + a2 E(x2 )+ anE(xn )

Var(y) = a12Var(x1 ) + a2

2Var(x2)+ an2Var(xn )

)(N~x 2i,i0i σµ


Example 2:

Two mating parts are assembled as shown below:

X Y

Assume µx = 2.010 cm., σx = .002 cm.µy = 2.004 cm., σy = .001 cm.

What fraction of parts will have positive clearance?


• Assembly process assessment – Calculate the nonconforming fraction of the final assembled

product when each component distribution is known (Example 7-9, P388 )

• Tolerance design for each component (Example 7-10, P389 )– To meet the specification limits of the final assemblies

• 2W is the width of SL for the final assemblies • NTL is defined so that no more than α% of the final assemblies will fall outside

SL ( the width of NTL= )• assume n components with the same variance σ2

• Design minimum clearance C (Example 7-11, P391)

– Pr{clearance<C}= = α

⎟⎟⎠

⎞⎜⎜⎝

⎛

σµ−

Φ−⎟⎟⎠

⎞⎜⎜⎝

⎛

σµ−

Φ=≤≤

σ=σ⇒µ=µ⇒= ∑∑∑===

y

y

y

y

n

1i

2i

2i

2y

n

1iiiy

n

1iii

LSLUSL}USLyLSLPr{

ccxcy

nZW 2

y2

2/y

σ≤σ⇒≤σ

α

W2Z2 y2/ =σα

Applications

αα− −==σ

µ−ZZ

C

y

y1⎟

⎟⎠

⎞⎜⎜⎝

⎛

σµ−

Φy

yC


7-7.2 Nonlinear Combinations

For nonlinear functions

y = g(x1, x2 , , xn )

We can use a second order Taylor series approximation to gand expand the solution around µ1, µ2, . . . . , µn

y = g(x1, x2 , , xn )

= g(µ1,µ 2, ,µ n ) + (xi – µ i)∂g∂xi µ 1, µ 2 , ,µ n

i =1

n

∑ + R

µ y = E(y) ≈ g(µ1,µ 2, ,µ n )

σY

2 = Var(y) ≈∂g∂xi µ 1, µ 2 , ,µ n

⎛

⎝ ⎜

⎞

⎠ ⎟

i =1

n

∑2

σ i2


Example 7-12, P392:

V = IR Expanded, V ≈ µI µR + (I – µI )µR + (R – µR ) µI µV ≈ µI µR σ2

V ≈ µR2 σI2+µI2 σR2


Example 2:

Consider a cylinder cut from sheet aluminum as illustrated:

X

Y

Assume X ~ N (20,2) and Y~N(10,1)

What is the mean and variance of the resultant volume?

Recall: Area of a circle = π r2

Circumference of a circle = 2 π r


7-8 Estimating Natural Tolerance Limits of a Process

In general we want an interval which contains (1-α)% of the process withprobability γ.

For a normal process distribution, Tables VII and Table VIII (see Appendix)provide values for K for two- and one-sided tolerance limits respectively;

X– ± K S

for 2 ≤ n ≤ 1000, γ = .90, .95, .99 and α = .10, .05, .01.


Ch14 Acceptance Sampling for Attributes

• What is acceptance sampling?• Why use acceptance sampling?• Types of Sampling Plans

– Single sampling plan– Double sampling plan– Multiple-sampling plan


What is acceptance sampling?

• Scenario 1– A company received a shipment of product from a vendor.

This product is often a component or raw material used in the company’s manufacturing process. A sample is taken from the lot, and some quality characteristic of the units in the sample is inspected. On the basis of the information in this sample, a decision is made regarding lot disposition. Usually, the decision is either to accept or to reject the lot. This procedure is called acceptance sampling.

• Scenario 2– A manufacturer will sample and inspect its own product at

various stages of production. Lots that are accepted are sent forward for further processing, while rejected lots may be reworked or scraped.


Why use acceptance sampling?

• Three methods for lot sentencing– Accept without inspection– 100% inspection– Acceptance sampling

• Advantages of acceptance sampling– It is usually less expensive because there is less inspection

(less personal, less inspection time, less inspection error)– There is less handling of product, hence reduced damage– Rejection of entire lots as opposed to the simple return of

defectives often provides a stronger motivation to the vendor for quality improvement.

• Disadvantages– Risks of accepting “bad” lots and rejecting “good” lots.– Less information is generated– Need planning and documentation


Types of Acceptance Sampling

• Single sampling plan– The lot disposition is determined by one single sample.

• Double-sampling plan– The decision from the first sample is to (1) accept the lot, (2)

reject the lot, (3) take second sample.– If we need to take the second sample, the lot disposition is

determined by both the first and the second sample.

• Multiple-sampling plan– It is an extension of double-sampling plan.– Sequential sampling is an ultimate extension of multiple

sampling.


Considerations in acceptance sampling

• Lot formation– Homogeneous– Larger lots are better– Lots should be conformable to the materials-handling systems

used in both the vendor and consumer facilities.• Random sampling

• An acceptance sampling plan is a statement of the sample size tobe used and the associated acceptance or rejection criteria for sentencing individual lots.


Single-Sampling Plans for Attributes

• Definition– A single-sampling plan is defined by the sample size n and the

acceptance number c.

Example:If the lot size is N=10000, then the sampling plan n=89, c=2 means that from a lot of size 10000 a random sample of n=89 units is inspected and the number of nonconforming or defective items d observed. If d is less or equal to c = 2, the lot will be accepted. If d is larger than 2, the lot will be rejected.


How to evaluate the acceptance sampling plan?The performance is evaluated by OC curve: a plot that shows the probability of accepting the lot versus the lot fraction defective.

OC curve shows the discriminatory power of the sampling plan.


How to get OC curve?

∑=

−

−

−−

=≤=

−−

==

c

d

dnda

dnd

ppdnd

ncdpP

ppdnd

ndfdP

0

)1()!(!

!}{

)1()!(!

!)(}defectives {


Draw OC curve for the single-sampling plan n=50,c=1. Assume lot is infinite large


The ideal acceptance sampling plan

It can only be achieved by 100% inspection if there is no inspection error.


Effects of sample size n and defective number c on OC curves


Characterization of OC curve

• AQL: represents the poorest level of quality for the vendor’s process that the consumer would consider to be acceptable as a process average.– It is defined by the consumer.– An acceptance sampling plan should give a high probability of

acceptance at the AQL.• LTPD: lot tolerance percent defective. It is the poorest level of

quality that the consumer is willing to accept in an individual lot.– It is also called rejectable quality level (RQL) or limiting quality

level (LQL).– An acceptance sampling plan should give a low probability of

acceptance at RQL


Type A and Type B OC curve

• Type A OC curve: isolated lot with finite (small) size.– Lot size N, sample size n, acceptance number c– The # of defective units in the sample is “Hypergeometric”

distributed.

• Type B OC curve: sample is from a infinite (large) lot– The # of defective units in the sample is binomial distributed.– Type B OC curve is always higher than type A curve– If n/N>0.1, type A and type B are almost the same.

Np is D,0

∑=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=c

da

nN

dnDN

dD

P


An example of type A and type B OC curve


Design a single-sampling plan with a specified OC curve

• Need two points to define a curve.– It does not matter which two points we select. It is customary

in industry to use AQL and RQL for this purpose.• Given Pr(accept|p1)=1-α where p1=AQL and Pr(accept|p2)=β where

p2=RQL, how to get n and c? Assume the lot size is large.

∑

∑

=

−

=

−

−−

=β

−−

=α−

c

d

dnd

c

d

dnd

ppdnd

n

ppdnd

n

022

011

)1()!(!

!

)1()!(!

!1


Solve the equation by Binomial nomograph

Example: p1=0.01, α=0.05, p2=0.06, β=0.1338J. Shi, the University of Michigan, [email protected], 734-763-5321(O)

Rectifying Inspection

• Acceptance-sampling programs usually require corrective action when lots are rejected. If the action is 100% inspection and to replace all nonconforming parts, such sampling programs are called rectifying inspection programs.


Average Outgoing Quality

• It is the average value of total quality that would be obtained over a long sequence of lots from a process.

• The calculation of AOQ– n items in the sample which, after the inspection, contain no

defectives.– N-n items which, if the lot is rejected, also contain no

defectives– N-n items which, if the lot is accepted, contain p(N-n)

defectives

pPAOQN

nNpPAOQ

a

a

≈

−=

)(


Example: suppose that N=10,000, n=89, and c=2, and that the incoming lots are of quality p=0.01. What is AOQ?


AOQ curve

The worst possible average quality that would result from the rectifying inspection program us called average outgoing quality limit (AOQL)


Average total inspection

• The average amount of inspection per lot will is the average total inspection (ATI).

• Example: N=10,000, n=89, c=2, and p=0.01. What is ATI?

))(1( nNPnATI a −−+=


Average total inspection curves


Summary of single-sampling plan for attributes

• Definition• Performance evaluation: OC curve

– AQL and RQL points– Type A and Type B OC curves– Design a single-sampling plan with a specified OC curve

• Rectifying Inspection– AOQ

• AOQL

– ATI


Double, Multiple, and Sequential Sampling• Definition of double

sampling plan– n1: sample size

on the first sample

– c1: acceptance number of the first sample

– n2: sample size on the second sample

– c2: acceptance number for both samples

• The operation of double sampling– Example:

n1=50,c1=1,n2=100,c2=3


Double-Sampling plan

• Advantages– Curtailment: reject a lot without complete the second sample– Inspection size is smaller than single sampling plan

• Disadvantages– If not applied correctly, it may need more inspections– Need more administration work


The OC curve of a double-sampling plan


Average Sample Number Curve• Average sample number (ASN)

• ASN for a double-sampling plan with curtailment

)1()1)(( 21211 III PnnPnnPnASN −+=−++=

∑+=

+−++−+−+=2

1 122222211 ]/)2,1()1(),()[,(

c

cjML pjcnPjcjcnPnjnPnASN


Design Double-Sampling Plans

• Required conditions– p1, 1-α– p2, β– Another relation between parameters. We often require that n2

is a multiple of n1.


Rectifying Inspection

IIa

Iaa

aII

aI

a

IIa

Ia

PPP

PNPnnPnATIN

pnnNPnNPAOQ

+=

−+++=

−−+−=

)1()(

)]()([

211

211

list of topics in lecture 1 ioe 466 statistical quality …. shi, the university of michigan,...

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