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Page 1

CHAPTER 1

INTRODUCTION TO STATISTICAL QUALITY CONTROL

Quality means doing it right when no one is
looking.

Henry Ford

1.1 WHAT IS QUALITY CONTROL?

Quality can mean different things to different people. Quality is often used to signify excellence of a manufactured
product or a service received. From the manufacturing standpoint quality is simply conformance to specifications.
Quality is also defined as meeting the requirements of the customer. Exceeding customer expectations has been ex-
tremely effective in building a loyal customer base. It is estimated to be five to seven times more costly to attain a new
customer that it is to retain a current one, so it makes a lot of sense to go that extra step [3].

There are various definitions attributed to historical leaders in the field of quality. Juran described quality as “fitness
for use”.

Quality control is a process employed to ensure a certain level of quality in a product or service, which may include
whatever actions a business deems necessary to provide for the control and verification of certain characteristics of a
product or service. The basic goal of quality control is to ensure that the products, services, or processes provided meet
specific requirements and are dependable, satisfactory, and fiscally sound.

Essentially, quality control involves the examination of a product, service, or process for certain minimum levels of
quality. The goal of a quality control team is to identify products or services that do not meet a companys specified
standards of quality. If a problem is identified, the job of a quality control team or professional may involve stopping
production temporarily. Depending on the particular service or product, as well as the type of problem identified,
production or implementation may not cease entirely.

Usually, it is not the job of a quality control team or professional to correct quality issues. Typically, other individ-
uals are involved in the process of discovering the cause of quality issues and fixing them. Once such problems are
overcome, the product, service, or process continues production or implementation as usual.

1.2 WHAT IS STATISTICAL QUALITY CONTROL?

Statistical quality control (SQC) is a term used to describe the activities associated with ensuring that goods and services
satisfy customer needs. SQC uses statistical analysis based on measurements taken from a process or from a sample
of products or services, to make decisions regarding the quality of goods and services. The statistical methods of SQC
may be divided into two main categories: Statistical process control (SPC) and acceptance sampling. SPC refers to the
use of statistical methods to measure and control the performance of a process to ensure that the output meets customer
needs. Acceptance sampling is a methodology of taking samples from lots of materials or products and inspecting the
items to determine if the items meet customer requirements. SPC may be used to to help control almost all processes
that can measured or monitored to ensure that the process performs within limits.

Deming, a statistician who gained fame by helping Japanese companies improve quality after the World War II,
believed that quality and productivity increase as variability decreases and, because all things vary, statistical methods

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of quality control must be used to measure and gain understanding of the causes of the variation. Many companies,
particularly those in the auto industry, have adopted Deming’s philosophy and approach to quality.

The causes of variations in a product quality characteristic may be broadly classified into two main categories: com-
mon causes of variation (variation due to the system itself) and special causes of variation (variation due to factors
external to the system).

• Chance or common causes: when only these causes of variations are present in a process, the process is considered
to be stable or in-control. Examples of such causes include atmospheric pressure or temperature changes in
the production area, worker fatigue, and fluctuations caused by hiring, training, and supervisory policies and
practices. Special causes of variation are the responsibility of management.

• Assignable or special causes: when these causes of variations are present in a process, variation will be excessive
and the process is considered to be unstable or out-of-control. Examples of such causes include tampering or
unnecessary adjusting the process when it is inherently stable, using a wrong tool or an incorrect procedure.
Special causes of variation are the responsibility of workers and engineers.

1.3 A BRIEF HISTORY OF QUALITY CONTROL

The quality movement can trace its roots back to medieval Europe, where craftsmen began organizing into unions
called guilds in the late 13th century. Until the early 19th century, manufacturing in the industrialized world tended to
follow this craftsmanship model. The factory system, with its emphasis on product inspection, started in Great Britain
in the mid-1750s and grew into the Industrial Revolution in the early 1800s.

In the early 20th century, manufacturers began to include quality processes in quality practices. After the United
States entered World War II, quality became a critical component of the war effort: Bullets manufactured in one state,
for example, had to work consistently in rifles made in another. The armed forces initially inspected virtually every
unit of product; then to simplify and speed up this process without compromising safety, the military began to use
sampling techniques for inspection, aided by the publication of military-specification standards and training courses
in Walter Shewharts statistical process control techniques.

The birth of total quality in the United States came as a direct response to the quality revolution in Japan following
World War II. The Japanese welcomed the input of Americans Joseph M. Juran and W. Edwards Deming and rather than
concentrating on inspection, focused on improving all organizational processes through the people who used them. By
the 1970s, U.S. industrial sectors such as automobiles and electronics had been broadsided by Japans high-quality
competition. The U.S. response, emphasizing not only statistics but approaches that embraced the entire organization,
became known as total quality management (TQM). By the last decade of the 20th century, TQM was considered a fad
by many business leaders. But while the use of the term TQM has faded somewhat, particularly in the United States,
its practices continue.

In the few years since the turn of the century, the quality movement seems to have matured beyond Total Quality.
New quality systems have evolved from the foundations of Deming, Juran and the early Japanese practitioners of
quality, and quality has moved beyond manufacturing into service, healthcare, education and government sectors.

1.4 WHAT IS STATISTICAL PROCESS CONTROL?

A process is the transformation of a set of inputs, and may include customer services, productions systems, and ad-
ministration activities. In each area or function of an organization there will be many processes taking place. Each
process may be analyzed by an examination of the inputs and outputs. This will determine the action necessary to
improve quality [2]. In order to produce an output which meets the requirements of the customer, it is necessary to
define, monitor and control the inputs to the process.

Statistical process control (SPC) is a procedure in which data is collected, organized, analyzed, and interpreted so
that a process can be maintained at its present level of quality or improved to a higher level of quality. SPC requires that
the process be improved continuously by reducing its variability. SPC refers to a number of different methods for mon-
itoring and assessing the quality of manufactured goods. Combined with methods from the Design of Experiments,
SPC is used in programs that define, measure, analyze, improve, and control development and production processes.
These programs are often implemented using “Design for Six Sigma” methodologies.

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1 lambda = 0 .5 ; % parameter
2 x = 0: .1 : 5 ;
3 f x = pdf ( ' exp ' , x , lambda ) ; % pdf
4 p l o t ( x , fx , ' LineWidth ' ,2 ) ; % V i sua l i ze the pdf
5 f i g u r e ;
6 Fx = cdf ( ' exp ' , x , lambda ) ; % cdf
7 p l o t ( x , Fx , ' LineWidth ' ,2 ) ; g r i d % Vi sua l i ze the cd f

0 1 2 3 4 5
0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

x

f
(x

)





λ=0.5

λ=1.0

λ=1.5

0 1 2 3 4 5
0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x
F

(x
)





λ=0.5

λ=1.0

λ=1.5

(a) (b)

FIGURE 2.19: Exponential distribution X ∼ exp(λ) for various values of λ: (a) pdf; (b)
cdf.

Example 2.26 Jobs are sent to a printer at an average of 3 jobs per hour.

(i) What is the expected time between jobs?

(ii) What is the probability that the next job is sent within 5 minutes?

Solution: Job arrivals represent rare events, thus the time X between them is exponentially distributed with rate 3
jobs/hour i.e. λ = 3.

(i) E(X) = 1/λ = 1/3 hours or 20 minutes

(ii) Using the same units (hours), we have 5 min=1/12 hours. Thus, the probability that the next job is sent within 5
minutes is

P(X < 1/12) = F(1/12) = 1 − e−(3)(1/12) = 0.2212.

Example 2.27 Suppose that the time in months between line stoppages on a production line follows an exponential distribution
with λ = 1/2.

(i) What is the probability that the time until the line stops again will be more than 15 months?

(ii) What is the probability that the time until the line stops again will be less than 20 months?

(iii) What is the probability that the time until the line stops again will be between 10 and 15 months?

(iv) Calculate the mean µ and standard deviation σ. Then, calculate the probability that the time until the line stops will be
between (µ − 3σ) and (µ + 3σ).

Solution: The time X between them line stoppages is exponentially distributed with λ = 1/2.

(i) The probability that the time until the line stops again will be more than 15 months is

P(X > 15) = e−15λ = e−(15)(1/2) = e−7.5 ≈ 0

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(ii) The probability that the time until the line stops again will be less than 20 months is

P(X < 20) = 1 − e−(20)(1/2) ≈ 1

(iii) The probability that the time until the line stops again will be between 10 and 15 months is

P(10 ≤ X ≤ 15) = F(15)− F(10) = e−(10)(1/2)− e−(15)(1/2) ≈ 0.0062

(iv) The mean and the standard deviation are given by µ = σ = 1/λ = 2. Thus, the probability that the time until the
line stops will be between (µ − 3σ) and (µ + 3σ) is

P(µ − 3σ ≤ X ≤ µ − 3σ) = P(−4 ≤ X ≤ 8) = P(0 ≤ X ≤ 8) = F(8) = 1 − e−(8)(1/2) ≈ 1.

2.4.5 CHI-SQUARED DISTRIBUTION

The chi-squared distribution with n degrees of freedom is the distribution of a sum of the squares of n independent
standard normal random variables. It is typically used in hypothesis testing and in determining confidence intervals.

Definition 2.11 Suppose that Z1, Z2, . . . , Zn ∼ N(0, 1). Then, the random variable X = Z21 + Z22 + . . . + Z2n is said to have a
chi-squared distribution, denoted X ∼ χ2(n), with n degrees of freedom. The density function of X is

f (x) =





1

2n/2Γ(n/2)
x(n−2)/2e−x/2 if x ≥ 0,

0 otherwise.
(2.4.12)

where the gamma function is defined by the integral Γ(ν) =
∫ ∞

0
e−ttν−1dt.

The mean and variance of X ∼ χ2(n) are E(X) = n and Var(X) = 2n

The pdf and cdf of the χ2-distribution for various values of degrees of freedom n are shown in Figure 2.20. The
distribution is asymmetric and as the degrees of freedom increase, the χ2-curve approaches a normal distribution. The
mean of the χ2-distribution is the degrees of freedom and the standard deviation is twice the degrees of freedom. This
implies that the χ2-distribution distribution is more spread out, with a peak farther to the right, for larger than for
smaller degrees of freedom.

1 n = 5; % degree of freedom
2 x = 0: .1 : 3 0 ;
3 p l o t ( x , pdf ( ' ch i2 ' , x , n ) , ' LineWidth ' ,2 ) ; % V i sua l i ze the pdf
4 f i g u r e ;
5 p l o t ( x , cd f ( ' ch i2 ' , x , n ) , ' LineWidth ' ,2 ) ; % V i sua l i ze the cd f

For α ∈ (0, 1), let χ2α,n be the percentage point of the χ2 distribution as shown in Figure 2.21. That is,

P(X > χ2α,n) = 1 − P(X ≤ χ2α,n) = 1 − F(χ2α,n) = α

Thus, χ2α,n = F
−1(1 − α) where F−1 denotes the inverse cdf of the χ2(n) distribution.

For α = 0.05 and n = 5, the value of χ2α,n is computed using:

MATLAB code
>> alpha = 0.05; n = 5;

>> icdf(’chi2’,1-alpha,n)

ans =

11.0705

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Example 3.19 The time it takes a central processing unit to process a certain type of job is normally distributed with mean 20
seconds and standard deviation 3 seconds. If a sample of 15 such jobs is observed, what is the probability that the sample variance
will exceed 12?

Solution: From the given information, we have n = 15 and σ2 = 9. Thus,

P(S2 > 12) = P

(
(n − 1)S

2

σ2
> (14)

12

9

)

= P(χ2(14) > 18.67)

= 1 − P(χ2(14) ≤ 18.67) = 0.1779,

where χ2(14) is a χ2-distribution with n − 1 = 14 degrees of freedom.

MATLAB code
>> n = 15; sigma2 = 9; x = 12;

>> chisq = (n-1)*x/sigma2;

>> P = 1-cdf(’chi2’,chisq,n-1)

Example 3.20 The Bravo Widget Company claims that their widgets last 5 years, with a standard deviation of 1 year. Assume
that their claims are true. If you test a random sample of 9 Acme widgets, what is the probability that the standard deviation in
your sample will be less than 0.95 years?

Solution: From the given information, we have n = 9 and σ = 1. Thus,

P(S < 0.95) = P

(
(n − 1)S

2

σ2
< (8)

(0.95)2

(1)2

)

= P(χ2(8) < 7.22) = 0.4869,

where χ2(8) is a χ2-distribution with n − 1 = 8 degrees of freedom.

If in Eq. (3.2.7) we replace σ by the sample standard deviation S, then the sampling distribution of the statistic

t =
X − µ
S/


n
=

X−µ
σ/


n

S/σ
∼ N(0, 1)√

1
n−1 χ

2(n − 1)
∼ t(n − 1) (3.2.9)

has a t-distribution with n − 1 degrees of freedom.

Example 3.21 Eureka Corporation manufactures light bulbs. The CEO claims that an average Eureka light bulb lasts 300 days.
A researcher randomly selects 15 bulbs for testing. The sampled bulbs last an average of 290 days, with a standard deviation of 50
days. If the CEO’s claim were true, what is the probability that 15 randomly selected bulbs would have an average life of no more
than 290 days?

Solution: From the given information, we have n = 15, µ = 300, x̄ = 260 and s = 50. Thus,

P(X ≤ 290) = P
(

X − µ
S/


n
≤ 290− 300

50/


15

)

= P(t(14) ≤ −0.7746) = 0.2257,

where t(14) is a t-distribution with n − 1 = 14 degrees of freedom.

MATLAB code
>> n = 15; mu = 300; xbar = 290; s = 50;

>> x = (xbar-mu)/(s/sqrt(n));

>> P = cdf(’t’,x,n-1)

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Let X1, . . . , Xn1 ∼ N(µX, σ2X) and Y1, . . . , Yn2 ∼ N(µY, σ2Y) be two samples from two normal populations. Then, the
sampling distribution of the statistic

F =
S2X/σ

2
X

S2
Y

/σ2
Y

∼ F(n1 − 1, n2 − 1) (3.2.10)

has an F-distribution with n1 − 1 and n1 − 1 degrees of freedom, where S2X and S2Y are the sample variances of the
corresponding populations.

Example 3.22 Consider two independent samples: the first of size 10 from a normal population having variance 4 and the second
of size 5 from a normal population having variance 2. Compute the probability that the sample variance from the second sample
exceeds the one from the first.

Solution: From the given information, we have n1 = 10, n2 = 5, σ
2
X = 4 and σ

2
Y = 2. Thus,

P(S2Y > S
2
X) = P

(
S2X
S2

Y

< 1

)

= P

(
S2X/σ

2
X

S2
Y

/σ2
Y

<
σ2Y
σ2X

)

= P(F(9, 4) < 1/2) = 0.1782,

where F(9, 4) is an F-distribution with n1 − 1 = 9 and n1 − 1 = 4 degrees of freedom.

MATLAB code
>> n1 = 10; n2 = 5; sigmax2 = 4; sigmay2 = 2;

>> x = sigmay2/sigmax2;

>> P = cdf(’F’,x,n1-1,n2-1)

3.3 PROBLEMS

¶ An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed,
with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a random sample
of 16 bulbs will have an average life of less than 775 hours.

· Tabarca Electronics is a company that manufactures circuit boards. The average imperfection on a board is µ = 5
with a standard deviation of σ = 2.34 when the production process is under statistical control. A random sample
of n = 36 circuit boards has been taken for inspection and a mean of X = 6 defects per board was found. What is
the probability of getting a value of X ≤ 6 if the process is under control?

¸ A product-fill operation produces net weights that are normally distributed with mean µ = 8.06 ounces and
standard deviation σ = 0.37 ounces.

a) Estimate the percent of the containers that have a net weight less than 7.08 ounces.

b) What is the probability that a sample of nine randomly selected containers will have an average net weight
less than 7.08 ounces?

¹ If 40 percent of the parts that come off a production line are defective, what is the probability of taking a random
sample of size 75 from the line and finding that 70 percent or less are defective?

º Forty percent of all the employees have signed up for the stock option plan. An HR specialist believes that this
ratio is too high. She takes a sample of 450 employees and finds that 200 have signed up. What is the probability
of getting a sample proportion larger than this if the population proportion is really 0.4?

» A manufacturer of car batteries guarantees that the batteries will last, on average, 3 years with a standard devi-
ation of 1 year. If five of these batteries have lifetimes of 1.9, 2.4, 3.0, 3.5, and 4.2 years, should the manufacturer
still be convinced that the batteries have a standard deviation of 1 year? Assume that the battery lifetime follows
a normal distribution.

55

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