control chart for variable and attribute

PREFACE
This e-book published with a goal to ease reference, strengthen understanding, and increase
achievement. Sources gathered and compiled to complete this e-book come from wide range of
area involving reference books, notes from internet and polytechnic past year final examination
paper. This e-book contains inquisition technique with various difficulty level and step by step
problem solution.
The material in this e-book could be incorporated in a polytechnic engineering syllabus especially
for Quality Control course.
I am constantly improving and refining this teaching resource for future. I would appreciate any
feedback or suggestion you may have. Finally, I hope you enjoy using this e-book.

Ts. Rozi binti Kamsi

.

CONTENTS 1–5
6 – 16
1 Statistical Quality Control 17 – 18
2 Variable control chart 19 – 23
3 State of control 24 – 35
4 Capability analysis 36 – 55
5 Attribute control chart 56
6 Exercises 57
7 Answers
8 References

1

1.0 STATISTICAL QUALITY CONTROL (SQC)
Definition: Use of statistical tools to observe performance of the production.
There are three methods of SQC:

1. Statistical Process Control (SPC)
2. Acceptance Sampling (AS)
3. Design Of Experiment (DOE)

STATISTICAL PROCESS CONTROL (SPC)
Statistical process control (SPC) is a collection of tools that when used together can
result in process stability and variance reduction.

There are two types of controls:

1. Process control: Its aim is to quickly detect the occurrence of assignable causes of
quality variation in the production process so that corrective action may be
undertaken before many nonconforming units are produced. Process control is
achieved through the technique of control charts.

2. Product control: Its aim is to ensure that lots of manufactured products do not
contain large proportion of defective items. Product control is achieved through the
technique of acceptance sampling inspection plans.

2

Seven (7) basic tools of SPC

GENERATING IDEA
1. CHECKSHEET
2. SCATTER DIAGRAM
3. CAUSE AND EFFECT DIAGRAM

ORGANIZE THE DATA
4. PARETO CHART
5. FLOW CHART

IDENTIFYING PROBLEM
6. HISTOGRAM
7. CONTROL CHART

CONTROL CHART

The control chart is a graph used to study how a process changes over time. Data
are plotted in time order. A control chart always has a central line for the average, an
upper line for the upper control limit, and a lower line for the lower control limit.
These lines are determined from historical data. By comparing current data to these
lines, you can draw conclusions about whether the process variation is consistent (in
control) or is unpredictable (out of control, affected by special causes of variation).

Control charts for variable data are used in pairs. The top chart monitors the
average, or the centering of the distribution of data from the process. The bottom
chart monitors the range, or the width of the distribution. If your data were shots in
target practice, the average is where the shots are clustering, and the range is how
tightly they are clustered. Control charts for attribute data are used singly.

3

Data
Types of data
1) Variable data - quality characteristics that are measurable values.
Example: Length, Temperature.
2) Attribute data - quality characteristics that are observed to be either present or
absent, conforming or nonconforming.
Example: Good, Bad.

Types of control chart
1) Variables charts

X bar and R chart
X bar and s chart

2) Attributes charts
p chart
100 p chart
np chart
c chart
u chart

Variation
Concept of variation

• No two natural items in any category are the same.
• Ability to measure variation is necessary before it can be controlled

There are 3 categories of variation in piece part production (example: Light bulbs,

washer, nuts)

1. Within piece - e.g. surface roughness

2. Piece to piece - e.g. diameter

3. Time to time - different outcomes e.g. morning & afternoon

4

Sources of variation

Many factors contribute to variation:

Equipment - tool wear.

Material - tensile strength.

Environment - temperature.

Operator - motivation level.

Inspection - inspection equipment.

Causes of variation
1. Chance causes / random causes/common causes
2. Assignable causes/special causes

Chance or random causes is natural process variation or system variation. A part
of process. Cannot eliminate but can be reduced.
Examples of Common Causes:

• Variations in environmental variables such as temperature and humidity
• Variations among machines
• Variations among worker’s skills
• Variations in raw materials

Figure 1
Based on the figure 1, notice that all points lie within the control limits.
The process above is in control.

5

Assignable or Special cause variation is caused by some problem or
extraordinary occurrence in the system. Can be eliminated.

Examples of Assignable Causes:
Temporary change to a new supplier
Temporary worker not properly trained

Figure 2

Based on Figure 2, notice that a single point can be found outside the control
limits(above them). The process above is out of control.

6

2.0 VARIABLE CONTROL CHART

Objectives Of Variable Control Chart:
• For quality improvement
• To determine process capability
• For decisions in product specifications
• Provide information on production processes for current decisions

Pairing X-Bar with R-Charts

X-Bar (average) charts and R (range) -charts are often paired together. The
X-Bar chart displays the centerline, which is calculated using the grand
average, and the upper and lower control limits, which are calculated using
the average range. Future experimental subsets are plotted compared to
these values. This demonstrates the centering of the subset values. The R-
chart plots the average range and the limits of the range. Again, the future
experimental subsets are plotted relative to these values. The R-chart
displays the dispersion of the subsets. X-Bar/R-Chart plot a subgroup
average. Note that they should only be used when subgroups really make
sense. For example, in a Gage R&R study, when operators are testing in
duplicates or more, subgrouping really represents the same group.

Pairing X-Bar with S-Charts

Alternatively, X-Bar charts can be paired with S-charts (standard deviation).
This is typically done when the size of the subsets are large. For larger
subsets, the range is a poor statistic to estimate the distributions of the
subsets, and instead, standard deviation is used. In this case, the X-Bar chart
will display control limits that are calculated using the average standard
deviation. The S-Charts are similar to the R-charts; however, instead of the
range, they track the standard deviation of multiple subsets.

7

Control chart techniques
1. Select quality characteristic
2. Choose rational subgroup
3. Collect data
4. Determine the central line and trial limits
5. Establish the central line and revised limits
6. Achieve the objective

1. Select quality characteristic
Measurable data (weight, length, mass, time, etc.)

2. Choose rational subgroup

Two ways selecting subgroup samples

1. Select period of time products are produced.

2. Select from homogeneous lot: same machine, same operator.

Lot Size Sample

91 – 150 Size
151 – 280 10
281 – 400 15
401 – 500 20
501 – 1200 25
1201 – 3200 35
3201 – 10,000 50
10,001 – 35,000 75
35,001 – 150,000 100
150

Table 1: Sample size
Source: ANSI/ASQ Z1.9, Normal Inspection, Level II.

8

The use of Table 1 can be a valuable aid in making judgements on the amount of
sampling required. If a process is expected to produce 15,000 pieces per day, then
100 pieces inspection are suggested.

3. Collect data
1. Use form or standard check sheet
2. Collect a minimum of 25 subgroups
3. Data can be vertically / horizontally arranged

Subgroup/ Time Measurements Average Range
Sample
1 8.30 X1 X2 X3 X4 XR
2 9.30
55 52 51 53 52.8 4

51 52 57 50 52.5 7

Sample or subgroup Sample or subgroup size is
number is symbolized with g. symbolized with n.

Table 2: Example of a method of reporting inspection results

Figure 3: Formation of subgroup

9

4. Determine trial central line and control limits
i. Average ( X ) and range ( R ) control chart
Central line

X =X R= R
g g

where, X = average of subgroup average (read “X double bar”)

X = average of subgroup

g = number of the subgroups
R = average of the subgroup ranges
R = range of the subgroups

Control limit X - chart Control limit R – Chart
Upper control limit
Upper control limit
UCLR = D4 R
UCL = X + A2 R
x

Lower control limit Lower control limit

LCL = X − A2 R LCL R = D3 R
x

Where, UCL = upper control limit
LCL = lower control limit

A2, D3, D4 = factors found in Table B: factors for computing central lines and

3σ control limit for X, s and R charts.

10

Example 1:
Using the data of Table 3.1, determine the central line and control limits for X and R
charts.

Subgroup Date Time MEASUREMENTS Average Range Comment

number X1 X2 X3 X4 R

1 26/12 8.30 6.35 6.40 6.32 6.37 6.36 0.08

2 11.30 6.46 6.37 6.36 6.41 6.40 0.10

3 1.45 6.34 6.40 6.34 6.36 6.36 0.06

4 3.45 6.69 6.64 6.68 6.59 6.65 0.10 New, temporary

5 4.20 6.38 6.34 6.44 6.40 6.39 0.10 operator

6 27/12 8.35 6.42 6.41 6.43 6.34 6.40 0.09

7 9.00 6.44 6.41 6.41 6.46 6.43 0.05

8 9.40 6.33 6.41 6.38 6.36 6.37 0.08

9 1.30 6.48 6.44 6.47 6.45 6.46 0.04

10 2.50 6.47 6.43 6.36 6.42 6.42 0.11

11 28/12 8.30 6.38 6.41 6.39 6.38 6.39 0.03

12 1.35 6.37 6.37 6.41 6.37 6.38 0.04

13 2.25 6.40 6.38 6.47 6.35 6.40 0.12

14 2.35 6.38 6.39 6.45 6.42 6.41 0.07

15 3.55 6.50 6.42 6.43 6.45 6.45 0.08

16 29/12 8.25 6.33 6.35 6.29 6.39 6.34 0.10

17 9.25 6.41 6.40 6.29 6.34 6.36 0.12

18 11.00 6.38 6.44 6.28 6.58 6.42 0.30 Damaged oil line

19 2.35 6.35 6.41 6.37 6.38 6.38 0.06

20 3.15 6.56 6.55 6.45 6.48 6.51 0.11 Bad material

21 30/12 9.35 6.38 6.40 6.45 6.37 6.40 0.08

22 10.20 6.39 6.42 6.35 6.40 6.39 0.07

23 11.35 6.42 6.39 6.39 6.36 6.39 0.06

24 2.00 6.43 6.36 6.35 6.38 6.38 0.08

25 4.25 6.39 6.38 6.43 6.44 6.41 0.06

SUM 160.25 2.19

Table 3: Data on the Depth of the Keyway (mm) Source: Besterfield 2013

11

Solution 1

Data given:

Subgroup / Sample size, n = 4

Number of subgroups / samples, g = 25
Based on Table B: factors for computing central lines and 3σ control limit for

X , s and R charts.

A2 = 0.729 D3 = 0 D4 = 2.282

Control chart – X Control chart – R

Central line Central line

X =X R= R
g
g
X = 160.25
25 R = 2.19
25
= 6.41
= 0.0876

Control limit Control limit
Upper control limit Upper control limit

UCL = X + A2 R LCL = X − A2 R
x x

= 6.41+ (0.729)(0.0876) = 6.41− (0.729)(0.0876)

= 6.47 = 6.35

Upper control limit Lower control limit

UCL R = D4 R LCLR = D3 R
= (2.282)(0.0876) = (0)(0.0876)
= 0.20 =0

12

Figure 4: X and R chart for preliminary data with trial control limits

Source: Besterfield 2013

Control chart in Figure 4 show that there are out of controlXpoints on chart at

subgroup 4,16 and 20 and an out of control point on R chart at subgroup 18.
When a point falls outside its control limits, the process is out of control. This
means that an assignable cause of variation is present.

13

ii. Average (X ) and sample standard deviation ( s ) control chart

Control chart – X Control chart – s

Central line (CL) Central line (CL)

X =X s = s
g
g

Upper control limit (UCL) Upper control limit (UCL)

UCL = X + A3 s UCL s = B4 s
x

Lower control limit (UCL) Lower control limit (LCL)

LCL = X − A3 s LCL s = B3 s
x

where,
s = sample standard deviation of subgroup values

s = average of the subgroup sample standard deviations

A3, B3, B4 = factors found in Table B: factors for computing central lines and

3σ control limit for X , s and R charts.

14

Example 2
Using the data of Table 4, determine the central line and control limits for X and s
charts.

Table 4: Data on the Depth of the Keyway (mm)
Source: Besterfield 2013

15

Solution 2

Data given:

Sample size, n = 4

Number of subgroups, g = 25
Based on Table B: factors for computing central lines and 3σ control limit for

X , s and R charts.

A3 = 1.628 B3 = 0 B4 = 2.266

4(6.352 + 6.402+ 6.322 + 6.372) − (6.352 + 6.402+ 6.322+ 6.372)
= √ 4(4 − 1)
= 0.034

Control chart – X Control chart – s
Central line (CL) Central line (CL)

X =X s = s
g
g = 0.975
= 160.25
25
25 = 0.039
= 6.41
Upper control limit (UCL)
Upper control limit (UCL)
UCL s = B4 s
UCL = X + A3 s = (2.266)(0.039)
x = 0.088

= 6.41+ (1.628)(0.039) Lower control limit (LCL)

= 6.47 LCL s = B3 s
= (0)(0.039)
Lower control limit (UCL) =0

LCL = X − A3 s
x

= 6.41− (1.628)(0.039)

= 6.35

16

Figure 5: X and s chart for preliminary data with trial control limits
Source: Besterfield 2013

Control chart in Figure 5 show that there are out of control points on X - chart at
subgroup 4, 16 and 20 and an out of control point on s- chart at subgroup 18.
Process out of control.

17

3.0 STATE OF CONTROL
Process Out Of Control
When a point (subgroup value) falls outside its control limits, the process is
out of control.
Out of control means a change in the process due to a special cause. A
process can also be considered out of control even when the points fall
inside the 3ơ limits:
❑ It is not natural for seven or more consecutive points to be above or
below the central line.
❑ Also when 10 out of 11 points or 12 out of 14 points are located on one
side of the central line, it is unnatural.
❑ Six points in a row are steadily increasing or decreasing indicate an out
of control situation

Analysis of Out-of-Control Patterns
1. Change or jump in level.
2. Trend or steady change in level
3. Recurring cycles
4. Two populations (also called mixture)
5. Mistakes

18

1. Change/Jump in level
 A new operator.

2. Trend or steady change in level
 tool or die wear.

3. Recurring cycles
• The seasonal effects of incoming material.

4. Two populations (mixture)
• Large difference in material quality.

19

4.0 CAPABILITY ANALYSIS

Capability analysis is about determining how well a process meets a set of

specification limits, based on a sample of data taken from a process. It can be

used to establish baseline for the process and measure the future state

performance of the process for comparison.

Process capability refers to the uniformity of the process. It is also measures
the “goodness of a process”.

Process capability is equal to 6σ when the process is in statistical control.

The process capability can be obtained by using the sample standard

deviation and the range.

Procedure (the sample standard deviation method)

1. Take 20 subgroups of size 4.

2. Calculate the sample standard deviation, s, for each subgroup s c4

3. Calculate the average sample standard deviation, s = s g

4. Calculate the estimate of population standard deviation, ˆo =
5. Calculate Process Capability = 6ˆ

Example 3

A new process is started and the sum of the sample standard deviation for 20

subgroups of size 4 is 84. Determine the process capability.

Solution 3:

Data given:

Sample size, n = 4

Number of subgroups, g = 20

The sum of the sample standard deviation, ∑ = 84

Based on Table B: factors for computing central lines and 3σ control

limit for X , s and R charts.

c4 = 0.9213

∑ 84
= = 20 = 4.2

4.2
= 4 = 0.9213 = 4.6
6 = 6(4.6) = 27.6

20

Procedure (the range method)
1. Take 20 subgroups of size 4.
2. Calculate the range, R, for each subgroup

3. Calculate the average range, R = R g

4. Calculate the estimate of population standard deviation, ˆ o = R d2
5. Calculate Process Capability = 6ˆ

Example 4
An existing process not meeting in the Rockwell-C specifications. Determine
the process capability based on the range values for 20 subgroups of size 4.
Data are 7, 5, 5, 3, 2, 4, 5, 9, 4, 5, 4, 7, 5, 7, 3, 4, 7, 5, 5, and 7.

Solution 4:
Data given:
Sample size, n = 4
Number of subgroups, g = 20
The sum of the range, ∑ = 103
Based on Table B: factors for computing central lines and 3σ control
limit for X , s and R charts.
d2 = 2.059
∑ 103
= = 20 = 5.15
5.15
= 2 = 2.059 = 2.50
6 = 6(2.5) = 15

21

Process Capability Index
Process capability is measured by the process capability index, Cp and Cpk.
Cp is used when the process variability is centered on the specification range.
If not, Cpk will be used to measure process capability.

Comment concerning Cp and Cpk are as follows:
1. The Cp value does not change as the process center changes.
2. Cp = Cpk when process is centered.
3. Cpk is always equal to or less than Cp
4. A Cpk value of 1.00 is a de facto standard. It indicates that the process is

producing product that conforms to specifications.
5. A Cpk value less than 1.00 indicates that the process is producing product

that does not conforms to specifications.
6. A Cp value less than 1.00 indicates that the process is not capable.
7. A Cpk value of zero indicates that the average is equal to one of the

specification limits.
8. A negative Cpk value indicates that the average is outside the

specifications.

A process capability index, Cp
A process capability index, Cp is formed from the combination of process
capability and the tolerance.

Cp = USL − LSL
6

where Cp = capability index
USL – LSL = upper specification limit – lower specification limit or tolerance

6σ = process capability
Notes
Specification limits are quantified performance standards determined by
customer requirements.

22

Example 5
A new process is started and the sum of the sample standard deviations for
20 subgroups of size 4 is 600. If the specifications are 700 ± 80, what is the
process capability index, Cp ? comment the process ?

Solution 5:

Data given:

Sample size, n = 4

Number of subgroups, g = 20

The sum of the sample standard deviation, ∑ = 600

Based on Table B: factors for computing central lines and 3σ control

limit for X , s and R charts.

c4 = 0.9213 LSL = 700 – 80 = 620
USL = 700 + 80 = 780 ∑ 600
= = 20 = 30

30
= 4 = 0.9213 = 32.56
6 = 6(32.56) = 195.36

− 780 − 620
= = 195.36 = 0.819
Comment: Cp value less than 1.00 indicates that the process is not capable.

23

A process capability index, Cpk

C pk = min USL −  , − LSL 
 3 3 

where,

µ= the mean of the process

σ = the population standard deviation of the process

Example 6
Calculate the Cpk measure of process capability for the following machine and
interpret the findings.
Process mean, µ = 60
Process standard deviation, σ = 10
Lower specification limit = 50
Upper specification limit = 110

Solution 6:

C pk = min USL −   − LSL 
 3 , 3 

= min 110− 60 , 60 − 50 
3(10) 3(10)

= min 50 , 10 
 30 30 

= min(1.67,0.33)

C pk = 0.33

This means that the process is not capable and not conform to the

specification because Cpk < 1.0.

24

5.0 ATTRIBUTE CONTROL CHART
Types of attribute control charts
a. p chart
This chart shows the fraction of nonconforming or defective
product produced by a manufacturing process.
b. np chart
This chart shows the number of nonconforming or defective
product produced by a manufacturing process. Almost same as the
p chart.
c. c chart
This chart shows the number of nonconformities or defects
produced by a manufacturing process.
d. u chart
This chart shows the nonconformities per unit or defects per unit
produced by a manufacturing process.

CONTROL CHART FOR NONCONFORMING UNIT
There are two types of charts:

1. The fraction of nonconforming unit – p chart
2. The number of nonconforming unit – np chart
1. p – CHART
The p-chart is used in quality control to report the fraction of nonconforming a
product, quality characteristic or group of quality characteristic. It is used to
report the performance of an operator, a group of operator or management as
a mean of evaluating their quality performance.

Fraction nonconforming, p: p = np
where; n

p = proportion or fraction nonconforming in the sample or subgroup

n = number in the sample or subgroup

np = number nonconforming in the sample or subgroup

25

Example 7
During the first shift, 450 inspection are made of book-of the month shipments
and 5 nonconforming units are found. Production during the shift was 15,000
units. What is the fraction nonconforming?

Solution 7:
Number of lot, N = 15000 units
Number in sample or subgroup, n = 450 units
Number nonconforming in sample or subgroup, np = 5 units
Fraction nonconforming, p = ?

p = (np)/n = 5/450 = 0.01

Calculation for the trial central line and control limits.

Central line =The average fraction nonconforming unit,

∑
p̅ = ∑

Upper control limit, Lower control limit,

UCL = p + 3 p(1− p) LCL = p −3 p(1− p)
n n

Example 8: Number Number Proportion
Subgroup Inspected, Nonconforming, Nonconforming,

n np p
12 0.040
1 300 3 0.010
9 0.030
2 300 4 0.013
0
3 300 0.0

4 300

5 300

26

6 300 6 0.020
7 300 6 0.020
8 300 1 0.003
9 300 8 0.027
10 300 11 0.037
11 300 2 0.007
12 300 10 0.033
13 300 9 0.030
14 300 3 0.010
15 300 0
16 300 5 0.0
17 300 7 0.017
18 300 8 0.023
19 300 16 0.027
20 300 2 0.053
21 300 5 0.007
22 300 6 0.017
23 300 0 0.020
24 300 3
25 300 2 0.0
Total 7500 138 0.010
0.007

Table 5: Inspection results of hair dryer blower motor, motor department, May

27

Solution 8:

CL = p =  np = 138 = 0.018
n 7500

UCL = 0.018+ 3 0.018(1− 0.018) LCL = 0.018− 3 0.018(1− 0.018)
300 300

= 0.041 = −0.005 = 0.0

Note:
Negative value of LCL is possible in a theoritical result, but not in practical.
Therefore, the LCL value of -0.005 is changed to zero.

p 0.053

0.041 UCL

0.0 p-bar
0.0183 LCL
0.0 2

1

0

5 10 15 20 25

Subgroup

Figure 6: p – control chart

28

2. np – CHART

The np chart is almost the same as the p chart.

Calculation for the trial central line and control limits.

p0 is the standard or reference value of the fraction nonconforming.

Central line = npo

Upper control limit, Lower control limit,

UCL = npo + 3 npo (1− po ) LCL = npo −3 npo (1− po )

Example 9
A government agency samples 200 documents per day from a daily lot of 6000.
From past records the standard or reference value for the fraction
nonconforming, p0, is 0.075. Calculate central line and control limit.

Solution 9:

lot, N= 6000 sample, n = 200, p0 = 0.075

Central line = npo = 200(0.075) = 15.0 LCL = npo −3 npo (1− po )
LCL =15 − 3 15(1 − 0.075)
UCL = npo + 3 npo (1− po ) LCL = 3.8

UCL =15 + 3 15(1 − 0.075)
UCL = 26.2

If po is unknown, it must be determined by collecting data, calculating UCL,
LCL.

29

̅ is the average fraction nonconforming of many subgroups.

p = np
n

Central line,

CL =np

Upper control limit, Lower control limit,

UCL = n p + 3 n p(1 − p) LCL = n p − 3 n p(1 − p)

Example 10
Based on table 5, calculate central line and control limit for np – chart.

Solution 10:

p =  np = 138 = 0.018
n 7500

Central line, CL = np = (300)(0.018) = 5.4

UCL = n p + 3 n p(1 − p) LCL = n p − 3 n p(1 − p)
UCL = 5.4 + 3 5.4(1 − 0.018) LCL = 5.4 − 3 5.4(1 − 0.018)
UCL =12.3 LCL = −1.5 = 0

Note:
p is the fraction nonconforming in a single subgroup
̅ is the average fraction nonconforming of many subgroups.
P0 is the standard or reference value of the fraction nonconforming.

30

Figure 7: np – control chart
Comment: Process is out of control

31

CONTROL CHART FOR COUNT OF NONCONFORMITIES
There are two types of charts:

1. Count of nonconformities – c chart
2. Count of nonconformities per unit – u chart
1. c – CHART
The c – chart is applicable where the subgroup size is an inspected unit of one
such as a canoe, a car and a ream of paper. The inspected unit can be any
size, however it must be constant.

c

Central line, CL = c =
g
Upper control limit, Lower control limit,

UCL = c + 3 c LCL = c −3 c

Example 11:

SERIAL COUNT OF SERIAL COUNT OF
NUMBER NONCONFORMITIES
NUMBER NONCONFORMITIES

MY102 7 MY198 3

MY113 6 MY208 2

MY121 6 MY222 7

MY125 3 MY235 5

MY132 20 MY241 7

MY143 8 MY258 2

MY150 6 MY259 8

MY152 1 MY264 0

MY164 0 MY267 4

MY166 5 MY278 14

MY172 14 MY281 4

MY184 3 MY288 5

MY185 1

Table 6: Count of Blemish Nonconformities by Canoe Serial Number

32

Solution 11:

CL = c =  c = 141 = 5.64

g 25
UCL =c +3 c = 5.64 + 3 5.64 = 12.76
LCL =c −3 c = 5.64 − 3 5.64 = −1.48 = 0.0

c-Chart
25

Count of Nonconformities 20
c
UCL
c-bar

15 LCL

10

5

0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Subgroup Num ber

Figure 8: c – control chart

Comment: process is out of control

2. u – CHART
The u – chart is mathematically equivalent to the c – chart. The u – chart can
be used when the subgroup size is vary.

Count of nonconformities per unit in a subgroup, u = c
n

Central line = average count of nonconformities per unit for many subgroup

u =  c
 n

33

Upper control limit, Lower control limit,

UCL =u + 3 u LCL =u −3 u
n n

Example 12:

Date n c Date nc
30-Jan 110 120 16-Feb 85 92

31-Jan 82 94 17-Feb 101 140
1-Feb 96 89 18-Feb 42 60
2-Feb 115 162 20-Feb 97 121

3-Feb 108 150 21-Feb 92 108
4-Feb 56 82 22-Feb 100 131
6-Feb 120 143 23-Feb 115 119

7-Feb 98 134 24-Feb 99 93
8-Feb 102 97 25-Feb 57 88
9-Feb 115 145 27-Feb 89 107

10-Feb 88 128 28-Feb 101 105
11-Feb 71 83 1-Mar 122 143
13-Feb 95 120 2-Mar 105 132

14-Feb 103 116 3-Mar 98 100
15-Feb 113 127 4-Mar 48 60

Table 7: count of nonconformities per unit for waybill

34

Solution 12:

For January 30:

u Jan 30 = c = 120 = 1.09
n 110

CL = u = c = 3389 = 1.20
n 2823

UCLJan30 =1.20 + 3 1.20 = 1.51 LCLJan30 =1.20 − 3 1.20 = 0.89
110 110

Date ncu UCL u-Bar LCL
30-Jan 0.89
31-Jan 110 120 1.091 1.51 1.20 0.84
1-Feb 0.87
2-Feb 82 94 1.146 1.56 1.20 0.89
3-Feb 0.88
4-Feb 96 89 0.927 1.54 1.20 0.76
6-Feb 0.90
7-Feb 115 162 1.409 1.51 1.20 0.87
8-Feb 0.88
9-Feb 108 150 1.389 1.52 1.20 0.89
10-Feb 0.85
11-Feb 56 82 1.464 1.64 1.20 0.81
13-Feb 0.86
14-Feb 120 143 1.192 1.50 1.20 0.88
15-Feb 0.89
16-Feb 98 134 1.367 1.53 1.20 0.84
17-Feb 0.87
18-Feb 102 97 0.951 1.53 1.20 0.69
20-Feb 0.87
21-Feb 115 145 1.261 1.51 1.20 0.86
22-Feb 0.87
88 128 1.455 1.55 1.20

71 83 1.169 1.59 1.20

95 120 1.263 1.54 1.20

103 116 1.126 1.52 1.20

113 127 1.124 1.51 1.20

85 92 1.082 1.56 1.20

101 140 1.386 1.53 1.20

42 60 1.429 1.71 1.20

97 121 1.247 1.53 1.20

92 108 1.174 1.54 1.20

100 131 1.310 1.53 1.20

35

23-Feb 115 119 1.035 1.51 1.20 0.89
24-Feb 99 93 0.939 1.53 1.20 0.87
25-Feb 57 88 1.544 1.64 1.20 0.77
27-Feb 89 107 1.202 1.55 1.20 0.85
28-Feb 101 105 1.040 1.53 1.20 0.87
1-Mar 122 143 1.172 1.50 1.20 0.90
2-Mar 105 132 1.257 1.52 1.20 0.88
3-Mar 98 100 1.020 1.53 1.20 0.87
4-Mar 48 60 1.250 1.67 1.20 0.73

Figure 9: u – control chart
Comment: process is in control

36

6. EXERCISES

QUESTION 1

A quality inspector intends to construct a mean ( X ) and range control chart to

monitor the quality of service in a bank. 20 sample of the length of time to process

loan application were taken.Each sample has three observations and all values are

in minutes as shown in table below.

Sample Observation

number X1 X2 X3

1 12 10 11

2 9 10 8

3 10 10 10

48 9 12

5 13 14 15

67 8 10

7 15 10 10

8 14 13 8

9 10 9 10

10 9 10 10

11 14 8 9

12 8 9 9

13 7 10 9

14 10 10 8

15 10 9 15

16 10 14 14

17 14 8 10

18 16 16 17

19 17 15 14

20 8 7 8

37

i. Determine the central line and control limits for control chart –X and R.
ii. Construct the control chart on a graph paper.
iii. Interpret the process based on the control chart.

SOLUTION 1:

Data given:

Sample size, n = ______

Number of subgroups, g = ______

Based on Table B: factors for computing central lines and 3σ control limit for

X , s and R charts.

A2 = ____________ D3 = _____________ D4 = ____________

Sample Observation Average Range
number R
X1 X2 X3 X
1
2 12 10 11
3
4 9 10 8
5
6 10 10 10
7
8 8 9 12
9
10 13 14 15
11
12 7 8 10
13
14 15 10 10
15
16 14 13 8
17
10 9 10

9 10 10

14 8 9

899

7 10 9

10 10 8

10 9 15

10 14 14

14 8 10

38

Sample Observation Average Range
number
X1 X2 X3 X R
18 16 16 17
19
20 17 15 14

878

SUM

Control chart X Control chart R
Central line (CL) Central line (CL)

X = X R= R

g g
= =
= =

Upper control limit(UCL) Upper control limit (UCL)

UCL = X + A2 R UCL R = D4 R
x =
=
=
Lower control limit (LCL)
=
LCL R = D3 R
Lower control limit (UCL) =
=
LCL = X − A2 R
x

=

=

39

QUESTION 2
The following table gives the average and range in kilograms for tensile test on an
improved plastic cord. The subgroup size is 4.

SUBGROUP SUBGROUP
NUMBER
1 X R NUMBER X R
2 23
3 476 32 16 496 25
4 466 24
5 484 24 17 478 23
6 466 25
7 470 32 18 484
8 494
9 486 26 19 506
10 496
11 488 24 20 476
12 482
13 498 24
14 464
15 484 28
482
506 23

24

26

25

24

24

22

23

iv. Determine the central line and control limits for control chart – X and R.
v. Construct the control chart on a graph paper.
vi. Interpret the process based on the control chart.

40

SOLUTION 2

Data given:

Sample size, n = ______

Number of subgroups, g = ______
Based on Table B: factors for computing central lines and 3σ control limit for

X , s and R charts.

A2 = ____________ D3 = _____________ D4 = ____________

SUBGROUP X SUBGROUP R
NUMBER 23
1 476 R NUMBER X 25
2 466 24
3 484 32 16 496 23
4 466 24 17 478 25
5 470 32 18 484
6 494 26 19 506
7 486 24 20 476
8 496 24 SUM
9 488 28
10 482 23
11 498 24
12 464 26
13 484 25
14 482 24
15 506 24
22
23

Control chart X 41
Central line (CL)
Control chart R
X = X Central line (CL)

g R= R
=
= g
=
Upper control limit(UCL) =

UCL = X + A2 R Upper control limit (UCL)
x
UCL R = D4 R
= =
=
=
Lower control limit (LCL)
Lower control limit (UCL)
LCL R = D3 R
LCL = X − A2 R =
x =

=

=

42

QUESTION 3
Children’s plastic cups are the products of Sarjana Sdn Bhd. Raw plastic materials
are initially poured and packed according to the desired shape. Recovery time (cure
time) during the casting process affects the quality of the side (edge) of the product
is given in a table below.

Subgroup Time (second)
T1 T2 T3
1 31 29 31
2 34 25 37
3 41 39 29
4 29 25 25
5 25 34 30
6 46 34 41
7 35 38 33
8 34 33 35
9 43 37 38
10 37 42 35
11 26 30 33
12 28 30 31
13 25 34 30
14 46 34 41
15 35 38 33

Calculate:
i. Average, X and sample standard deviation, s for each of subgroup.
ii. Determine the central line and control limits for control chart – X and s.
iii. Construct the control chart on a graph paper.
iv. Interpret the process based on the control chart.

43

SOLUTION 3:
Data given:
Sample size, n = ______
Number of subgroups, g = ______
Based on Table B: factors for computing central lines and 3σ control limit for

X , s and R charts.

A3 = ____________ B3 = _____________ B4 = ____________

Subgroup Time (second) X s
T1 T2 T3
1 31 29 31
2 34 25 37
3 41 39 29
4 29 25 25
5 25 34 30
6 46 34 41
7 35 38 33
8 34 33 35
9 43 37 38
10 37 42 35
11 26 30 33
12 28 30 31
13 25 34 30
14 46 34 41
15 35 38 33

SUM

Control chart – X 44
Central line (CL)
Control chart – s
X = X Central line (CL)

g s = s
=
= g
=
Upper control limit(UCL) =

UCL = X + A3 s Upper control limit (UCL)
x
UCL s = B4 s
= =
=
=
Lower control limit (LCL)
Lower control limit (UCL)
LCL s = B3 s
LCL = X − A3 s =
x =

=

=

45

QUESTION 4
Control charts for and s are to be established on the Brinell hardness of hardened
tool steel in kilograms per square millimeter. Data for subgroup sizes of 8 are shown
in the table below.

subgroup X subgroup s
number s number X 26
540 26 16 579 28
1 534 23 17 549 23
2 545 24 18 508 22
3 561 27 19 569 28
4 576 25 20 574
5 523 50
6 571 29
7 547 29
8 584 23
9 552 24
10 541 28
11 545 25
12 546 26
13 551 24
14 522 29
15

Calculate:
i. Determine the central line and control limits for control chart – X and s.
ii. Construct the control chart on a graph paper.
iii. Interpret the process based on the control chart.

46

SOLUTION 4

Data given:

Sample size, n = ______

Number of subgroups, g = ______
Based on Table B: factors for computing central lines and 3σ control limit for

X , s and R charts.

A3 = ____________ B3 = _____________ B4 = ____________

subgroup X subgroup s
number 26
540 X 28
1 534 23
2 545 s number 22
3 561 26 16 579 28
4 576 23 17 549
5 523 24 18 508
6 571 27 19 569
7 547 25 20 574
8 584 50 SUM
9 552 29
10 541 29
11 545 23
12 546 24
13 551 28
14 522 25
15 26
24
29

Control chart – X 47
Central line (CL)
Control chart – s
X = X Central line (CL)

g s = s
=
= g
=
Upper control limit(UCL) =

UCL = X + A3 s Upper control limit (UCL)
x
UCL s = B4 s
= =
=
=
Lower control limit (LCL)
Lower control limit (UCL)
LCL s = B3 s
LCL = X − A3 s =
x =

=

=