Sampling 4 pieces of precision cut wire

Intro To Operations Management

Chapter 6s: Control Chart

Formulas

For X Bar Charts when we know 

UCLx̅ = x̿ + z σx̅

LCLx̅ = x̿ - z σx̅

σx̅ = σ/√n

Where

x̿ = mean of the sample means or a target value set for the process

z = number of normal standard deviations 𝜎�̅� = standard deviation of the sample means

 = population (process) standard deviation

n = sample size (subgroup number)

When we don’t know 

X Bar Chart

UCLX̅ = x̿ + A2 R̅

LCLX̅ = x̿ - A2 R̅

R Chart

UCLR = D4 R̅

LCLR = D3 R̅

Factor for Computing Control Chart

Sample Size (n) Mean Factor A2 Upper Range D4 Upper Range D3

2 1.88 3.268 0

3 1.023 2.574 0

4 0.729 2.282 0

5 0.577 2.115 0

6 0.483 2.004 0

7 0.419 1.924 0.076

8 0.373 1.864 0.136

9 0.337 1.816 0.184

10 0.308 1.777 0.223

Chapter 6s: Control Chart

Question 1. Boxes of chocolates are produced to contain average of 14 ounces

(target value), with a standard deviation of 0.1 ounce. Set up the 3-sigma �̅�

chart for a sample size of 36 boxes. Determine the upper and lower control

limits

Solution:

�̿� = 14 ounces z = 3

𝜎�̅� = 𝜎/√𝑛

𝜎�̅� = 0.1/√36 = 0.0167

𝑈𝐶𝐿�̅� = �̿� + z 𝜎�̅�

𝑈𝐶𝐿�̅� = 14 + (3 * 0.0167) = 14.05 oz

𝐿𝐶𝐿�̅� = �̿� - z 𝜎�̅�

𝐿𝐶𝐿�̅� = 14 – (3 * 0.0167) = 13.95 oz

Question 2. The overall average on a process you are attempting to monitor is

50 kg. The process standard deviation is 1.72. Determine the upper and lower

control limits for a mean chart, if you choose to use a sample size of 5.

a) Determine the upper and lower control limits for the x-bar chart Set z=3

b) Determine the upper and lower control limits for the x-bar chart Set z=2.

Solution

n= 5 �̿� = 50  = 1.72 z = 3

a) 𝜎�̅� = (1.75/√5) = 0.78

𝑈𝐶𝐿�̅� = 50 + (3 * 0.78) = 52.34

𝐿𝐶𝐿�̅� = 50 - (3 * 0.78) = 47.66

b) 𝜎�̅� = (1.75/√5) = 0.78

𝑈𝐶𝐿�̅� = 50 + (2 * 0.78) = 51.56

𝐿𝐶𝐿�̅� = 50 - (2 * 0.78) = 48.44

Chapter 6s: Control Chart

Question 3. For the past 7 hours, seven samples of size 3 were taken from a

process, and their weights measured. The sample averages and sample ranges

are in the following table.

a) Use the R chart to compute the UCL & LCL

b) Use X bar chart to calculate the UCL & LCL X bar chart

c) Is the process in control?

d) What additional steps should the quality assurance team take?

Note: Round up/down to the nearest whole number

Hour sample 1 sample 2 sample 3

1 32 28 30

2 40 29 31

3 32 30 42

4 31 30 29

5 29 27 28

6 32 30 27

7 42 40 39

Solution

n = 3

From the Factor for Computing Control Chart Table:

A2 = 1.023

D4 = 2.574

D3 = 0

Day sample 1 sample 2 sample 3 R x̅

1 32 28 30 32-28= 4 �̅� = 32+28+30

3 = 30

2 40 29 31 40-29 =11 �̅� = 40+29+31

3 = 33

3 32 30 42 42- 30 = 12 �̅� = 32+30+42

3 = 35

4 31 30 29 31- 29 = 2 �̅� = 31+30+29

3 = 30

5 29 27 28 29- 27 = 2 �̅� = 29+27+28

3 = 28

6 32 30 27 32- 27 = 5 �̅� = 32+30+27

3 = 30

7 42 40 39 42- 39 = 3 �̅� = 42+40+39

3 = 40

�̅� = 4+11+12+2+2+5+3

7 = 6

x̿ = 30+33.33+34.67+30+28+29.67+40.33

7 = 32

Chapter 6s: Control Chart

a) R chart

UCLR = D4 R̅

UCLR = (2.574 * 6) = 15

LCLR = D3 R̅

LCLR = (0 * 6) = 0

b) X bar chart

UCLX̅ = x̿ + A2 R̅

UCLX̅ = 32 + (1.023 * 6) = 38

LCLX̅ = x̿ - A2 R̅

LCLX̅ = 32 + (1.023 * 6) = 26

c) The process is out of control because x̅ for sample 7 is greater than the UCL in

x-bar chart. For R chart indicates all of the samples are within control limit. We

can conclude that the process is out of control.

d) The quality assurance team should investigative sample 7 to find the root cause of

the problem.

Question 4. Sampling 4 pieces of precision-cut wire (to be used in computer

assembly) every hour for the past 12 hours has produced the following results:

Hour x̅ R

1 3.5 1.25

2 3 1

3 1.25 1.5

4 3.5 1.25

5 3 1.5

6 2.75 0.25

7 3 0.5

8 2.5 1.5

9 3 0.75

10 2.75 1.5

11 2.25 1

12 3 0.5

Chapter 6s: Control Chart

a. Determine the upper and lower control limits for the x-bar chart

b. Determine the upper and lower control limits for the R-chart

c. Is the process in control? Why?

Note: use 2 decimal places

Solution

First we need to find the x̿, you have to add all of the x̅ and divide them by the sample

number 12:

x̿ = 3.5+3+1.25+3.5+3+2.75+3+2.5+3+2.75+2.25+3

12

x̿ = 2.79

Now we need to find the R̅, (R = Highest – Lowes value for each sample) in this example R

is already given so all we need to do is to add all of the Rs and divide them by 12 (number

of samples)

R̅= 1.04

n = 4

From the Factor for Computing Control Chart Table:

A2 = 0.729

D4 = 2.282

D3 = 0

a. UCL & LCL For x-BAR CHART

UCLX̅ = x̿ + A2 R̅

= 2.79 + (0.729 * 1.04) = 3.55

LCLX̅ = x̿ - A2 R̅

= 2.79 – (0.729 * 1.04) = 2.03

b. UCL & LCL For R CHART

UCLR = D4 R̅

= 2.282 * 1.04 = 2.38

LCLR = D3 R̅

= 0 * 1.04 = 0

Chapter 6s: Control Chart

c. For x bar chart: x̅ for sample 3 is out of control, for R chart: all of the samples are within the

control limits. We can conclude that the process is currently out of control.

Question 5. A process that is considered to be in control measures an

ingredient in ounces. Below are the last 10 samples (each of size n=5) taken.

The population process standard deviation is 1.36

Samples

1 2 3 4 5 6 7 8 9 10

10 9 13 10 12 10 10 13 8 10

9 9 9 10 10 10 11 10 8 12

10 11 10 11 9 8 10 8 12 9

9 11 10 10 11 12 8 10 12 8

12 10 9 10 10 9 9 8 9 12

a) What is the standard deviation of the sample means (𝜎�̅� ) ?

b) If z = 3, what are the control limits for the mean chart

c) If  isn’t given, what are the control limits for x bar chart?

d) What are the control limits for the range chart?

e) Is the process in control?

f) What additional steps should the quality manager take?

Note: use 2 decimal places

Solution:

a) Process (population) standard deviation () = 1.36,

b) Using  x̅

 

  x

x

  

 

UCL 10 3 0.61 11.83

LCL 10 – 3 0.61 8.17

Standard deviation of the sampling means

1.36 5

0.61

x





Chapter 6s: Control Chart

c) X bar chart First you have to find the x ̅ and R for each sample, then X double bar and R bar.

Sample

x̅ 1 x̅ 2 x̅ 3 x̅ 4 x̅ 5 x̅ 6 x̅ 7 x̅ 8 x̅ 9 x̅ 10

10 10 10.2 10.2 10.4 9.8 9.6 9.8 9.8 10.2

x̿ = 10+10+10.2+10.2+10.4 +9.8+9.6+9.8+9.8+10.2

10 = 10

Sample

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R 10

3 2 4 1 3 4 3 5 4 4 R̅ = 3.3

Since n = 5, A2 = 0.577 (check the chart)

 

  x

x

  

 

UCL 10 3.3 0.577 11.90

LCL 10 – 3.3 0.577 8.10

d) R Chart

UCLR = 2.115(3.3) = 6.98

LCLR = 0(3.3) = 0

e) Yes, both mean and range charts indicate process is in control.

f) As the process is in control no further action is needed.

Chapter 6s: Control Chart

Question 6. Auto pistons at Wemming Chung’s plant in Shanghai are

produced in a forging process, and the diameter is a critical factor that must be

controlled. From sample sizes of 10 pistons produced each day, the mean and

the range of this diameter have been as follows:

Day Mean (mm) Range (mm)

1 156.9 4.2

2 153.2 4.6

3 153.6 4.1

4 155.5 5

5 156.6 4.5

a. What is the value of the mean of �̿� ?

b. what is the �̅� ?

c. What are the 𝐔𝐂𝐋�̅� and 𝐋𝐂𝐋�̅�?

d. What are the UCLR and LCLR?

(c) -chart:X

2

2

155.16 mm from the sample data

UCL 155.16 (0.308 4.48) 156.54 mm

LCL 155.16 (0.308 4.48) 153.78 mm.

x

x

X

X A R

X A R



     

     

(d) UCLR = 1.777 x 4.48 = 7.96

LCLR = 0.223 x 4.48 = 0.99

156.9 153.2 153.6 155.5 156.6 (a) 155.16 mm

5

4.2 4.6 4.1 5.0 4.5 (b) 4.48 mm

5

X

R

     

     