control chart for variable and attribute

48

QUESTION 5
Three bottling machines at Cocoa Fizz are being evaluated for their capability:

Bottling Machine Standard deviation
A 0.05
B 0.1
C 0.2

If specifications are set between 15.8 and 16.2 ounces, determine which of
the machines are capable of producing within specifications.

QUESTION 6
Three bagging machines at the Crunchy Potato Chip Company are being
evaluated for their capability. The following data are recorded:

Bagging Machine Standard Deviation
A 0.2
B 0.3
C 0.5

If specifications are set between 12.35 and 12.65 ounces, determine
which of the machines are capable of producing within specification.

QUESTION 7
A new process is started and the sum of range for 20 subgroups of size 4 is
1.30 mm. If the upper specification limit is 25.30 mm and the lower
specification limit is 25.22 mm. Determine the process capability index, Cpk
when the average is 25.25 mm and comment the process.

49

QUESTION 8
A production line supervisor decides to count the number of nonconforming
units ffrom a process. the supervisor then makes an inspection on 23 samples
and records the data as shown in table below. Each sample contains 250 units
of products.
i. Calculate the fraction of non conforming for each sample
ii. Determine the value of center line, upper control limit and lower control
limit for a p – chart.
iii. Construct the p – chart of the process
iv. Based on the graph in question (b) iii, state whether the process is in
control or out of control. Explain your answer.

number number of 11 20
of nonconforming 12 45
13 32
sample 14 56
15 43
1 30 16 44
17 43
2 24 18 43
19 44
3 26 20 33
21 32
4 34 22 23
23 30
5 40

6 33

7 32

8 22

9 26

10 21

SOLUTION 8:
Data given:
Sample size, n = ______
Number of subgroups, g = ______

50

Fraction nonconforming:

p = np
n

i. The fraction of nonconforming unit

Number of Number of Fraction of
Nonconforming,,p
sample Nonconforming,np
∑ =
1 30

2 24

3 26

4 34

5 40

6 33

7 32

8 22

9 26

10 21

11 20

12 45

13 32

14 56

15 43

16 44

17 43

18 43

19 44

20 33

21 32

22 23

23 30

∑ =

ii. Central line, 51

CL = p =  np = Lower control limit,
n
LCL = p −3 p(1− p)
Upper control limit, n

UCL = p +3 p(1− p)
n

52

QUESTION 9
Use data in table below, to plot the np-chart and show all the calculations
involved. Interpret the process based on the chart.

Day Defectives Sample 13 16 100
size 14 21 100
1 10 100 15 20 100
2 12 100 16 12 100
3 10 100 17 11 100
4 11 100 18 6 100
56 100 19 10 100
67 100 20 10 100
7 12 100 21 11 100
8 10 100 22 11 100
96 100 23 11 100
10 11 100 24 6 100
11 9 100 25 9 100
12 14 100

SOLUTION 9:
Data given:
Sample size, n = ______
Sum of sample size, ∑ = _______
Sum of defective unit, ∑ = _______
Number of subgroups, g = ______

p = np
n

CL =np =

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

53

QUESTION 10
i. Calculate the value of central line (CL), upper control limit (UCL) and
lower control limit (LCL) for c – chart.
ii. Construct a c-chart on a graph paper.
iii. Interpret the process control based on the chart

Plate Number of Plate Number of
number defects number defects
0 2
1 2 11 2
2 4 12 3
3 1 13 0
4 2 14 2
5 3 15 3
6 0 16 4
7 0 17 0
8 1 18 1
9 2 19 1
10 20

SOLUTION 10:
Data given:
Sample size, n = ______
Sum of defect, ∑ = _______
Number of subgroups, g = ______

i. Central line, Lower control limit,

CL = c =  c = LCL = c −3 c

g

Upper control limit,

UCL =c + 3 c

54

QUESTION 11
i. Calculate the value of central line (CL), upper control limit (UCL) and
lower control limit (LCL) for u – chart.
ii. Construct a c-chart on a graph paper.
iii. Interpret the process control based on the chart

Lot Sample Total Lot Sample Total
number size nonconformities number
10 size nonconformities
1 10 45 11
2 10 51 12 12 25
3 9 36 13
4 10 48 14 12 35
5 10 42 15
6 10 5 16 12 32
7 8 33 17
8 8 27 18 10 43
9 8 31 19
10 21 20 10 48

11 35

10 39

10 29

10 37

10 33

SOLUTION 11:
Data given:
Sample size, n = ______
Sum of sample size, ∑ = _______
Sum of defect, ∑ = _______

Count of nonconformities per unit in a subgroup, u=c
n
i. Central line,

CL = u = c =
n

Upper control limit, Lower control limit,

UCL = u + 3 u LCL =u −3 u
n n

55

Lot Sample Total u=c CL UCL LCL
number size Nonconformities n

1 10 45
2 10 51
3 10 36
49 48
5 10 42
6 10 5
7 10 33
88 27
98 31
10 8 21
11 12 25
12 12 35
13 12 32
14 10 43
15 10 48
16 11 35
17 10 39
18 10 29
19 10 37
20 10 33

56

7. ANSWER

1. X bar chart, CL = 10.78 UCL =13.80 LCL = 7.77
R chart, CL = 2.95 UCL = 7.59 LCL = 0

2. X bar chart, CL = 484.1 UCL =502.4 LCL = 465.8
R chart, CL = UCL = 7.59 LCL = 0

3. X bar chart, CL = 33.8 UCL = 40.9 LCL = 26.5
s chart, CL = 3.7 UCL = 9.5 LCL = 0

4. X bar chart, CL = 550.9 UCL = 580.6 LCL = 521.2
s chart, CL = 27 UCL = 49 LCL = 0

5. Cp = 1.33, machine A is capable.

6. No machine is capable.

7. Cpk = 0.33, process is producing product does not conform to specification.

8. CL = 0.13 UCL = 0.19 LCL = 0.07

9. CL = 11 UCL = 20.39 LCL = 1.01

10. CL = 1.65 UCL = 5.50 LCL = 0

57

8. REFERENCES

Dale H. Besterfield,(1994). Quality Control.4th ed. Prentice - Hall

Dale H. Besterfield, (2001). Quality Control, 6th edition. Prentice Hall.

Grant, E. L., and R. S. Leavenworth. Statistical Quality Control.6th ed. New
York:McGraw-Hill, 1998.

Jay Heizer & Barry Render, (1999). Operation Management, 5th edition. Prentice
Hall.

Neil A. Weiss, (1995). Introductory Statistics, 4th edition. Addison - Wesley.

http://www.wiley.com

http://asq.org/index.aspx

http:// www.wps.prenhall.com/wps/media/objects/4947/.../heizer_supp_06.ppt

Final examination paper of Quality Control, Politeknik Malaysia.