# P Cha r t of N onconf or mi ng Assembl i es ( Ex 6 - 1 N um)

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## Transcript Of P Cha r t of N onconf or mi ng Assembl i es ( Ex 6 - 1 N um)

Chapter 6 Exercise Solutions

Notes: 1. New exercises are denoted with an “”. 2. For these solutions, we follow the MINITAB convention for determining whether a
point is out of control. If a plot point is within the control limits, it is considered to be in control. If a plot point is on or beyond the control limits, it is considered to be out of control. 3. MINITAB defines some sensitizing rules for control charts differently than the standard rules. In particular, a run of n consecutive points on one side of the center line is defined as 9 points, not 8. This can be changed under Tools > Options > Control Charts and Quality Tools > Define Tests. Also fewer special cause tests are available for attributes control charts.

6-1. n  100;

m  20;

m
 Di  117;
i 1

m
p  i1 Di  117  0.0585 mn 20(100)

UCL p  p  3 p(1 p)  0.0585  3 0.0585(1 0.0585)  0.1289

n

100

LCL p  p  3 p(1 p)  0.0585  3 0.0585(1 0.0585)  0.0585  0.0704  0

n

100

MTB > Stat > Control Charts > Attributes Charts > P

P Chart of Nonconforming Assemblies (Ex6-1Num)

0.16

1

0.14
UCL=0.1289 0.12

0.10

Proportion

0.08

_

0.06

P=0.0585

0.04

0.02

0.00

LCL=0

2

4

6

8 10 12 14 16 18 20

Sample

Test Results for P Chart of Ex6-1Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 12

6-1

Chapter 6 Exercise Solutions

6-1 continued

Sample 12 is out-of-control, so remove from control limit calculation:

m

m

 Di 102

n  100; m  19;  Di  102; p  i1 

 0.0537

i 1

mn 19(100)

UCL p  0.0537  3 0.0537(1 0.0537)  0.1213 100

LCL p  0.0537  3 0.0537(1 0.0537)  0.0537  0.0676  0 100

MTB > Stat > Control Charts > Attributes Charts > P

Proportion

P Chart of Nonconforming Assemblies (Ex6-1Num)
Sample 12 removed from calculations

0.16

1

0.14

0.12

UCL=0.1213

0.10

0.08

0.06

_

P=0.0537

0.04

0.02

0.00

LCL=0

2

4

6

8 10 12 14 16 18 20

Sample

Test Results for P Chart of Ex6-1Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 12

6-2

Chapter 6 Exercise Solutions

6-2. n  150;

m  20;

m
 Di  69;
i 1

m
p  i1 Di  69  0.0230 mn 20(150)

UCL p  p  3 p(1 p)  0.0230  3 0.0230(1 0.0230)  0.0597

n

150

LCL p  p  3 p(1 p)  0.0230  3 0.0230(1 0.0230)  0.0230  0.0367  0

n

150

MTB > Stat > Control Charts > Attributes Charts > P

P Chart of Nonconforming Switches (Ex6-2Num)
1
0.10

0.08
1
0.06

UCL=0.0597

Proportion

0.04

_

0.02

P=0.023

0.00

LCL=0

2

4

6

8 10 12 14 16 18 20

Sample

Test Results for P Chart of Ex6-2Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 9, 17

6-3

Chapter 6 Exercise Solutions

6-2 continued Re-calculate control limits without samples 9 and 17:
MTB > Stat > Control Charts > Attributes Charts > P
P Chart of Nonconforming Switches (Ex6-2Num)
Samples 9 and 17 excluded from calculations
1
0.10

Proportion

0.08
1
0.06 1
0.04

UCL=0.0473

0.02

_

P=0.0163

0.00

LCL=0

2

4

6

8 10 12 14 16 18 20

Sample

Test Results for P Chart of Ex6-2Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 1, 9, 17

6-4

Chapter 6 Exercise Solutions

6-2 continued

Also remove sample 1 from control limits calculation:

m

m

 Di 36

n  150; m  17;  Di  36; p  i1 

 0.0141

i 1

mn 17(150)

UCL p  0.0141 3 0.0141(1 0.0141)  0.0430 150

LCL p  0.0141 3 0.0141(1 0.0141)  0.0141 0.0289  0 150

MTB > Stat > Control Charts > Attributes Charts > P

P Chart of Nonconforming Switches (Ex6-2Num)
Samples 1, 9, 17 excluded from calculations
1
0.10

Proportion

0.08
1
0.06 1
0.04

UCL=0.0430

0.02

_

P=0.0141

0.00

LCL=0

2

4

6

8 10 12 14 16 18 20

Sample

Test Results for P Chart of Ex6-2Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 1, 9, 17

6-5

Chapter 6 Exercise Solutions

6-3. NOTE: There is an error in the table in the textbook. The Fraction Nonconforming for Day 5 should be 0.046.

m

m

m

m

m  10;  ni  1000;  Di  60; p   Di  ni  60 1000  0.06

i 1

i 1

i 1

i 1

UCLi  p  3 p(1 p) ni and LCLi  max{0, p  3 p(1 p) ni }

As an example, for n = 80:

UCL1  p  3 p(1 p) n1  0.06  3 0.06(1 0.06) 80  0.1397

LCL1  p  3 p(1 p) n1  0.06  3 0.06(1 0.06) 80  0.06  0.0797  0

MTB > Stat > Control Charts > Attributes Charts > P

P Chart of Nonconforming Units (Ex6-3Num)
0.16

0.14 UCL=0.1331
0.12

0.10

Proportion

0.08

_

0.06

P=0.06

0.04

0.02

0.00

LCL=0

1

2

3

4

5

6

7

8

9 10

Sample

Tests performed with unequal sample sizes

The process appears to be in statistical control.

6-6

Chapter 6 Exercise Solutions

6-4. (a)
n  150;

m  20;

m
 Di  50;
i 1

m
p   Di i 1

mn  50 20(150)  0.0167

UCL  p  3 p(1 p) n  0.0167  3 0.0167(1 0.0167) 150  0.0480

LCL  p  3 p(1 p) n  0.0167  3 0.0167(1 0.0167) 150  0.0167  0.0314  0

MTB > Stat > Control Charts > Attributes Charts > P
P Chart of Nonconforming Units (Ex6-4Num)
0.05 UCL=0.04802

0.04

Proportion

0.03

0.02

_

P=0.01667

0.01

0.00

LCL=0

2

4

6

8 10 12 14 16 18 20

Sample

The process appears to be in statistical control.
(b) Using Equation 6-12, n  (1 p) L2
p  (1 0.0167) (3)2
0.0167  529.9 Select n  530.

6-7

Chapter 6 Exercise Solutions

6-5. (a) UCL  p  3 p(1 p) n  0.1228  3 0.1228(1 0.1228) 2500  0.1425
LCL  p  3 p(1 p) n  0.1228  3 0.1228(1 0.1228) 2500  0.1031

MTB > Stat > Control Charts > Attributes Charts > P

0.200 0.175 0.150 0.125

P Chart of Nonconforming Belts (Ex6-5Num)

1

1

1

1

1

UCL=0.1425
_ P=0.1228

Proportion

0.100

1

1

1

0.075

LCL=0.1031
1 1

0.050

1

2

4

6

8 10 12 14 16 18 20

Sample

Test Results for P Chart of Ex6-5Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 1, 2, 3, 5, 11, 12, 15, 16, 17, 19, 20
(b) So many subgroups are out of control (11 of 20) that the data should not be used to establish control limits for future production. Instead, the process should be investigated for causes of the wild swings in p.

6-8

Chapter 6 Exercise Solutions

6-6. UCL  np  3 np(1 p)  4  3 4(1 0.008)  9.976 LCL  np  3 np(1 p)  4  3 4(1 0.008)  4  5.976  0
MTB > Stat > Control Charts > Attributes Charts > NP
NP Chart of Number of Nonconforming Units (Ex6-6Num)
1
12

10

UCL=9.98

Sample Count

8

6

__

4

NP=4

2

0

LCL=0

1

2

3

4

5

6

7

8

9

10

Ex6-6Day

Test Results for NP Chart of Ex6-6Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 6

6-9

Chapter 6 Exercise Solutions 6.6 continued Recalculate control limits without sample 6:

Sample Count

NP Chart of Number of Nonconforming Units (Ex6-6Num)
Day 6 excluded from control limits calculations
1
12

10

8

UCL=8.39

6

4

__

NP=3.11

2

0

LCL=0

1

2

3

4

5

6

7

8

9

10

Ex6-6Day

Test Results for NP Chart of Ex6-6Num
TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 6
Recommend using control limits from second chart (calculated less sample 6).

6-10
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