Chapter 6 Revised

HomeworkProblems 6, 8, 10, 21

Problem 6 (a)

• Given• Target = 12 ounces• If process in control, R-bar = .6• Is process in control at these levels?• n = 6• k = 5

288.12

)6.0(48.012

712.11

)6.0(48.012

2

2

RAXUCL

RAXLCL

Problem 6 (a) – X-bar Chart

LCL UCL Mean Inside Limits?11.712 12.288 12.1 Yes11.712 12.288 11.8 Yes11.712 12.288 12.3 No11.712 12.288 11.5 No11.712 12.288 11.6 No

Problem 6 (a) – X-bar Chart

Problem 6 (a) – X-bar Chart

X-bar Chart

11.0

11.5

12.0

12.5

1 2 3 4 5

LCL

UCL

CL

Mean

X-bar Chart

• The means of the last three samples fall outside of the control limits.

• Therefore, the X-bar chart is not in control.• The process mean is unstable and,

consequently, not predictable.• Consequently, the mean is not on target. • The operators are responsible for

identifying and removing the special causes responsible for the instability.

X-bar Chart

• Once the special causes are removed and the mean is in control, the mean will be stable but it may or may not be on target.

• If it is not, management action is required to correct the mean to the target value.

• The interpretation of the X-bar chart assumes that R chart is in control.

Problem 6 (a) – R Chart

2.1

)6.0(2

0

)6.0(0

UCL

LCL

2

0

6

4

3

D

D

n

2.1

)6.0(2

0

)6.0(0

UCL

LCL

2

0

6

4

3

D

D

n

R Chart

R Chart

0

0.5

1

1.5

1 2 3 4 5

LCL

UCL

R-bar

Range

R Chart

R Chart

• The R chart is in control• The process variance is therefore stable

and predictable.• The variance can be estimated as

24.03

6)6.0)(48.0(3

ˆ 2 nRA

Fraction Defective (Extra)

• Assume that USL = 12.5 ounces and LSL = 11.5 ounces.

• If the mean was in control and centered on the target of 12 ounces, the fraction defective would be

0.0376.4812]-2[.5 defectiveFraction

4812.)08.2(

08.224.

125.12ˆ

Area

XUSLz

Fraction Defective (Extra)

Sample n X p=X/n1 20 1 0.052 20 3 0.153 20 2 0.14 20 1 0.055 20 4 0.26 20 1 0.057 20 2 0.18 20 0 09 20 3 0.1510 20 1 0.05

Sum 0.9p-bar 0.09

Problem 8 (a)

10

20

k

n

09.01090.0

kp

p

npppLCL

npppUCL

)1(3

)1(3

0102.0.20

)09.1(09.309.

282.020

)09.1(09.309.

Problem 8 (a)

p Chart

00.05

0.10.15

0.20.25

0.3

1 2 3 4 5 6 7 8 9 10

Sample

p

p=X/nLCLUCLp-bar

Problem 8 (a)

Problem 8 (a)

• All the sample fraction defectives fall within the control limits and form a random pattern.

• The process appears in control.• We can therefore estimate the process

fraction defective.• Our best estimate is the p-bar of .09.

Problem 8 (a)

• Thus, 9% of the tires produced are defective.

• Since the process is stable, management action is required to improve the process by reducing the fraction defective.

• On Day 8, no defective tires were found.• Since this point is on the LCL, it should be

investigate for a special cause, which may have a favorable impact on the fraction defective.

Day n X p=X/n LCL UCL p-bar11 20 6 0.3 0 0.282 0.0912 20 3 0.15 0 0.282 0.0913 20 3 0.15 0 0.282 0.0914 20 4 0.2 0 0.282 0.09

Problem 8 (b)

The sample fraction defective on Day 11 falls above the UCL. The process fraction defective isTherefore out of control. The underlying special cause has an unfavorable affect on the processbecause it shifted the process fraction defective upward.

Problem 10

044.01253.0

ku

u

Week n c u=c/n1 100 4 0.042 100 5 0.053 100 6 0.064 100 6 0.065 100 3 0.036 100 2 0.027 100 6 0.068 100 7 0.079 100 3 0.03

10 100 4 0.0411 100 3 0.0312 100 4 0.04

Sum u 0.53

107.0100044.3044.0

3

0100044.3044.0

3

nuuUCL

nuuLCL

u-Chart – Control Limits

u-Chart for Billing Errors

0.000

0.020

0.040

0.060

0.080

0.100

0.120

1 2 3 4 5 6 7 8 9 10 11 12

Week

c, n

umbe

r of d

efec

ts

LCL

UCL

u

u-bar

Interpretation of Chart

• The process is in control.• Each billing statement contains on

average .044 errors, or less than an average of one billing error per statement.

• Since there should be no billing errors, management action is required to achieve further reductions in the average number of errors.

39.1

)12.0(65.95.10

ˆ6

LSLUSLCp

Problem 21

Since the capability index is greater than 1, the process is capable.

17.4

)39.1(3

3

Cpz

Problem 21

Process fraction defective = .00003, or 3 outof every 100,000 packages will be outside of the specification limits.

.000015

0-4.17 z

.000015

4.17

NORMSDIST(-4.17) = Area under curve to left of z

Problem 21

USLLSL

Problem 21

• The Cp index assumes that the process is on target.

• However, the process is not on target. • The mean is 9.8 and the target is 10.• The fraction defection will therefore be

greater than 3 out of 100,000.• Therefore, we should compute CT.

71.0

)108.9()12.0(6

5.95.10

Target)(ˆ6

2

2

X

LSLUSLCT

Problem 21

Problem 21

• Since the capability index is less than 1, the process is not capable.

• The actual process fraction defective is .006, or 6 out of 1,000 packages.

• If process is center on target, the capability index would increase to 1.39, and the process fraction defective would decrease to 3 out of 100,000 packages.

.006 0 .006 defectiveFraction

0)83.5(

83.512.

105.10ˆ

0006.)5.2(

5.2.12.

105.9ˆ

Area

XUSLz

Area

XLSLz

U

L

Problem 21