Chapter 6

Homework

Problems 6, 8, 10, 21

Problem 6

Sample Sample Mean Range1 12.1 0.72 11.8 0.43 12.3 0.64 11.5 0.45 11.6 0.9

Sum 59.3 3.0Mean 11.9 0.6

9.115

3.55

5

6.115.113.128.111.12

5,6

k

XX

kn

X-bar Chart

1.12

)6.0(48.09.11

6.11

)6.0(48.09.11

2

2

RAXUCL

RAXLCL

X-bar Chart

X-bar Chart

X-bar Chart

11

11.5

12

12.5

1 2 3 4 5

LCL

UCL

CL

Mean

X-bar Chart

• A special cause occurred on the 3rd and 4th samples.

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

consequently, not predictable.• Consequently, it is impossible to get a

valid estimate of the mean.• The operators are responsible for finding

and correcting or institutionalizing special causes.

6.0

5

3

5

9.04.06.04.07.0

k

RR

R Chart

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(

3ˆ 2

nRA

Fraction Defective

• 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

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

09.010

90.0

k

pp

n

pppLCL

n

pppUCL

)1(3

)1(3

0102.0.20

)09.1(09.309.

282.020

)09.1(09.309.

Problem 8 (a)

p Chart

0

0.05

0.1

0.15

0.2

0.25

0.3

1 2 3 4 5 6 7 8 9 10

Sample

p

p=X/n

LCL

UCL

p-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.

Week c c-bar1 4 4.42 5 4.43 6 4.44 6 4.45 3 4.46 2 4.47 6 4.48 7 4.49 3 4.410 4 4.411 3 4.412 4 4.4

Sum c 53c-bar 4.4

Problem 10

4.4

12

53

k

cc

72.10

4.434.4

3

0or 89.1

4.434.4

3

ccUCL

ccLCL

c-Chart – Control Limits

c-Chart

0

2

4

6

8

10

12

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

Week

c, n

um

be

r o

f d

efe

cts

c

LCL

UCL

c-bar

Interpretation of Chart

• The process is in control.

• Each billing statement contains on average 4.4 errors, which is unacceptably high.

• Since the process is in control, management action is required to improve the process by reducing the average number of errors.

39.1

)12.0(6

5.95.10

ˆ6

LSLUSL

Cp

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 bottles will be out of spec.

Problem 21

The index assumes that the process is on target. However, the process is not on target. Therefore,we should compute CT.

z was computed using the NORMSDIST in Excel.

.

=2*NORMSDIST(G26)

-4.17 0.000030526

G

.00003

0-4.17 z

.00003

4.17

NORMSDIST(G26) = Area under curve to left of z

Problem 21

USLLSL

71.0

)108.9()12.0(6

5.95.10

Target)(ˆ6

2

2

X

LSLUSLCT

Problem 21

Since the capability index is less than 1, the processis not capable. If process is center on target, the capability index would increase to 1.39, and the process would be capable.