Mont4e Sm Ch16 Supplemental

8/2/2019 Mont4e Sm Ch16 Supplemental

1/21

Supplementary Exercises

16-57 a)X-bar and Range - Initial Study

--------------------------------------------------------------------------------

X-bar | Range

---- | -----

UCL: + 3.0 sigma = 64.0181 | UCL: + 3.0 sigma = 0.0453972

Centerline = 64 | Centerline = 0.01764

LCL: - 3.0 sigma = 63.982 | LCL: - 3.0 sigma = 0

|

out of limits = 0 | out of limits = 0

--------------------------------------------------------------------------------

Chart: Both Normalize: No

Estimated

process mean = 64

process sigma = 0.0104194

mean Range = 0.01764

0Subgroup 5 10 15 20 25

63.98

63.99

64.00

64.01

64.02

SampleMean

Mean=64.00

UCL=64.02

LCL=63.98

0.00

0.01

0.02

0.03

0.04

0.05

SampleRange

R=0.01764

UCL=0.04541

LCL=0

Xbar/R Chart for C1-C3

The process is in control.

b) = =x 64 0104.0693.1

01764.0

2

===d

R

c) 641.0)0104.0(6

98.6302.64

6 =

=

= LSLUSL

PCR

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The process does not meet the minimum capability level of PCR 1.33.

d)

[ ]

641.0

641.0,641.0min

)0104.0(398.6364,

)0104.0(36402.64min

3,

3min

=

=

=

=

LSLxxUSLPCRk

e) In order to make this process a six-sigma process, the variance 2 would have to be decreased such that

PCRk= 2.0. The value of the variance is found by solving PCRk=x LSL

=3

2 0

. for :

0033.0

6

98.630.64

98.630.646

0.23

98.6364

=

=

=

=

Therefore, the process variance would have to be decreased to 2 = (0.0033)2 = 0.000011.

f) x = 0.0104

8295.00020.08315.0

)88.2()96.0()96.088.2(

0104.0

01.6402.64

0104.0

01.6498.63)02.6498.63(

==


3/21

16-58 a)

0Subgroup 5 10 15 20 25

63.98

63.99

64.00

64.01

64.02

SampleMean

Mean=64.00

UCL=64.02

LCL=63.98

0.00

0.01

0.02

Sa

mpleStDev

S=0.009274

UCL=0.02382

LCL=0

Xbar/S Chart for C1-C3

b) = =x 64 0104.08862.0

009274.0

4

===c

s

c) Same as 16-57 641.0)0104.0(6

98.6302.64

6=

=

=

LSLUSLPCR

The process does not meet the minimum capability level of PCR 1.33.

d) Same as 16-57

[ ]

641.0

641.0,641.0min

)0104.0(3

98.6364,

)0104.0(3

6402.64min

3,

3min

=

=

=

=

LSLxxUSLPCRk

e) Same as 16-57 e). In order to make this process a six-sigma process, the variance 2 would have to be

decreased such that PCRk= 2.0. The value of the variance is found by solving PCRk=x LSL

=3

2 0

. for :

0033.0

6

98.630.64

98.630.646

0.23

98.6364

=

=

=

=

Therefore, the process variance would have to be decreased to 2 = (0.0033)2 = 0.000011.

16-39


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f) Same as 16-57 x = 0.0104

8295.00020.08315.0

)88.2()96.0()96.088.2(

0104.0

01.6402.64

0104.0

01.6498.63)02.6498.63(

==


5/21

0 10 20

0.04

0.09

0.14

0.19

Sample Number

Proportion

P Chart for def2

P=0.1063

UCL=0.1717

LCL=0.04093

There are no further points beyond the control limits.

c) A larger sample size with the same percentage of defective items will result in more narrow control limits.The control limits corresponding to the larger sample size are more sensitive to process shifts.

16-60 a)

0 5 10 15 20 25

0

1

2

Sample Number

SampleCount

U Chart for Defects

1

1

U=0.528

UCL=1.503

LCL=0

Points14 and 23 are beyond the control limits. The process is out of control.

16-41


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b) After removing points 14 and 23, the limits are narrowed.

0 10 20

0.0

0.5

1.0

1.5

Sample Number

SampleCount

U Chart for Defects

U=0.4261

UCL=1.302

LCL=0

c) The control limits are narrower for a sample size of 10

2520151050

1.0

0.5

0.0

Sample Number

SampleCount

U C hart for defects n=10

1

1

U=0.264

UCL=0.7514

LCL=0

20100

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Sample Number

SampleCount

U Chart for defects n=10

U=0.2130

UCL=0.6509

LCL=0

16-42


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16-61 a) Using I-MR chart.

Observa t ion

IndividualValue

2018161412108642

60.3275

60.3270

60.3265

60.3260

60.3255

_X=60.32641

UC L=60.327362

LCL=60.325458

Observa t ion

MovingRange

2018161412108642

0.00100

0.00075

0.00050

0.00025

0.00000

__MR=0.000358

UC L=0.001169

LCL=0

I -MR Chart of C1

b) The chart is identical to the chart in part (a) except for the scale of the individuals chart.

Observa t ion

IndividualValue

2018161412108642

0.0015

0.0010

0.0005

0.0000

-0.0005

_X=0.00041

UC L=0.001362

LCL=-0.000542

Observa t ion

MovingRange

2018161412108642

0.00100

0.00075

0.00050

0.00025

0.00000

__MR=0.000358

UC L=0.001169

LCL=0

I -MR Chart of C1

c)The estimated mean is 60.3264. The estimated standard deviation is 0.0003173.

0.0021.0505

6 6(0.0003173)

USL LSLPCR

= = =

0.0009 0.0011min , 0.9455

3 3kPCR

= =

16-43


8/21

16-62 a)

Observat ion

IndividualValue

30272421181512963

7000

6000

5000

4000

_

X=5832

UC L=6669

LCL=4995

Observat ion

MovingRange

30272421181512963

1000

750

500

250

0

__MR=315

UC L=1028

LCL=0

11

1

1

111

1

I -MR Chart of enery

b) The data does not appear to be generated from an in-control process. The average tends to drift

to larger values and then drop back off over the last 5 values.

16-63 a)Trial control limits :

S chart

UCL= 170.2482

CL = 86.4208

LCL = 2.59342

X bar chart

UCL= 670.0045

CL = 558.766

LCL = 447.5275

5 10 15 20 25

400

600

800

Index

xbar

5 10 15 20 25

0

50

150

Index

s

16-44


9/21

b) An estimate of is given by 8259.909515.0/4208.86/ 4 ==cS

PCR=500/(6*90.8259)=0.9175 and PCRk=

)8259.90(3

33077.558,

)8259.90(3

77.558830min =0.8396

Based on the capability ratios above (both


10/21

d) In-control distribution X ~ N(551.9477, )234.6

Out-of-control distribution X ~ N(600, )234.6

P[448.1 X 655.8, when =600]

= ]6.346008.655

6.346001.448[ ZP

= ]6124.139.4[ ZP=0.9463

Out-of-control ARL= = 9463.01

118.6 19.

16-64 X Control Chart with 2-sigma limits

nLCL

nUCL

CL

2,2 =+=

=

02275.097725.01)2(1)2(2/

2

02275.097725.01)2(1)2(2/

2

==

ZPZPn

XP

nXP

and

ZPZPn

XPn

XP

The answer is 0.02275 + 0.02275 = 0.0455. The answer for 3-sigma control limits is 0.0027. The 3-sigma

control limits result in many fewer false alarms.

16-65

a) The following control chart use the average range from 25 subgroups of size 3 to estimate the process

standard deviation. Minitab uses a pooled estimate of variance as the default method for an EWMA control

chart so that the range method was selected from the options. Points are clearly out of control.

Sample

EWMA

24222018161412108642

64.7

64.6

64.5

64.4

64.3

64.2

64.1

64.0

63.9

__X=64.1334

UCL=64.2759

LCL=63.9910

EWMA Char t of C3, ..., C5

16-46


11/21

b) The following control chart use the average range from 25 subgroups of size 3 to estimate the process

standard deviation. There is a big shift in the mean at sample 10 and the process is out of control at

this point.

Sample

EWMA

24222018161412108642

65.75

65.50

65.25

65.00

64.75

64.50

64.25

64.00

__X=64.133

UCL=64.380

LCL=63.887

EWMA Chart of C3, . .., C5

16-66

a) The data appears to be generated from an out-of-control process.

Sample

EWMA

30272421181512963

6600

6400

6200

6000

5800

5600

5400

5200

5000

__

X=5832

UCL=6111

LCL=5553

EWMA Chart of C1

b) The data appears to be generated from an out-of-control process.

16-47


12/21

Sample

EWMA

30272421181512963

7000

6500

6000

5500

5000

4500

__X=5832

UCL=6315

LCL=5349

EWMA Chart of C1

16-67 a) The process appears to be in control.

Sample

EWMA

2018161412108642

60.3268

60.3267

60.3266

60.3265

60.3264

60.3263

60.3262

60.3261

__X=60.32641

UCL=60.3267273

LCL=60.3260927

EWMA Chart of C1

b) The process appears to be in control.

Sample

EWMA

2018161412108642

60.32700

60.32675

60.32650

60.32625

60.32600

__X=60.32641

UCL=60.326960

LCL=60.325860

EWMA Chart of C1

16-48


13/21

16-68

Sample

CumulativeSum

30272421181512963

5.0

2.5

0.0

-2.5

-5.0

0

UCL=5.27

LCL=-5.27

CUSUM Chart of C1

Process standard deviation is estimated using the average moving range of size 2 with MR/d2, where d2 =

1.128. The estimate is 1.05. Recommendation forkand h are 0.5 and 4 or 5, respectively forn =1. For this

chart h = 5 was used.

16-69 The process is not in control.

K= k = 1, so that k= 0.5

H= h = 10, so that h = 5

40

30

20

10

0

-10

10

-10

20100

Subgroup Number

CumulativeSum

Upper CUSUM

Lower CUSUM

CUSUM Chart for hardness

16-49


14/21

16-70

-80

0

80

-85.8529

85.8529

0 10 20

Subgroup Number

CumulativeSum

Upper CUSUM

Lower CUSUM

CUSUM Chart for Viscosity

Process standard deviation is estimated using the average moving range of size 2 with MR/d2, where d2 =

1.128 for a moving range of 2. The estimate is 17.17. Recommendation forkand h are 0.5 and 4 or 5,

respectively, forn = 1.

16-71 a)

Sample

CumulativeSum

24222018161412108642

0.020

0.015

0.010

0.005

0.000

-0.005

-0.010

0

UCL=0.01152

LCL=-0.01152

CUSUM Char t of x

is estimated using the moving range: 0.0026/1.128=0.0023. H and K were computed using

k=0.5 and h=5. The process is not in control.

b) EWMA gives similar results.

16-50


15/21

Sample

EWMA

24222018161412108642

0.404

0.403

0.402

0.401

0.400

0.399

__X=0.401184

UCL=0.403489

LCL=0.398879

EWMA Chart of x

16-72 a) Letp denote the probability that a point plots outside of the control limits when the mean has shifted from

0 to = 0 + 1.5. Then,

5.0]0[5.0

)6()0()06(

3/

5.1

/3

/

5.1

33

)( 00

==


16/21

16-73. ARL = 1/p wherep is the probability a point falls outside the control limits.

a) and n = 1= +0

02275.0

]0[97725.01

1)4()2(

)3()3(

/

3

/

3

)()(

0000

=

+=

==

=

=

=

nwhenZPZP

nZPnZP

n

n

ZPn

n

ZP

LCLXPUCLXPp

Therefore, ARL = 1/p = 1/0.02275 = 43.9.

b) = +0 2

15866.0

]0[84134.01

1)5()1(

)23()23(

/

23

/

23

)()(

0000

=

+=

==

=

=

nwhenZPZP

nZPnZP

n

nZP

n

nZP

LCLXPUCLXP

Therefore, ARL = 1/p = 1/0.15866 = 6.30.

c) = +0 3

50.0

]0[50.01

1)6()0(

)33()33(

/

33

/

33

)()(

0000

=+=

==

=

=

nwhenZPZP

nZPnZP

n

nZP

n

nZP

LCLXPUCLXP

Therefore, ARL = 1/p = 1/0.50 = 2.00.

d) The ARL is decreasing as the magnitude of the shift increases from to 2 to 3. The ARL decrease asthe magnitude of the shift increases since a larger shift is more likely to be detected earlier than a smaller

shift.

16-52


17/21

16-74 a) Because ARL = 370, on the average we expect there to be one false alarm every 370 hours. Each 30-day

month contains 30 24 = 720 hours of operation. Consequently, we expect 720/370 = 1.9 false alarms each month

0027.0)00135.0(2)3()3()3()3( === zPzPXXPXXP ARL=1/p=1/0.0027=370.37

b) With 2-sigma limits the probability of a point plotting out of control is determined as follows, when

= +0

P X UCL P X LCL

PX

PX

P Z P Z

P Z P Z

( ) ( )

( ) ( )

( ) [ ( )]

. .

.

> + +

+

<

= > + <

= < + 12.24 when = 16) =

>

416

1624.12ZP

= P(Z > 1.88) = 1 P(Z < 1.88)= 1 0.03005= 0.96995

P( U < 3.78) =

10.68 when = 16) = P Z >

10 68 1616

10

. = P(Z > 4.22) = 1

So the probability is 1.

16-80 u = 10 a) n = 1

51.01

103103

49.191

103103

===

=+=+=

n

uuLCL

n

uuUCL

P( U > 19. 94 when = 14) = P Z >

19 94 14

14

.

= P(Z > 1.47)= 1 P(Z < 1.47) = 1 0.9292 = 0.0708and

P( U < 0.51) = 014

1451.0=

14.74 when = 14) =

>

4

14

1474.14ZP = P(Z >0.40) = 1 0.6554 = 0.3446

P( U < 5.26 when = 14) =

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Mont4e Sm Ch16 Supplemental