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CHAPTER 10:QUALITY CONTROL

Teaching Notes

As a result of increased global competition, a rapidly growing number of companies of all sizes arepaying much more attention to issues involving quality and productivity. Many statistical techniques areavailable to assist organizations in improving the quality of their products and services. It is important forcompanies to use these techniques in the context of an overall quality system (Total Quality Management)which requires quality awareness, careful planning and commitment to quality at all levels of theorganization. Many companies are not only utilizing these statistical techniques themselves, but are alsorequiring their suppliers to meet certain standards of quality based on various statistical measures. Thischapter covers the statistical applications of quality control. Control charts are given the primaryemphasis, but other quality control topics such as process capability analysis is also important.

Through the use of control charts, the nonrandom (special) causes of variation will be controlled, and

random (common) causes of variation will be analyzed through process capability.

Answers to Discussion & Review Questions

1. The elements in the control process are:

a. Defineb. Measurec. Compare to standardd. Evaluatee. Take corrective action if neededf. Evaluate corrective action to ensure it is working

2. Control charts are based on the premise that a process which is stable will reflect onlyrandomness. Statistics of samples taken from the process (means, number of defects, etc.)will conform to asampling distribution with known characteristics. From this, controllimits are determined. Observing a sample statistic outside the control limits or a trend insample statistic indicates the process has significantly changed, hence there must be anassignable cause.

3. Control charts are used to judge whether the sample data reflects a change in theparameters (e.g., mean) of the process. This involves a yes/no decision and not anestimation of process parameters.

4. Order of observation of process output is necessary if patterns (e.g., trends, cycles) inthe output are to be detected.

5. a. x chart - A control chart used to monitor process by focusing on thecentral tendency of a process.b. Range control charts are used to monitor process, focusing on the dispersion of a

process.

c. p-chart - is a control chart used to monitor the proportion of defectives in aprocess.

d. c-chart - is a control chart used to monitor the number of defects per unit.

6. Specifications are limits on the range of variation of output which are set by design(e.g., engineering, customers). Control limits are statistical bounds on a sampling

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distribution. They indicate the extent to which summary values such as sample means orsample ranges will tend to vary solely on a chance basis.Process variability refers to the

inherent variability of a processthe extent to which the output of a process will tend tovary due to chance. Control limits are a function of process variability as well as samplesize and confidence level. Both are essentially independent of tolerances.

7. The problem is that even when the machine is functioning as well as it can,unacceptable output will result. Among the possible options that should be considered are:

a. Use 100 percent inspection to weed out the defectives. If destructive testing isrequired, this may not be feasible.

b. Attempt to convince customer (offer a lower price?) to widen tolerances orengineering (communicate the cost of 100 percent inspection if relevant). The problem

is that customers/engineers may resent this suggestiondepending on how it ishandled. Moreover, it may be that the tolerance is necessary for proper functioning ofthe final product or service.

c. Attempt to substitute a different machine (e.g., a newer one) which has thecapability to handle the job.

8. This problem often goes undetected since there are no complaints fromcustomers about output being within specs. However, it is quite possible to realizedecreased costs or more profits by taking certain actions. A marketing approach to thisproblem might be to see if the customer is willing to pay more for output that meetsnarrower tolerances. If not, perhaps the job could be shifted to another, less capablemachine, freeing up this equipment for more demanding work. Still a third option would beto cut back on inspection since virtually 100 percent of the output will be acceptable.

9. a. An optimum level of inspection is one where the cost and effort ofinspection equals the benefits derived from inspection, or the point (number of unitsinspected) at which the marginal cost of inspection equals the marginal benefit frominspection.b. Cost of product or service, volume, cost of inspection, cost of letting undetected

defects slip through, degree of human involvement, stability of process, and thenumber and size of lots.

c. The main issues in the decision of whether to inspect on site or in a centrallocation are the situation (size & mobility), inspection time, cost of processinterruption, need for quick decision, importance to avoid extraneous factors affectingsamples or tests, need for specialized equipment, and the need for a more favourabletesting environment.

d. Raw materials & purchased parts, finished products, before a costly operation,before an irreversible process, and before a covering process.

10. a. Type I error

b. Type II error

Memo Writing Exercise

A p-chart is used to monitor the proportion of defective units generated by a process, while an x chart is

used to monitor the central tendency of a process (i.e. change in the mean or the nominal value of aprocess). A p chart classifies the observations into one of two mutually exclusive categories (good vs.

bad, pass vs. fail, etc.). An x chart usually requires taking measurements in data to monitor the average

of a process. Examples of characteristics requiring an x chart include measurement of a diameter of a

tire, length of a bolt, tensile strength of a rubber product, and weight of a cereal box. In using a p chart

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data collection is usually easier because instead of taking actual measurements, we would simply recordwhether the item is conforming or not conforming. In addition, the p chart requires a considerably larger

sample size than the x chart. On the other hand, if workers do the charting on the line, the training

required for p chart is simpler than the training required for x and R chart.

x chart is usually preferred over p chart for characteristics that require taking actual

measurements because the lower number of observations and higher information content will outweighthe extra cost of measurement. However, when the characteristic in question is a dichotomous

classification (defective vs. nondefective, on vs. off) the x chart is not applicable and p chart should be

used.

Solutions

1. specs: 24 kg. to 25 kg.

= 24.5 kg. [assume = ] = .2 kg.a.

z =+.5

= 2.5 2(.0062) = .0124 or 1.24%.2

b. 2 = 24.5 2 .2 = 24.5 .1 or 24.4 to 24.6n 16

2. = 2.0 litres = .01 litren = 5

a. Control limits: 2n

[z = 2.0 for 95.5%]

UCL is 2.0 + 2.01

= 2.009 litres5

LCL is 2.0 - 2.01

= 1.991 litres5

Yes, the process is in control.3. n = 10

A2 = 0.31 = 3.10 a. Mean Chart: A2 = 3.1 0.31(0.45)

D3= 0.22 = 0.45 = 3.1 .14

D4 = 1.78 Hence, UCL is 3.24and LCL is 2.96

Range Chart: UCL is D4= 1.78(0.45) = .801

LCL is D3= 0.22(0.45) = .099b. In control since all points are within these limits.

4. Sample Mean Range

1 79.48 2.6Mean Chart: A2= 79.96 0.58(1.90)

2 80.14 2.3 = 79.96 1.1

3 80.14 1.4 UCL = 81.06, LCL = 78.86


.0062.0062

24 24.5 25 16

-2.5 0 +2.5 z-scale

b.

FromAppendix B,Table B

2.015

2.010

2.005

2.000

1.9951.990

1.895

UCL

LCL

(litres)Mean

*

*

*

*

**

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4 79.60 1.7 Range Chart: UCL = D4= 2.11(1.90) = 4.0095 80.02 2.0 LCL = D3= 0(1.90) = 06 80 .38

79.961 .41.9

[Both charts suggest that the process is in control: Neither has anypoints outside the limits.]


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Solutions (continued)

Embedded Excel Worksheet

5.Octane Moving

obs # rat ing Range1 89.20

2 86.50 2.70

3 88.40 1.90 avg= 88.56

4 91.80 3.40 std dev= 2.91

5 90.30 1.50

6 87.50 2.80 UCLx= 97.29

7 92.60 5.10 LCLx= 79.83

8 87.00 5.60

9 89.80 2.80 All observat ions fal l wi thin above limits

10 92.20 2.40

11 85.40 6.80 avg moving range= 4.11

12 91.60 6.20

13 87.70 3.90 UCLmr= 13.42

14 85.00 2.70 LCLmr= 0

15 91.50 6.5016 90.30 1.20 All moving ranges fall wi thin above l imits

17 85.60 4.70

18 90.90 5.30 Therefore, process is in cont rol.

19 82.10 8.80

20 85.80 3.70

6.Octane Moving

obs # rat ing Range

1 16.20

2 13.80 2.40

3 17.00 3.20 avg= 15.23

4 15.80 1.20 std dev= 1.50

5 13.50 2.30

6 14.70 1.20 UCLx= 19.73

7 14.00 0.70 LCLx= 10.748 14.80 0.80

9 13.20 1.60 All observat ions fal l wi thin above limits

10 16.80 3.60

11 14.90 1.90 avg moving range= 1.92

12 13.00 1.90

13 12.50 0.50 UCLmr= 6.29

14 16.70 4.20 LCLmr= 0

15 15.90 0.80

16 14.60 1.30 All moving ranges fall wi thin above limits

17 16.50 1.90

18 18.40 1.90 Therefore, process is in cont rol.

19 15.20 3.20

20 14.60 0.60

21 17.20 2.60

22 16.10 1.1023 14.40 1.70

24 17.00 2.60

25 13.80 3.20

26 15.50 1.70


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7. n = 200a. 1 2 3 4

4= .020

12= .010

5= .025

9= .045

200 200 200 200

b. (0.02 + 0.01 + 0.025 + 0.045)/4 = 0.025c. mean = .025

011.200

)975(.025.)1(.devStd. ==

=

n

pp

d. = 0.03 confidence = .97 z = 2.17.025 2.17(0.011) = .025 .024 = .001 to .049.

e. .025 + z(.011) = .047Solving, z = 2, leaving .0228 in each tail. Hence, alpha = 2(.0228)

= .0456.f. Yes, all sample proportions in part (a) fall within these control limits.g. mean = .02

.02(.98)Std. dev. .0099200= =

h. .02 2(.0099) = 0.0002 to .0399.

No, the last sample is beyond the upper limit.

8. n = 200 Control Limits =n

ppp

)1(2

Thus, UCL is .0234 and LCL is 0 (because it can't be negative).Since n = 200, the fraction represented by each data point is half theamount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is too large. Omitting that value and recomputing

limits with = 18 = .0075 yields12(200)

UCL = .0075 + .0122 = .0197 and LCL = 0.9. = 110

14= 7.857

Control Limits: 3 = 7.857 8.409

UCL is 16.266, LCL, if negative, should be changed to 0.All numbers of daily compaints are within the limits.

10. = 2114 = 1.5 Control Limits: 3 = 1.5 3.67

UCL is 5.17, LCL becomes 0.All values are within the limits.


0096.)200(13

25==p

0138.0096.

200

)9904(.0096.20096.

=

=

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11. = total number of defectives =87

= .054Total number of observations 16(100)

Control limits are z (1 - ) = .054 1.96.054(.946)

n 100

= .054 .044. Hence, UCL = .098LCL = .01

Note that observations must be converted to fraction defective, or control limits must beconverted to number of defectives. In the latter case, the upper control limit would be 9.8defectives and the lower control limit would be 1 defective. Even though all points are withinthese limits, the process error rate can be shown statistically to be different from 4% error ratebecause 5.4% is too far from 4% given 1600 observations.

12. There are several slightly different ways to solve this problem. The most straightforward seems tobe the following:

(1) Observe that the upper control limit is six standard deviations above the lowercontrol limit.

(2) Compute the value of the upper control limit at the start:

cm.06.15

1

01.615 =+

(3) Determine how many pieces can be produced before the upper control limit justtouches the upper tolerance, given that the upper limit increases by .004 cm. per piece:

15.2cm. - 15.06cm.= 35 pieces.

.004 cm./piece

13. a. Out of the 30 observations, only one value exceeds the tolerances, or 3.3%. [This case isessentially the one portrayed in the text in Figure 10-11A.] Thus, it seems that the tolerances

are being met: approximately 97 percent of the output will be acceptable.b. = 1.90 from problem 4.A2 = .58

2n 5

A R (.58)(1.90) . 823 3

= =

P

max spec min spec 81 78C .61

6 6(.82)

= = = < 1 => not capable

No, part (a) and (b) give different results. It is best to use CP Index because it is based onpopulation, not just a sample of 30.

14. a. = .146

n = 6

.85.339

15.150

39===

xx

Control limits are.146

x 2 .385 2 3.85 .119n 6

= =

So UCL is 3.97, LCL is 3.73. Sample 29 is outside the UCL, so the process isnot in control.


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15.

(i) The upper control limit is 6 standard deviations above the lower control limits.(ii) When UCL = 3.5 Cm, the LCL = 3.5 - 6 0.01 = 3.5 - 0.06 = 3.44 Cm.

1

(iii) Determine how many pieces can be produced before the LCL just crosses the lowertolerance of 3 Cm.

3.44 - 3.00=

0.44=

440= 440 pieces

0.001 0.001 1

16. It is necessary to see if the process variability is within 9.96 and 10.35. Two observations havevalues above the specified limits, i.e., 10% of the 20 observations fall outside the limits.

2n

A R3

from (10-4)4

(.73)(.52) .2533

= =

P

max spec min spec 10.35 9.65C .46

6 6(.253)

= = = < 1 => process is not capable

One should reduce the process variability, e.g., using experimental design or use more accuratemachines.

17. a. 1 2 3 44.3 4.5 4.5 4.7

b. = (4.3 + 4.5 + 4.5 + 4.7)/4 = 4.5std. dev. (of data set) = .192 using Excel

c. mean = 4.5, std. dev. = .192/ 5 = .086d. 4.5 3(.086) = 4.5 .258 = 4.242 to 4.758

The risk is 2(.0013) = .0026.e. 4.5 + z(.086) = 4.86

Solving, z = 4.19, so the risk is close to zero.

f. None.g. = (.3 + .4 + .2 + .4)/4 = .325n = 5Means: A2 = 0.58 A2= 4.5 0.58(.325) = 4.3115 to 4.6885.

The first mean is below the lowest limit, and the last mean is above theupper limit.

Ranges: D3 = 0 0 to 2.11(.325) = 0 to .68575 All ranges are within the limits.


3.5Cm

3.44

3.0 Cm

3 sigma

n = 1

= 0.01 cmUCL

Mean

LCL

3 sigma

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h. Two different measures of dispersion are being used, the standard deviation and the range, andthe process distribution is not quite Normal.

i. 4.4 3 0.18 = 4.4 .241 = 4.16 to 4.64. The last sample mean is above the upper limit.

5


4.5 4.6 4.5 4.7

4.2 4.5 4.6 4.6

4.2 4.4 4.4 4.8

4.3 4.7 4.4 4.5

4.3 4.3 4.6 4.9

mean= 4.5

std dev= 0.191943

Bin Frequenc y

4.1 4.1 04.2 4.2 2

4.3 4.3 3

4.4 4.4 3

4.5 4.5 4

4.6 4.6 4

4.7 4.7 2

4.8 4.8 1

4.9 4.9 1

More 0

18. Process mean = 0.03 cm, = 0.003 cm, tolerance = (0.02, 0.04) cm.

a. Cp =specification width

=.04 - .02 = .02

=.02

= 1.11process width 6(.003) .018

b. In order to be capable, the process capability ratio must be at least 1.00. In this instance, the indexis 1.11, so the process is capable. (If 1.33 were used, the process would not be capable.)

19.Machine Standard Deviation (cm) Job Specification (cm) Cp Capable ?

001 0.02 0.05 0.833 No

002 0.04 0.07 0.583 No

003 0.10 0.18 0.600 No

004 0.05 0.15 1.000 Yes

005 0.01 0.04 1.333 Yes

20. Machine Cost per unit ($) Standard Deviation (mm.) Cp

A 20 0.079 1.013

B 12 0.080 1.000

C 11 0.084 0.952

D 10 0.081 0.988


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You can narrow the choice to machines A and B because they are the only ones with a capabilityratio of at least 1.00. You would need to know if the slight additional capability of machine A isworth an extra cost of $8 per unit.

21. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,

X = Process mean, = Process standard deviation

For process H:

}{pk

X LSL 15 14.1.9375

3 (3)(.32)

USL X 16 151.04

3 (3)(.32)

C min .9375, 1.04 .9375

.9375 1.0, not capable

= =

= =

= =

==

=

=

=

=

For process T:

}{

capable.isprocesstheSince

C

XUSL

LSLX

pk

,0.106.1

06.106.1,33.1min

06.1)5.0)(3(

5.181.20

3

33.1)5.0)(3(

5.165.18

3

>

==

=

=

=

=


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22. No, the maximum value of CPKoccurs when the process mean is centred in the specificationrange, and then CPKwill equal CP because

CPK=max spec process mean process mean min spec 1 max spec min spec

3 3 2 3

= =

= CP.

Therefore CPK CP.

23. Let USL = Upper Specification Limit

X = Process mean, = Process standard deviation.

USL= 45 minutes,

38 min., 3 min .

37 min., 2.5 min .

37.5 min., 2.5 min .

Armand Armand

Jerry Jerry

Melissa Melissa

X

X

X

= =

= =

= =

For Armand:

45 38.78

3 (3)(3)

.78 1.0, .

= = =

pk

USL X C

Since Jerry is capable

For Melissa:

45 37.51.0

3 (3)(2.5)

1.0 1.0, .

pk

USL X C

Since Melissa is capable

Jerry is most capable.

= = =


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24. a.Stroh Brewery

Sample 1 2 3 4 5 6 7

0.6 0.4 0.5 0.4 0.65 0.6 0.7

0.55 0.45 0.3 0.4 0.45 0.5 0.55

0.5 0.35 0.4 0.5 0.45 0.55 0.55

Process m 0.49

Process st 0.10

Bin Freq.

0.3 0.3 1

0.4 0.4 5

0.5 0.5 7

0.6 0.6 6

0.7 0.7 2

0.8 0.8 0

More 0

Tolerance (0,1.45) cc

Cpk = 3.18 >1 Capable

0

2

4

6

8

0.3 0.4 0.5 0.6 0.7 0.8 More

Bin

Frequency

b.Sample 1 2 3 4 5 6 7

0.6 0.4 0.5 0.4 0.65 0.6 0.7

0.55 0.45 0.3 0.4 0.45 0.5 0.55

0.5 0.35 0.4 0.5 0.45 0.55 0.55 avg.

Sample m 0.55 0.40 0.40 0.43 0.52 0.55 0.60 0.49

Sample R 0.10 0.10 0.20 0.10 0.20 0.10 0.15 0.14

Sample Mean control chart:

UCL = 0.63

LCL = 0.35

Sample Range control chart:

UCL = 0.35

LCL = 0.00

= grand mean

= average range

Mean chart

0.00

0.20

0.40

0.60

0.80

1 2 3 4 5 6 7

Range chart

0.00

0.10

0.20

0.30

0.40

0.50

1 2 3 4 5 6 7


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c.Time period

Sample 1 2 3 4 5 6 70.6 0.4 0.5 0.4 0.65 0.6 0.7

0.55 0.45 0.3 0.4 0.45 0.5 0.55

0.5 0.35 0.4 0.5 0.45 0.55 0.55 avg.

Sample m 0.55 0.40 0.40 0.43 0.52 0.55 0.60 0.49

Sample R 0.10 0.10 0.20 0.10 0.20 0.10 0.15 0.14

Mean chart

0.00

0.50

1.00

1.50

2.00

1 2 3 4 5 6 7 8

Range chart

0.00

0.10

0.20

0.30

0.40

0.50

1 2 3 4 5 6 7 8

Out of control

In control

= grand mean

= average range

Case: Toys Inc.

A consultant must consider the long-term implications of decisions suggested.

1. Cutting cost in design and product development may not be beneficial to the company in the longrun.

2. The trade-in and repair program, while appeasing customers in the short run, may be too costlyand will not be correcting the root cause of the problem.

3. Since the company thrives on its reputation of high quality products, it needs to continue todesign products of high quality that fulfils the needs of the market place. 100% inspection may betoo expensive. Manufacturing needs to place greater emphasis on preventive qualitymanagement/control (e.g., use of control charts) rather than inspecting already completed parts.The company may want to consider investing more in R&D.

Case: Tiger Tools

Answers to Questions:

1. For the first data set = .873. From Table 10-2 , for n = 20, A2 = .18. Using Equation 10.4, theestimated standard deviation is:


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2

n 20A R (.18)(.873) .234

3 3 = =

The process capability is1.20

= .854. Because this is less than 1.00, the process is not capable6(.234)

2. For the second data set, = .401. From Table 10-2 , A2=.58. Performing the samecalculations as in #1, we obtain an estimated standard deviation of:

2A R .58(.401)n 5 .1733 3

= =

The process capability is1.2

= 1.153. Because this is more than 1.00, the process is capable.6(.173)

The process seems to be cycling, as indicated by the plot of the smaller sample size below.Taking large samples probably resulted in combining the results of several different processmeans, and therefore resulted in a large estimate for .

Even though process is capable, it is not in control (because of the wave-like pattern of theaverage lengths of prybars). The cause for this should be investigated and fixed.


For n = 5

Sample number2 4 6 8 10 12 14 16 18 20 22 24 26

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Case: Canadian Springs


1.

2. Quality for Canadian Springs means that the product, water, is free of any impurities or most bacteria.Also, the water's taste, smell, and colour should be acceptable.

Canadian Springs draws the water from closed aquifers that have almost pure water. By usingsanitized tanker trucks and equipment in the plants, contamination of water is kept to a minimum. Inaddition, water goes through filteration processes. Also, the returned bottles are cleaned andsterilized. Finally, regular hourly quality tests are performed on the water in the holding tanks, andbottled water is kept for up to 30 days and tested to see how the water keeps over time.

Case: In the Chips at Jays


1.


Conveyorbelt

SortSemi-trailersof potatoes

Holdingbins

Holding

bins

big

small

Washing SkinningInspectionfor rottenpotatoes

Inspectionfor burnt

chips

Chippers

Cooking incorn oil,

circulated,salted,

flavoured

Laser check(opti-sort

scanner) fordark spots,

holes

Brokenchips fallthroughScalesBagging

. . .aquifer

Tankertrucks

Carbonfilters &

membranes

Springwater

Premiumwater

Holdingtanks

ReverseOsmosis Holding

tanks

Filling Capping

Returnedbottles

Conveyorbelt

VisualInspection

Cleansing &Sterilization

Plant

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2. Chips should taste good (i.e., from good-tasting, not rotten, potatoes), be pure (withoutskin), be whole (not broken), not burnt, and consistently flavoured. First, good quality, NorthDakota potatoes are purchased, then washed and skinned, and inspected to remove the rottenones. While frying, the oil and flavours are circulated to provide consistency. After cooking,burnt and broken ones are separated too.


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