INSTRUCTORS MANUAL

Chapter 10:

Quality control

Teaching Notes

As a result of increased global competition, a rapidly growing number of companies of all sizes are paying much more attention to issues involving quality and productivity. Many statistical techniques are available to assist organizations in improving the quality of their products and services. It is important for companies 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 the organization. Many companies are not only utilizing these statistical techniques themselves, but are also requiring their suppliers to meet certain standards of quality based on various statistical measures. This chapter covers the statistical applications of quality control. Control charts are given the primary emphasis, but other quality control topics such as process capability and inspection are also discussed.

When covering the material in this chapter, we need to stress that through the use of control charts, the nonrandom (special) causes of variation must be controlled before random (common) causes of variation and process capability can be analyzed.

Answers to Discussion and Review Questions

1.The elements in the control process are:

a. Define

b. Measure

c. Compare to standard

d. Evaluate

e. Take corrective action if needed

f. Evaluate corrective action to insure it is working

2.Control charts are based on the premise that a process which is stable will reflect randomness: statistics of samples taken from the process (means, number of defects, etc.) will conform to a sampling distribution with known characteristics, so that statistical significance tests can be performed on sample statistics; and successive samples will not reveal any patterns which will enable prediction of future values other than specification of range of variability.

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

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

5.a.

chartA control chart used to monitor process variables by focusing on the central tendency of a process.

b. Range control charts are used to monitor process variables, focusing on the dispersion of a process.

c. p-chartis a control chart for attributes, used to monitor the proportion of defectives in a process.

d. c-chartis a control chart for attributes, used to monitor the number of defects per unit.

6.A run is a sequence of observations with a given characteristic. Run tests are helpful in detecting patterns in time series (e.g., control chart) data.

7.All points can be within control limits but with certain patterns developing in the data which would suggest the output is not random, and hence, not in control for long.

8.It is usually desirable to use both an up/down and a median run test on a given set of data because the tests are sensitive to different things. For example, one test can be more sensitive to trend and the other to bias.

9.No, there is always the possibility of a Beta or Type II error which is the probability of calling something random when in fact it is non-random or concluding that non-randomness is not present, when it actually is.

10.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 distribution. They indicate the extent to which summary values such as sample means or sample ranges will tend to vary solely on a chance basis. Process variability refers to the inherent variability of a process(the extent to which the output of a process will tend to vary due to chance. Control limits are a function of process variability as well as sample size and confidence level. Both are essentially independent of tolerances.

11.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 is required, this may not be feasible.

b. Attempt to convince customers (offer a lower price?) to widen tolerances or engineering (communicate the cost of 100 percent inspection if relevant). The problem is that engineering may resent this suggestion(depending on how it is handled. Moreover, it may be that the tolerance is necessary for proper functioning of the final product or service.

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

d. Hope for a miracle.

12.a. This problem often goes undetected since there are no complaints from customers about output not within specs. However, it is quite possible to realize decreased costs or more profits by taking certain actions.

b. A marketing approach to this problem might be to see if the customer is willing to pay more for output that meets narrower tolerances. If not, perhaps the job could be shifted to another, less capable machine, freeing up this equipment for more demanding work. Still a third option would be to cut back on inspection since virtually 100 percent of the output will be acceptable, and even a slight out of control situation will not warrant corrective action.

13.a.An optimum level of inspection is one where the cost and effort of inspection equals the benefits derived from inspection, or the point (number of units inspected) at which the marginal cost of inspection equals the marginal benefit from inspection.

b. Cost of product or service, volume, costs of inspection, cost of letting undetected defects slip through, degree of human involvement, stability of process, and the number and size of lots.

c. The main issues in the decision of whether to inspect on site or in a central location are the situation (size and mobility), inspection time, costs of process interruption, need for quick decision, importance to avoid extraneous factors affecting samples or tests, need for specialized equipment, and the need for a more favorable testing environment.

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

14.Two basic assumptions that must be satisfied in order to use a process capability index are:

a.The process is stable (Non-random causes of variation have been identified and corrected);

b.The process distribution is normal.

15.It is very important. The companys (and managers) reputation is at stake, and there may be cost, liability, legal and safety issues. Although the risks may differ substantially for different products and services, ethical standards should be maintained across the board.

16.a.Type I error

b.Type II error

c.Eating a dirty cookie is a Type II error

Not eating a clean cookie is a Type I error

d. Type II error

Taking Stock

1. a. In deciding whether to use 2-sigma or 3-sigma limits, the quality control people should be involved as well as the accounting /record keeping personnel because it will be critical to determine the cost of unnecessarily stopping the process vs. cost of not correcting a special cause of variation. In addition, we may want to involve the customers quality control personnel since they will be ultimately affected by this decision.

b. The quality control department should make this decision. However, input from the production control department may be very useful to estimate the cost of sampling.

c. Increasing the capability of the process is a significant and potentially costly adventure. Therefore, the upper management in consultation with the quality and production departments should determine what type of systematic improvements need to be made to improve the capability of the process.

2. In setting the quality standards, customers should definitely be involved since they will ultimately be using the product. In consultation with the upper management, the quality and production departments should work as a team in establishing the quality standards.

3. The technology had a profound impact on quality. Improvement in measurement systems drastically improved the measurement of quality. The computer technology has enabled many companies to perform on-line, real-time statistical process control, which enabled companies to respond to quality problems faster. Due to technological improvements in computerized design, the products are designed better, thus have significantly fewer quality problems. The artificial intelligence systems forewarn potential problems before they occur.

Critical Thinking

1.If the analysis of the output of a process suggests that there is an unusual occurrence, but the result of the investigation cannot pinpoint or determine the assignable causes, the limits may be set too tight. Loosening the limits will allow us to determine the possible assignable causes of variation. If the record keeping is poor it may be very difficult to identify the assignable causes of variation. Improving the record keeping may assist us in identifying non-random or assignable causes of variation.

2.A single standard would be easier to work with, and everyone would know what the standard was. Multiple standards might be used if the cost of errors differed significantly across products or services. For instance, not meeting the specifications for a product such as paper clips wouldnt have the same degree of importance as meeting the specifications for heart-monitoring equipment. Also, differing standards might reflect progress in continuous improvement; as processes are improved, the standard would be increase to reflect that, and perhaps be used as a competitive strength in B2B dealings.

Additional Experiential Learning Exercises

Sampling Demonstrations. Bowls of colored beads (e.g., 1,000 beads, 40% white, 30% green, 20% red, 5% black, 3% yellow, etc.) are available with paddles that have indentations :that facilitate quickly obtaining samples of various sizes. Use to demonstrate sampling variability take repeated small samples (or have a student do it) to demonstrate that different percentages of a color appear in different samples. Then increase the sample size, focusing on a specific color, and have students recognize that there is less variability (i.e., sampling variability distribution) as the sample size increases. Afterward, explain why in process sampling, small samples are used even though large samples are more accurate. One is the cost, time, and disruption caused by taking large samples. Even more important is the ability to capture process changes that might occur between small samples (e.g. 6 samples of n = 10 taken over time could reveal a trend, whereas 1 sample of 60 could not).

Process Control Demonstrations. Obtain 30 clear plastic zipper bags and an ample supply of colored beads, marbles, or jelly beans.

a.

b.

c.

Place 10-20 beads/beans in each of 10 bags, focusing, say on red beads/beans. Number the bags 1 to 10. Put 0, 1, or 2 red beads/beans in each bag, along with other colors of beads/beans. Pass the bags out to the class and ask them what kind of control chart would be appropriate (p-chart or c-chart). Then say, lets suppose the red beads/beans are defects, so we shall construct a c-chart. Have then report (in order), the number of defects they have counted. Next, have the class calculate 2s control limits, and then see that all are within the control limits.

In a second set of bags numbered 11-21, arrange it so that bag #14 has too many red beads/beans, making it out of control. Pass out those bags and have the students report the numbers of red they find. When the student with bag #14 reports a large number of reds, tell the class that the process would be halted while efforts were made to find and correct the problem. Then resume sampling with the process now back in control.

Arrange the third set so that a trend begins to appear at about the fifth bag, but with all results still within the limits. Point out that the process doesnt appear to be random, so it would be halted at the ninth or tenth sample to find and correct the problem.

Memo Writing Exercises (included on the DVD)1.A p-chart is used to monitor the proportion of defective units generated by a process, while an chart is used to monitor the central tendency of a process (i.e. change in the mean or the nominal value of a process). A p-chart classifies the observations into one of two mutually exclusive categories (good vs. bad pass vs. fail, etc.). An chart usually requires taking measurements in data collection to monitor the average of a process. Examples of characteristics requiring an 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 data collection is usually easier because instead of taking actual measurements, we would simply record whether the item is conforming or not conforming. In addition, the p-chart requires a considerably larger sample size than the 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 and R chart.

chart is usually preferred over p-chart for characteristics that require taking actual measurements because the lower cost of sampling and higher information content will outweigh the extra cost of measurement. However, when the characteristic in question is a dichotomous classification (defective vs. nondefective, on vs. off) the chart is not applicable and p-chart should be used.

2.In order to monitor the capability of the process to control the number of defective units, we must first make sure that the special (assignable) causes of variation are eliminated with the use of control charts. However, control charts will not pinpoint the cause of defective units because the natural process variability may exceed the specification limits (tolerances). In other words, even if a process is in control using control charts it may still be producing defective units. Therefore, after the process is deemed to be in control using control charts, we need to determine whether the process is capable by making sure that the natural variation of the process is within the specification limits, and the percentage of expected number of defective units is acceptable.

Solutions

1. specs: 24 oz. to 25 oz.

( = 24.5 oz. [assume ( = EQ \O(=,x) ]

( = .2 oz.

a. [refers to population]

b.

2. ( = 1.0 liter

( = .01 liter

n = 25

a.

[z = 2.17 for 97%]

3. a.n = 20

A2 = 0.18 EQ \O(=,X) = 3.10 Mean Chart: EQ \O(=,X) ( A2 EQ \O((,R) = 3.1 ( 0.18(0.45)

D3 = 0.41 EQ \O((,R) = 0.45 = 3.1 ( .081

D4 = 1.59Hence, UCL is 3.181

and LCL is 3.019. All means are within these limits.

Range Chart: UCL is D4 EQ \O((,R) = 1.59(0.45) = .7155

LCL is D3 EQ \O((,R) = 0.41(0.45) = .1845

In control since all points are within these limits.

4.SampleMeanRange

179.482.6Mean Chart: EQ \O(=,X) ( A2 EQ \O((,R) = 79.96 ( 0.58(1.87)

280.142.3= 79.96 ( 1.08

380.141.2UCL = 81.04, LCL = 78.88

479.601.7Range Chart: UCL = D4 EQ \O((,R) = 2.11(1.87) = 3.95

580.022.0LCL = D3 EQ \O((,R) = 0(1.87) = 0

680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]

Solutions (continued)

5. n = 200

a.1234

.020.010.025.045

b. (2.0 + 1.0 + 2.5 + 4.5)/4 = 2.5%

c. mean = .025

d. z = 2.17

.025 ( 2.17(0.011) = .025 ( .0239 = .0011 to .0489.

e..025 + z(.011) = .047

Solving, z = 2, leaving .0228 in each tail. Hence, alpha = 2(.0228)

= .0456.

f. Yes.

g. mean = .02

h..02 ( 2(.01) = 0 to .04

The last sample is beyond the upper limit.

6.n = 200

Control Limits =

Thus, UCL is .0234 and LCL becomes 0.

Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.

Sample 10 is too large. Omitting that value and recomputing limits with

UCL = .0197 and LCL = 0.

7.

Control limits:

UCL is 16.266, LCL becomes 0.

All values are within the limits.

Solutions (continued)

8.

Control limits:

UCL is 5.17, LCL becomes 0.

All values are within the limits.

9.

Hence, UCL = 0.10

LCL = 0.01

Note that observations must be converted to fraction defective, or control limits must be converted to number of defectives. In the latter case, the upper control limit would be 7.10 defectives and the lower control limit would be .11 defective. Even though all points are within these limits, the process appears to be out of control because 75% of the values are above 4%.

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

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

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

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

15.2cm. 15.06cm.= 35 pieces.

.004 cm./piece

11. Out of the 30 observations, only one value exceeds the tolerances, or 3.3%. [This case is essentially the one portrayed in the text in Figure 109A.] Thus, it seems that the tolerances are being met: approximately 97 percent of the output will be acceptable.

12. a. ( = .146

n = 14

Control limits are

So UCL is 3.97, LCL is 3.73. Sample 29 is outside the UCL, so the process is not in control.

Solutions (continued)

b. [median is 3.85]

SampleA/BMeanU/DSampleA/BMeanU/D

1A3.86(21B3.84D

2A3.90U22B3.82D

3B3.83D23A3.89U

4B3.81D24A3.86D

5B3.84U25A3.88U

6B3.83D26A3.90U

7A3.87U27B3.81D

8A3.88U28A3.86U

9B3.84D29A3.98U

10B3.80D30A3.96D

11A3.88U31A3.88D

12A3.86D32B3.76D

13A3.88U33B3.83U

14B3.81D34B3.77D

15B3.83U35A3.86U

16A3.86U36B3.80D

17B3.82D37B3.84U

18A3.86U38B3.79D

19B3.84D39[B]3.85U

20A3.87U

Testobs.exp.( Z

Conclusion

Median1820.53.08.81random

Up/down2925.72.571.28random

13.a.

AAB BA BA BB ABAABAAA B BBABABAB

DD DU DU DDUD UUDU UD D DU UD UDUD

b.

AAAABAABBBBBABBBAAAABBBBBB

U D U DU DD UD UD UD UD U DU DDUD UDD

Solutions (continued)

Summary:obs.exp.(zConclusion

a. median18142.501.6random

up/down17172.070.0random

b. median8142.502.40nonrandom

up/down22172.072.41nonrandom

14. a. [Data from Chapter 10, Problem 8]

Median is 1.5 A = Above, B = Below, U = Up, D = Down.

Sample:1234567891011121314

Median:AABBBAABABABAB

Data:23101320213120

Up/down:UDDUUDDUDUDUD

b. [Data from #171] Median is 7.5.

Day:1234567891011121314

Median:BAAAABBAABBBBA

Data:4101489651213764210

Up/downUUDUDDUUDDDDU

For part a and b:

E(r)med =

E(r)u/d =

For part a:

Zmed =

For part b:

Zmed =

Since the absolute values of all Z statistics calculated above are less than 2, all patterns appear to be random.

Solutions (continued)

Summary:

Testobs.Exp.(zConclusion

a.median1081.801.11random

up/down1091.47.68random

b.median681.801.11random

up/down791.471.36random

15.

DayAmountDayAmountDayAmount

1B27.69(21B28.60U41B26.76D

2B28.13U22B20.02D42B30.51U

3A33.02U23B26.67U43B29.35D

4B30.31D24A36.40U44B24.09D

5A31.59U25A32.07D45B22.45D

6A33.64U26A44.10U46B25.16U

7A34.73U27A41.44D47B26.11U

8A35.09U28B29.62D48B29.84U

9A33.39D29B30.12U49A31.75U

10A32.51D30B26.39D50B29.14D

11B27.98D31A40.54U51A37.78U

12A31.25U32A36.31D52A34.16D

13A33.98U33B27.14D53A38.28U

14B25.56D34B30.38U54B29.49D

15B24.46D35A31.96U55B30.81U

16B29.65U36A32.03U56B30.60D

17A31.08U37A34.40U57A34.46U

18A33.03U38B25.67D58A35.10U

19B29.10D39A35.80U59A31.76D

20B25.19D40A32.23D60A34.90U

Summary:

Testobs.exp.(zConclusion

median22313.842.34non-random

up/down3539.673.221.45random

Since one of the tests suggests non-randomness, the conclusion must be that the process is not in control. In other words, the variation in daily expenses is not random. Further investigation would be necessary in order to determine what sort of pattern is present.

Solutions (continued)

16.

(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

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

3.44 3.00=0.44=440= 440 pieces

0.0010.0011

17. It is necessary to see if the process variability is within 9.96 and 10.35. Two observations have values above the specified limits, i.e., 10% of the 20 observations fall outside the limits. Perhaps the process mean should be set a bit lower.

18. 1 Step 10% scrap, 2nd 6%, and 3rd 6%.

a. Let x be the number of units started initially at Step 1. With a scrap rate of 10% in Step 1 the input to Step 2 is 0.9x. The input to Step 3 is (1 0.06) (1 0.10)x. With a scrap rate of 6% at Step 3 the number of good units after Step 3 = (1 0.06) (1 0.06) (1 0.10)x = (0.94)2(0.90)x.

The required output is 450 units

(0.94)2(0.90)x = 450 units

x = 565.87 ( 566 units

b. (1 0.03)2(1 0.05)x = 450 units

(0.97)2(0.95)x = 450 units

x = 503.44 ( 504 units

Savings of 566 504 = 62 units

c. From (a)

The scrap = 566 450 = 116 units

@ $10 per unit, The Total Cost = $1,160.00

Solutions (continued)

19. Sample #

1234567891011121314151617181920

32451241213424213134

Median = 2.5 A/BABAABBABBBAABABBABAA

Up/DownDUUDUUDUDUUDUDDUDUU

N = 20

Standard

ObservedExpectedDeviationZConclude

Median13112.17940.9177Random

Up/down14131.79810.5561Random

20.a.1234

4.34.54.54.7

b. EQ \O(=,x) = (4.3 + 4.5 + 4.5 + 4.7)/4 = 4.5

std. dev. (of data set) = .192

c. mean = 4.5, std. dev. =

d. 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. EQ \O((,R) = (.3 + .4 + .2 + .4)/4 = .325

n = 5

Means: A2 = 0.58 EQ \O(=,x) ( A2 EQ \O((,R) = 4.5 ( 0.58(.325) = 4.3115 to 4.6885.

The last mean is above the upper limit.

Ranges: D3 = 0 0 to 2.11(.325) = 0 to .6875 All ranges are within the limits.

h. Two different measures of dispersion are being used, the standard deviation and the range.

i.

to 4.64. The last value is above the upper limit.

21. Solution

a.

b. In order to be capable, the process capability ratio must be at least 1.33. In this instance, the index is 1.11, so the process is not capable.

Solutions (continued)

22.MachineStandard Deviation (in.)Job Specification ((in.)CpCapable ?

0010.020.050.833No

0020.040.070.583No

0030.100.180.600No

0040.050.151.000No

0050.010.041.333Yes

23.MachineCost per unit ($)Standard Deviation (mm.)Cp

A200.0591.355

B120.0601.333

C110.0631.27

D100.0611.311

You can narrow the choice to machines A and B because they are the only ones with a capability ratio of at least 1.33. You would need to know if the slight additional capability of machine A is worth an extra cost of $8 per unit.

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

= Process mean, ( = Process standard deviation

For process H:

For process K:

Since 1.0 < 1.33, the process is not capable.

Solutions (continued)

For process T:

Since 1.33 = 1.33, the process is capable.

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

= Process mean, ( = Process standard deviation.

USL = 90 minutes, LSL = 50 minutes,

= 74 minutes,

= 4.0 minutes

= 72 minutes,

= 5.1 minutes

For the first repair firm:

Since 1.333 = 1.333, the firm 1 is capable.

For the second repair firm:

Since 1.18 < 1.33, the firm 2 is not capable.

Solutions (continued)

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

= Process mean, ( = Process standard deviation.

USL = 30 minutes, LSL = 45 minutes,

Armand = 38 minutes, ( Armand = 3 minutes

Jerry = 37 minutes, ( Jerry = 2.5 minutes

Melissa = 37.5 minutes, ( Melissa = 2.5 minutes

For Armand:

Since .78 < 1.33, Armand is not capable.

For Jerry:

Since .93 < 1.33, Jerry is not capable.

For Melissa, since, the process is centered, therefore we will use Cp to measure process capability.

Since 1.39 > 1.33, Melissa is capable.

27.a. Cp = spec width = 20 = 1.33. Solving, ( = 2.506.

6( 6(b. Box variance = 3.84; box ( = 1.96. Average box weight = 6(1.01) = 6.06 ounces. Students can

deduce that 1 ounce = 28.33 grams, making the box weight in grams: 6.06(28.33) = 171.70 gr.

Cpk =

Upper spec ( 171.70

3(=

180 ( 171.70

3(1.96)

= 1.41 (capable).

c. The lowest setting implies a mean that is less than average:

Cpk =

Mean ( Lower spec

3(1.96)

= 1.33.

Solving, mean = 167.82. Mean/6 = 27.97 gr. = .987 ounces.

28. Note that all points are within the control limits, so the process is apparently in control.

Run tests:

TestObservedExpectedStd. dev. ZConclusion

Median15 122.291.31 random

Up/down19 14.331.892.46 non-random

Because one of the run tests indicated that the output is not random, the process is probably not random, and should be investigated to determine the cause.

Case: Toys Inc.

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

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

2.The trade-in and repair program, while appeasing customers in the short run, may be too costly and 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 to design products of high quality that fulfils the needs of the market place. Manufacturing needs to place greater emphasis on preventive quality management/control rather than inspecting already completed parts. The company may want to consider investing more in R&D.

4.If implemented well, this strategy will enable the company to become more competitive in the long run.

Case: Tiger Tools

1. For the first data set = .873. From Table 102, for n = 20, A2 = .18. Using the hint, the estimated standard deviation is .234:

Rearranging terms, we have

Solving, we obtain

The process capability is Because this is less than 1.33, the process is not capable.

2. The process seems to be cycling, as indicated by the control chart for the smaller sample size. Taking large samples probably resulted in combining the results of several different process means, and therefore did not reveal the changes that were occurring. By taking smaller samples more frequently, the pattern was easier to discern.

Control chart for n = 20:

Control chart for n = 5:

3.If the cycling can be removed. The true process standard deviation is probably much smaller than the apparent process standard deviation. For the second data set, R = .411. From Table 102, A2=.58. Performing the same calculations as in #1, we obtain an estimated standard deviation of .178.

The process capability is Because this is more than 1.33, the process is capable.

4.Small samples tend to be less reliable than large samples (the standard deviation of the sampling distribution of means decreases as the sample size increases). Also, a manager must weigh the cost of inspecting each item and cost of taking a sample. If the cost to obtain a sample is high, but the cost to inspect an item is low, larger samples might be the better choice.

*

(

(

(

(

(liters)

Mean

out

out

LCL

UCL

1.006

1.0043

1.002

1.000

.998

.9957

.994

-2.5 0+2.5z-scale

24 24.52516

.0062

.0062

b.

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3.44

3.0 Cm

3 sigma

3 sigma

n = 1

( = 0.01 cm

UCL

Mean

LCL

(

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24 6 8 101214 1618

LCL = 4.86

sample number

UCL = 5.16

UCL = 5.24

LCL = 4.76

Sample number

246 81012 141618 20 22 2426

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

(

EMBED Equation.3

EMBED Equation.3

1274Operations Management, 9/e

273Instructors Manual, Chapter 10

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