1. Quality and Performance Chapter 5
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What is Quality?
Quality
A term used by customers to describe their general satisfaction with a service or product.
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Costs of Quality
Prevention Costs
Appraisal Costs
Internal Failure Costs
External Failure Costs
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Ethics and Quality
Balancing the traditional measures of quality performance and the overall benefits to society.
Identifying deceptive business practices.
Developing a culture around ethics.
Training employees to understand how ethics interfaces with their jobs.
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5. Total Quality Management
TQM
A philosophy that stresses principles for achieving high levels of process performance and quality.
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6. Customer Satisfaction
Conformance to Specifications
Value
Fitness for Use
Support
Psychological Impressions
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Employee Involvement
Cultural Change
Teams
Employee Empowerment
Problem-solving teams
Special-purpose teams
Self-managed teams
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8. Continuous Improvement
Kaizen
Problem-solving
tools
PDSA Cycle
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What is Six Sigma?
Six Sigma
A comprehensive and flexible system for achieving, sustaining, and maximizing business success by minimizing defects and variability in processes.
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10. Six Sigma Approach Process average OK; too much variation
Process variability OK; process off target X X
Process on target with low variability
Reduce spread
Center process X
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11. Six Sigma Improvement Model
Six Sigma Certification
Master Black Belts
Black Belts
Green Belts
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Acceptance Sampling
Acceptance Sampling
The application of statistical techniques to determine if a quantity of material from a supplier should be accepted or rejected based on the inspection or test of one or more samples.
Acceptable Quality Level
A statement of the proportion of defective items that the buyer will accept in a shipment.
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13. Acceptance Sampling Interface
Firm A uses TQM or Six Sigma to achieve internal process performance
Buyer
Manufactures furnaces
Fan motors
Supplier uses TQM or Six Sigma to achieve internal process performance
Firm A
Manufacturers furnace fan motors
TARGET: Buyer’s specs
Motor inspection
Fan blades
Accept motors?
Supplier
Manufactures fan blades
TARGET: Firm A’s specs
Blade inspection
Yes No
Accept blades?
Yes No
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14. Statistical Process Control (SPC)
The application of statistical techniques to determine whether a process is delivering what the customer wants.
Performance Measurements
Variables - Characteristics that can be measured.
Attributes - Characteristics that can be counted.
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Sampling
Sampling Plan
Size of the sample
Time between successive samples
Decision rules that determine when action should be taken
Complete Inspection
Inspect each product at each stage
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Sampling Statistics
Sample Mean
Sum of the observations divided by the total observations
Sample Range
Difference between the largest and smallest observation in a sample
where
xi = observation of a quality characteristic (such as time)
n = total number of observations
= mean
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Sampling Statistics
Standard deviation– The square root of the variance of a distribution. An estimate of the process standard deviation based on a sample is given by:
where
σ = standard deviation of a sample
xi = observation of a quality characteristic (such as time)
n = total number of observations
= mean
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18. Sampling Statistics
The sample mean is the sum of the observations divided by the total number of observations.
where
xi = observation of a quality characteristic (such as time)
n = total number of observations
x = mean
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19. Sampling Statistics
The range is the difference between the largest observation in a sample and the smallest. The standard deviation is the square root of the variance of a distribution. An estimate of the process standard deviation based on a sample is given by
where
σ = standard deviation of a sample
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20. Sampling Distribution
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Types of Variation
Common cause
Variation that is random, unidentifiable and unavoidable
Assignable cause
Variation that can be identified and eliminated
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22. Effects of Assignable Cause Variation on the Process Distribution
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Control Charts
Time-ordered diagram used to determine whether observed variations are abnormal
Mean
Upper control limit
Lower control limit
Steps for a control chart
Random sample
Plot statistics
Eliminate the cause, incorporate improvements
Repeat the procedure
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24. Control Limits and Sampling Distribution
UCL
Nominal
LCL 1 2
Assignable
causes likely 3
Samples
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25. Control Charts UCL Variations Nominal LCL Sample number
(a) Normal – No action
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26. Control Charts UCL Variations Nominal LCL Sample number
(b) Run – Take action
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27. Control Charts UCL Variations Nominal LCL Sample number
(c) Sudden change – Monitor
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28. Control Charts UCL Variations Nominal LCL Sample number
(d) Exceeds control limits – Take action
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Control Chart Errors
Type I error – Concluding that a process is out of control when it is in control
Type II error –
Concluding that a process is in control when it is out of control
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Control Chart Types
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31. Variable Control Charts
R-Chart
UCLR = D4R and LCLR = D3R
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32. Variable Control Charts
UCLx = x + A2R and LCLx = x – A2R
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33. Calculating Control Chart Factors
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34. Steps for x- and R-Charts
Collect data.
Compute the range.
Use Table 5.1 to determine R-chart control limits.
Plot the sample ranges. If all are in control, proceed to step 5. Otherwise, find the assignable causes, correct them, and return to step 1.
Calculate x for each sample.
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35. Steps for x- and R-Charts
Use Table 5.1 to determine x-chart control limits
Plot the sample means. If all are in control, the process is in statistical control. Continue to take samples and monitor the process. If any are out of control, find the assignable causes, correct them, and return to step 1.
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Example 5.1
The management of West Allis Industries is concerned about the production of a special metal screw used by several of the company’s largest customers. The diameter of the screw is critical to the customers. Data from five samples appear in the accompanying table. The sample size is 4. Is the process in statistical control?
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Example 5.1
Compute the range for each sample and the control limits
UCLR = D4R =
2.282(0.0021) = in.
LCLR = D3R =
0(0.0021) = 0 in.
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38. Example 5.1 Process variability is in statistical control.
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Example 5.1
Compute the mean for each sample and the control limits.
(0.0021) = in.
– 0.729(0.0021) = in.
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40. Example 5.1 Process average is NOT in statistical control.
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41. An Alternate Form
If the standard deviation of the process distribution is known, another form of the x-chart may be used:
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Example 5.2
For Sunny Dale Bank the time required to serve customers at the drive-by window is an important quality factor in competing with other banks in the city.
After several weeks of sampling, two successive samples came in at 3.70 and 3.68 minutes, respectively. Is the customer service process in statistical control?
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43. Example 5.2
For Sunny Dale Bank the time required to serve customers at the drive-by window is an important quality factor in competing with other banks in the city.
After several weeks of sampling, two successive samples came in at 3.70 and 3.68 minutes, respectively. Is the customer service process in statistical control?
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44. Example 5.2 x = 5 minutes σ = 1.5 minutes n = 6 customers z = 1.96
The process variability is in statistical control, so we proceed directly to the x-chart. The control limits are
UCLx = x + zσ/n =
(1.5)/6 = 6.20 minutes
LCLx = x – zσ/n =
5.0 – 1.96(1.5)/6 = 3.80 minutes
The new process is an improvement.
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Application 5.1
Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped.
Webster is concerned whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes are taken and each tube is weighed in ounces.
Tube Number
Sample 1 2 3 4 5 6 7 8
Avg
Range
7.98
8.34
8.02
7.94
8.44
7.68
7.81
8.11
8.040
0.76
8.23
8.12
8.41
8.31
8.18
7.99
8.06
8.160
0.43
7.89
7.77
7.91
8.04
8.00
7.93
8.09
7.940
0.32
8.24
7.83
8.05
7.90
8.16
7.97
8.07
8.050
0.41
7.87
8.13
7.92
8.10
8.14
7.88
7.980
0.33
8.130
0.03
Avgs
0.38
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46. Application 5.1
Assuming that taking only 6 samples is sufficient, is the process in statistical control?
1.864(0.38) = 0.708
0.136(0.38) = 0.052
The range chart is out of control since sample 1 falls outside the UCL and sample 6 falls outside the LCL.
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47. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Application 5.1
Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures.
Tube Number
Sample 1 2 3 4 5 6 7 8
Avg
Range
7.98
8.34
8.02
7.94
8.44
7.68
7.81
8.11
8.040
0.76
8.23
8.12
8.41
8.31
8.18
7.99
8.06
8.160
0.43
7.89
7.77
7.91
8.04
8.00
7.93
8.09
7.940
0.32
8.24
7.83
8.05
7.90
8.16
7.97
8.07
8.050
0.41
7.87
8.13
7.92
8.10
8.14
7.88
7.980
0.33
Avgs
8.034
0.45
What is the conclusion on process variability and process average?
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Application 5.1
1.864(0.45) = 0.839
0.136(0.45) = 0.061
(0.45) = 8.202
8.034 – 0.373(0.45) = 7.866
The resulting control charts indicate that the process is actually in control.
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49. Control Charts for Attributes
p-charts are used to control the proportion defective
Sampling involves yes/no decisions so the underlying distribution is the binomial distribution
The standard deviation is
p = the center line on the chart
and
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Example 5.3
Hometown Bank is concerned about the number of wrong customer account numbers recorded. Each week a random sample of 2,500 deposits is taken and the number of incorrect account numbers is recorded
Using three-sigma control limits, which will provide a Type I error of 0.26 percent, is the booking process out of statistical control?
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51. Example 5.3 Sample Number Wrong Account Numbers 1 15 7 24 2 12 8 3 1910 4 17 5 11 6
Total
147
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52. Total number of observations
Example 5.3
p =
Total defectives
Total number of observations
147
12(2,500)
= =
σp = p(1 – p)/n
= 0.0049(1 – )/2,500 =
UCLp = p + zσp
= (0.0014) =
LCLp = p – zσp
= – 3(0.0014) =
Calculate the sample proportion defective and plot each sample proportion defective on the chart.
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53. Example 5.3 The process is NOT in statistical control. X .0091
Fraction Defective
Sample
Mean
UCL
LCL
.0091
.0049
.0007
| | | | | | | | | | | | X
The process is NOT in statistical control.
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54. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Application 5.2
A sticky scale brings Webster’s attention to whether caulking tubes are being properly capped. If a significant proportion of the tubes aren’t being sealed, Webster is placing their customers in a messy situation. Tubes are packaged in large boxes of 144. Several boxes are inspected and the following numbers of leaking tubes are found:
Sample
Tubes 1 3 8 6 15 5 2 9 4 16 10 17 11 18 12 19 13 20 7 14
Total = 72
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Application 5.2
Calculate the p-chart three-sigma control limits to assess whether the capping process is in statistical control.
The process is in control as the p values for the samples all fall within the control limits.
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56. Control Charts for Attributes
c-charts count the number of defects per unit of service encounter
The underlying distribution is the Poisson distribution
UCLc = c + zc and LCLc = c – zc
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Example 5.4
The Woodland Paper Company produces paper for the newspaper industry. As a final step in the process, the paper passes through a machine that measures various product quality characteristics. When the paper production process is in control, it averages 20 defects per roll.
a. Set up a control chart for the number of defects per roll. For this example, use two-sigma control limits.
b. Five rolls had the following number of defects: 16, 21, 17, 22, and 24, respectively. The sixth roll, using pulp from a different supplier, had 5 defects. Is the paper production process in control?
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58. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Example 5.4
a. The average number of defects per roll is 20. Therefore
= (20) = 28.94
LCLc = c – zc
= 20 – 2(20) = 11.06
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59. Example 5.4 b.
The process is technically out of control due to Sample 6. However, Sample 6 shows that the new supplier is a good one.
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Application 5.3
At Webster Chemical, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Even when the process is in control, there will still be an average of 4 lumps per tube of caulk. Testing for the presence of lumps destroys the product, so Webster takes random samples. The following are results of the study:
Tube #
Lumps 1 6 5 9 2 4 10 3 7 11 8 12
Determine the c-chart two-sigma upper and lower control limits for this process.
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Application Problem 5.3
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62. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Process Capability
Process Capability – The ability of the process to meet the design specification for a service or product
Nominal Value
Tolerance
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63. Process Capability Nominal value Process distribution Lower Upper20 25 30
Minutes
Upper
specification
Lower
Nominal
value
Process distribution
(a) Process is capable
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64. Process Capability Nominal value Process distribution Upper Lower20 25 30
Minutes
Upper
specification
Lower
Nominal
value
Process distribution
(b) Process is not capable
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65. Process Capability Nominal value Six sigma Four sigma Two sigma Upper
Lower
specification
Mean
Upper
Nominal value
Six sigma
Four sigma
Two sigma
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66. Process Capability Index
Measures how well a process is centered and whether the variability is acceptable
Cpk = Minimum of ,
x – Lower specification 3σ
Upper specification – x
where
σ = standard deviation of the process distribution
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67. Process Capability Ratio
A test to see if the process variability is capable of producing output within a product’s specifications.
Cp =
Upper specification – Lower specification 6σ
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68. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Example 5.5
The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.35 minutes.
The nominal value for this service is 25 minutes + 5 minutes.
Is the lab process capable of four sigma-level performance?
Upper specification =
30 minutes
Lower specification
20 minutes
Average service
26.2 minutes
 =
1.35 minutes
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Cpk = Minimum of ,
Example 5.5
Cpk = Minimum of – 20 , 30 – 26.2
3 ( 1.53) ( 1.53)
Cpk = Minimum of , 0.94
Process does not meets 4-sigma level of 1.33
Cpk = 0.94
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Example 5.5
Cp = Upper Specification – Lower Specification 6
Cp = 30 – = 1.23
6 (1.35)
Process did not meet 4-sigma level of 1.33
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71. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Example 5.5
New Data is collected:
Upper specification =
30 minutes
Lower specification
20 minutes
Average service
26.1 minutes
 = 1.20 minutes
Cp = Upper - Lower 6
Cp = 30 – = 1.39
6 (1.20)
Process meets 4-sigma level of 1.33 for variability
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72. x – Lower specification Upper specification – x
Example 5.5
Cpk = Minimum of ,
x – Lower specification 3σ
Upper specification – x
Cpk = Minimum of – , 30 – 26.1
3 ( 1.20) ( 1.20)
Cp = 1.08
Process does not meets 4-sigma level of 1.33
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Application 5.4
Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces ± 0.60 ounces. The target process capability ratio is 1.33, signifying that management wants 4-sigma performance. The current distribution of the filling process is centered on ounces with a standard deviation of 0.192 ounces. Compute the process capability index and process capability ratio to assess whether the filling process is capable and set properly.
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74. x – Lower specification Upper specification – x
Application 5.4
a. Process capability index:
Cpk = Minimum of ,
x – Lower specification 3σ
Upper specification – x
= Minimum of = 1.135, = 0.948
8.054 – 7.400
3(0.192)
8.600 – 8.054
The value of is far below the target of Therefore, we can conclude that the process is not capable.
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75. Upper specification – Lower specification
Application 5.4
b. Process capability ratio:
Cp =
Upper specification – Lower specification 6σ
= =
8.60 – 7.40
6(0.192)
The value of Cp is less than the target for four-sigma quality.
Therefore we conclude that the process variability must be addressed first, and then the process should be retested.
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76. Quality Loss Function Loss (dollars) Lower Nominal Upper
specification value specification
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77. International Quality Documentation Standards
ISO 9001:2008 – Quality Standards
ISO 14000:2004 – Environmental Management Standards
ISO 26000:2010 – Social Responsibility Guidelines
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78. Baldridge Performance Excellence Program
Leadership
Strategic Planning
Customer Focus
Measurement, Analysis, and Knowledge Management
Workforce Focus
Operations Focus
Results
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79. Solved Problem 1
The Watson Electric Company produces incandescent light bulbs. The following data on the number of lumens for 40-watt light bulbs were collected when the process was in control.
Observation
Sample 1 2 3 4
604
612
588
600
597
601
607
603
581
570
585
592
620
605
595 5
590
614
608
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80. Solved Problem 1 604 + 612 + 588 + 600 4 = 601 x = R = 612 – 588 = 24 4
= 601
x =
R =
612 – 588 = 24
Sample R 1
601 24 2
602 10 3
582 22 4 32 5
604
Total
2,991
112
Average
x = 598.2
R = 22.4
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81. Solved Problem 1 The R-chart control limits are 2.282(22.4) = 51.12
2.282(22.4) = 51.12
0(22.4) = 0
The x-chart control limits are
(22.4) =
598.2 – 0.729(22.4) =
b. First check to see whether the variability is still in control based on the new data. The range is 53 (or 623 – 570), which is outside the UCL for the R-chart. Since the process variability is out of control, it is meaningless to test for the process average using the current estimate for R. A search for assignable causes inducing excessive variability must be conducted.
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82. Solved Problem 2
The data processing department of the Arizona Bank has five data entry clerks. Each working day their supervisor verifies the accuracy of a random sample of 250 records. A record containing one or more errors is considered defective and must be redone. The results of the last 30 samples are shown in the table. All were checked to make sure that none was out of control.
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83. Solved Problem 2 Sample Number of Defective Records 1 7 16 8 2 5 17 123 19 18 4 10 6 11 20 21 22 9 23 24 13 25 26 27 14 28 29 15 30
Total 300
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84. Number of Defective Records
Solved Problem 2
a. Based on these historical data, set up a p-chart using z = 3.
b. Samples for the next four days showed the following:
Sample
Number of Defective Records
Tues 17
Wed 15
Thurs 22
Fri 21
What is the supervisor’s assessment of the data-entry process likely to be?
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85. Solved Problem 2 SOLUTION
a. From the table, the supervisor knows that the total number of defective records is 300 out of a total sample of 7,500 [or 30(250)]. Therefore, the central line of the chart is
= 0.04
300
7,500
p =
The control limits are
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86. Number of Defective Records
Solved Problem 2
b. Samples for the next four days showed the following:
Sample
Number of Defective Records
Proportion
Tues 17
0.068
Wed 15
0.060
Thurs 22
0.088
Fri 21
0.084
Samples for Thursday and Friday are out of control. The supervisor should look for the problem and, upon identifying it, take corrective action.
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87. Solved Problem 3
The Minnow County Highway Safety Department monitors accidents at the intersection of Routes 123 and 14. Accidents at the intersection have averaged three per month.
a. Which type of control chart should be used? Construct a control chart with three sigma control limits.
b. Last month, seven accidents occurred at the intersection. Is this sufficient evidence to justify a claim that something has changed at the intersection?
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88. Solved Problem 3 SOLUTION
a. The safety department cannot determine the number of accidents that did not occur, so it has no way to compute a proportion defective at the intersection. Therefore, the administrators must use a c-chart for which
UCLc = c + z c
= = 8.20
LCLc = c – z c
= 3 – = –2.196
There cannot be a negative number of accidents, so the LCL in this case is adjusted to zero.
b. The number of accidents last month falls within the UCL and LCL of the chart. We conclude that no assignable causes are present and that the increase in accidents was due to chance.
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89. Solved Problem 4
Pioneer Chicken advertises “lite” chicken with 30 percent fewer calories. (The pieces are 33 percent smaller.) The process average distribution for “lite” chicken breasts is 420 calories, with a standard deviation of the population of 25 calories. Pioneer randomly takes samples of six chicken breasts to measure calorie content.
a. Design an x-chart using the process standard deviation.
b. The product design calls for the average chicken breast to contain 400 ± 100 calories. Calculate the process capability index (target = 1.33) and the process capability ratio. Interpret the results.
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90. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Solved Problem 4
SOLUTION
a. For the process standard deviation of 25 calories, the standard deviation of the sample mean is
(10.2) = calories
420 – 3(10.2) = calories
Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.

91. Upper specification – Lower specification
Solved Problem 4
b. The process capability index is
Cpk = Minimum of ,
= Minimum of = 1.60, = 1.07
420 – 300
3(25)
500 – 420
The process capability ratio is
Cp =
Upper specification – Lower specification 6σ
= = 1.33
500 – 300
6(25)
Because the process capability ratio is 1.33, the process should be able to produce the product reliably within specifications. However, the process capability index is 1.07, so the current process is not centered properly for four-sigma performance. The mean of the process distribution is too close to the upper specification.
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Printed in the United States of America.
Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.