Statistical Quality Control Test Bank Chapter 3

Chapter 3:
Statistical Quality Control

True/False

1. ____________involves monitoring and controlling a process to prevent poor quality.
   1. Process capability measures
   2. Statistical process control
   3. Pattern tests
   4. Control charts
2. Statistical process control is based on a philosophy of inspection as opposed to prevention.
3. One goal of statistical process control is to prevent a process from producing items that have to be scrapped or reworked.
4. Two types of variation associated with the output of a process are common cause (random) and special cause (nonrandom).
5. Control limits are based on the special cause (nonrandom) variation inherent in a process.
6. A process that is determined to be in control contains no variation.
7. Common cause (random) variation provides evidence that the process is not in control.
8. After ______________ is detected, the focus changes to identifying the root cause of the variation and eliminating it.
   1. a C_pk > 1
   2. a C_pk < 1
   3. common cause variation
   4. special cause variation
9. Process control is achieved by taking periodic samples from a process and plotting the sample points on a chart to determine if the process is within control limits.
10. When a control chart detects no special cause (nonrandom) variation in a process, the upper and lower control limits are the same value.
11. It is sometimes not necessary to determine new control limits after special cause (nonrandom) variation has been identified if the source has been eliminated without changing the process.
12. When Cpk differs from Cp it indicates the
    1. tolerance has changed.
    2. control limit has changed.
    3. process mean is centered.
    4. the process is mean is off center.
13. With a c-chart, the sample size is small and may contain only one item.
14. A ______ is used to monitor the proportion defective in the output of a process.
    1. c-chart
    2. p-chart
    3. -chart
    4. range chart
15. control charts are used to monitor descriptive characteristics of the output of a process rather than measurable characteristics.
    1. Attribute
    2. Variable
    3. Pattern test
    4. Range
16. The formula used to determine the upper and lower control limits are based on product specification limits.
17. Variable control charts are used to monitor measurable characteristics of a process’s outputs rather than descriptive characteristics.
18. An x-bar and R-chart constructed to monitor and control a process use the same raw data.
19. Variable control charts are used for quantitative measures such as weight or time.
20. Construction and use of an -chart is based on an assumption that the sample points are normally distributed around the centerline.
21. The _______ is the difference between the smallest and largest values in a sample.
    1. attribute
    2. mean
    3. range
    4. variation
22. The range measures the variation within samples versus the variation between samples.
23. It is possible to have low variation within samples while at the same time having high variation between sample means.
24. One advantage of using a pattern test is that special cause variations may be identified before any points are plotted outside the control limits.
25. A pattern test can identify an out-of-control process even if all sample points are within control limits.
26. A control chart is in control when the plot of the sample points exhibits a pattern.
27. If the points plotted on a control chart display a pattern, it is called a
    1. pattern test.
    2. run.
    3. Trend.
    4. slope.
28. When constructing a control chart for the first time, all points should be within the control limits indicating the process is in control.

1. Process control charts are often used at a critical point after which it is difficult to correct or rework the process output.
2. Control charts visually show when a process is not within statistical control limits.
3. The popularity of Excel and other data analysis software has been a major factor in the increased use of statistical process control.
4. Tolerances or specification limits are allowable variation prescribed in a product design.
5. ________ reflect the amount of common cause variation allowed in a process.
   1. Control limits
   2. Tolerances
   3. Process capability
   4. Statistical process control
6. For a given process, the process capability ratio is not related to its specification limits.
7. A process capability ratio __________ shows that a process is capable of producing output within its specification limits.
   1. of zero
   2. less than one
   3. equal to one
   4. greater than one
8. All processes contain a certain amount of variation in their output.
9. A sequence of sample points that display a pattern is known as a run.
10. Statistical process control can prevent poor quality before it occurs if a pattern is evident in the plotted points.
11. The ___________ indicates how much a process mean differs from the target specification value.
    1. -chart
    2. range chart
    3. process capability ratio
    4. process capability index

Multiple Choice Questions

1. If a sample point plotted on a control chart is outside the control limits
2. the evidence indicates the process is in control.
3. the evidence indicates the process is out of control.
4. the evidence is inconclusive.
5. None of these answer choices is correct.
6. Which of the following could be responsible for variability that is special cause (nonrandom)?
7. Broken machinery.
8. Defective parts and materials.
9. Operator error.
10. All these answer choices are correct.
11. Common (random) variation of a process depends on all the following except
12. errors due to lack of training.
13. the equipment and machinery used.
14. the operator.
15. system used for measurement.
16. An attribute measure is a product characteristic such as
17. weight.
18. color.
19. length.
20. time.
21. A variable measure is a product characteristic such as
22. color.
23. smoothness.
24. temperature
25. good taste
26. Which of the following services can be measured and monitored with control charts?
27. Hospitals
28. Airlines
29. Banks
30. All these answer choices are correct.
31. Control charts are typically used at the ___________ of a process.
32. beginning.
33. middle.
34. end.
35. All these answer choices are correct.
36. Which of the following is not a primary purpose of statistical process control?

a. to establish control limits

b. to detect special cause variations

c. to identify specification limits

d. to determine when a process is not in control

1. Four common types of control charts include all of the following except:

a. -chart

b. t-chart

c. p-chart

d. c-chart

1. Which of the following is not a characteristic of a control chart?

a. the centerline is determined using special cause (nonrandom) variations.

b. the upper and lower control limits are based on special cause (nonrandom) variation.

c. the centerline is determined by using the target value.

d. None of these answer choices is correct.

1. Special cause (nonrandom) variation in a process is more likely to be detected with

a. wider control limits

b. narrow control limits

c. wider specification limits

d. narrow specification limits

1. Which of the following statements concerning control chart limits is true?
2. the smaller the value of z, the more narrow the control limits are and the more sensitive the chart is to changes in the production process
3. the larger the value of z, the more narrow the control limits are and the more sensitive the chart is to changes in the production process
4. the smaller the value of z, the wider the control limits are and the less sensitive the chart is to changes in the production process
5. the larger the value of z, the more narrow the control limits are and the less sensitive the chart is to changes in the production process
6. The basic purpose of control charts include(s)
7. establishing control limits for a process.
8. monitoring the process to indicate when it is out of control.
9. both establishing control limits for a process and monitoring the process to indicate when it is out of control are basic purposes.
10. None of these answer choices is correct.
11. The formulas for determining the upper and lower control limits are based on the number of standard deviations, z, from the process average. Management usually selects a z value of _______.
12. one
13. two
14. three
15. six

1. When a control chart is first developed, if the process is found to be out of control,
2. the control chart can be utilized.
3. the control chart should not be utilized until more samples are taken,
4. the process should be examined and corrections made before a new control chart is constructed.
5. the process should be replaced by a new process.

55. A control chart that uses the actual number of defects per item to monitor a process is known as a

1. p-chart
2. c-chart
3. R-chart
4. -chart

56. If a sample of 40 units of output found 500 defects, then the center line for monitoring the average number of defects per unit of output would be

a. = 40

b. = 0.08

c. = 12.5

d. = 20,000

Solution: centerline = 500/40

57. If a sample of 40 units of output found 500 defects, then the 3-sigma upper control limit for the chart would be

1. 12.50
2. 23.11
3. 37.50
4. 75.00

Solution: UCL = 12.5 + 3 × SQRT(12.5) = 23.11

58. A company randomly selects 100 light bulbs every day for 40 days from its production process. If 600 defective light bulbs are found in the sampled bulbs then the estimate for the average percent defective would be

1. 6.667
2. 0.167
3. 0.150
4. 0.250

Solution: process average = 600/(100 × 40) = 0.15

59. A company randomly selects 100 light bulbs every day for 40 days from its production process. If 600 defective light bulbs are found in the sampled bulbs then the 3-sigma lower control limit would be

1. 0.0357
2. 0.0429
3. 0.1500
4. 0.1857

Solution: LCL = .15 – 3 × SQRT(.15 × .85/100) = .0429

60. Which of the following control charts is based on the number of defects within a sample?

a.

b. R

c. c

d. p

61. Which of the following control charts is used to monitor the percent of defective items within a sample?

a.

b. R

c. c

d. p

62. If the quality of a process’s output is determined by classifying the output as being defective or not defective, use a(n) ________control chart.

a.

b. R

c. c

d. p

63. Which of the following control charts are based on sample sizes as small as one?

a.

b. R

c. c

d. p

64. Which of the following control charts are often based on sample sizes equal to or larger than one hundred?

a.

b. R

c. c

d. p

1. Consider a production process that produces batteries. A quality engineer has taken 20 samples each containing 100 batteries. The total number of defective batteries observed over the 20 samples is 200. The centerline for the control chart constructed using z equal to two is
2. 0.03
3. 0.04
4. 0.05
5. 0.10

Solution: 2100/(20 × 100) = 0.10

1. Consider a production process that produces batteries. A quality engineer has taken 20 samples each containing 100 batteries. The total number of defective batteries observed over the 20 samples is 200. The sample standard deviation is
2. 0.03
3. 0.04
4. 0.05
5. 0.10

Solution: SQRT(.10 × .90/100) = .03

1. Consider a production process that produces batteries. A quality engineer has taken 20 samples each containing 100 batteries. The total number of defective batteries observed over the 20 samples is 200. The UCL for the control chart constructed using two sigma is
2. 0.088
3. 0.094
4. 0.104
5. 0.160

Solution: UCL = .10 + 2 × .03 = .160

1. Consider a production process that produces batteries. A quality engineer has taken 20 samples each containing 100 batteries. The total number of defective batteries observed over the 20 samples is 200. The LCL for the control chat constructed using two sigma is
2. 0.01
3. 0.04
4. 0.12
5. 0.16

Solution: LCL = .10 – 2 × .03 = .04

1. Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The centerline for the control chart is
2. 0.08
3. 0.11
4. 0.16
5. 0.21

Solution: 340/(15 × 200) = 0.11

1. Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The sample standard deviation is
2. 0.005
3. 0.011
4. 0.022
5. 0.028

Solution: SQRT(.11 × .89/200) = .022

1. Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The UCL for the control chart constructed using two sigma is
2. 0.025
3. 0.094
4. 0.122
5. 0.154

Solution: UCL = .11 + 2 × .022 = .154

1. Consider a production process that produces tires. A quality engineer has taken 15 samples, each containing 200 tires. The total number of defective tires over the 15 samples is 340. The LCL for the control chat constructed using two sigma is
2. 0.022
3. 0.036
4. 0.048
5. 0.066

Solution: LCL = .11 – 2 × .022 = .066

1. Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. What type of control chart would you recommend?
   1. p-chart
   2. c-chart
   3. chart
   4. R chart
2. Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The centerline for the control chart constructed is
3. 5
4. 10
5. 15
6. 20

Solution: centerline=200/20=10

1. Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The standard deviation for the control chart is
2. 2.2
3. 3.2
4. 3.9
5. 4.5

Solution: standard deviation = Sqrt(10) = 3.2

1. Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The UCL for the control chart constructed using three sigma is
2. 19.6
3. 26.7
4. 33.5
5. 50.0

Solution: UCL = 10 + 3 × 3.2 = 19.6

1. Easy Tax is a service company that prepares tax returns. An outside auditor has examined 20 samples each containing one completed tax return. The total number of defects observed over the 20 samples is 200. The LCL for the control chart constructed using three sigma is
2. -1.6
3. 0.4
4. 3.3
5. 6.5

Solution: LCL = 10 − 3 × 3.2 = 0.4

1. Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below:

Given the sample information above, the average number of defects per unit for this process would be

• 1. 160
  2. 80
  3. 16
  4. 10

Solution: average number of defects = (17 + 19 + 5 + 18 + 16 + 14 + 15 + 16 + 14 + 15 + 15)/10 = 16

1. Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below:

Given the sample information above, the standard deviation of the number of defects for this process would be

1. 16
2. 10
3. 4
4. 0

Solution: Standard deviation = SQRT(16)

1. Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below:

Given the sample information above, the UCL using sigma = 3 for this process would be

1. 36
2. 32
3. 30
4. 28

Solution: UCL = 16 + 3 × 4 = 28

1. Marble Inc. makes countertops from a variety of high-end materials. To monitor the quality of its production processes the company randomly selects one countertop and counts the number of blemishes. The results for ten samples are shown below:

Given the sample information above, the LCL using sigma = 3 for this process would be

1. 12
2. 8
3. 4
4. 0

Solution: UCL = 16 − 3 × 4 = 4

1. A control chart that reflects the amount of variation, or spread, present within each sample is known as a(n)
2. p-chart
3. c-chart
4. R-chart
5. -chart
6. Which of the following control charts is used to control the variation within samples?

a. -chart

b. R chart

c. c-chart

d. p-chart

1. Which of the following control charts is used to control the variation between samples?

a. -chart

b. R-chart

c. c-chart

d. p-chart

1. Which of the following charts are frequently used together to monitor and control quality?

a. p and

b. R and p

c. c and R

d. R and

1. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, what is the centerline of an R-chart for this process?
2. .50
3. .85
4. 1.20
5. 1.55

Solution: centerline = 12.68/15 = .85

1. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the UCL for an R-chart of this process would be
2. 0.0
3. 1.0
4. 1.7
5. 2.4

Solution: UCL = 2 × .85 = 1.7

1. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the UCL for an R-chart of this process would be
2. 0.0
3. 1.0
4. 1.7
5. 2.4

Solution: LCL = 0 × .85 = 0.0

1. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the centerline for an X-bar chart of this process would be
2. 4.75
3. 5.00
4. 5.35
5. 5.69

Solution: centerline = 80.20/15 = 5.35

1. 90. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the UCL for an X-bar chart of this process would be5.00
2. 5.35
3. 5.76
4. 6.45

Solution: UCL = 5.35 + .48 × .85 = 5.76

1. Pizazz manufactures a 5.0 oz. energy drink of the same name. Because the cans are so small, consumers are concerned that they are not receiving the full 5 ounces in each can. A quality engineer at the company is charged with analyzing the filling process and ensuring accurate readings. On 15 different occasions over the past month, she has taken a sample of 6 energy drinks off the production line and recorded their weight. If the sum of the sample means is 80.20 ounces and the sum of the sample ranges is 12.68 ounces, the LCL for an X-bar chart of this process would be
2. 4.50
3. 4.65
4. 4.79
5. 4.94

Solution: LCL = 5.35 − .48 × .85 = 4.94

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/- 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data from one restaurant appears below. If DTG were to construct an X-bar chart from this data, the centerline would be

1. 2.87
2. 3.70
3. 3.81
4. 4.28

Solution: Centerline = (4.38 + 4.18 + 3.46 + 4.13 + 2.83 + 3.93)/6 = 3.81

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/- 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data from one restaurant appears below. If DTG were to construct an X-bar chart from this data, the 3-sigma UCL would be

• 1. 6.44
  2. 5.87
  3. 3.50
  4. 2.82

Solution: UCL = 3.81 + .73 × 2.82 = 5.87

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/− 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data from one restaurant appears below. If DTG were to construct an X-bar chart from this data, the 3-sigma LCL would be

1. 0.00
2. 1.00
3. 1.76
4. 2.82

Solution: LCL = 3.81 – (.73 × 2.82) = 1.76

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/- 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data from one restaurant appears below. If DTG were to construct an R-chart from this data, the centerline would be

1. 1.00
2. 2.82
3. 3.54
4. 3.81

Solution: Centerline = (4.5 + 2.6 + 3.54 + 2.5 + 2.8 + 1)/6 = 2.82

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/- 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data from one restaurant appears below. If DTG were to construct an R-chart from this data, the 3-sigma UCL would be

• 1. 6.42
  2. 5.87
  3. 5.63
  4. 5.01

Solution: UCL = 2.28 × 2.82 = 6.42

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/- 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data from one restaurant appears below. If DTG were to construct an R-chart from this data, the 3-sigma LCL would be

1. 2.82
2. 1.41
3. 0.00
4. −1.41

Solution: UCL = 0 × 2.82 = 0

1. Dumplings –To-Go (DTG) provides take-out dumplings and noodle dishes to customers at its chain of drive-through restaurants. The target for a customer’s waiting time is 3.0 minutes +/- 1 minute. Each month, one of the managers observes the drive-through process and collects a sample of 4 waiting times a day over a 6 day period. The data for one restaurant appears below. Calculate the process capability ratio. Is this restaurant capable of meeting DTG standards?

1. yes
2. no
3. Cannot be determined from the above information

Solution: Cp = (4 − 2)/(5.87 − 1.76) = .486; No

99. In general, a process is considered to be in control for all the following conditions except

a. no points are outside the control limits

b. all points are above the centerline

c. the points are randomly distributed following a normal population

d. no pattern exists in the plotted points

100. A process is generally considered to be in control when

1. there are no sample points outside the control limits
2. most points are near the center line, without many being close to the control limits
3. sample points are randomly distributed equally above and below the center line
4. All of these answer choices are correct.

101. For the process to be capable of meeting design specification the process capability index must be

1. less than one (1.0)
2. equal to or greater than one (1.0)
3. less than zero (0.0)
4. equal to or greater than zero (0.0)
5. 102. A company produces a product which is designed to weigh 10 oz., with a tolerance of + 0.5 oz. The process produces products with an average weight of 9.95 oz. and a standard deviation of 0.10 oz. The process capability ratio for this process is 1.67
6. 0
7. 0.8333
8. -1.67

Solution: process capability ratio = 1.0/(6 × 0.1) = 1.67

103. A company produces a product which is designed to weigh 10 oz., with a tolerance of + 0.5 oz. The process produces products with an average weight of 9.95 oz. and a standard deviation of 0.10 oz. According to the process capability ratio is the process capable of meeting design specifications?

1. No, the process capability ratio is less than 1.0
2. Yes, the process capability ratio is less than 1.0
3. No, the process capability ratio is greater than 1.0
4. Yes, the process capability ratio is greater than 1.0

Solution: process capability ratio=1.0/(6 × 0.1) = 1.67

104. A company produces a product which is designed to weigh 10 oz., with a tolerance of + 0.5 oz. The process produces products with an average weight of 9.95 oz. and a standard deviation of 0.10 oz. The process capability index for this process is

1. 1.50
2. −1.50
3. 1.83
4. −1.83

Solution: Process capability index = min{(.55/(3 × .1), (45/(3 × .1)} = min{1.83,1.50}=1.50

105. A company produces a product which is designed to weigh 10 oz., with a tolerance of + 0.5 oz. The process produces products with an average weight of 9.95 oz. and a standard deviation of 0.10 oz. According to the process capability index

1. the process mean is off center having shifted to the right
2. the process mean is off center having shifted to the left
3. the process mean is centered on the design target
4. None of these answer choices is correct.

Solution: Process capability index=min{(.55/(3*.1),(45/(3*.1)}=min{1.83,1.50}=1.50

1. XYZ manufacturing has received an order to produce a rod 5 inches in diameter + .04 inch. In sample runs, the machine tool that will be making the rod has been able to produce rods with a mean diameter of 4.99 inches and a standard deviation of 0.011 inch. The process capability ratio for this process is
2. −1.44
3. −1.21
4. 1.21
5. 1.44

Solution: process capability ratio=0.08/(6*.011)=1.21

1. XYZ manufacturing has received an order to produce a rod 5 inches in diameter + .04 inch. In sample runs, the machine tool that will be making the rod has been able to produce rods with a mean diameter of 4.99 inches and a standard deviation of 0.011 inch. The process capability index for this process is
2. 0.91
3. 1.51
4. −0.91
5. −1.51

Solution: process capability index=min{0.03/(3*.011),(0.05/(3*.011)}=min{0.91,1.51}=1.5

1. XYZ manufacturing has received an order to produce a rod 5 inches in diameter + .04 inch. In sample runs, the machine tool that will be making the rod has been able to produce rods with a mean diameter of 4.99 inches and a standard deviation of 0.011 inch. Which of the following statements is true?
   1. The process is capable of meeting design spec but is off-center.
   2. The process is capable of meeting design specs and is on-center.
   3. The process in incapable of meeting design specs.

Solution: Cp < 1, so incapable.

Short Answer Questions

1. Briefly discuss attribute and variable quality measures.

Solution:

The quality of a product or service can be evaluated using either an attribute of the product or service or a variable measure. An attribute is a product characteristic such as color, surface texture, or perhaps smell or taste. Attributes can be evaluated quickly with a discrete response such as good or bad, acceptable or unacceptable, or yes or no. Even if quality specifications are complex and extensive, a simple attribute test might be used to determine if a product or service is or is not defective. A variable measure is a product characteristic that is measured on a continuous scale such as length, width, time, or temperature. Because a variable evaluation is the result of a measurement it is sometimes referred to as a quantitative classification method. An attribute evaluation is sometimes referred to as a qualitative classification, since the response is not measured. Because it is a measurement, a variable typically provides more information about the product that does an attribute.

1. Using control charts, how do we evaluate whether a process is in control?

Solution:

Sample points are plotted on the control chart and the chart is examined to determine if the process is in control. Generally, a process will be considered to be in control if (a) there are no sample points outside the control limits, (b) most points are near the process average or center line, without too many close to the control limits, (c) approximately equal numbers of sample points occur above and below the center line, and (d) the points appear to be randomly distributed around the center line. If any of these conditions are violated, the process may be out of control.

1. What is a c-chart and when is it used?

Solution:

A c-chart is a type of attribute control chart. A c-chart uses the actual number of defects per item in a sample. A c-chart is used when it is not possible to compute a proportion defective and the actual number of defects must then be used. For example, it is possible to count the number of blemishes on a painted surface but we cannot compute a proportion because the total number of possible blemishes is not known. In such a situation the number of blemishes would be monitored using a c-chart.

1. Why are and R-charts used together?

Solution:

When monitoring a variables characteristic, that is one that can be measured, it is possible for the process to lose control in terms of its central tendency (mean) and in terms of its variability (range). IN order for the process to be in control it must be in control with respect to its average and its variability. The two charts measure the process differently. It is possible for samples to have very narrow ranges, suggesting little process variability, but the sample averages might be beyond the control limits. Conversely, it is possible for sample averages to be in control, but the ranges might be very large. IN order to monitor both the mean and the variability of a process using a variable measured on a continuous scale the two charts must be used together.

1. What is the process capability ratio and how is it calculated?

Solution:

Process capability refers to the natural variation of a process relative to the variation allowed by the design specifications. In other words, how capable is the process of producing acceptable units according to the design specifications? Process control charts are used for process capability to determine if an existing process is capable of meeting design specifications. There are three main elements associated with process capability—process variability (the natural range of variation of the process), the process center line (mean), and the design specifications.

114. For a p-chart, UCL = 0.58 and LCL = 0.26. Then p = ______.

a) There is not enough information to determine p

b) 0.84

c) 0.58

d) 0.42

Solution: p = (UCL + LCL)/2 = 0.42

115. For a p-chart, UCL = 0.58 and LCL = 0.26. Then the proportion of compliant products is = ______.

a) There is not enough information to determine q

b) 0.16

c) 0.42

d) 0.58

Solution:1 − p = 1 − (UCL + LCL)/2 = 1 − 0.42 = 0.58

116. For a p-chart, UCL = 0.58 and LCL = 0.26. Then the overall percentage of compliant items is ______%.

a) There is not enough information to determine q

b) 58

c) 42

d) 16

Solution:1 − p = 1 − (UCL + LCL)/2 = 1 − 0.42 = 0.58 = 58%

117. For a p-chart, UCL = 0.58 and LCL = 0.26. The standard deviation of the proportions is ______.

a) There is not enough information to determine the standard deviation of the proportions

b) 0.16

c) 0.11

d) 0.05

Solution: σ = (UCL − LCL)/6 = = 0.05

118. For a p-chart, the standard deviation of the proportions is 0.066933. The number of items in each sample is 45. Then p is ______.

a) There is not enough information to determine p

b) 0.25

c) 0.28

d) 0.32

Solution: σ = 0.066933 = SQRT(p × (1 – p)/45) => 0.066933^2 × 45= p × (1 – p) → p = 0.28

119. For a p-chart, the standard deviation of the proportions is 0.084439. If p is 0.31, then the number of items in each sample is ______.

a) There is not enough information to determine the number of items in each sample

b) 28

c) 30

d) 35

Solution: σ = 0.084439 = SQRT(0.31 × (1 – 0.31)/n) → 0.084439^2 × n= 0.2139→ n = 30

120. For a p-chart, the number of items in each sample is 30. The maximum width (UCL – LCL) possible is ______.

a) There is not enough information to determine the number of items in each sample

b) 0.32

c) 0.43

d) 0.55

Solution: UCL − LCL = 6 × SQRT(p × (1 – p)/30), which is largest for p = 0.5 → max width = 6 × SQRT(0.25/30) = 0.55

121. A company produces electronic components designed to have a collector-emitted voltage of 30 V (volts), with a tolerance of + 0.2 V. The production process produces components with an average collector-emitted voltage of 30.1 V and a standard deviation of 0.15 V. The process capability index for this production process is ______.

a)

b)

c)

d)

Solution: Process capability index =
min{(30.1 – 29.8)/(3 × 0.15), (30.2 – 30.1)/(3 × 0.15)} = min{0.67, 0.22} = 0.22

122. A company produces electronic components designed to have a collector-emitted voltage of 30 V (volts), with a tolerance of + 0.2 V. The production process has a standard deviation of 0.15 V and a process capability index of 0.4. Then the average collector-emitted voltage of the units produced is ______.

a) 29.98

b) 30.01

c) 30.02

d) 30.04

Solution: Process capability index =
min{(x̅ – 29.8)/(3 × 0.15), (30.2 – x̅)/(3 × 0.15)} = 0.4 → x̅ = 30.02

123. A company produces electronic components designed to have a collector-emitted voltage of 30 V (volts), with a tolerance of + 0.2 V. The production process produces components with an average collector-emitted voltage of 30.04 V. The process capability index is 0.6. Then the standard deviation of the collector-emitted voltages of the units produced is ______.

a) 0.0889

b) 0.0079

c) 0.0071

d) 0.0069

Solution: Process capability index =
min{(30.04 – 29.8)/(3 × σ), (30.2 – 30.04)/(3 × σ)} =

min{(30.04 – 29.8), (30.2 – 30.04)}/(3 × σ) =  0.6 → 0.16/0.6 = 3 × σ → σ = 0.0889