Statistical Quality Control Pune University MCQs
Statistical Quality Control Pune University MCQs
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “A Brief History of Quality Control and Improvement – 1”.
1. Quality is ______ variability.
a) opposite of
b) proportional to
c) reciprocal of
d) synonym of
Answer: c
Explanation: Quality is defined as the fitness of a product for use. In normal circumstances, the fitness of a product for use increases with a reduction in variability. So, Quality is reciprocal of Variability.
2. Which one of these is a dimension of quality?
a) Performance
b) Hazard rate
c) Process Capability
d) Control limits
Answer: a
Explanation: Performance is a key component of quality. Hence, it is counted in the dimensions of quality by Garvin .
3. Performance of a product is _____
a) how long the product lasts
b) how easy it is to repair the product
c) how often the product fails
d) whether the product is capable of doing the intended job
Answer: d
Explanation: Performance of a manufactured product is defined as the capability of a product to do the job, for which it is designed.
4. Which one of these is not a component of quality?
a) Reliability
b) Durability
c) Acceptance sampling
d) Serviceability
Answer: c
Explanation: There are 8 components of quality which include reliability, durability, and serviceability. Acceptance sampling is not one of them.
5. How is the aesthetics of a product defined as?
a) How good the product performs its job
b) How good it looks
c) How fast the job of the product is completed
d) Whether the product is made exactly as the designer intended
Answer: b
Explanation: The visual appeal of the product is regarded as the aesthetics of the product. The looking of a product is the aesthetic dimension of quality.
6. What does reduced variability result in?
a) Increased failure rate
b) Decreased reliability
c) Fewer repair claims
d) Increased repair costs
Answer: c
Explanation: Reduction in variability removes harmful differences between product units. This means fewer failures, hence lesser repair claims.
7. Quality of a product is at its lowest when ____ quality component is neglected while it’s manufacturing.
a) performance
b) reliability
c) aesthetics
d) serviceability
Answer: a
Explanation: If a product does not carry out its job efficiently , it will be of no use. So performance is a key factor of quality of any product.
8. Quality characteristics are classified into variables and ________
a) constants
b) attributes
c) standards
d) specifications
Answer: b
Explanation: Quality characteristics are broadly divided into two parts. Variables and Attributes are the two parts of Quality characteristics.
9. Read the following sentences and choose the correct option.
Variables can only take discrete values while Attributes can take continuous values.
Variables can only take continuous values while Attributes take discrete value.
Attributes and variables are dependent on each other.
Width of an impeller blade is a variable but no. of defective blades in a lot is an attribute.
a) and are correct
b) and are incorrect
c) Only is correct
d) , , and
Answer: b
Explanation: Variable can only take continuous values. Attributes can only take discrete values. Plus, variables are independent of attributes.
10. Which one of these is physical CTQ characteristic?
a) Length
b) Appearance
c) Reliability
d) Color
Answer: a
Explanation: Length is a physical CTQ characteristic, whereas appearance and color are sensory CTQ characteristics. Reliability is a Time-Orientation CTQ characteristic.
11. What does CTQ characteristic mean?
a) Close to quality characteristic
b) Comparison to quality characteristic
c) Consumer to quality characteristic
d) Critical to quality characteristic
Answer: d
Explanation: CTQ characteristic is termed as Critical to Quality characteristics. All the other options are not any relevance to Statistical Quality Control.
12. Specifications are defined as ______
a) desired measurements of CTQ characteristics of whole product
b) actual measurements of CTQ characteristic of whole product
c) difference from desired measurements of CTQ characteristic of the components of product
d) desired measurements of CTQ characteristics of the product components
Answer: d
Explanation: For any manufactured product, the specifications are the desired measurements for the quality characteristics of the components, which make up the product.
13. Length of a rectangular box is _______
a) a continuous measurement, i.e. a variable
b) a discrete measurement, i.e. an attribute
c) a continuous measurement, i.e. an attribute
d) a discrete measurement, i.e. a variable
Answer: a
Explanation: Length of any object is a continuous value. As all continuous values are a variable, so will be length of the rectangular box.
This set of Statistical Quality Control Interview Questions and Answers focuses on “A Brief History of Quality Control and Improvement – 2”.
1. What is the “value of a measurement that corresponds to the desired value for that quality characteristic” called?
a) Nominal value
b) USL value
c) LSL value
d) Original value
Answer: a
Explanation: A target or desired value, for a product component to be manufactured, is said to be the Nominal value, which is lower than USL and higher than LSL.
2. USL is defined as _____
a) the highest positive allowable difference from the nominal value of a CTQ characteristic
b) the highest allowable value of a CTQ characteristic
c) the lowest difference between the highest allowable value and the nominal value of a CTQ characteristic
d) the highest difference between the highest allowable value and the lowest allowable value of a CTQ characteristic
Answer: b
Explanation: USL is defined as the highest tolerable value of a Critical-to-Quality characteristic. There is no mention of difference in the definition of USL.
3. Full form of LSL is _____
a) lowest safe limit
b) largest safe limit
c) lowest specification limit
d) lowest specified limit
Answer: c
Explanation: The smallest permissible value of a particular quality characteristic is described as the lowest specification limit or LSL of that particular quality characteristic.
4. Nonconformity is described as _______
a) a client not confirming a product deal
b) a defect in manufacturing process
c) a difference from the Nominal value of a specification
d) a specific type of failure
Answer: d
Explanation: Nonconforming products are defined as the products, which fail to meet one or more of its specifications. Nonconformity is defined as a specific type of failure.
5. Who gave the principles of “Scientific Management”?
a) Henry Ford
b) Mayo
c) Fayol
d) F. W. Taylor
Answer: d
Explanation: The principles of “Scientific Management” were given by Frederick W. Taylor, in the year of 1875. This affected the production standards the most.
6. Who introduced the Assembly Line concept?
a) W.S. Gosset
b) Henry Ford
c) R.A. Fisher
d) A.V. Feigenbaum
Answer: b
Explanation: The concept of assembly line was first introduced by Henry Ford. This refined the work methods. The goal of this concept was to improve productivity and quality.
7. Who was the man who first introduced the concept of Control charts in the arena of quality control?
a) Frank Gilbreth
b) Henry Grant
c) W. A. Shewhart
d) G. Taguchi
Answer: c
Explanation: In 1924, W. A. Shewhart introduced the control chart concept in a Bell Laboratories technical memorandum. This was an avant-garde step that made the producers thinking to improve their quality standards to the modern world.
8. Acceptance sampling methodology was first developed by _______
a) H. F. Dodge and H. G. Romig
b) Motorola
c) U.S. War department
d) W. E. Deming
Answer: a
Explanation: In 1928, acceptance sampling methodology was developed and introduced by H. F. Dodge and H. G. Romig at Bell Laboratories.
9. By whom was Six-sigma methodology first developed?
a) Motorola
b) Bell Laboratories
c) Ford Motors
d) The ministry of Supply Advising Service on Statistical methods and Quality Control, UK
Answer: a
Explanation: Motorola developed Six-Sigma technique, and started implementing the initiative, after the year of 1987.
10. When was first ISO quality system standards published?
a) 1977
b) 1989
c) 1987
d) 1981
Answer: c
Explanation: In the year of 1987, ISO published its first quality system criterion was published. The goal of these was to improve quality system standards.
11. What does ASQC stand for?
a) All-India Standards for Quality Control
b) Attribute Specified Quality Control
c) American Society for Quality Control
d) All-India Society for Quality
Answer: c
Explanation: ASQC stands for American Society for Quality Control, formed in 1946. This organization promotes the use of quality improvement techniques, used in all kinds of production industries.
12. There are two types of inputs to a process besides the raw material. That are controllable inputs and _____ inputs.
a) noise
b) voice
c) monitoring
d) service
Answer: a
Explanation: There are 3 types of input to every process viz. raw materials, controllable inputs and uncontrollable inputs. Uncontrollable inputs are also called Noise inputs.
13. ASQC and ASQ stand for the same object.
a) True
b) False
Answer: a
Explanation: ASQC stands for American Society for Quality control formed in 1946. This company was renamed, American Society of Quality, back in the year 1998.
This set of Statistical Quality Control Questions and Answers for Freshers focuses on “Management Aspects of Quality Improvement – 2”.
1. PDSA is expanded as ____________
a) Plan-Define-Study-Analyze
b) Plan-Define-Study-Act
c) Plan-Do-Study-Act
d) Plan-Do-Study-Analyze
Answer: c
Explanation: PDSA cycle is termed as Plan-Do-Study-Act cycle. Sometimes, study step is called check step. So, PDSA cycle becomes PDCA cycle.
2. Which of the following is not a part of Juran Trilogy?
a) Planning
b) Control
c) Check
d) Improvement
Answer: c
Explanation: Juran Trilogy consists of 3 points viz. planning, control, improvement. The check step is not included in the Juran trilogy.
3. According to Armand V. Feigenbaum, Which of these is not the 3 tools of Quality improvement?
a) Quality leadership
b) Quality Technology
c) Organizational commitment
d) Quality commitment
Answer: d
Explanation: According to Feigenbaum, the 3 tools of quality improvement are Quality leadership, quality technology, and organizational commitment.
4. ISO was founded in the year of ______
a) 1943
b) 1946
c) 1949
d) 1956
Answer: b
Explanation: ISO was founded in the year of 1946 in Geneva, Switzerland for development of standards for quality systems. It issued its first standards in 1987.
5. The Malcolm Baldrige National Quality Award was created by ______
a) U.S. Congress
b) British Government
c) ISO
d) ASQ
Answer: a
Explanation: The Malcolm Baldrige National Quality Award was created by U.S. Congress in 1987. It is given to recognize U.S. organizations for performance excellence.
6. In 3-σ quality performance, the probability of producing a conforming product is _______
a) 0.9973
b) 0.9500
c) 1
d) 0.9467
Answer: a
Explanation: In 3-σ quality performance, the probability of getting a product, conforming to standards, is exactly 0.9973 which is almost 2700ppm defective.
7. Using 3 sigma quality performances, we get 2700 products produced which are nonconforming. If we operate on the 3- σ performance, and we produce 100 independent product components, what is the probability of finding a product not defective?
a) 0.7891
b) 0.27
c) 0.7631
d) 0.007631
Answer: c
Explanation: Probability of finding an single product not defective= 1- = 1 – 0.0027= 0.9973
Probability of finding a single product not defective chosen from 100 independent product components
= 0.9973 x 0.9973 x 0.9973 x … x 0.9973 = 0.9973 100 = 0.7631.
8. For any stable process, if the process follows 6 – sigma quality performance, how many products it will produce defective?
a) 1.9946 ppm
b) 3.4000 ppm
c) 1200 ppm
d) 934 ppm
Answer: b
Explanation: By reaching till the level of six-sigma quality level, we get almost 3.4 ppm defective products.
9. “Process performance is not predictable unless the process behavior is stable”.
a) True
b) False
Answer: a
Explanation: If the process behavior is variable, the process performance will vary accordingly. So, process performance can only be predicted when process is stable.
10. If the value-add time for a process is 10% of the process cycle time, what will be the PCE or Process Cycle efficiency of the mentioned process?
a) 10%
b) 12%
c) 23%
d) 14%
Answer: a
Explanation: PCE= Value-add time / Process Cycle Time = 10% Process cycle time / Process cycle time = 10%.
11. Consider a mortgage refinance operation at a bank. If the average completion rate for submitted applications is 100 completions per day, and there are 1500 applications waiting for processing, the process cycle time is ___________
a) 15 hrs
b) 15 days
c) 150 days
d) 15.33 days
Answer: b
Explanation: Process Cycle Time = Work-in-progress / Average completion rate = 1500/100 = 15 days.
12. Which of these is an appraisal cost?
a) Inspection and testing of incoming material
b) New product review
c) Complaint adjustment
d) Rework
Answer: a
Explanation: Appraisal costs are the costs, which are associated with measuring, evaluating, or auditing products, and purchased material, to insure the conformance to standards that have been imposed.
13. Which one of them is an External failure cost?
a) Burn-in
b) Materials and the services consumed
c) Scrap
d) Indirect costs
Answer: d
Explanation: External failure costs are the costs occurring when the product does not perform satisfactorily after it is delivered to customers. Indirect cost is one of them.
14. Automations, computers and new machinery will solve problems.
a) True
b) False
Answer: a
Explanation: According to Deming, the belief that automations, computers and new machinery could be solutions to the new problems; is not good for business quality development.
15. Failure analysis is a ______ cost.
a) prevention cost
b) internal failure cost
c) appraisal cost
d) external failure cost
Answer: b
Explanation: The cost incurred to determine the causes of product failures is defined as failure cost and it is counted in the internal failure costs.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Overview of DMAIC – 1”.
1. DMAIC stands for ______
a) Do-Measure-Act-Implement-Check
b) Define-Measure-Act-Implement-Control
c) Define-Measure-Analyze-Improve-Control
d) Do-Measure-Analyze-Improve-Control
Answer: c
Explanation: DMAIC is a problem solving process. It is a collection of 5 steps. D stands for Define, M stands for Measure, A stands for Analyze, I stands for Improve and C stands for control.
2. What DMAIC process does is, to _______
a) Manufacture any product
b) Define specification limits for a product
c) Solve root cause of quality and process problems
d) Define quality system standards
Answer: c
Explanation: DMAIC is a problem solving tool. Hence, its main objective is to find root causes of problems in a process, which create variability. It attempts to remove those root origins.
3. Which of these is not a part of the Define step of DMAIC process?
a) Verify and gain approval for final solutions
b) Define critical customer requirements
c) Establish project charter and build team
d) Document process
Answer: a
Explanation: Define step of DMAIC, which consists of identifying or validating the improvement opportunity, knowing customer requirements, mapping process, and establishing project charter.
4. DMAIC is often associated with ______
a) Six-sigma activities
b) Kaizen board
c) 5-s
d) Acceptance sampling
Answer: a
Explanation: DMAIC is often associated with six-sigma activities. The main objective of this is to give the maximum output with good quality. Almost all six-sigma implementation use DMAIC PSP for project management and completion.
5. PSP is expanded as ____
a) Problem Solving Process
b) Product safeguarding process
c) Product Sampling Process
d) Project Settling Procedure
Answer: a
Explanation: PSP is expanded as Problem Solving Process. There are many types of PSP’s available. DMAIC PSP is one of them.
6. The Cause and Defect Diagram is a part of _____ step of DMAIC process.
a) Define
b) Analyze
c) Control
d) Measure
Answer: d
Explanation: The cause and defect diagram is an diagram which shows defects in a product, and the corresponding cause for it. This is counted in the Measure step of DMAIC.
7. Tollgate reviews are presented to managers and “owner” of a process __________ in the six-sigma organizations.
a) The project champions and the black belts not working directly on the project
b) Industrial engineering team
c) Quality control manager
d) Developers of the factory
Answer: a
Explanation: Tollgate reviews are the reviews of a report developed by the project team. In a six sigma organization, the project champion and the black belts also review this report, besides the manager and the owner.
8. Improve step of DMAIC can also be changed by Design step in the case of some processes.
a) True
b) False
Answer: a
Explanation: When the original process is operating very badly, or if there is a new product or service is required, then the Improvement step of DMAIC process actually becomes Design step.
9. DFSS refers to _____
a) Development of Six-Sigma
b) Design for Six-Sigma
c) Define for Six-Sigma
d) Development for Six-Sigma
Answer: b
Explanation: For a six-sigma organization, when it is determined that a new product is required, the improve step in DMAIC becomes Design step. This is sometimes called, Design for Six-Sigma.
10. Tollgate reviews are done _______
a) Before each step of DMAIC
b) After DMAIC process ending
c) Before the whole DMAIC process
d) After each step of DMAIC process
Answer: d
Explanation: Tollgate reviews are done after each step of DMAIC process. They contain a review by the manager and owner, of a report of the step which is developed by the project team. This offers evaluation of the work of the team.
11. Process capability tests are a part of _____ step of DMAIC process.
a) Define
b) Improvement
c) Measure
d) Control
Answer: c
Explanation: Measure step contains the evaluation of the current state of the production process.
12. A DMAIC process is implemented on a process only when ______
a) There are signs of potential breakthrough
b) It is estimated that quality of products will decrease in near future
c) Exceptionally bad quality review of a product
d) When the project impact of DMAIC is estimated low
Answer: a
Explanation: DMAIC process is put into practice only when there are chances of improvement in the final quality of product/service. This improvement or potential change is measured on the basis of financial improvement.
13. Projects with high potential negative impact are _____
a) Most desirable
b) Least desirable
c) Having highest risk
d) Have a low risk
Answer: c
Explanation: Projects which have a chance of positive impact on the final product quality are most suitable for DMAIC application. But if the impact is predicted negative, they become the least desirable of all projects.
14. DMAIC can be carried out without tollgate reviews.
a) True
b) False
Answer: b
Explanation: DMAIC is a problem solving process carried out by project team. If this team makes an mistake in any one step, and it is not reviewed by experts like manager and the owner, it can result in negative impact on the process.
15. In SIPOC diagram, P stands for _______
a) Process
b) Product
c) Progress
d) Profit
Answer: a
Explanation: SIPOC is a high level map of a process where SIPOC stands for Supplier, Input, Process, Output, and Customer. This is an important part of Define step of the DMAIC process.
This set of Statistical Quality Control Interview Questions and Answers for freshers focuses on “Overview of DMAIC – 2”.
1. Input is defined as _____
a) Material provided to a process
b) Changes made to controllable variables
c) Development of in the process
d) Desired product specification
Answer: a
Explanation: The input to a process is defined as the material and the information provided to a process, before it starts producing any product. This is an important factor to the quality of the product.
2. KPIV’s and KPOV’s are measured in _______
a) Define step
b) Measure step
c) In both Define and Measure step
d) Improve step
Answer: c
Explanation: KPIV’s and KPOV’s are the variables, which matter the most on the output of a process. They are identified at least tentatively in the define step, and mostly identified during the measure step.
3. A gauge repeatability and reproducibility study can also be said _____
a) Formal gauge capability
b) Gauge metrology
c) Gauge potential
d) Gauge capacity
Answer: a
Explanation: For evaluation of current state of the process, we need to find the process capability. This is found via using formal gauge capability study, which is also called gauge repeatability and reproducibility study.
4. Repeatability is defined as ____
a) Repeating ability of a process to give almost same products in short interval of time
b) Repeating ability of a measuring instrument to give almost same measurement in short interval of time
c) Repeating ability of a process to give almost same products in long interval of time
d) Repeating ability of a measuring instrument to give almost same measurement in long interval of time
Answer: b
Explanation: Repeatability is generally described as, “The ability of a measuring instrument, to give almost or exactly same measurement of one dimension of an object, when the measuring person and the outer conditions are same.”
5. Which of these is not necessary in a tollgate report after measure step of DMAIC?
a) A list of identified KPIV’s and KPOV’s
b) Measurement system capability
c) SIPOC diagram
d) Assumptions made during the data collection
Answer: c
Explanation: In a tollgate review, all the important things used in measure step like KPIV’s and KPOV’s, capability of measurement system, and the assumptions made during the data collection etc are listed. SIPOC is the part of define step.
6. “The data collected in measure step may have a reference from the historical records.” Which of these options is correct bearing the above statement in mind?
a) Data from the historical records is satisfactory
b) Data from the historical records may not always be satisfactory
c) Data must always be collected only from historical records
d) Historical records are most reliable for data collection
Answer: b
Explanation: The statement says we can only have a reference from historical records, but we cannot always rely on it. This is because history may be incomplete or the methods of record keeping may have changed.
7. RPN stands for _____
a) Reliability priority number
b) Risk priority number
c) Recalled precautions number
d) Reliability precautions number
Answer: b
Explanation: The scores of the 3 criteria of the FMEA give information about variability, failure, and error score. They are multiplied together to find the RPN.
8. In FMEA, M stands for _____
a) Modes
b) Measurement
c) Manufacture
d) Material
Answer: a
Explanation: FMEA is expanded as Failure Modes and Effect Analysis which is used as a very important tool during measurements of Measure step of DMAIC.
9. The 3 criterion for FMEA are scored on _____ scale.
a) 0 to 1
b) 0 to 100
c) 0 to 10
d) 0 to 5
Answer: c
Explanation: The 3 criterion for FMEA are scored between o and 10. 0 stands for least possible and 10 stands for highest possible.
10. Which one of these is not a part of the 3 criteria for FMEA?
a) Likelihood of a process going wrong
b) Ability of detecting a failure
c) Cause for a failure
d) The severity of a failure
Answer: c
Explanation: The 3 criteria for FMEA are the likelihood that something will go wrong, the ability to detect a failure, and the severity of a failure.
11. Which of these is not a part of tollgate review after Improve step?
a) Adequate documentation of how the solution was obtained
b) Documentation on alternative solutions that were considered
c) Complete results of pilot tests
d) Results of FMEA
Answer: d
Explanation: The tollgate review after Improve step includes documentation about the appropriate solution and other alternative solutions considered, and the results of pilot tests, whereas, results of FMEA are part of the Measure step tollgate review.
12. Pilot tests and mistake proofing are parts of _____
a) Measure step
b) Define step
c) Improve step
d) Analyze step
Answer: c
Explanation: Mistake proofing is done to ensure the process improvement method is correct. Pilot test is also a form of confirmation experiment. So they both are included in Improve step of DMAIC.
13. Six-Sigma systems do not require DMAIC process at all.
a) True
b) False
Answer: b
Explanation: There may be chances of potential improvement in the six-sigma systems also. So, it cannot be predicted that six-sigma systems do not require DMAIC process at all.
14. Regression analysis and other multivariate methods are a part of which step of DMAIC process?
a) Define
b) Analyze
c) Improve
d) Control
Answer: c
Explanation: Regression analysis and other multivariate methods are a collection of tools required to analyze the data obtained from measure step. So they fall in the category of Analyze step of DMAIC problem solving process.
15. Designed experiments are a part of both, Analyze step and Improve step of DMAIC process.
a) True
b) False
Answer: a
Explanation: Design of experiments need the data from measure step to analyze and to make an improvement in a process. So, they are counted in both Analyze and Improve step of DMAIC process.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Modeling Process Quality – Describing Variation – 1”.
1. Which of these is not a tool to describe variation in product units?
a) The box plot
b) The histogram
c) Stem-and-Leaf plot
d) Acceptance Sampling
Answer: d
Explanation: The box plot, the histogram, and the Stem-and-Leaf plot, all are used to illustrate variation among the product units. But, Acceptance sampling cannot be used to describe variation as; it is not a variation describing tool.
2. Descriptive statistics is used _____
a) To develop information regarding the product sample using the measured data
b) To measure the data for a sample
c) To draw conclusions about the population
d) To control the variation
Answer: a
Explanation: Descriptive statistics is used to show how simple graphical and numerical techniques can be used, to summarize the information about the product. The evaluation of this information is done, by using Statistical Inference.
3. Which of these shows the ordered Stem-and-Leaf Plot?
a) Stems arranged by magnitude
b) Stems not arranged by magnitude
c) Stems and the leaves, both arranged by magnitude
d) Neither stems nor leaves are arranged by magnitude
Answer: c
Explanation: A Stem-and-Leaf plot is said to be ordered if it has its leaves arranged by magnitude. If it has stems also arranged with leaves, it surely is the ordered Stem-and-Leaf plot.
4. The median for an odd number of observations is _________ where n is the number of observations.
a) [/2 + 1] rank on the ascending order of observations
b) [/2 + 1] rank on the ascending order of observations
c) Average of [ / 2] and rank on the ascending order of observations
d) rank on the ascending order of observations
Answer: a
Explanation: Median is the fiftieth percentile of the data distribution. For the odd number of observations , the median is [/2 + 1] rank on the ascending order of observations. This can also be written as [/2] rank on the ascending order of observations.
5. For even number of observations of a data distribution, what is the median?
a) [/2 + 1] rank on the ascending order of observations
b) rank on the ascending order of observations
c) Average of th observation and th observation
d) Average of [ / 2] and rank on the ascending order of observations
Answer: c
Explanation: For even number of observation of a data variation, the median is Average of th observation, and th observation. This is because median of a data distribution is the fiftieth percentile of the data distribution.
6. IQR is defined as ____________
a) difference between fourth and first quartile
b) difference between fourth and third quartile
c) difference between third and first quartile
d) difference between second and first quartile
Answer: c
Explanation: IQR is expanded as interquartile range. It is defined as the difference between third and first quartile. IQR = Q3-Q1; where Q1 = First Quartile; Q3 = Third Quartile.
7. In stem-and-leaf plot, the measure of variability is _____________
a) IQR
b) Mean
c) Median
d) Third quartile
Answer: a
Explanation: IQR is the interquartile range, which is equal to the value of difference between third and first quartile. It is used as a measure of variability, in the case of stem-and-leaf plot.
8. Which of these, does not take time order of observations into account?
a) Time series plot
b) Run chart
c) Marginal plot
d) Stem-and-leaf plot
Answer: d
Explanation: Although Stem-and-leaf plot is an excellent way to visually show the variability in the observations. But it does not take time order of observations into account. So, time series plot or marginal plot or run chart is used.
9. Which of these is not the name of the intervals created for the construction of the histogram?
a) Class intervals
b) Cells
c) Bins
d) Boxes
Answer: d
Explanation: To construct histogram for continuous data, we must divide the range of the data into intervals, which are usually called class intervals, cells, or bins.
10. For a set of data having 100 observations, how many bins must be created for satisfactory results of histogram?
a) 25
b) 10
c) 28
d) 4
Answer: b
Explanation: The bin number, in the case of creating a histogram, must be between 5 and 20, to give satisfactory results. If we choose no. of bins approx. equal to the square root of the number of observations, it results well. √n=√100=10
11. The histogram has the same meaning as the box plot, i.e. there are two names of same thing.
a) True
b) False
Answer: b
Explanation: The histogram is a plot between the frequency of an observation in a data, and the bins of data created. But, the box plot is a graphical display that simultaneously displays many features of data, like variability and departure from symmetry.
12. Relative frequency is defined as ____________
a) difference of frequency of an observation from the highest one
b) frequency of an observation
c) highest known frequency among all frequencies of the data
d) frequencies of each bin divided by the number of observations
Answer: d
Explanation: Relative frequencies are defined as the frequencies of each bin of the histogram, divided by the total number of observations in the data. They are shown on vertical scale of histogram.
13. Cumulative frequency plot is defined as ____________
a) height of each bar less than or equal to upper limit of bin
b) height of each bar more than or equal to upper limit of bin
c) height of each bar less than or equal to lower limit of bin
d) height of each bar more than or equal to upper limit of bin
Answer: a
Explanation: Increasing frequency plot or Cumulative frequency plot is defined as, the histogram in which, the height of each bar is less than or equal to upper limit of the bin or class interval.
14. Frequency of a bin in the histogram must be higher than 0.
a) True
b) False
Answer: b
Explanation: In a histogram, frequency of an observation in a particular bin, depends on the data given. It is not necessary that frequency of a bin cannot be zero.
16. Which of these give an expression for the mean of a discrete distribution?
a) \
dx\)
b) \
\)
c) \
^2 p
\)
d) \
^2 fdx\)
Answer: b
Explanation: The expression for the mean of a discrete distribution is given by,
\
\)
This set of Statistical Quality Control Questions and Answers for Experienced people focuses on “Modeling Process Quality – Describing Variation – 2”.
1. The most important measure of central tendency in a sample is _____
a) Sample Average
b) Sample variance
c) Frequency of highest observation
d) Frequency of lowest observation
Answer: a
Explanation: For a sample data of n observations, sample average is the most important measure of central tendency. It is arithmetic mean of all the observations of the sample data.
2. Which of these is called the center of mass of the sample data?
a) Sample variance
b) Median of the data
c) Arithmetic mean of the data
d) The number having the highest frequency
Answer: c
Explanation: The histogram exactly “balances” at the point of the sample average, which is arithmetic mean of the whole data. So it is called the center of mass of the data.
3. The variability in the sample data is measured by _____
a) Sample Data
b) Sample mean
c) Sample variance
d) Range of data
Answer: c
Explanation: The sample variance is the measure of the variability of the sample data, because it shows deviation of the observations from the mean of the data.
4. The sample mean/average of the data is expressed by _____
a) \
\
\)
c) \
\)
d) \( \sum_{i=1}^n x_i \div n \)
Answer: d
Explanation: The mean of a sample data with n number of observations is expressed by,
\( \overline{x} = \sum_{i=1}^n x_i \div n \)
5. Which of the expression/inequality is always true?
a) \
sample standard deviation = \ sample standard deviation > sample variance
d) \( \overline{x} = s^2\)
Answer: b
Explanation: Sample variance is always the square of the sample standard deviation. This clarifies that,
sample standard deviation = \(\sqrt{Sample \,Variance}\).
6. The standard deviation of the data is expressed by _____
a) \
\
\
\(s = \frac{\sum_{i=1}^n
^2}{n}\)
Answer: b
Explanation: The data sample standard deviation is expressed as,
\(s = \left\{\frac{\sum_{i=1}^n
^2}{n-1}\right\}^{1/2}\)
7. The standard deviation does not reflect the magnitude of the sample data, only the scatter about the average.
a) True
b) False
Answer: a
Explanation: The standard deviation of a sample data is the square root of the sample variance. As variance is also calculated with respect to the mean, the above statement is true.
8. The sample variance is always greater than or equal to the sample standard deviation.
a) True
b) False
Answer: b
Explanation: Although sample variance is the square of sample standard deviation, but it can never be predicted that it will always be greater than or equal to the sample standard deviation. For example, for S.D. = 0.2, Variance = 0.04.
9. Which of these cannot be displayed by the box plot?
a) Location or central tendency
b) Spread
c) Departure from symmetry
d) Mean of the data
Answer: d
Explanation: The box plot graphically displays simultaneously several features of data, such as location, spread/variability, departure from symmetry, and identification of outliers.
10. Line at the either end of the box plot shows the _____
a) Extreme values
b) First quartile
c) Third quartile
d) Median
Answer: a
Explanation: A line at the either end of the box of the box plot shows the extreme values, i.e. maximum and minimum values. This line is also called whisker, bases on what, the box plot is also called box and whisker plot.
11. Probability distribution relates the value of a variable to ____
a) Its frequency
b) Its probability of occurrence
c) Random variable
d) Probability of occurrence of values other than that
Answer: b
Explanation: Probability distribution is a mathematical model, which relates the value of the variable with the probability of occurrence of that value in the population.
12. Which of these statements give exact definition of a random variable?
a) A random value from the set of integers
b) A random value from the set of natural numbers
c) A random value from the set of whole numbers which can take many values
d) A variable which takes on different values in population according to some random mechanism
Answer: d
Explanation: A random variable is generally described as a variable, which can take different values in population, i.e. the set of data, and follows no calculated mechanism.
13. Which of these gives a correct definition of continuous distributions?
a) Probability distribution of the variable being measured which takes on random values
b) Probability distribution of the variable being measured which takes on values dependent over a calculated mechanism
c) Probability distribution of the variable which takes some certain values such as integers
d) Probability distribution of the variable being measured that takes values on continuous scale
Answer: d
Explanation: When the parameter being measured can be expressed only on a continuous scale, its probability distribution is called a continuous distribution. E.g. length of a chip.
14. Which of these expressions is correctly describing a continuous distribution?
a) P(x=x i )=1-P(x≠x i )
b) P{a≤x≤b} = \
dx\)
c) P{x=x i }=p(x i )
d) P(x=x i )≤1-p(x≠x i )
Answer: b
Explanation: For a continuous distribution, probability of a value x lying in between a and b is expressed as,
P{a≤x≤b} = \
dx\)
15. Which of these give an expression for the mean of a continuous distribution?
a) \
dx\)
b) \
\)
c) \
^2 fdx\)
d) \
^2 p
\)
Answer: a
Explanation: The expression for the mean of a continuous distribution is given by,
\
dx\)
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Modeling Process Quality – Discrete Distributions – 1”.
1. What will be the variance of the sample following hyper geometric distribution and having 5 items defective in 100 lot items, if 10 samples are taken without replacement?
a) 0.431818
b) 0.410023
c) 0.483838
d) 0.568898
Answer: a
Explanation: variance of hyper geometric distribution is given by,
\
\) Putting, N=100, n=10, D=5; we get variance=0.431818.
2. For a Poisson distribution, parameter λ is having a value of 7. What is the variance for the distribution?
a) 2.64
b) 2.78
c) 7
d) 14
Answer: c
Explanation: For the Poisson probability distribution, the variance is equal to the parameter λ of the probability distribution. So for the above stated distribution, the variance will be equal to 7.
3. For 9 Bernoulli trials, what will be the mean of trials when the probability of failure is 0.56?
a) 5.04
b) 1.74
c) 3.96
d) 2.82
Answer: c
Explanation: For n=9 and probability of failure=0.56, we get probability of success;
p = 1 – 0.56 = 0.44.
Using this, we get;
mean= np=9*0.44=3.96.
4. In sample fraction defective, the ̂ symbol represents that ______
a) p ̂ is an estimate of the true value of binomial parameter p
b) It is the “raised to power” operator
c) It is having no significance
d) It is the true value of binomial parameter p
Answer: a
Explanation: The ̂ symbol represents the estimate of the true, unknown value of the binomial parameter p.p is the value of sample fraction defective or sample fraction nonconforming.
5. For a Pascal distribution, the probability of occurrence of a event x is given by _____
a) \=\leftMissing or unrecognized delimiter for \right p^x ^{x-r}\)
b) \=\leftMissing or unrecognized delimiter for \right p^x ^{x-r}\)
c) \=\leftMissing or unrecognized delimiter for \right p^x ^{1-r}\)
d) \=\leftMissing or unrecognized delimiter for \right p^x ^{x-1}\)
Answer: a
Explanation: The Pascal probability distribution is given by the following equation,
\=\leftMissing or unrecognized delimiter for \right p^x ^{x-r}\)
Where x=r, r+1, r+2… and r≥1 is an integer.
6. When is Pascal distribution, called negative binomial distribution?
a) When r>0 but not necessarily an integer
b) When r<0 but an integer
c) When r>3 and an integer
d) When r=0 and an integer
Answer: a
Explanation: When in a Pascal probability distribution, r is greater than 0 but not necessarily an integer, the Pascal distribution is called, negative binomial distribution. This is a special case of Pascal distribution.
7. When r=1 in a Pascal distribution, what is this case called?
a) Negative binomial distribution
b) Positive binomial distribution
c) Hyper geometric distribution
d) Geometric distribution
Answer: d
Explanation: The special case of Pascal distribution when r=1, we get geometric distribution. This represents the number of Bernoulli trials until the first success occurs.
8. Lognormal distributions are a part of discrete distributions.
a) True
b) False
Answer: b
Explanation: Lognormal distributions are an important part of continuous distributions, as they represent probability distribution of variables having continuous values. They are not counted in discrete distributions.
9. Which of the expression represents the mean of Pascal distribution?
a) r/p
b) np
c) nD/N
d) np
Answer: a
Explanation: The mean of the Pascal distribution is given by the following equation,
μ=r/p
Where r > 1 and an integer and p is probability of success.
10. The equation of variance, i.e. σ 2 =\Missing open brace for subscript Hyper geometric distribution
b) Pascal Distribution
c) Normal distribution
d) Binomial distribution
Answer: b
Explanation: The expression \(\frac{r}{p^2}\) is useful for finding out variance of the Pascal probability distribution. We can identify it as it has “r” in the expression of variance.
11. Bernoulli trials are a part of _____ probability distribution.
a) Poisson
b) Binomial
c) Hyper geometric
d) Normal
Answer: b
Explanation: Bernoulli trials are the independent trials having a result either a “success” or a “failure”. They are an important part of binomial probability distribution. The probability distribution is based upon the Bernoulli trials.
12. \Missing or unrecognized delimiter for \right\) is expanded as __________
a) \
\
\
\(\frac{b!}{a!}\)
Answer: b
Explanation: \Missing or unrecognized delimiter for \right\) is generally acronym for the combination \(a \atop b \)C or \(\frac{a!}{b!!}\)
13. For a Pascal distribution; r=3 and the probability of success is 0.7895. What will be the value of “σ”?
a) 1.0131
b) 3
c) 1.0065
d) 1.7320
Answer: c
Explanation: We know, for a Pascal distribution,
σ 2 =r/p 2
Evaluation of the above equation using the values r=3 and p=0.7895, we get, σ=1.0063.
14. For evaluation probability of measurement of length of a part being a certain value, discrete probability distributions can be used.
a) True
b) False
Answer: b
Explanation: The length of a part is always measured as a continuous value. Taking this fact into account, we cannot use discrete probability distributions for evaluation of probability. The continuous probability distributions are used for the above mentioned purpose.
This set of Statistical Quality Control Interview Questions and Answers for Experienced people focuses on “Modeling Process Quality – Discrete Distributions – 2”.
1. Which of these is not a discrete probability distribution?
a) Hyper geometric Distribution
b) Binomial Distribution
c) Normal Distribution
d) Poisson Distribution
Answer: c
Explanation: Hyper geometric distribution, Binomial distribution, and Poisson distribution are all part of discrete probability distribution family. But, Normal distribution is a Continuous distribution.
2. The mean of hyper geometric probability distribution is expressed as _____
a) μ=nD/N
b) μ=np
c) μ=r/p
d) μ=nN/D
Answer: a
Explanation: The mean of hyper geometric distribution is given by,
μ=nD/N.
3. Probability function for hyper geometric probability distribution is _____
a) \ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
b) \ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
c) \ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
d) \ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
Answer: a
Explanation: The probability function for the hyper geometric probability distribution is evaluated by following equation,
\ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
Where D is a number of interest, N is total no. of items in population, x is the no. of sample falling in the class of interest and n is no. of items in a sample.
4. For N=100, D=5, n=10, and x = 2; probability according to hypergeometric distribution will be _____
a) 0.58375
b) 0.92314
c) 0.99336
d) 0.99975
Answer: c
Explanation: According to hyper geometric probability distribution,
\ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
This gives P=0.99336 for the mentioned values.
5. The independent trials which have either a “success” or a “failure” as an outcome, are called ____
a) Normal trials
b) Bernoulli trials
c) Poisson’s trials
d) Hyper geometric trials
Answer: b
Explanation: When a process contain a sequence of independent trials which don’t depend on each other’s result, and which have outcome either a “success” or a “failure”, the trials are called Bernoulli trials.
6. The binomial distribution is given by _____
a) \ = \leftMissing or unrecognized delimiter for \right p^x ^{n-x}\)
b) \ = \leftMissing or unrecognized delimiter for \right p^{n-x} ^x\)
c) \ = \leftMissing or unrecognized delimiter for \right p^x ^{n-x}\)
d) \ = \leftMissing or unrecognized delimiter for \right p^{n-x} ^x\)
Answer: a
Explanation: The binomial distribution is based on Bernoulli trials, which gives the probability distribution as,
\ = \leftMissing or unrecognized delimiter for \right p^x ^{n-x}\).
7. Let us take Bernoulli trial of tossing a coin. If there are 70 independent trials, and finding a head is a success, what will be the variance?
a) 17.5
b) 18.5
c) 4.8
d) 4.18
Answer: a
Explanation: Variance of the binomial distribution,
σ 2 =np
For tossing a coin and finding a head, p = 0.5, which gives variance = 17.5.
8. Poisson distribution is given by ____
a) \ = \frac{e^{-λ} λ^x}{x!}\)
b) \ = \frac{e^{-x} λ^x}{x}\)
c) \ = \frac{e^{-λ} λ^x}{x}\)
d) \ = \frac{e^{-λ} λ^x}{λ!}\)
Answer: a
Explanation: The Poisson distribution probability is given by the following equation,
\ = \frac{e^{-λ} λ^x}{x!}\)
Where λ is a parameter and λ > 0.
9. The mean of a Poisson distribution is _____
a) Greater than λ
b) Lesser than λ
c) Equal to λ
d) Having no relation with λ
Answer: c
Explanation: According to Poisson distribution, the mean and the variance, both are always equal to the parameter used in the Poisson probability distribution, i.e. λ.
10. For a company, which uses Poisson distributions to describe all the defects in its product, which has the parameter λ=4, what will be the probability of finding the selected product from a sample that will have 2 or less than 2 defects?
a) 0.2381
b) 0.2561
c) 0.2104
d) 0.3310
Answer: a
Explanation: The Poisson probability distribution which has λ=5, we can determine the above stated probability using,
P{x≤2} = \(\sum_{x=2}^2 \frac{e^{-4} 4^x}{x!}\) = 0.018316 + 0.73263 + 0.146525 = 0.238104.
11. For a experiment following binomial distribution, the mean is 54.3, and the number of independent trials is 121. What will be the probability of failure?
a) 0.55123
b) 0.51000
c) 0.44877
d) 0.45700
Answer: a
Explanation: μ=54.3, n=121. We know that, p=μ/n
p=0.44876.
So probability of failure = 1-p = 0.55123.
12. A lot follows hyper geometric distribution for defects found in its items. It contains 100 items out of which, 5 are defective. If 10 items are selected without replacement in a random order, what is the probability of finding 0 defective items?
a) 0.5837521
b) 0.4162479
c) 0.5708992
d) 0.4522222
Answer: a
Explanation: According to hyper geometric probability distribution,
\ = \frac{\leftMissing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right}{\leftMissing or unrecognized delimiter for \right} \)
Putting values from the question, we get, p = 0.5837521.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Modeling Process Quality -Continuous Distributions – 1”.
1. Which of these cannot be shown on the continuous distributions?
a) Length dimension measurement of a box
b) Volume measurement of the box
c) Area measurement of one face of the box
d) Number of defects on the surface of the box
Answer: d
Explanation: Continuous distributions are used to describe the variation in the values of variables which are continuous, i.e. which take values on continuous scale. Number of defects is discrete not continuous parameter.
2. Which of these is a continuous distribution?
a) Pascal distribution
b) Lognormal distribution
c) Binomial distribution
d) Hyper geometric distribution
Answer: b
Explanation: Pascal, binomial, and hyper geometric distributions are all part of discrete distribution which are used to describe variation of attributes. Lognormal distribution is a continuous distribution used to describe variation of the continuous variables.
3. Which of these distributions has an appearance of bell-shaped or unimodal curve?
a) Lognormal distributions
b) Normal distribution
c) Exponential distribution
d) Cumulative exponential distributions
Answer: b
Explanation: Out of all continuous distributions, Normal distributions are the only distributions, which have a shape of curve as a bell. The curves of them are mostly unimodal.
4. The rule of multiplication of probability is possible only ____
a) When events are independent
b) When events are mutually exclusive
c) When events are Bayesian
d) When events are Empirical
Answer: a
Explanation: The rule of multiplication of probability is possible only when the events are independent.
P=P.P
5. Which of these equations describe the normal continuous distribution?
a) \=\frac{1}{\sigma \sqrt{2π}} e^{-0.5
^2}, -\infty < x < -\infty\)
b) \=\frac{1}{\sqrt{2π}} e^{-0.5
^2}, -\infty < x < -\infty\)
c) \=\frac{1}{\sigma \sqrt{π}} e^{-0.5
^x}, -\infty < x < -\infty\)
d) \=\frac{1}{\sigma \sqrt{2π}} e^{-0.5
^x}, -\infty < x < -\infty\)
Answer: a
Explanation: Normal distribution is a part of continuous distributions, which is described by the following equation,
\=\frac{1}{\sigma \sqrt{2π}} e^{-0.5
^2}, -\infty < x < -\infty\)
6. “ϕ” is said to be cumulative distribution function of _________
a) Standard binomial distribution
b) Standard normal distribution
c) Standard exponential distribution
d) Standard gamma distribution
Answer: b
Explanation: For the standard normal distribution, “ϕ” is described as the cumulative distribution function, which has the value of mean equal to zero and the corresponding standard deviation equal to unity.
7. The probability that the normal random variable x is less than or equal to some value a is said to be the _____
a) Standard normal distribution
b) Lognormal distribution
c) Exponential distribution
d) Cumulative normal distribution
Answer: d
Explanation: The normal distribution has several important special cases, out of which, the cumulative normal distribution is defined as the probability that the normal random variable x≤a.
8. Which of these is true for the normal distributions?
a) \
\
\
\(σ_y^2 = a_1^2 σ_1^2 + a_2^2 σ_2^2 +⋯+ a_n^2 σ_n^2\)
Answer: a
Explanation: From property, for any normal distribution which has,
\(y = a_1^2 x_1 + a_2^2 x_2 +⋯+ a_n^2 x_n\)
We have,
\(μ_y = a_1 μ_1 + a_2 μ_2 +⋯+ a_n μ_n\)
9. The central limit theorem is true for _______ distribution.
a) Normal distribution
b) Lognormal distribution
c) Exponential distribution
d) Gamma distribution
Answer: a
Explanation: The central limit theorem uses an assumption that, the normal distribution is an appropriate distribution for a random variable. That’s why; the central limit theorem is true only for Normal distribution.
10. If a variable x is having its value equal to the exponential function of another variable w, i.e. “x=exp”, and w has normal distribution; the distribution of x is called _______
a) Normal distribution
b) Exponential distribution
c) Lognormal distribution
d) Gamma distribution
Answer: c
Explanation: For a random variable x, which has “x=exp” where w is having a normal distribution; the distribution of such a random variable x, is said to be the Lognormal distribution. It is a continuous distribution.
11. For a lognormal random variable, what is the value of mean?
a) \
\
\
\(μ=e^{\theta+w^2}\)
Answer: a
Explanation: if for x, which has x=e w , and w has a normal distribution with mean θ and variance w 2 ; x is a lognormal random variable with mean expressed as
μ=e (θ+ w 2 /2) .
12. The exponential distribution is given by _____
a) f=xe λ
b) f= e -λx
c) f= λe -λx
d) f=e w
Answer: c
Explanation: The continuous probability distribution is said to be exponential if the distribution follows this equation,
f= λe -λx .
13. The cumulative distribution function for the lognormal distribution is given by _____
a) \
\
\
\(\phi[\frac{ln-\theta}{\theta}]\)
Answer: a
Explanation: The lognormal cumulative distribution function defines the probability that the variable x is less than or equal to a is given by,
\(\phi[\frac{ln-\theta}{\omega}]\)
14. If a card is chosen from a deck of cards, what is the probability that it is either 7 or 9?
a) 4/52
b) 7/52
c) 9/52
d) 8/52
Answer: d
Explanation: There are 8 cards in the deck which are either 7 or 9. There are total 52 cards in the deck, so the probability that the card is either a 7 or a 9 is 8/52; based upon the outcomes of interest divided by the total possible outcomes.
15. The rule of multiplication of probability is possible only _____________
a) When events are independent
b) When events are mutually exclusive
c) When events are Bayesian
d) When events are Empirical
Answer: a
Explanation: The rule of multiplication of probability is possible only when the events are independent.
P=P.P
This set of Statistical Quality Control test focuses on “Modeling Process Quality – Continuous Distributions – 2”.
1. The display of possible outcomes of an event with their corresponding probabilities is called ____________
a) Probability Plot
b) Contingency table
c) Bayesian Table
d) Frequency Plot
Answer: a
Explanation: The graphical representation of possible outcomes of an event with their corresponding probabilities; is called Probability plot.
2. Which one of these is not one of the conditions for the binomial distribution?
a) Only two outcomes must be there for an event
b) Independent trials must be there
c) Trials must be more than 4
d) The probability of failure must be constant
Answer: c
Explanation: For a binomial distribution, Bernoulli trials are necessary, i.e. the outcomes of an event must be either a success or a failure, and this means the probability of failure must be constant.
3. For mean=7 and standard deviation=2, what will be probability that a variable, which follows normal distribution, will have a value less than or equal to 10?
a) 0.93319
b) 0.94520
c) 0.96409
d) 0.91924
Answer: a
Explanation: Here μ=7, σ=2, we know that, P[x≤a] = P[z≤\(\frac{a-\mu}{σ}\)]
So, P[x≤10]=P[z≤\(\frac{10-7}{2}\)]= P[z≤1.5]=φ[1.5]=0.93319.
4. The variance of lognormal distribution is given by ____
a) σ 2 = e 2θ+w
b) σ 2 = e 2θ+w 2 (e w -1)
c) σ 2 = e 2θ+w 2
d) σ 2 = e 2θ+w 2
Answer: c
Explanation: The variance of a variable having a lognormal distribution is given by,
σ 2 = e 2θ+w 2 (e w 2 -1).
5. The relationship between the mean and variance of the exponential distribution is expressed by _____
a) μ = λσ 2
b) μ = σ 2 /λ
c) σ 2 =λ 2 μ
d) σ 2 =μ/λ 2
Answer: a
Explanation: For an exponential distribution, μ= 1 ⁄ λ and σ 2 = 1/λ 2
So, for an exponential distribution, μ = λσ 2 .
6. (1-e -λa ) is the cumulative ______ distribution.
a) Gamma
b) Normal
c) Lognormal
d) Exponential
Answer: d
Explanation: The cumulative distribution for standard exponential distribution is given by following equation,
F=(1-e -λa ).
7. The exponential distribution is used in reliability engineering as a model of the time to failure of a system. The parameter λ is called _____ in this application.
a) Failure rate
b) MTF
c) Hazard rate
d) MTBF
Answer: a
Explanation: The parameter λ in the reliability engineering application of the exponential distribution is called the failure rate of the system. The application uses it as a model of time to failure.
8. The mean of the exponential distribution used in reliability engineering is used as mean time to failure.
a) True
b) False
Answer: a
Explanation: In reliability engineering, the exponential distribution is used as a model of the time to failure of a system. The mean of the distribution is called MTF or mean time to failure.
9. The mean of the gamma distribution is given by _________
a) μ=1/λ
b) μ=r/λ
c) μ=r 2 /λ
d) μ=1/r
Answer: b
Explanation: For a gamma distribution,
μ=r/λ
Where r and λ are the shape parameter and the scale parameter, respectively.
10. Exponential distribution is a special case of gamma distribution.
a) True
b) False
Answer: a
Explanation: When the shape factor r=1, in the case of a gamma distribution, gamma distribution reduces to an exponential distribution which has only the scale parameter λ.
11. The relationship between the mean and the standard deviation of the gamma distribution is given as _____
a) \
\
\
\(\frac{\mu}{\sigma} = \sqrt[3]{r}\)
Answer: a
Explanation: The mean of a gamma distribution = \(\frac{r}{λ}\) and standard deviation σ = \(\frac{\sqrt{r}}{λ}\); so,
\(\frac{\mu}{\sigma} = \sqrt{r}\).
12. If the shape parameter of the gamma distribution is twice the scale parameter, what will be the value of the mean of the distribution?
a) 1.5
b) 2
c) 4
d) 1
Answer: b
Explanation: As for a gamma distribution, mean μ=r/λ; putting the value r=2λ from the question, we get,
μ= 2λ ⁄ λ =2.
13. For an exponential distribution, what is the probability that the value of the exponential random variable with parameter λ=3.2; will be having a value higher than 0.2353?
a) 0.5290
b) 0.4710
c) 0.2213
d) 0.3452
Answer: b
Explanation: P[x>0.2352]=1-P[x≤0.2352]=1-{1-exp}=0.4710.
14. Which of these will not be distributed on a discrete distribution?
a) The number of houses in a colony
b) The number of bedrooms in a house
c) The number of scratches on a car hood
d) The diameter of a car tire
Answer: d
Explanation: The number of houses, bedrooms, and scratches, are all incremented by a discrete number, which is an integer, whereas, the diameter of a car tire will be increased on a continuous scale. Thus, it must be plotted on a continuous distribution.
15. For a variable distributed log-normally with θ=6, ω=1.2; what is the probability that the variable exceeds a value of 500?
a) 0.4290
b) 0.5710
c) 0.4990
d) 0.4937
Answer: a
Explanation: For a lognormal distribution,
P[x > a]=1-P[x≤a]=1-P[z≤\(\frac{ln-θ}{ω}\)]
Putting the values of θ and ω we get,
P[x>500]=0.4290.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “SPC Methods and Philosophy – Statistical Basis of the Control Chart – 1”.
1. The average value of the quality characteristic corresponding to in-control state is represented by _____ in the control charts.
a) CL
b) UCL
c) LCL
d) Sample number
Answer: a
Explanation: The control chart contains a centre line that represents the average value of the quality characteristic corresponding to in-control state. This may be the desired value of that quality characteristic.
2. The highest value that a quality characteristic can take before the process becomes out-of-control, is called ______
a) Center line
b) Upper control limit
c) Lower control limit
d) Control limit
Answer: b
Explanation: The UCL is the highest value of a quality characteristic at which the process is in-control, i.e. it produces products of desired specifications.
3. The horizontal line in the control chart which shows the minimum value of a quality characteristic, before the process gets out-of-control, is called the _____
a) Upper control limit line
b) Lower control limit line
c) Desired value
d) Center line
Answer: b
Explanation: The lowest value for a Critical-to-quality characteristic corresponding to in-control state is shown on the control chart with the lowest horizontal line, which is called the lower control limit line.
4. In the horizontal axis of a control chart displays _______
a) Sample number
b) Time
c) Either sample number or time
d) Neither sample number or time
Answer: c
Explanation: The horizontal axis of a control chart displays either sample numbers or time elapsed from a certain time or from the time of process starting.
5. If for a process, 18 out of 20 points are plotted above the CL but below the upper control limit, and only 2 of 20 are plotted between the center line and the lower control limit, what can we say about the process state?
a) In-control
b) Out-of-control
c) Data is not enough to predict
d) Process state is not dependent over this data
Answer: b
Explanation: The above question concludes a systematic pattern. For a process to be perfectly in-control state, the plotted points on the corresponding control chart should have an essentially random pattern.
6. Shifting of mean from the current value to a new value and remaining there, is called _____ shift of the mean.
a) Continuous
b) Abrupt
c) Random
d) Sustained
Answer: d
Explanation: Due to some reason, it may be possible that the process mean shifts from one point to another, and remain there for quite some time. This shifting of mean is called the sustained shift of the mean.
7. A process is said to be in the statistical control if it operates with ______ of variation.
a) Chance causes
b) Assignable causes
c) Both chance and assignable causes
d) Neither chance nor assignable causes
Answer: a
Explanation: Chance causes of variation produce the natural variability of the stable process. As this natural variability cannot be avoided, so a process is said to be in statistical control if it operates only with chance causes of variation.
8. Which of these is not a part of SPC’s 7 tools?
a) Pareto chart
b) Histogram
c) Design of Experiments
d) Check sheet
Answer: c
Explanation: Pareto chart, check sheet, histogram, cause-and-effect diagram, scatter diagram, and control chart are all part of SPC . Design of Experiments is not one of them.
9. If a process has its operation with _________ causes of variation, it is said to be an out-of-control process.
a) Chance
b) Assignable
c) Neither chance nor assignable
d) Out-of-control
Answer: b
Explanation: Only assignable causes of variation are able to shift the process from in-control to out-of-control state. So if a process possesses assignable causes of variation, it is stated to be out-of-control.
10. Chance causes of variation are also called ________ and assignable causes are also called _______
a) Common causes, special causes
b) Special causes, common causes
c) Variability causes, non-variability causes
d) Non-variability causes, variability causes
Answer: a
Explanation: Chance causes of variation, and assignable causes of variation are also called the common causes, and special causes of variation, respectively.
11. The chance and assignable cause terminology was developed by _______
a) Deming
b) Hawthorne
c) ISO
d) Shewhart
Answer: d
Explanation: The chance and assignable cause terminology was developed by W. A. Shewhart. Now, common and special cause terminology is used.
12. OC curve can be used to determine the probability of _______
a) Type I error
b) Type II error
c) Both type I and type II error
d) Neither type I nor type II error
Answer: b
Explanation: OC curve for any process is a very useful tool to find out the probability of a type II error which means, concluding the process in-control when it is actually out-of-control.
13. Control charts can be used very effectively in the process of hypothesis testing.
a) True
b) False
Answer: a
Explanation: Control charts and hypothesis testing have a really close relation as control charts can be used to check if the assumptions of hypothesis testing hold good in the process operation.
14. Type I error is described as the situation of the conclusion of the process state as out-of-control when it is in-control.
a) True
b) False
Answer: a
Explanation: Type I error is actually concluding the process state being out-of-control when the process state is in-control. The opposite conclusion gives the conditions for Type II error.
This set of Statistical Quality Control Quiz focuses on “SPC Methods and Philosophy – Statistical Basis of the Control Chart – 2”.
1. In the name of the OC curve, OC stands for ________
a) Operation Characteristic
b) Operating Characteristic
c) Operator Characteristic
d) Operated Characteristic
Answer: b
Explanation: The OC curve stands for Operating Characteristic curve which is used to develop an understanding of probability variation of the Type II error in the quality control.
2. The general theory of Control charts was first developed by ________
a) Walter A. Shewhart
b) Deming
c) ISO
d) ASQC
Answer: a
Explanation: W.A. Shewhart was the first person to develop an applicable theory for quality control for the processes used in manufacturing a product. This was known as the general theory of Control Charts.
3. Control charts are the part of _______ step of DMAIC process.
a) Define
b) Measure
c) Act
d) Control
Answer: d
Explanation: The control charts use the data from the measure step to analyze the process situation and then apply the steps to control the quality of the process. So they are used in the control step, analyze step of DMAIC process.
4. The size of sample is 5 for a process. If the process standard deviation is 0.15 micron, and the mean of process is 1.5 micron, the standard deviation of the sample average will be _______
a) 0.0500
b) 0.1000
c) 0.6710
d) 0.0671
Answer: d
Explanation: The standard deviation of the sample average is calculated by,
σ x =σ/√n
Putting σ=0.15, n=5, we get, σ x =0.0671.
5. For a process, sample size=5, process standard deviation=0.15, mean=1.5, and the std. deviation of mean is 0.0671. What will be the value of 3σ upper control limit for the construction of control chart?
a) 1.70
b) 1.29
c) 1.92
d) 0.170
Answer: a
Explanation: As we know, UCL 3σ =;n=5; σ=0.15, \(\overline{x}\) = 1.5; σ x =0.0671{from question}
So, UCL 3σ =1.5+3=1.7013.
6. The correct expression for UCL for construction of a control chart is given by _____
a) UCL = \
\)
b) UCL = \
\)
c) UCL = \
\)
d) UCL = \
\)
Answer: a
Explanation: The UCL is defined as the upper control limit for the control chart and it is evaluated as,
UCL = \
\); where \(\overline{x}\) is the mean of process, σ x is the std. deviation of mean and \(Z_{\frac{\alpha}{2}}\) is a arbitrary constant.
7. The general model for the lower control limit for a value of quality characteristic “w” will be _____
a) LCL = μ w + Lσ w
b) LCL = μ w – Lσ
c) LCL = μ w – Lσ w
d) LCL = μ w + Lσ
Answer: c
Explanation: The general expression for LCL of a quality characteristic “w” is given by,
LCL = μ w – Lσ w ;
Where “L” denotes the distance of control limits from the center line.
8. The center line of a control chart will be having a value ______
a) Higher than mean of quality characteristic
b) Lower than mean of quality characteristic
c) Equal to mean of quality characteristic
d) Which is higher than UCL
Answer: c
Explanation: The center line of the control chart denotes the value of the process mean of the quality characteristic. It is always the desired value.
9. Which of these is a part of corrective action process associated with a control chart?
a) OCAP
b) DMAIC
c) OC curve
d) LCL
Answer: a
Explanation: OCAP is an important part of the corrective action process associated with the control charts which uses results of control charts to control the process.
10. Which of these can be used to estimate capability of the process?
a) Control charts
b) Process mean
c) Acceptance Sampling
d) Designed Experiments
Answer: a
Explanation: Control charts are also used as estimating device as their results can be used to estimate the process capability of a certain process.
11. Control charts for central tendency and the variability are called _________ control charts.
a) Variables
b) Attributes
c) Acceptance
d) Rejections
Answer: a
Explanation: It is convenient to describe the variables with a measure of central tendency and measure of variability. So the control charts for central tendency and variability are called Variable control charts.
12. The control charts formed for judgment of conformities and non-conformities are called ______ control charts.
a) Variables
b) Attributes
c) Acceptance
d) Rejections
Answer: b
Explanation: Attributes are increased on the discrete scale and they give results as either conformity or non-conformity. So the charts plotted for non-conformities and conformities are called Attributes control charts.
13. Control charts with points around mean and in predicted or fixed manner indicate ________
a) Stationary variability
b) Non-stationary variability
c) Auto correlated variability
d) Process out of control
Answer: a
Explanation: The control charts which have points around the mean and in a predicted manner indicate that the process has stationary variability. This explains the nature of the process.
14. Auto correlated stationary process data is dependent on each other.
a) True
b) False
Answer: a
Explanation: In the auto correlated data, one data observations influence the next point, which means, if one point is above the mean in the control chart, the next one will also be above the mean.
15. Control charts are not effective in defect prevention.
a) True
b) False
Answer: b
Explanation: The control charts will help in the “Do it right from the first time” philosophy. This means, it helps to control the quality of the process, which means we get more conforming products than from out-of-control process.
This set of Statistical Quality Control MCQs focuses on “SPC Methods and Philosophy – Statistical Basis of the Control Chart – 3”.
1. Moving the UCL and LCL line far from the Center line means ____
a) Decreasing the possibility of type I error
b) Increasing the possibility of type II error
c) Increasing the possibility of type I error
d) Decreasing the possibility of type II error
Answer: a
Explanation: Type I error is, “concluding the process to be out-of-control when it is actually in-control”. So, more distance between UCL and LCL means, a point without an assignable cause will lie in between the UCL and LCL, i.e. in-control.
2. If we decrease the distance between LCL and UCL, what will happen?
a) Decreasing the possibility of type I error
b) Increasing the possibility of type II error
c) Increasing the possibility of type I error
d) Decreasing the possibility of type II error
Answer: d
Explanation: Type II error is, “concluding the process to be in-control when it is actually out-of-control”. This means the decreased distance between LCL and UCL will cause a point with assignable cause, lie not in between them, i.e. out-of-control.
3. If “two sets of limits” approach is taken to construct a control chart, what are the outer set of limits called?
a) Action Limits
b) Warning Limits
c) Variable Limits
d) Constant Limits
Answer: a
Explanation: The outer set of limits used in a “two sets of limits” approach of constructing a control chart, are generally called Action limits as action is to be done to correct the assignable cause if any point out-lies the action limits.
4. The inner limits in the “Two sets of limits” approach of the construction of control chart are called _____
a) Action Limits
b) Variable Limits
c) Constant Limits
d) Warning Limits
Answer: d
Explanation: The “two sets of limits” approach utilizes two different sets of UCL and LCL. In those two sets, the inner set of LCL and UCL is called the warning limits.
5. ARL is termed as ____________
a) Average Run Length
b) Allocating Run Length
c) Affected Run Length
d) Assumed Run Length
Answer: a
Explanation: The Average Run Length has an acronym of ARL. It provides useful way to make effective decisions for the sample size to be selected for the construction of the control chart.
6. The probability of a point to exceed the control limits for a control chart is 0.0040. What will be the ARL for this case?
a) 250
b) 278
c) 210
d) 216
Answer: a
Explanation: We define ARL as the reciprocal of the probability of a point falling out of the area between control limits. Here,
ARL=\(\frac{1}{p}=\frac{1}{0.0040}\)=250
7. For an average run length of 370, what will be the probability of a point falling out of the area between the control limits?
a) 0.0027
b) 0.0013
c) 0.0045
d) 0.0040
Answer: a
Explanation: As we know,
ARL = 1 ⁄ p
Putting value of ARL=370, we get, p=0.0027.
8. What is the full-form of ATS?
a) Average time to Stop
b) Average Time to Start
c) Average Time to Signal
d) Average Time to Select
Answer: c
Explanation: ATS is the abbreviation used for the Average Time to Signal. It is a useful way to express the performance of a control chart very effectively and easily.
9. If for a process, the samples are taken 5 hrs apart and its ARL is 24.24, What will be ATS for it?
a) 121.2
b) 110.0
c) 113.3
d) 137.8
Answer: a
Explanation: We know that,
ATS=ARL.h
Putting values h=5,ARL=24.24; we get,
ATS=121.2.
10. Decreasing sample size will ________
a) Decrease the slope of OC curve
b) Increase the slope of OC curve
c) First decrease then increase the slope of OC curve
d) Makes OC curve have 0 slope
Answer: b
Explanation: The slope of OC curve varies inversely with the sample size taken for the control chart construction. The decrease in the size of the sample taken to construct a control chart will increase the slope of OC curve.
11. Which of these statements, correctly describes the concept of rational subgroups?
a) If assignable causes are present, the chances of difference between subgroups should be largest
b) If assignable causes are present, the chances of difference between subgroups must be lowest
c) If assignable causes are present, the chances of difference between the subgroups must be moderate
d) If assignable causes are present, the chances of difference between the subgroups must be zero
Answer: a
Explanation: Rational Subgroups concept says that, for subgroups, if assignable causes are present, the chances of differences between subgroups should be maximized, while the chance for differences due to these causes within a subgroup must be lowest.
12. If the ARL is 34.8 for a process which has ATS=69.6 hrs, what is the time interval before taking samples?
a) 0.5 hrs
b) 8 hrs
c) 2 hrs
d) 4 hrs
Answer: c
Explanation: The time interval before taking samples is given by,
h=\(\frac{ATS}{ARL}\)
Putting values of ATS and ARL, we get h=2 hrs.
13. Which of these is not one of the sensitizing Western Electric rules that indicate “action needed” for Shewhart control charts?
a) One or more points near a warning limit/control limit
b) One point plots outside 3-sigma control limits
c) Eight consecutive points plot on one side of the control line
d) Two of 3 consecutive points plot beyond the 2-sigma warning limits
Answer: a
Explanation: The Western Electric rules which indicate “action needed” for Shewhart control charts do not state that, if one or more points are plotted near a warning limit or control limit will need an action to control the process.
14. 14 points in a row alternating up and down the mean will not indicate action needed for any process for which a control chart is plotted.
a) True
b) False
Answer: b
Explanation: If 14 points in a row, on a control chart, plot alternating up and down the mean, will surely indicate that that process is giving warning signals. This means the process is presently in-control but will go out-of-control in near future.
15. If an assignable cause is expected in the process, the subgroups taken subsequently should have a maximum difference.
a) True
b) False
Answer: a
Explanation: If a process is estimated to have an assignable cause, from the concept of Rational Subgroups; the chances of difference between the subgroups or samples should be maximum possible.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “SPC Methods and Philosophy – Rest of Magnificent Seven”.
1. Which of these is not a part of magnificent seven of SPC?
a) Pareto chart
b) Check Sheet
c) Scatter Diagram
d) 2 k factorial design
Answer: d
Explanation: Pareto chart, check sheet, and scatter diagram are all parts of the 7 problem solving tools of SPC or magnificent 7 of SPC . 2 k factorial design is a part of design of experiments.
2. Check sheet shows ___________
a) If a process part is completed
b) If the process components are all checked
c) If the samples of the process are checked
d) Time oriented summary of defects
Answer: d
Explanation: A check sheet is very easy way, to show the time oriented summary of defects in the number of samples taken from a lot of products from a certain process.
3. The graphical representation of the total frequencies of occurrence of each type of defects type against the various defect types will be called as ____________
a) Check sheet
b) Pareto chart
c) Histogram
d) Control charts
Answer: b
Explanation: The graphical representation of frequencies of defects with the information of the names of defects is generally done using a Pareto chart. It is quite useful in analysis of bigger data.
4. Pareto chart identifies the ____________ defects not the ____________ defects.
a) The most important, the most frequent
b) The most frequent, the most important
c) The smallest defects, the largest defects
d) The largest defects, the smallest defects
Answer: a
Explanation: It is property of a Pareto chart that, it does not automatically identify the most important defects but only the most frequent. There is no information about severity of the defect.
5. Pareto charts are a main part of _____ steps of DMAIC.
a) Measure and Analyze
b) Define and Measure
c) Define and Improve
d) Analyze and Control
Answer: a
Explanation: Pareto charts are useful to store a data about frequencies of occurrence of a number defects. This means it is first used in measure step, then in the analyze step to analyze data from the measure step.
6. Pareto chart is invented by from Italian economist ________
a) Alfred Pareto
b) Vilfred Pareto
c) Jon Pareto
d) Paulo Pareto
Answer: b
Explanation: Pareto charts have their name derived from the name of the Italian economist, Vilfred Pareto who theorized that in certain economies where the majority of the wealth was held by a disproportionately small segment of population.
7. In Cause and Effect diagram, what procedure is adopted?
a) First defects are identified and then the corresponding effects on working of product is determined
b) First defects in a product are identified and then, the corresponding causes are discovered
c) First causes of defects are plotted then the effects of them, i.e. defects are identified
d) Causes and their effects are identified simultaneously
Answer: b
Explanation: To plot a cause and effect diagram, first defects in a product are identified and then, the corresponding causes which are grounds of the origin of those defects.
8. Defect concentration diagrams are an important part of ____________ step of DMAIC.
a) Analyze
b) Define
c) Improve
d) Control
Answer: a
Explanation: Defect concentration diagrams are an important part of the Analyze step of the DMAIC problem solving process as it aims to analyze the defects present in different locations of units.
9. Phase II of control chart applications contain __________
a) Monitoring the process by comparing each sample statistic to control limits
b) Comparison of a set of data from control limits
c) Corrective action only to get data in control the process
d) Initial measurement of data
Answer: a
Explanation: Phase II of control chart application contains monitoring of the process by comparing each sample statistics data of the process production to some finalized control limits.
10. Which of these is a useful plot for identifying a potential relationship between two variables of a process?
a) Pareto chart
b) Defect concentration diagram
c) Scatter diagram
d) Stem and Leaf plot
Answer: c
Explanation: A scatter diagram is a very useful plot between two variables which affects the manufacturing process. To identify any potential relationship between the two variables, scatter diagram is a very useful tool.
11. Which of these is having an important role in regression modeling?
a) Stem and Leaf plot
b) Pareto chart
c) Defect concentration diagram
d) Scatter diagram
Answer: d
Explanation: A Scatter diagram is playing an important role in regression modeling.
12. Scatter diagram is a technique used in ______ step of DMAIC process.
a) Define
b) Analyze
c) Improve
d) Control
Answer: b
Explanation: Scatter diagram have an important part in regression modeling which is an important procedure, used in analyze step of DMAIC PSP.
13. Defect concentration diagram is a ___________
a) Picture of the unit showing all relevant views
b) Graph of defects with their frequency
c) Time oriented summary of defects
d) Graph to determine underlying causes of any defect
Answer: a
Explanation: Defect concentration diagram is a very useful tool to analyze the whole body of product to find out areas of high defect intensity. It is a picture of the unit showing all relevant views.
14. Correlation in scatter diagram necessarily implies causality.
a) True
b) False
Answer: b
Explanation: Correlation in data can be caused by something quite different than the two variables. They may be, both, dependent on a third variable which denies the above mentioned statement.
15. Most frequent defects are the most dangerous/severe.
a) True
b) False
Answer: b
Explanation: We cannot say that the above statement is true because some low frequency defects could also be very dangerous. For example, casting voids occur very infrequently but they are very dangerous as only a little scratch exposing void could elevate failure possibility.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Variable Charts – Control Charts for x̅ and R – 1”.
1. Quantities that can be numerically measured, can be plotted on a _________ control chart.
a) X bar
b) P chart
c) C chart
d) np chart
Answer: a
Explanation: A quantity that is measured on a continuous scale, i.e. it is numerically measured, can be plotted on a control chart type named, X bar control chart.
2. A single measureable quality characteristic, such as dimension, weight, or volume, is called ________
a) Variable
b) Attribute
c) Variable and an Attribute
d) Mean and variablility
Answer: a
Explanation: A single measurable quality characteristic is that CTQ that can be measured on a continuous scale. So it is also called a variable, be it a dimension or a weight measurement of a product unit.
3. A variable quality characteristic will have both __________
a) Mean and variability
b) Discrete and continuous values
c) Zero and infinite value
d) One or zero
Answer: a
Explanation: A variable quality characteristic will vary on a continuous scale. So it will have both, a mean and its variability.
4. Control of the process average or mean quality level is usually done with the ___________ control chart.
a) X bar control chart
b) S control chart
c) R chart
d) P chart
Answer: a
Explanation: If a quality characteristic has a mean and variability, it is called a variable. If the control of process average or mean quality level is to be done for a variable, an x bar variable control chart is used.
5. S chart is used to monitor _________ of a quality characteristic.
a) Mean
b) Range
c) Variability
d) Attributes
Answer: c
Explanation: S chart is a control chart used to monitor the variability of a process quality characteristic. It is variable chart as it is used for monitor variability of variables.
6. Toughness of a bolt mount on a tank is __________
a) An attribute
b) A variable
c) Variable and an attribute
d) Variability
Answer: b
Explanation: A quality characteristic having a numerical measurement on a continuous scale is called a variable. As the toughness of a bolt is a continuous quality characteristic, it is called a variable.
7. X chart is a ____________
a) Attribute control chart
b) Variable control chart
c) Neither a variable control chart nor an attribute control chart
d) Falls in the category of both variable and attribute control charts
Answer: b
Explanation: Mean is only an asset of a variable quality characteristic. As x denotes mean of a variable, x chart is counted in the variable control chart category.
8. If a process is said to be in control, what can we say about the variation?
a) Random
b) Normal
c) Attribute
d) Assignable
Answer: a
Explanation: If a process is said to be in control, its sample data will have a random pattern. So when plotted on a control chart, they will appear to have a random distribution.
9. Tolerances are said to be ________
a) limits of natural variability
b) Statistical limits of variability
c) Limits determined by the customers of the product
d) Limits of inherent process variability
Answer: a
Explanation: The natural variability varies from a certain highest value to some lowest value. These limits of natural variability are also called as Tolerance.
10. Upper control limit for a x chart is expressed by ________
a) \
\
\
\(\bar{\bar{x}} + A_3 R\)
Answer: a
Explanation: The expression for the upper control limit of the x chart is \(\bar{\bar{x}} + A_2 R \)
11. The center line for a \Missing open brace for subscript Mean of any sample
b) Mean of means of the sample
c) Mean of any sample + 0.5
d) / 0.5
Answer: b
Explanation: The center line for a \(\bar{\bar{x}}\) chart is generally mean of means of all the samples taken from a process. This gives more accurate results than choosing the mean of any sample to be taken as Center line.
12. Specifications have the same meaning as ____________
a) Control limits
b) UCL
c) LCL
d) Tolerances
Answer: d
Explanation: The tolerance for a variable quality characteristic is generally having the same meaning as specifications of a product, i.e. the desired values for that quality characteristic.
13. For a random variable having a normal distribution, the ratio of its range to the standard deviation is called ____________
a) Relative range
b) Absolute range
c) Major range
d) Minor range
Answer: a
Explanation: For a normal random variable, the ratio of its range to the standard deviation is called the relative range for that random variable.
14. Process variability can only be described and monitored by the s control chart.
a) True
b) False
Answer: b
Explanation: The process variability can be monitored with either a control chart for standard deviation, called the s control chart, or a control chart for the range, called an R control chart.
15. The lower control limit for an x bar control chart is lesser than the mean of means of the samples taken.
a) True
b) False
Answer: a
Explanation: The lower control limit for any control chart is expressed by
LCL = \(\bar{\bar{x}} – A_2 R\)
So it is lower than process mean or the mean of means of samples taken.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Variable Charts – Control Charts for x̅ and R – 2”.
1. LCL for the R chart is given by __________
a) D 3 R
b) D 2 R
c) R – D 3 R
d) d 2 R
Answer: a
Explanation: LCL for an R chart is always given by the following equation,
LCL = D 3 R .
2. In the general equation of UCL of a control chart, for any x chart, which of these is used as the estimator of μ?
a) \
\
\
\(\bar{\bar{R}}\)
Answer: c
Explanation: UCL=μ+ Z α/2 σ is the general equation for UCL for any control chart. For any \(\bar{x}\) chart, \(\bar{\bar{x}}\) is used as the estimator of μ in the above mentioned equation.
3. Which of these gives the correct value of A 2 used in the equation for control limits of a x control chart?
a) \
\
\
\(\frac{3}{\sqrt{d_2}}\)
Answer: a
Explanation: The value of A 2 used in the equation of control limits of a x control chart is given by following equation,
\(\frac{3}{d_2 \sqrt{n}}\)
4. In phase I application of x and R chart, the control limits obtained from the equations are treated as ____________
a) Final limits
b) Trial limits
c) Warning limits
d) Pattern limits
Answer: b
Explanation: The obtained limits from the equation of control limits for a x and R chart, are generally treated as Trial limits. They allow us to determine whether the process was in control when the m initial samples were taken.
5. Which term is having a closest meaning as Sampling Distributions?
a) Control charts
b) On site inspection
c) Whole lot inspection
d) Acceptance sampling
Answer: a
Explanation: The term “control charts” is having a closest meaning to “sampling distribution” because, control charts are also plotted on the data obtained from the sample inspection and also, they show variation in sample data.
6. Process capability generally uses __________
a) Specifications
b) Control Limits
c) Process standard deviation
d) Mean of any one sample
Answer: b
Explanation: Process capability studies make use of the specifications of any certain Critical-to-quality characteristic or quality characteristic to estimate the performance of any process.
7. The process standard deviation is given by __________
a) R /d 2
b) R d 2
c) 1/d 2
d) R /d
Answer: a
Explanation: The process standard deviation may be estimated by using the following equation,
\(\hat{σ} = \frac{\bar{R}}{d_2}\)
8. For any process, the sample ranges are, 1.2,1.5,1.1,1.4,1.5. The subgroup size is 5. What will be the process standard deviation? Given: d 2 =2.326 and A 2 =0.577
a) 0.576
b) 2.322
c) 0.511
d) 2.463
Answer: a
Explanation: We know that,
\(\bar{R} = \frac{\sum_{i=0}^{i=n} R_i}{n}\) and “process standard deviation = \(\frac{\bar{R}}{d_2}\)“, by using the values of R and d 2 in the question, we get process standard deviation=0.576
9. A tolerance diagram is also called ____________
a) Scatter diagram
b) Defect concentration diagram
c) Histogram
d) Tier chart
Answer: d
Explanation: The run chart of individual observations in each sample is called the tolerance diagram for any process. The tolerance diagram is also called Tier chart of the process.
10. Is there any relationship between specification limits and control limits of x and R charts?
a) Yes, Specification limits = Control limits
b) Yes, Control limits=Specification limits/2
c) No
d) Yes, Control limits*0.5 = Specification limits
Answer: c
Explanation: There is no certain relationship defined; between the control limits of x and R charts and the specification limits of any quality characteristic.
11. Control limits are ___________
a) Limits defined by customers
b) Limits driven by the natural variability of the process
c) Limits driven by the inherent variability of the process
d) Statistical limits
Answer: b
Explanation: The control limits are the limits for a quality characteristic for a process to be in-control. They are driven by the natural variability of the process.
12. The natural variability of the process is measured by ____________
a) Process mean
b) Sample standard deviation
c) Process standard deviation
d) Sample mean
Answer: c
Explanation: The natural variability of any process is the main factor affecting the control limits of any quality characteristic while plotting a control chart. They are measured by process standard deviation, σ.
13. What type of chart will be used to plot the number of defectives in the output of any process?
a) x bar chart
b) R chart
c) c chart
d) p chart
Answer: d
Explanation: The number of defectives in the samples of the output of a process is monitored by the p chart and the number of defects is monitored by a “c chart”.
14. Process standard deviation is necessarily equal to the sample standard deviation of the same process.
a) True
b) False
Answer: b
Explanation: It is not necessary that the sample standard deviation will always be equal to the process standard deviation, for a same process. This is because; there may be assignable causes of variation while the sample is being produced by the process.
15.“There is no need of revision of control limits once calculated by the equations of control limits.
a) True
b) False
Answer: b
Explanation: In the phase I application of control charts, the limits calculated from equations are used as trial limits because it gives us an idea about the variation, both, inherent and natural. So with time the variation changes, to compensate this, limits must be revised after a certain time.
This set of Statistical Quality Control online test focuses on “Variable Charts – Control Charts for x̅ and R -3”.
1. Once a set of reliable control limits is obtained, we use the control chart for monitoring future production. This is called __________
a) Phase I control chart usage
b) Phase II control chart usage
c) Phase III control chart usage
d) Phase IV control chart usage
Answer: b
Explanation: The use of reliable control limits to monitor the future production from a process, is generally mentioned as the Phase II application of control chart while, setting the trial control limits to monitor the process is called the Phase I application.
2. When R chart is out of control, we __________
a) Eliminate the out-of-control points and recalculate the control limits
b) Take one more sample and recalculate the control limits
c) Eliminate the out-of-control points and the nearest two points, and recalculate the control limits
d) Take no action
Answer: a
Explanation: When R chart is out of control, we often eliminate the out-of-control points and recompute a revised value of R bar. This will help us recalculate the control limits.
3. When the upper and lower natural tolerance limits are equal to the upper and lower specification limits, the process capability ratio, c p is ________
a) Greater than 1
b) 0
c) Less than 1
d) Equal to 1
Answer: d
Explanation: When the natural tolerance limits are having a equal value to the specification limits for a quality characteristic, the process capability ratio, c p is having a value equal to 1.
4. X bar chart monitors __________
a) Between-sample variability
b) Within-sample variability
c) Neither between-sample nor within-sample variability
d) Both between-sample variability and within-sample variability
Answer: a
Explanation: A x chart is only used to monitor the variability of a whole process over time. It does not detect variability within a sample but variability between all the samples of a process output.
5. Unlike x chart, which measures between-sample variability only, an R chart is used to monitor ____
a) Both between-sample variability and within-sample variability
b) Within-sample variability only
c) Between-sample variability only
d) Neither between-sample variability nor within-sample variability
Answer: b
Explanation: A R chart is a chart which uses sample ranges to calculate its center line and control charts. So it only measures within-sample variability, i.e. the instantaneous variability at a given time.
6. When using standard values of process mean and standard deviation, the equation of UCL for a x chart is given as, UCL = μ+Aσ
What is the value of A here?
a) 6/√n
b) 3/√n
c) √n/6
d) √n/3
Answer: b
Explanation: When using the standard values of process mean and standard deviation, A in the equation of UCL for x chart is given by
A=\(\frac{3}{n^{0.5}} = \frac{3}{\sqrt{n}}\)
7. For standard values of mean and standard deviation used, what does the center line of the R chart represent?
a) R bar
b) d 2 σ
c) D 2 σ
d) d 2 R
Answer: b
Explanation: For the standard values of mean and standard deviation used, the center line for a R chart is determined by the following formula, d 2 σ.
8. The control limits obtained by specifying the type I error level for the test, are called ________
a) Probability limits
b) Trial limits
c) Error limits
d) Unreliable limits
Answer: a
Explanation: It is possible to define the control limits for a control chart by specifying the type I error level. The limits obtained by this way, are called the Probability limits for the control chart.
9. Which chart should be interpreted first when both, x chart, and R chart are indicating a non-random behavior?
a) x chart
b) R chart
c) X and R chart
d) Trial Limits
Answer: b
Explanation: It is a property of control charts for x bar and R, that if they both indicate non-random behavior, and if R chart is interpreted first to delete the assignable causes in it, it will automatically delete assignable causes in x bar chart.
10. Which of these is a cause of trend patterns on a control chart?
a) Gradual wearing out of some critical process component
b) Operator fatigue
c) Environmental changes
d) Over-control
Answer: a
Explanation: A trend or continuous movement in one direction on a control chart is generally caused by gradual wearing out of some critical process component like deterioration of a tool.
11. Shift in process level can be seen on the control charts when __________
a) Operator fatigue occurs
b) Temperature changes
c) Over-control of process
d) New workers introduction
Answer: d
Explanation: Shift in the process level is a phenomenon seen on the control chart patterns. It occurs when there is introduction of new workers, or there is a change of methods.
12. Stratification is defined as ________
a) Tendency for the points to cluster artificially around the center line
b) Shift in the process level
c) Continuous movement of points in one direction
d) When the points fall near or slightly outside the control limits
Answer: a
Explanation: Stratification is defined as the tendency of the points on a control chart, to cluster around the center line of the control chart. It occurs very frequently while applying control charts to a process.
13. Stratification of points on a control chart indicates __________ of natural variability of the process.
a) Lack
b) Increase
c) Constancy
d) Randomness
Answer: a
Explanation: Stratification makes the points in a control chart to be plotted around the control chart center line; which indicates that the randomness is gone and there is a lack in natural variability of the process.
14. Never attempt to interpret the x chart when the R chart indicates the out of control condition.
a) True
b) False
Answer: a
Explanation: It is easy to eliminate the assignable causes in the x chart by eliminating the assignable causes in the R chart. This implies, if there is indication of out-of-control condition of process from the R chart and x chart; then first, R chart must be interpreted.
15. The 3 sigma limits on x bar control charts imply that the type I error probability is __________
a) 0.0012
b) 0.0072
c) 0.0027
d) 0.0037
Answer: c
Explanation: The 3 sigma control limits on x bar control charts will imply that the type I error probability is 0.0027 or 0.27%. It will give 2700 defectives per million units of production.
This set of Statistical Quality Control online quiz focuses on “Variable Charts – Control Charts for x̅ and R – 4”.
1. The assumption that links normality to the control charts is ___________
a) The underlying distribution of the quality characteristic is normal
b) The normal mean will be equal to the process mean
c) The normal distribution is not the correct distribution for all the quality characteristic
d) Every distribution is a part of normal distribution
Answer: a
Explanation: There is an assumption that links the normality to the control charts in the development of the performance properties of x bar and R charts; that is, the underlying distribution of the quality characteristic is normal.
2. The probability of not detecting “an in-control shift” of the mean of the process is said to be the _____
a) α – Risk
b) β – Risk
c) γ – Risk
d) δ – Risk
Answer: b
Explanation: If the mean of a process shifts from one value to another value with process being in-control, and shift is not detected on the control chart, the probability of this event is called the β – Risk.
3. If the sample size is 7 and the Average run length is 122 for a process, what will be the expected number of the individual units sampled?
a) 850
b) 854
c) 867
d) 844
Answer: b
Explanation: We know,
I=nARL
Putting n=7 and ARL=122, we get, I = 854.
4. If β- risk of any process is 0.75, what will be the ARL for that process?
a) 4
b) 1.33
c) 0.86
d) 2
Answer: a
Explanation: ARL is given by,
ARL=\(\frac{1}{1-β}\)
Putting β=0.75, we get ARL=4.
5. Slope of OC curve for x bar chart will _________ when sample size is increased.
a) Decrease
b) Increase
c) Remain same
d) Decrease then increase
Answer: b
Explanation: The OC curve is the curve between the β- risk against the magnitude of the shift in mean expressed in std. deviation units for different sample size. Its slope decreases as the sample size is increased.
6. If the probability of one point plotting out of control limits of a X bar control charts is 0.0143, what will be its expected number of individual units sampled with subgroup size 8?
a) 559
b) 544
c) 530
d) 580
Answer: a
Explanation: We know, ARL= 1 ⁄ α and I=nARL. From question we get, α=0.0143 and n=8.
Using this data we get; I≈559.
7. A mixture pattern in the control chart points is exhibited when __________ occurs.
a) Operator fatigue
b) Change of raw materials
c) Continuous wear of tool
d) Adjustments too often responding to random variation
Answer: d
Explanation: When the “Over-control” occurs, i.e. the operator adjusts the processes very frequently and recurrently, the random pattern of the process data vanishes and a mixed pattern is obtained.
8. If the means for sample 1 to 4 for a process are, 12.67, 22.32, 14.53, 12.11; what value will be the center line of x chart indicating?
a) 15.2
b) 15.4
c) 14.8
d) 14.9
Answer: b
Explanation: The center line of the x chart, indicates the mean of all means of samples; i.e.,
CL=\(\bar{\bar{x}} = \frac{\sum \bar{x}}{N}\)=15.4
9. A permanent change in the sample size is made only because of cost or because has exhibited good stability and fewer resources are allocated for process monitoring.
a) True
b) False
Answer: a
Explanation: The permanent change in sample size is not a frequent activity. So it is done because of some major reasons like cost or extraordinary good performance of process on control chart.
10. Which of these should be plotted on the x bar chart?
a) Counts
b) Defects
c) Problems Solved
d) Measurements
Answer: d
Explanation: Counts, defects and problems solved are all quantities which increase in a discrete amount. Measurements increase in a continuous way. Only Continuous quantities are plotted on the x bar chart.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Variable Charts – Control Charts for x̅ and S – 1”.
1. What is the estimator of standard deviation in the x bar and R charts?
a) Mean of one sample
b) Mean of whole process
c) Range
d) Process capability ratio
Answer: c
Explanation: In x bar and R charts, process standard deviation is estimated indirectly through the use of the range R. x bar is used as an estimator of mean.
2. What does “s” denote in x bar and s charts?
a) Sample
b) Sample standard deviation
c) Process standard deviation
d) Statistics
Answer: b
Explanation: Process standard deviation in the x bar and s charts, is estimated directly instead of indirectly through the use of Range as in x bar and R charts. Here “s” denotes the sample standard deviation.
3. What is an unbiased estimator of unknown variance of a probability distribution?
a) Sample mean
b) Sample standard deviation
c) Sample variance
d) Sample range
Answer: c
Explanation: If σ 2 is the unknown variance of a probability distribution, then an unbiased estimator must be used to estimate σ 2 . In this case, sample variance is used as the required estimator.
4. What is the standard formula of sample variance?
a) \
\
\
\(\frac{\sum_{i=1}^n
^{2}}{n}\)
Answer: a
Explanation: The sample variance is given by the following formula.
s 2 = \(\frac{\sum_{i=1}^n
^{1/2}}{n-1}\).
5. Which of these formulas gives the exact equation for the UCL of s chart with a std. value for σ given?
a) B 6 σ
b) B 5 σ
c) c 4 σ
d) c 3 σ
Answer: a
Explanation: The UCL parameter of the s chart with a std. value for σ given, is expressed by
UCL=B 6 σ.
6. The center line of the s chart with a standard value for σ given, denotes the value of _____
a) B 6 σ
b) c 4 σ
c) B 5 σ
d) c 5 σ
Answer: b
Explanation: The center line of the s chart with a standard value for σ given, denotes the value equal to,
CL=B 6 σ
7. If the sample standard deviations for a process are 1.567, 1.429, 1.323, 1.525, 1.989, 1.457, what will be the mean standard deviation?
a) 1.548
b) 1.858
c) 1.327
d) 1.967
Answer: a
Explanation: The mean standard deviation of the sample standard deviations is given by,
\(\bar{s} = \frac{1}{m} \sum_{i=1}^m s_i\)
Where s i denotes the standard deviation of ith sample. Calculating the mean using the above formaula gives, s =1.548.
8. What is the value of B 5 in the terms of c 4 ?
a) \
\
\
\(c_4-3\sqrt{
}\)
Answer: a
Explanation: The value of B 5 in the terms of c 4 is given by,
B 5 = \(c_4-3\sqrt{
}\)
9. The center line of the s chart denotes ____
a) Standard deviation of the process
b) Mean of m number of standard deviations, where m is the number of samples
c) c 4 s
d) B 5 s
Answer: b
Explanation: The center line of the s chart denotes the mean of m number of standard deviations, where m is the number of samples. This is the desired value of the sample standard deviation for the process to be in control.
10. What is the value of LCL for the s chart when the standard value for σ is not given?
a) B 5 s
b) B 4 s
c) B 6 s
d) B 3 s
Answer: d
Explanation: The LCL of the s chart gives the value equal to, B 3 s when the standard value for σ is not given. This is the lowest the value of s can be, for the process to be in-control.
11. What is the value of B 3 in the terms of c 4 ?
a) \
\
\
\(1-\frac{c_4}{3\sqrt{
}}\)
Answer: c
Explanation: The value of B 3 in the terms of c 4 is given by,
B 3 = \(1-\frac{3}{c_4} \sqrt{
}\)
12. What is the formula for UCL for x bar chart when s is known?
a) \
\
\
\(UCL = \bar{\bar{x}} + A_2 \bar{s}\)
Answer: a
Explanation: The formula of UCL for x bar and s chart construction when s is known, is given by
\(UCL = \bar{\bar{x}} + A_3 \bar{s}\)
13. For mean of all sample standard deviations=0.0094 and the sample size= 5, what will be the estimate of process standard deviation?
a) 100
b) 0.01
c) 0.0094
d) 94
Answer: b
Explanation: We know that estimate of the process standard deviation,
\(\hat{\sigma} = \frac{\bar{s}}{c_4} \)
Here for sample size=5, c 4 =0.94, and s = 0.0094, we get σ = 0.01.
14. Process standard deviation is the mean of all sample standard deviations.
a) True
b) False
Answer: b
Explanation: It is not necessary that process standard deviation is the mean of all sample deviations. This is because there is some inherent and natural variability in the process. This may or may not appear in every sample.
15. X bar and R charts are highly favorable when the sample size is high.
a) True
b) False
Answer: b
Explanation: X bar and R charts are not used for high sample sizes- say, n>10 or 12. This is because the range method for estimating σ loses its efficiency for moderate to large samples.
16. Which of this is a situation when x bar and s charts should be utilized instead of x bar and R charts?
a) When sample size is constant
b) When sample standard deviation is less than 1
c) When sample range is more than 1
d) When sample size is variable
Answer: d
Explanation: It is favorable to use the x bar and s charts over x bar and R charts when the sample size is variable. This is because when sample size is variable, it leads to a changing center line of R chart, which is difficult to interpret.
This set of Statistical Quality Control Question Bank focuses on “Variable Charts – Control Charts for x̅ and S – 2”.
1. When the sample size is variable, which one of these can be used to evaluate the value of x double bar?
a) \
\
\
\( \frac{\sum_{i=1}^m n_i \bar{x_i}}{\sum_{i=1}^m n_i}\)
Answer: d
Explanation: When sample size is variable, the formula to evaluate the process mean is given by,
\( \bar{\bar{x}} = \frac{\sum_{i=1}^m n_i \bar{x_i}}{\sum_{i=1}^m n_i}\)
2. Which one of these is correct to evaluate the mean standard deviation of the process samples?
a) \
\
\
\(\bar{s} = \frac{\sum_{i=1}^m
s_i^2}{\sum_{i=1}^m n_i-m}\)
Answer: c
Explanation: The mean of sample standard deviations in the case of variable sample size, is evaluated by following formula,
\(\bar{s} = \left[\frac{\sum_{i=1}^m
s_i^2}{\sum_{i=1}^m n_i-m}\right]^{1/2}\)
3. Which of these is taken as the sample size while estimating process standard deviation, where sample size is variable?
a) n i ; Which is the highest among all sample sizes
b) n i ; Which is the lowest among all sample sizes
c) n i =5
d) n i ; Which is the most frequently occurring among all the sample sizes
Answer: d
Explanation: When there are samples of variable sample sizes and we have to estimate process standard deviation, we use the sample size which is most frequenty occurring among all sample sizes.
4. The control charts based directly on the sample variance are called ____________
a) s Control charts
b) σ 2 Control charts
c) s 2 Control charts
d) x Charts
Answer: c
Explanation: The variable control charts are based on the range method or standard deviation method, but some practitioners use charts based directly on the sample variance, which are called the s 2 control charts.
5. What is the value of the center line of the sample variance control chart?
a) \
s 2
c) \
\(\overline{s^2}\)
Answer: a
Explanation: The center line in the case of sample variance control chart or s 2 control chart denotes the value of \(\bar{s}^2\).
6. Which of these is a name of s control chart?
a) s 2 Chart
b) Process standard deviation chart
c) σ Chart
d) σ 2 Chart
Answer: c
Explanation: The s chart or sample standard deviation chart is also called σ chart by some practitioners. This is based on the control limits set by calculations using the sample standard deviations.
7. X bar chart should be interpreted before s chart if both are indicating out of control situations.
a) True
b) False
Answer: b
Explanation: The base rule of x bar and s chart is that if both charts indicate out-of-control situation then the s chart should be interpreted first as deleting assignable causes in the s chart will automatically delete assignable causes in x bar chart.
8. X bar and S chart are more accurate in predicting out-of-control situations than the x bar and R charts, in the case of high sample size.
a) True
b) False
Answer: a
Explanation: X bar and R charts have a high level of β- risk when the sample size is high. This means they lose their efficiency at high or moderate sample sizes. So x bar and s charts are more accurate than the former ones.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Variable Charts – Shewhart Control Chart for Individual Measurements”.
1. The Shewhart control charts for sample size n=1 are called _________
a) Single Sample control charts
b) Stationary control charts
c) Control charts for zero variance
d) Control charts for individual measurements
Answer: d
Explanation: The Shewhart control charts for sample size n=1 are called, Control charts for individual measurements. It has no basis of rational sub grouping in this case.
2. In the case of automatic manufacturing, the sample size is ___________
a) 1
b) 2
c) 4
d) Greater than 5
Answer: a
Explanation: In the case of automated inspection and measurement technology, every unit manufactured is analyzed and so there is no basis of rational sub grouping. So the sample size here, is 1.
3. In the case of individuals control charts, which of these is used?
a) Relative range
b) Process standard deviation
c) Mean of the highest observations
d) Moving range
Answer: d
Explanation: In the case of individual measurements, we use the moving range of two successive observations as the basis of estimating the process variability.
4. MR is defined as
a) The difference of highest observation and the lowest observation
b) The difference between any two successive observations
c) The difference between the highest observation and mean of observations
d) The difference between the lowest observation and mean of the observations
Answer: b
Explanation: The difference between any two observations in the data is called the Moving range or MR of the two observations. This helps in estimation of the process variability in the case of the individuals control chart.
5. What is the correct expression for the UCL for an Individuals Control Chart?
a) UCL=\
UCL=\
UCL=\
UCL=\(\bar{x}+\frac{\overline{MR}}{d_2}\)
Answer: a
Explanation: The UCL for any individuals control chart is given by the following expression,
UCL=\(\bar{x}+\frac{3\overline{MR}}{d_2}\)
6. The center line for any individuals control chart represents the value equal to __________
a) The process mean
b) The moving range
c) The mean of moving ranges
d) The process standard deviation
Answer: a
Explanation: The center line of any individuals control chart represents the value equal to x or the process mean for that particular process.
7. The value of ARL for any x bar control chart having 3-σ limits is __________
a) 482
b) 310
c) 370
d) 270
Answer: c
Explanation: For any shewhart x bar control chart having 3 – sigma limits, we have p=0.0027. The average run length is calculated by formula,
\(ARL=\frac{1}{p}=\frac{1}{0.0027}=370\)
8. The average number of points that must be plotted before an out-of-control point is plotted on an Individuals control chart?
a) ARL
b) ATS
c) MTBF
d) Hazard rate
Answer: a
Explanation: The average number of points that must be plotted before a point indicating an out-of-control situation is plotted, is called the Average run length of any control chart.
9. The ARL for any individuals control chart with the 1 sigma limits, and the β=0.972, is ____
a) 2
b) 6.3
c) 43.96
d) 13
Answer: c
Explanation: We know that for any individuals control chart,
ARL=\(\frac{1}{1-\beta}\)
Putting the values of β=0.9772, we get ARL=43.96.
10. Which of this stands correct for the relation between the ARL for a shewhart control chart and the individual control chart ARL?
a) ARL individual control chart =1/ ARL shewhart control chart
b) ARL individual control chart = ARL shewhart control chart
c) ARL individual control chart ≪ARL shewhart control chart
d) ARL individual control chart ≫ ARL shewhart control chart
Answer: c
Explanation: The Average run length for any individuals control chart with 3 sigma limits is considerably lower than the value of Average run length for any Shewhart control chart.
11. What is the correct expression for average moving range for any individuals control chart?
a) \
\
\
\(\overline{MR} = \sum_{i=1}^m \frac{MR_i}{m-1}\)
Answer: a
Explanation: The average moving range for any individuals control chart is given by,
\(\overline{MR} = \sum_{i=2}^m \frac{MR_i}{m-1}\)
12. The estimator of the process standard deviation for any individuals control chart is written as?
a) \
\
\
\(1.128\overline{MR}\)
Answer: c
Explanation: The estimation of the process standard deviation for any individuals control chart is written as,
\(\widehat{σ_1}=\overline{MR}/d_2\)
For individuals control chart, n=2. For n=2, we have d 2 =1.128. So we get,
\(\widehat{σ_1}=\overline{MR}/1.128\).
13. The estimation of the process standard deviation for any individuals control chart is expressed as ____ in the terms of sample standard deviation.
a) \
\
\
\(\widehat{σ_2}=\frac{s}{c_3}\)
Answer: a
Explanation: The estimation of the process standard deviation for any individuals control chart is expressed as \(\widehat{σ_2}=\frac{s}{c_4}\) in the terms of sample standard deviation, s.
14. Moving range can only be calculated by the two successive observations.
a) True
b) False
Answer: b
Explanation: The moving range is mostly called as MR of span two because it uses 2 successive observations difference for MR calculation. The span can be increased, MR can be calculated by more than 2 observations.
15. “\
True
b) False
Answer: a
Explanation: The estimation for the process standard deviation is expressed in the terms of the median of the span-two moving ranges by the following expression,
\(\widehat{σ_3} = 1.047\overline{MR}\)
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Charts – Control Charts for Fraction Nonconforming – 1”.
1. Defectives word has almost same meaning as __________
a) Conforming
b) Nonconforming
c) Non-defective
d) Un-conforming
Answer: b
Explanation: A nonconforming item is an item which does not conform to the specifications defined for its quality characteristics. A defective is having almost same as the nonconforming item.
2. Quality characteristics which are related to only conforming or non conforming products, are called _________
a) Attributes
b) Continuous characteristics
c) Discrete characteristics
d) Variables
Answer: a
Explanation: The quality characteristics which are related to only conforming or non conforming products; are called attributes.
3. The control chart which relates to the fraction of defective product produced by a manufacturing process, is called _________
a) The control chart for nonconformities
b) Control charts for fraction nonconforming
c) Control charts for conformities per unit
d) Control chart for process mean
Answer: b
Explanation: The control charts which are related to the fraction of nonconforming or defective product produced by a manufacturing process is called the control chart for fraction nonconforming.
4. The control chart for fraction nonconforming is also called __________
a) u chart
b) c chart
c) p chart
d) R chart
Answer: c
Explanation: The control chart for fraction nonconforming is also called p chart. This chart is plotted for the fraction of nonconforming products produced by a manufacturing process.
5. The control chart designed to deal with the defects or nonconformities of a product, is called __________
a) p chart
b) c chart
c) R chart
d) s chart
Answer: b
Explanation: In some situations, it is more convenient to deal with the no of defects rather than the fraction nonconforming. The control charts designed for this purpose are called the c charts.
6. The c charts are also called _________
a) The control chart for nonconformities
b) Control charts for fraction nonconforming
c) Control charts for conformities per unit
d) Control chart for process mean
Answer: a
Explanation: The control charts which are made for the analysis of the defects in a product, are generally called the control charts for fraction nonconforming, or c charts.
7. The control charts used for the analysis of nonconformities per unit of a product, are called _________ charts
a) p
b) c
c) x
d) u
Answer: d
Explanation: The control charts useful in the situations where the average number of defects or conformities per unit, is more convenient basis for process control; are called the u charts.
8. The u charts are also called _________
a) The control chart for nonconformities
b) Control charts for fraction nonconforming
c) Control charts for conformities per unit
d) Control chart for process mean
Answer: c
Explanation: The control charts plotted for the conformities per unit are called u charts. They are helpful in the situations where average number of defectives is very important to determine process state.
9. The ratio of the number of nonconforming items in a population to total number of items in that population, is called _________
a) Fraction nonconforming
b) Fraction of nonconformities
c) Fraction of conformities per unit of product
d) Fraction of variability
Answer: a
Explanation: The fraction nonconforming is delineated by the ratio of nonconforming items in a population to the total number of items in the population.
10. The sample fraction nonconforming is expressed as __________
a) \
\
\
\(\hat{p}=\frac{D}{n}\)
Answer: d
Explanation: The sample fraction nonconforming is expressed by,
\(\hat{p}=\frac{D}{n}\)
Where D is nonconforming units and n is the sample size.
11. The center line of control chart of fraction nonconforming represents the value equal to ____________
a) Fraction nonconforming
b) Process mean
c) Process standard deviation
d) Sample mean
Answer: a
Explanation: The center line of the control chart of fraction nonconforming represents the value equal to the fraction nonconforming, i.e. the ratio between the number of nonconforming units to the total sample size.
12. If there are 9 items defective in the sample size of 28, what will be the value that the fraction nonconforming chart, will represent?
a) 0.2971
b) 0.3214
c) 0.6328
d) 0.8172
Answer: b
Explanation: The fraction nonconforming chart center line when standards are given, represents the value of fraction nonconforming which in this case, where m=1, D=9, n=28, is,
\(\bar{p}=\frac{[∑_{i=1}^m D_i]}{mn}=\frac{9}{28}\)=0.3214
13. If standards are not given, the estimate of the unknown fraction nonconforming of the process, is evaluated by formula ___________
a) \
\
\
\(\bar{p}=\frac{[∑_{i=1}^{n} D_i]}{mn}\)
Answer: c
Explanation: The estimation of the process’ unknown fraction nonconforming is done by taking average of all the sample fraction nonconforming, i.e. \(\bar{p}\)
\(\bar{p}=\frac{[∑_{i=1}^{m} D_i]}{mn}\)
14. The fraction nonconforming says the same thing as the term defects per unit.
a) True
b) False
Answer: b
Explanation: Fraction nonconforming is the ratio of the number of defectives in a sample to its sample size. “Defects per unit” counts the defects in a single unit.
15. Variable control charts can also be applied on attributes but their efficiency will be less as compared to the attribute control charts.
a) True
b) False
Answer: b
Explanation: Variables are incremented on a continuous scale whereas, attributes do not increment on a continuous scale. Attribute data is about conformity or nonconformity of a sample. As variable charts cannot be used for conforming and nonconforming data, they are not applied on attributes data.
This set of Statistical Quality Control Questions and Answers for Entrance exams focuses on “Attribute Charts – Control Charts for Fraction Nonconforming – 2”.
1. Which of these gives a correct equation for the general model for Shewhart control chart for a “w” quality characteristic statistic?
a) UCL = μ w + Lσ w
b) UCL = μ w – Lσ w
c) LCL = μ w + Lσ w
d) LCL = μ w + σ w
Answer: a
Explanation: The Upper control limit for constructing a Shewhart control chart for a quality characteristic w, we have;
UCL = μ w + Lσ w .
2. The value of L=3 in the genera model of control limits for a Shewhart control chart, explains ____________
a) There are 3 sigma limits taken
b) There are 3 quality characteristics
c) There are 6 quality characteristics
d) There are 6 sigma limits taken
Answer: a
Explanation: The value of L=3 in the general model of control limits for a Shewhart control chart signifies that there are 3 sigma limits taken, i.e. the 3 times shift of the standard deviation, of the mean will be measured on either side.
3. When the Lower control limits for a p chart are less than zero, what is done?
a) The value LCL<0 is utilized in control chart
b) There is a certain constant C added to both UCL and LCL, i.e. UCL new =UCL old + C and LCL new =LCL old +C
c) There is certain constant value added to LCL only, i.e. LCL new =LCL old +C
d) LCL=0 is taken and assumed that control chart only has the upper control limit
Answer: d
Explanation: The value of LCL is taken to be zero when there is a value less than 0 found for LCL. It is also assumed that the control chart only has the upper control limit.
4. LCL for any p chart when the standard values are given are ________
a) LCL=\
LCL=\
LCL=\
LCL=\(p-\sqrt{\frac{p}{n}}\)
Answer: d
Explanation: The LCL for any p chart or control chart for fraction nonconforming, when the standard values are given, are written as,
LCL=\(p-\sqrt{\frac{p}{n}}\)
5. The UCL for any p chart, when the standard values are not given, for 3 sigma limits, is written as ____
a) UCL=\
UCL=\
UCL=\
UCL=\(\bar{p}-\sqrt[3]{\frac{\bar{p}
}{n}}\)
Answer: c
Explanation: The upper control limit for a 3 sigma limit p-chart is expressed by
UCL=\(\bar{p}+\sqrt[3]{\frac{\bar{p}
}{n}}\)
when the standard values are not given.
6. What is done when there is a sample plotted out of control limits for a p-chart?
a) The sample is investigated for an assignable cause and then the sample data is eliminated to develop a new p-chart
b) The sample is only investigated for an assignable cause
c) The sample is not investigated at all
d) All the samples are investigated
Answer: a
Explanation: When a sample plots out of control in p-chart, it is investigated for an assignable cause and then the sample data is eliminated to develop a new p-chart.
7. Even if the process is in control according to p-chart and the fraction nonconforming is too high, it states that _________
a) The process is not stable
b) The process is stable but there are no operator controllable problems
c) The process is stable but there are some operator controllable problems
d) The process is out of control
Answer: b
Explanation: If the process is in control according to p chart but the fraction nonconforming is too high, it states that the process is stable with no operator controllable problems.
8. If the fraction nonconforming for 7 samples are 0.11,0.24,0.21,0.14,0.24,0.21,0.17, what is the value for the center line for a p-chart?
a) 0.19
b) 0.21
c) 0.12
d) 0.13
Answer: a
Explanation: The center line of a p chart represents the average of the all the fraction nonconforming. This means,
\(CL=\bar{p}=\frac{\sum_{i=0}^n p_i}{n}\)
This gives, CL=0.19 for the above question.
9. If the sample size for a p-chart is 50 and the value for the center line of the chart is 0.2313, what will be the value of the LCL of the chart?
a) 0.4108
b) 0.0524
c) 0.0762
d) 0.0389
Answer: b
Explanation: The LCL of the p chart is given by expression,
LCL=\(p-\sqrt{\frac{p}{n}}\)
When we put the value of CL= p =0.2313, we get LCL=0.0524.
10. The control limits for p-chart found from the use of estimated unknown fraction non conforming, are regarded as __________
a) Final limits
b) Concluded limits
c) Trial limits
d) Absolute limits
Answer: c
Explanation: The control limits for p chart, which are not found using the standard values of p, are generally regarded as trial limits. The finalized limits are calculated by first implementing the trial limits.
11. Which of these is not one of the parameters which need to be specified for fraction nonconforming control charts?
a) Sample Size
b) Frequency of sampling
c) Width of control limits
d) Units to be produced
Answer: d
Explanation: the fraction nonconforming chart has 3 parameters which need to be specified: the sample size, the frequency of sampling, and the width of control limits.
12. If δ is the magnitude of the process shift, n must specify __________
a) \
\
\
\(\delta=L\sqrt{\frac{np}{n}}\)
Answer: c
Explanation: The magnitude of process shift and the sample size n must satisfy,
\(\delta=L\sqrt{\frac{p}{n}}\)
for a process to be in control.
13. Which of these is true?
a) \
^2 p\)
b) \
^2 p\)
c) \
^2 \)
d) \
^2 p\)
Answer: a
Explanation: The value for sample size of a p-chart is determined by the following expression,
\
^2 p\)
14. In the p-chart, even if only one point is out of control, we should conclude that the process is out-of-control.
a) True
b) False
Answer: b
Explanation: It is generally not an intelligent decision to conclude the process state as out-of-control by obtaining only one point out of control in p-chart; since for any p>0, there is a probability of producing some defectives.
15. Non-defective term has the same meaning as Nonconforming.
a) True
b) False
Answer: b
Explanation: Non-defective product means a product, which is not defective, or which conforms to specifications defined for it. A nonconforming product does not conform to the specifications defined for it.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Charts – Control Charts for Fraction Nonconforming – 3”.
1. For LCL for the p-chart to be higher than zero, what condition should the sample size fulfill?
a) \
\
\
\(n<\left[\frac{ L^2}{p}\right] \)
Answer: a
Explanation: For the LCL of the p-chart to be higher than zero, the sample size n should fulfill the following condition.
\(n>\left[\frac{ L^2}{p}\right] \)
2. For what value of sample size, will the p-chart will have a LCL value higher than 0 when p=0.05 and 3 sigma limits are used?
a) n>121
b) n>100
c) n>125
d) n>171
Answer: d
Explanation: We know for a p chart having LCL value higher than zero, we have,
\(n>\frac{1-p}{p}*L^2 \)
For p=0.05 and L=3, we get, n>171.
3. For narrower limits of control, which of this is true?
a) Chart becomes insensitive to small shift
b) Chart becomes more sensitive to small shift but gives false alarms
c) Chart becomes more sensitive to small shift but gives accurate alarms
d) Chart gives no alarms at all
Answer: b
Explanation: If the narrower limits like 1-sigma limits or 2 – sigma limits are chosen for p-chart, the chart becomes more sensitive to small shifts. But in addition, it also gives more numbers of false alarms.
4. If the probability of a unit being conforming or nonconforming, depends on the previous unit being conforming or nonconforming, can p-chart be applied on the process?
a) Yes
b) No
c) This depends on the sample size
d) This depends on the operator changes
Answer: b
Explanation: In processes where nonconforming units are clustered, i.e. the probability of one unit to be conforming or nonconforming depends on the previous unit being defective or not, p-chart can’t be applied to them, as p-charts are designed on the basis of binomial probability or independent trials.
5. The number nonconforming chart is also called ___________
a) np-chart
b) p-chart
c) c-chart
d) s-chart
Answer: a
Explanation: The number nonconforming chart is also called the np-chart. This uses the number of nonconforming items in a sample instead of the fraction nonconforming.
6. The center line of the np-control chart represents the value equal to ____________
a) p
b) p
c) np
d) 1-np
Answer: c
Explanation: The center line of the np chart or the number nonconforming chart represents the value equal to np or n p .
7. If the p =0.2313 for a np chart, and the number of items in a sample are 50, what will be the center line value of the np chart?
a) 10.34
b) 11.56
c) 10.11
d) 13.21
Answer: b
Explanation: The center line of the np chart represents the value equal to,
CL=n p
Here, n=50 and p =0.2313, putting these values in the CL value, we get, CL=11.565.
8. For a control chart data having the average sample fraction nonconforming= 0.2313 and the sample size=50, what will be the value of the UCL for the np control chart?
a) 2.62
b) 20.510
c) 11.56
d) 11.892
Answer: b
Explanation: We know that for a np chart,
UCL=\(n\bar{p}+3\sqrt{n\bar{p}
}\)
Putting the values, we get, UCL=20.51.
9. What is the expression to calculate the variable-width control limits for the p-chart?
a) \
\
\
\(\bar{p} \pm \sqrt{\bar{p}
/n_i}\)
Answer: c
Explanation: The variable-width control limits of the p-chart, when the sample size is varying, are,
\(\bar{p} \pm 3\sqrt{\bar{p}
/n_i}\)
10. For a process, there are 25 samples taken from its output of variable sample size. There are total 234 defects in all the samples combined. If the total of all the sample sizes is 2450, what will be the value of the UCL in the case of variable-width control limits of p-chart?
a) \
\
\
\(0.087-3\sqrt{0.087*\frac{0.913}{n_i}}\)
Answer: a
Explanation: The UCL of variable-width control limits of p-chart is written as,
\(\bar{p} + 3\sqrt{\bar{p}
/n_i}\)
Here \(\bar{p} = \frac{234}{2450}\)=0.096. Putting this value in UCL formula, we get,
\(0.096+3\sqrt{0.096*\frac{0.904}{n_i}}\)
11. Which one of these is not a method to plot the variable sample size data on p-chart?
a) Variable-width control limits
b) Control limits based on average sample size
c) Tolerance diagram
d) Standardized Control Chart
Answer: c
Explanation: The three approaches to deal with variable sample size are: variable-width control limits, control limits based on average sample size, and standardized control chart.
12. In the case of average sample size based control limits, which of these states the correct expression for average sample size?
a) \
\
\
\(\bar{n}=\frac{∑_{i=1}^m n_i}{1-m}\)
Answer: b
Explanation: The average sample size in the case of control limits based on average sample size, is expressed by,
\(\bar{n}=\frac{∑_{i=1}^m n_i}{m}\)
13. The approximate upper control limit calculated for control chart limits based on an average sample size, is expressed by _____
a) UCL=\
UCL=\
UCL=\
UCL=\(\bar{p} + 3\sqrt{\bar{p}
/\bar{n}}\)
Answer: d
Explanation: The approximate upper control limit for control limits of p-chart based on an average sample size, is expressed by,
UCL=\(\bar{p} + 3\sqrt{\bar{p}
/\bar{n}}\)
14. Even if 100% inspection is done, control chart can have a variable sample size.
a) True
b) False
Answer: a
Explanation: In some applications of the p-chart, the sample is 100% inspection of process output over a time period. Since different number of units could be produced in each period, the control chart would then have a variable sample size.
15. The points plotting below the lower control limit of p-chart do not always represent the out-of-control process or an assignable cause.
a) True
b) False
Answer: a
Explanation: The point can be plotted below the LCL of p-chart because of errors in inspection process resulting from inadequately trained or inexperienced inspectors. Sometimes, inspectors pass deliberately the nonconforming units too.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Charts – Control Charts for Fraction Nonconforming – 4”.
1. The standardized control chart has the center line at ____________
a) 1 σ
b) 2 σ
c) Zero
d) -1σ
Answer: c
Explanation: The standardized control chart used for control charts for fraction nonconforming with variable sample size, has its center line at the value zero.
2. To deal with the variable sample size, the standardized chart is used, which has its upper limit at ___________
a) 3
b) -3
c) 1
d) 0
Answer: a
Explanation: The standardized control chart, which has been used over quite some time to handle the situations of variable sample size, has its upper limit at +3.
3. A type II error is also called a ___________
a) α Error
b) δ Error
c) β Error
d) ∅ Error
Answer: c
Explanation: A type II error is concluding a process in-control when it is actually in out-of-control state. A type II error is also called an β error.
4. An OC curve is a display of ___________
a) β Error against defectives in a process output over a limited period of time
b) β Error against time
c) β Error against the process fraction nonconforming
d) Fraction nonconforming against β Error
Answer: c
Explanation: An OC curve or Operation characteristic curve for a p-chart is a graphical representation of β Error against the process fraction nonconforming computed.
5. ARL for the fraction nonconforming control chart, is calculated by ___________
a) ARl=1/β
b) ARL=β/α
c) ARL=1/βn
d) ARL=\(\frac{1}{probability\, of \,sample \,point\, plots\, out\, of\, control}\)
Answer: d
Explanation: The average run lengths for the fraction nonconforming control charts are calculated by the following formula,
ARL=\(\frac{1}{probability\, of \,sample \,point\, plots\, out\, of\, control}\).
6. If the process is in control, the ARL for the fraction nonconforming chart, is calculated by __________
a) \
\
\
\(ARL_0=\frac{1}{\alpha \beta}\)
Answer: a
Explanation: The average run length, for the process of Fraction nonconforming control chart for processes in-control, is expressed by,
ARL 0 =1/α.
7. If the for a fraction nonconforming chart, the probability of a sample point plotting out of control is 0.0531, what will be the ARL for this?
a) 12.31
b) 8.16
c) 18.83
d) 22.78
Answer: c
Explanation: The fraction nonconforming chart which has the probability of a sample point plotting out of control as p, the ARL is given by,
ARL = 1 ⁄ p
Putting p=0.0531, ARL=18.83.
8. If a process is out of control, what will be its ARL for the fraction nonconforming chart?
a) \
\
\
\(ARL_1=\frac{1+β}{1-β}\)
Answer: a
Explanation: When the process is out of control, it has possibility of β error. The Average run length for the fraction nonconforming chart of this process, is calculated by following formula,
\(ARL_1=\frac{1}{1-β}\)
9. If type II error probability for a process is 0.8594, its ARL will be ____
a) 8
b) 1
c) 3
d) 7
Answer: d
Explanation: We know,
\(ARL_1=\frac{1}{1-β}\)
We have, β=0.8594, so ARL 1 ≅7
10. We cannot analyze abnormal runs or patterns directly, on p-chart with variable sample size.
a) True
b) False
Answer: a
Explanation: We cannot analyze abnormal runs or patterns directly on p chart with variable sample size. It’s because, the sample nonconforming data can indicate poorer quality for a sample with less defects or variability.
11. The gap between the UCL and LCL is double the gap between either one of them and the Center line, for the standardized chart.
a) True
b) False
Answer: a
Explanation: The Center line for standardized chart is at 0, and UCL and LCL for standardized chart are at ±3 respectively. So the gap between the UCL and LCL is 6, which is double the gap between the gap between either, one of them and the center line, which is 3 units.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Charts – Control Charts for Nonconformities – 1”.
1. The event of a specification not satisfied at one point of a product, is called ____________
a) Nonconformity
b) Non-defectiveness
c) Un-specification
d) Non-specification
Answer: a
Explanation: Each event, in which a specification is not satisfied or an error occurs in approaching the specification value, is called nonconformity or a defect.
2. A nonconforming item has at least ____________ conformities.
a) 4
b) 3
c) 1
d) 2
Answer: c
Explanation: A nonconforming item is defined as the item or product, which has at least one defect. It is necessary that a nonconforming item has one or more than one defects.
3. The c-control chart is plotted for _____________
a) Fraction nonconforming
b) Number of nonconformities per unit
c) Number of defects observed
d) Deviation from median of the defects in samples
Answer: c
Explanation: The c-control chart is based on the total number of nonconformities in a product unit of a process output. So a c-chart is also called control chart for nonconformities.
4. Which of these is an assumption made for designing a control chart for noncorformities?
a) Normal Distribution
b) Poisson Distribution
c) Lognormal Distribution
d) Weibull Distribution
Answer: b
Explanation: The control charts for nonconformities, make an assumption that the sample taken is well modeled by the Poisson distribution. This is a necessary assumption to deal with these control charts.
5. Which of these is not an assumption of the c-chart?
a) Number of potential locations for nonconformities should be large
b) Probability of occurrence of nonconformities should be small
c) Probability of occurrence of nonconformities should be variable
d) Probability of occurrence of nonconformities should be constant
Answer: c
Explanation: The assumption of Poisson distribution of nonconformities in the samples of constant size, for c-charts, makes it necessary that, number of potential locations for nonconformities should be large, and the probability of occurrence of nonconformities should be small and constant.
6. What will be the value of the 3-sigma Upper control limit for the c chart when standard is given?
a) UCL = c + 3√c
b) UCL = c – √c
c) UCL = c + √c
d) UCL = c + 2√c
Answer: a
Explanation: The 3 – sigma upper control limit for the c-chart is expressed algebraically by
UCL = c + √c.
7. The center line for the control chart for nonconformities is representing the value equal c, which is ____________
a) The total number of nonconformities
b) The average number of conformities in a preliminary sample
c) The total number of nonconforming products
d) The total number of conforming products
Answer: b
Explanation: If there is no standard given for the c-chart, the center line represents the estimation of standard, c which is equal to the average number of nonconformities in a preliminary sample.
8. When there is no standard given, the value of LCL for the c-chart is given by ___________
a) LCL=\
LCL=\
LCL=\
LCL=\(\bar{c}+3\sqrt{\bar{c}}\)
Answer: d
Explanation: When there is no standard given for the number of conformities , for the c chart, the lower control limit is given by,
LCL=\(\bar{c}+3\sqrt{\bar{c}}\)
9. If the average number of nonconformities in a preliminary sample of a process is 19.67, which of these represents the value of UCL for a c-chart for this process output?
a) 19.67
b) 6.37
c) 32.97
d) 25.77
Answer: c
Explanation: The UCL for the c-chart is given by,
UCL=\(\bar{c}-3\sqrt{\bar{c}}\)
Now UCL=32.97.
10. For c =13.11, what will be the value of LCL for the c-chart?
a) 23.97
b) 10.86
c) 8.84
d) 2.24
Answer: d
Explanation: The LCL for c-chart is expressed by
LCL=\(\bar{c}+3\sqrt{\bar{c}}\)
Putting c =13.11, we get LCL for the control chart for nonconformities = 2.24.
11. With c-chart, which of these is used to analyze nonconformity data?
a) X bar charts
b) R charts
c) Cause and Effect diagram
d) Tolerance Diagram
Answer: c
Explanation: The c-chart is usually used to analyze the nonconformity data to get more information about the process state. Cause and effect diagram are also a good way to analyze nonconformity data.
12. The control chart used to inspect the process state by using the average number of nonconformities per unit data, is called _______________
a) u-chart
b) c-chart
c) p-chart
d) R-chart
Answer: a
Explanation: The control chart, which uses the average number of nonconformities per unit data to analyze the process state, is generally termed as the Control chart for Average number of Nonconformities per Unit.
13. If x i =total number of nonconformities in a sample of n i inspection units, and there are N samples of different sample size, and x varies according to a Poisson distribution, what will be the value of center line of the u-chart?
a) \
\
\
\(\bar{u} = \frac{\sum_{i=1}^N \frac{x_i}{n_i}}{2N}\)
Answer: c
Explanation: As u-chart uses the value of the average number of nonconformities per unit, the center line of the u-chart is given by,
\(\bar{u} = \frac{\sum_{i=1}^N \frac{x_i}{n_i}}{N}\)
14. For a control chart for nonconformities, the sample size can only be 1.
a) True
b) False
Answer: b
Explanation: For a control chart for nonconformities, sample size can be greater than 1. This increases the area of opportunity for the occurrence of nonconformities. This can be used to obtain a positive lower control limit also.
15. The sample for the control chart for nonconformities is assumed to have, a variable sample size, and the sample conformities distributed according to Normal distribution.
a) True
b) False
Answer: b
Explanation: The control charts for nonconformities, usually assume that the occurrence of nonconformities in samples of constant size, is well modeled by the Poisson distribution.
16. Count rate of a defect is defined as _____________
a) Defects per unit
b) Average number of defects per unit
c) Total number of nonconformities
d) Occurrence of n defects in N number of units
Answer: d
Explanation: Defect level or Count Rate is defined as the number of units to be produced before occurrence of a certain number of defects, say n.
This set of Statistical Quality Control Questions and Answers for Campus interviews focuses on “Attribute Charts – Control Charts for Nonconformities – 2”.
1. The 3 – sigma upper limit for a control chart for average number of nonconformities per unit, is expressed by ___________
a) UCL=\
UCL=\
UCL=\
UCL=\(\bar{u} – \frac{3}{2} \sqrt{\frac{\bar{u}}{n}}\)
Answer: a
Explanation: The 3 – sigma upper control limit for a control chart for average number of nonconformities per unit, is expressed by the following expression.
UCL=\(\bar{u} + 3\sqrt{\frac{\bar{u}}{n}}\)
2. If the average number of nonconformities per unit per sample is 0.074, what will be the lower control limit for the u-chart?
a) 0.0414
b) 0.0221
c) 0.0
d) -0.0513
Answer: c
Explanation: The lower control limit for the u-chart is expressed by,
LCL = \(\bar{u} – 3\sqrt{\frac{\bar{u}}{n}}\)
Putting \(\bar{u}\)=0.074 we get, LCL=-0.0414. As LCL is less than 0, we use LCL=0.
3. The control chart for total number of events is called ____________
a) g-charts
b) R-charts
c) h-charts
d) c-charts
Answer: a
Explanation: The control charts, which are developed for the total number of events, are called the g-charts. They were first developed by Kaminski et al in 1992.
4. The control charts developed for the average number of events is generally called ______________
a) g-chart
b) h-chart
c) s-chart
d) s-square-charts
Answer: b
Explanation: The control charts, which are based on the average number of events data, are called the h-charts. They were also developed by Kaminski et al.
5. What will be the center line value for a Kaminski et al h-chart when standards are given?
a) \
\
\
\(\frac{1-p}{p}+a\)
Answer: d
Explanation: The center line value for the Kaminski et al h-chart is given by the equation,
CL = \(\frac{1-p}{p}+a\)
6. What is the value of the Upper control limit for the g-chart?
a) \
+L\sqrt{n}{p^2}\)
b) \
+L\sqrt{n}{p^2}\)
c) \
+L\sqrt{n}{p^2}\)
d) \
+L\sqrt{}{p^2}\)
Answer: c
Explanation: The value of upper limit for the g chart given by Kaminski et al is,
\
+L\sqrt{n}{p^2}\)
7. What is the value of the center line of g-chart?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: b
Explanation: The center line value for the g-chart construction is calculated by following formula,
\
\)
8. What is the value of center line of g-chart when there is no standard given?
a) \
\
\
\(\bar{t}+0.5\sqrt{\bar{t}}\)
Answer: b
Explanation: The center line value for the g-chart, when there is no standard given is given by,
Cl=\(\bar{t}=\frac{t_1+t_2+t_3+⋯+t_m}{m}\)
where t 1 , t 2 , t 3 ,…,t m are the total numbers of events in m subgroups each.
9. What is the center line value for the average number of events control chart, when there is no standard given?
a) t /n
b) 2t /3n
c) 2 t /n
d) t /2n
Answer: a
Explanation: The center line value for the average number of events control chart or h-chart, when there is no standard given, is expressed by,
CL= t /n.
10. Which of these is used in the standardized chart for u-chart with variable sample size?
a) Z i
b) M i
c) U i
d) A i
Answer: a
Explanation: For the standardized control chart for a u-chart with variable sample size, the construction of the standardized control chart requires plotting a standard statistic written as Z i .
11. What is the value of Z i in the standardized control chart for u-chart with variable sample size?
a) \
\
\
\(\frac{u_i-\bar{u}}{\sqrt{\frac{\bar{u}}{\bar{n}}}}\)
Answer: c
Explanation: The value of a standard statistic used to plot the standardized chart for the u-chart with a variable sample size, is written as,
\(\frac{u_i-\bar{u}}{\sqrt{\frac{\bar{u}}{n_i}}}\)
12. What is the Upper control limit for the standardized control chart for u-chart with variable sample size?
a) 0
b) 1
c) 1.5
d) 3
Answer: d
Explanation: The upper control limit for the standardized control chart, which is designed for u-chart with variable sample size, is generally set to value 3.
13. The center line of the standardized control chart designed for u-chart, is given by _____
a) 1
b) -3
c) 6
d) 0
Answer: d
Explanation: The center line for the standardized control chart is set to zero. That is why this control chart is named the standardized control chart.
14. The Lower control limit has a 6 unit distance from the upper control limit of the standardized control chart.
a) True
b) False
Answer: a
Explanation: The Upper and lower control limits of the standardized control chart designed for u-chart with variable sample size, have a value of 3 and -3, respectively. So the distance between them is 6 units.
15. Defect or nonconformity data are always lesser informative than fraction nonconforming.
a) True
b) False
Answer: b
Explanation: The nonconformity data are always more informative than the fraction nonconforming data, because there will usually be several different nonconformity types. By analyzing all their types, we get more insight into the defect cause which helps in making OCAPs.
16. The OC curve of the c-chart is a curve which plots β-risk against _______
a) Number of defectives per sample
b) True mean number of defects
c) Total number of defects
d) Demerits per unit
Answer: b
Explanation: The OC curve of the c chart is a curve which plots the type II error probability against the true mean number of defects in a process output.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Charts – Control Charts for Nonconformities – 3”.
1. If there is variable sample size to be used with the u-chart or average number of nonconformity per unit chart, which of these methods, is used?
a) DMAIC
b) Designing of Experiments
c) Acceptance sampling
d) Standardized control chart
Answer: d
Explanation: When the sample size is variable, for a u-chart, the method of using the standardized chart is used to deal with the variable sample size. This makes it easy to use all the sample size data with the nonconformity data.
2. The method of average sample size is used when there is _____
a) Variable sample size
b) Constant sample size
c) Sample size=1
d) Sample Size=10
Answer: a
Explanation: When the sample size is variable and the u-chart or the control chart for nonconformity data is to be used, the approach of average sample size is used.
3. In the average sample size approach to design u-chart with variable sample size, which of these is used to calculate the average sample size?
a) \
\
\
\(\bar{n}=\frac{\sum_{i=1}^n u_i}{m}\)
Answer: c
Explanation: The average sample size for the u-chart with variable sample size is calculated using the following formula,
\(\bar{n}=\frac{\sum_{i=1}^m n_i}{m}\)
4. In a demerit system for defects, how many classes are defined for defects to be classified?
a) 3
b) 4
c) 5
d) 8
Answer: b
Explanation: The demerit system for attribute data; classifies the defects in 4 classes; class A, class B, Class C, and Class D.
5. Serious defects are put into which class according to the Demerit system for attribute data?
a) Class C
b) Class D
c) Class A
d) Class B
Answer: d
Explanation: Serious Defects are put into the class B. In serious defects reduce the effective life of the component or increase the maintenance costs to be applied for the use of that component.
6. The Class A defects are ____________
a) Moderate
b) Minor
c) Serious
d) Very Serious
Answer: d
Explanation: The class A defects contains all the defects which are very serious to be present in the part. These defects either make the part totally unfit for use or will fail in service causing property damage with it.
7. The minor defects are put into class ____________ of the demerit system.
a) A
b) B
c) C
d) D
Answer: a
Explanation: The minor defects are the defects which do not affect the service of part effectively. They are classified in the Class D of the Demerit system for attribute data.
8. The minor finish, appearance, or quality of work defects are generally classified in _____ of the demerit system.
a) Class A
b) Class D
c) Class C
d) Class B
Answer: b
Explanation: The minor defects in finish, quality of work, and appearance are generally not serious. They do not affect the life of product effectively. So they are classified into class D of the demerit system.
9. Moderate defects are classified into which class of demerit system?
a) Class B
b) Class C
c) Class D
d) Class A
Answer: a
Explanation: The moderate defects are the defects which cause trouble that is less serious than operating failure.
10. The number of demerits in a system having c iA ,c iB ,c iC ,and c iD defects in classes A, B, C, and D respectively is expressed by ______ It is assumed that the probability of each defect varies according to a Poisson distribution.
a) d i = 100c iD + 50c iB + 10c iC + c iA
b) d i = 100c iA + 50c iB + 10c iC + c iD
c) d i = 100c iB + 50c iA + 10c iC + c iD
d) d i = 100c iA + 50c iC + 10c iB + c iD
Answer: b
Explanation: The number of demerits in a system is expressed by the following expression.
d i = 100c iA + 50c iB + 10c iC + c iD .
11. If there are 8, 12, 11, 3, 6 demerits in sample number 1, 2, 3, 4, and 5 respectively, what will be the demerits per units for the mentioned output?
a) 8
b) 11
c) 3
d) 5
Answer: a
Explanation: The demerits per unit are calculated by using the following formula,
\(u_i=\frac{D}{n}\)
Where D=\(\sum_{i=1}^n d_i\) = 40 and n=5. We get u i =8
12. The upper control limit for the demerits per unit chart is given by _____
a) UCL = \
UCL = \
UCL = \
UCL = \(\bar{u}-\widehat{σ_u}\)
Answer: b
Explanation: The upper control limit for the demerits per unit chart is given by the following formula,
UCL = \(\bar{u}+3\widehat{σ_u}\)
Where \(\bar{u}=100\overline{u_A}+50\overline{u_B}+10\overline{u_C}+\overline{u_D}\)
13. What is the value of the center line of the demerits per unit?
a) \
\
\
\(\bar{u}-3\sqrt{\bar{u}}\)
Answer: b
Explanation: The value of the center line of the demerits per unit chart is given by,
CL = u
14. A serious defect is defined as the defect which makes the unit totally unfit for use.
a) True
b) False
Answer: b
Explanation: The defect which makes the unit totally unfit for use, is called a very serious defect, and it is categorized in Class A demerit of the demerit system of the attribute data.
15. For construction of the control charts for nonconformities, the inspection unit must be same for each sample.
a) True
b) False
Answer: a
Explanation: It is necessary that the inspection unit must be same for each sample for the construction of the control chart for nonconformities. Thus, there will be a same area of opportunity for the occurrence of nonconformities for each inspection unit.
16. The defects which affect the proper working of the unit in the service arena, are called ____________
a) Appearance Defects
b) Functional Defects
c) Specification Defects
d) Aspect defects
Answer: b
Explanation: The defects or nonconformities which affect the proper functioning or working of the part in its area of service, are called the functional defects.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Charts – Choice between Attributes and Variable Control Charts – 1”.
1. The number of defects per unit of a casting process output, is plotted on _____________ chart.
a) x bar
b) s-chart
c) u-chart
d) p-chart
Answer: c
Explanation: The average number of defects or nonconformities per unit of a process output is always plotted on a u-chart or the control chart for average number of nonconformities per unit.
2. What type of control chart can be used to plot “number of defectives in the output of a process for making a machine part” data?
a) p-chart
b) c-chart
c) u-chart
d) s-chart
Answer: a
Explanation: The p-chart or the Control Chart for Fraction Nonconforming is used to plot “the number of defectives in the output of any manufacturing process” data, on a control chart.
3. Which of the control chart is used to find out variability in the data?
a) s-chart
b) x -chart
c) p-chart
d) c-chart
Answer: a
Explanation: s-chart or the control charts based on the sample standard deviation, are a useful method to find out the variability in the data. They are only used for Variable data, not for attribute data.
4. Which of the charts are more efficient to find out variability in the data when the sample size is more than 10?
a) R-chart
b) c-chart
c) s-chart
d) p-chart
Answer: c
Explanation: To find out variability in the data, R-charts and s-charts are used. But when the sample size is more than 10 or 12, the efficiency of s-charts to find variability in the data, is more than the R-charts.
5. Which of these is an advantage of attribute control chart?
a) Much useful information about the process performance can be gathered
b) Mean and variability is obtained directly
c) One quality characteristic is observed at a time
d) Several quality characteristics can be considered jointly
Answer: d
Explanation: The attribute control charts have a main advantage of consideration of many quality characteristics to be jointly analyzed. This is because a defect can be produced by one or more quality characteristics, which are out of control.
6. Which of these is an advantage of variable control chart?
a) Numerous quality characteristics considered at a time
b) To achieve the information very easily about the mean and variability
c) To have analyses of units nonconforming
d) To analyze the defects in one unit
Answer: b
Explanation: The main advantage of a variable control chart is that they provide much useful information about the process performance. They give more functional information about the process mean and variability.
7. X bar and R charts are __________ indicators of trouble.
a) Trailing
b) Inferior
c) Leading
d) Secondary
Answer: c
Explanation: x and R charts are leading indicators of trouble in the process. We conclude this because; they indicate the out-of-control process before the process has made defective products, unlike p-chart, which indicates the out-of control process after the process has made defectives.
8. The efficiency of variable control charts is _________ the efficiency of the p-charts when p is small and far away from 0.5.
a) Lesser than
b) More than
c) Equal to
d) Non-predictable
Answer: b
Explanation: The efficiency of variable control charts is considerable higher than the efficiency of the p-charts when the value of p is small or away from 0.5.
9. The efficiency of p-charts is __________ the efficiency of the x bar and R charts when p is closer to 0.5.
a) Equal to
b) More than
c) Less than
d) Non-predictable
Answer: b
Explanation: The efficiency of p-charts become higher or just more than the efficiency of x-bar and R charts, when the value of p comes closer to 0.5 or higher than 0.5.
10. Which of these is most economical in long term?
a) x And R charts
b) c-chart
c) p-chart
d) u-chart
Answer: a
Explanation: The variable charts are more economical in long term, because they indicate process state out-of-control, before we have paid to make a nonconforming product. Attribute charts become costly because the same indication from them comes, when defectives are already produced.
11. When a change in product specification is desired, which of these control charts should be used?
a) p-charts
b) x And R charts
c) c-charts
d) u-charts
Answer: b
Explanation: The variable control charts should be used whenever a new process is coming on stream, or a new product is being manufactured by an existing process, or a change in product specification is desired.
12. When we want to troubleshoot the out-of-control process, we use _________ control charts.
a) x
b) p
c) c
d) u
Answer: a
Explanation: Whenever a process is out-of-control and it is known to us, we use variable control charts to troubleshoot the process. It is because; we can focus on each quality characteristic individually using variable control charts.
13. In regulated industries, where process stability and process capability must be continually demonstrated __________ charts are used.
a) x And s
b) p
c) c
d) u
Answer: a
Explanation: A regulated industry needs continuous information about the process stability and capability. As variable control charts are more efficient to find out these, they are used in the regulated industries.
14. Before choosing which CTQ-characteristic to be controlled, control chart must be chosen.
a) True
b) False
Answer: b
Explanation: We cannot decide whether to use a variable control chart or an attribute control chart before choosing a quality characteristic. This is because attribute charts are only plotted for attribute data.
15. Proper type of control charts to be prepared, must be chosen after choosing the point in the process at which the control charts will be implemented.
a) True
b) False
Answer: b
Explanation: It is necessary to point out the location where the charts will be implemented, before choosing the control charts to be prepared. This is because the location of implementing charts can reveal more important factors to be considered while choosing the proper control chart.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Control Charts – Choice between Attributes and Variable Control Charts – 2”.
1. Which of these is a condition where attributes control charts should be used?
a) New product manufacturing from existing process
b) Troubleshooting a out-of-control process
c) When it is necessary to reduce process fallout
d) Change in product specification is required
Answer: c
Explanation: When it is necessary to reduce process fallout, and when the operators control the assignable causes, Attributes control charts are chosen over the Variables control chart.
2. When the process is a complex assembly operation, and the product quality can only be judged in terms of occurrence of nonconformities, the ______ are more helpful.
a) Only x Charts
b) R-charts
c) s-charts
d) p-charts
Answer: d
Explanation: When the process is a complex assembly operation and the product quality is assessed only on basis of occurrence of defects, the p-charts are more helpful.
3. When process needs to be controlled but the measurement data cannot be obtained, the control charts to be formed are _____________
a) p-charts
b) Only x Charts
c) s-charts
d) R-charts
Answer: a
Explanation: When process needs to be controlled but the measurement data can’t be acquired, the attributes charts are the better choice than the variables control chart.
4. When there are very tight specifications, overlapping assembly tolerances problems, which of the control charts are used?
a) Attributes control charts
b) Variables control charts
c) Both, attributes control charts and the variables control charts can be used
d) Neither one of attributes or variables control charts can be used
Answer: b
Explanation: Variable control charts are more favored when there are very tight specifications, overlapping assembly tolerances problems. This is because variables control charts deal better with specifications of quality characteristics than the attributes control chart.
5. When destructive testing is required, which type of control should be used?
a) p-chart
b) c-chart
c) x And R charts
d) u-charts
Answer: c
Explanation: When destructive testing is required or any other expensive testing procedure is needed, variable control charts are more suitable than the attributes control chart.
6. When attributes charts are used when _______
a) Destructive testing is required
b) Process must be continuously stabilized and monitored for its capability
c) Historical summary of process performance is needed
d) Existing process but in chronic trouble
Answer: c
Explanation: When historical summary of process performance is needed for the variables control charts, attributes control charts are used to develop the summary data.
7. When there is a need to reduce acceptance-sampling or other downstream testing to a minimum when the process can be operated in-control, which type of control charts should be preferred?
a) Attributes control chart
b) Variable control chart
c) Both can be used
d) None of the both types can be used
Answer: b
Explanation: When the acceptance sampling is to be reduced to minimum, the variables charts reveal better results and efficiency than the attributes control chart.
8. Which of these does not require SPC implementation?
a) When Specifications change without notice
b) When customer demands both control and capability
c) When process is stable
d) When the process is estimated to reduce its capability in near future
Answer: c
Explanation: SPC implementation is required when the specifications change without notice, and customer demands both control and capability, and when the process is predicted to reduce its capability in near future.
9. When it is inconvenient to obtain more than one measurement per sample output of any process, which of these is the best choice?
a) x bar and R charts
b) x bar and s charts
c) c-charts
d) Control charts for individuals
Answer: d
Explanation: The times when, it is inconvenient or impossible to obtain more than one measurement per sample, or repeat measurements will only be affected by lab or analysis error, the control charts for individuals are the best choice, e.g. Often in chemical processes.
10. Which of these should be carried out if the process is in-control and capable?
a) SPC only
b) Experimental design
c) Change process
d) Investigate specifications
Answer: a
Explanation: If the process currently studied is in control and also capable, the control and capability of the process are maintained as it is, by using the statistical process control only.
11. Computer should not be used with SPC.
a) True
b) False
Answer: b
Explanation: Some SPC consultants don’t recommend the use of computers with SPC. But now it is necessary to use computers with SPC as manual control charting cannot be transmitted to whole over company branches. Company-wide database can be updated using computers only.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Attribute Control Charts – Process Capability Analysis using a Histogram or a Probability Plot – 1”.
1. Process capability refers to ___________
a) Changes in process
b) Variability in process
c) Uniformity in process
d) Unevenness in process
Answer: c
Explanation: Process capability refers to the uniformity of the process. This uniformity in the process is always measured in the terms of the variability in the processs.
2. The process of usage of statistical methods to determine the variability in a process and to reduce or completely eliminating the variability, is called ________________
a) Process capability
b) Process capability Analysis
c) Process variability
d) Process variability analysis
Answer: b
Explanation: The method of using the statistical methods to determine the variability in a process and to reduce it to the minimum level, is generally known as the Process Capability Analysis.
3. Determining process capability is an important part of _____________ step of DMAIC process.
a) Analyze
b) Define
c) Control
d) Measure
Answer: a
Explanation: Determining process capability is an important part of analyze step of DMAIC process. It is generally done in the analyze step majorly but it is also used in improve step.
4. Which of these is used as the measure of process capability?
a) Process mean
b) Process standard deviation
c) Sample standard deviation
d) Six-sigma spread in the distribution of the product quality characteristic
Answer: d
Explanation: It is customary to take the six – sigma distribution of the product quality characteristic, as the measure of the process capability, in the process capability analysis.
5. For a six–sigma spread in the distribution of the product quality characteristic, the upper natural control limit will fall at _____________
a) μ + 2σ
b) μ + 3σ
c) μ + 4σ
d) μ + 6σ
Answer: b
Explanation: For six sigma distribution of the product quality characteristic, the upper natural tolerance control limit will fall at,
UNTL = μ + 3σ.
6. μ – 3σ, is the LNTL for ______________
a) 3-sigma spread in distribution of CTQ characteristic
b) 1-sigma spread in distribution of CTQ characteristic
c) 6-sigma spread in distribution of CTQ characteristic
d) 4-sigma spread in distribution of CTQ characteristic
Answer: c
Explanation: The six-sigma spread in the distribution of the CTQ characteristic of a process, is generally having its lower natural tolerance limit at,
LNTL = μ – 3σ.
7. 0.27 percent outside the normal tolerances can be obtained using ____________
a) 6-sigma both sides of mean
b) 3-sigma both sides of mean
c) 2-sigma both sides of mean
d) 8-sigma both sides of mean
Answer: b
Explanation: When there is NTL of μ±3σ, i.e. 3-sigma both sides of the mean of the variable, the 99.73% products are between the specification limit. So there are 0.27% outside the normal tolerances.
8. Which of these is not necessary to find the process capability?
a) Mean
b) Standard deviation
c) Spread
d) Design of Experiments
Answer: d
Explanation: In the process capability analysis, we need to have the following to start the study; mean of the process, spread or the standard deviation of the process, and the specified shape of the probability distribution.
9. When we don’t have the mean, spread and the shape of probability distribution of one CTQ characteristic of a process, we can use ___________ for process capability analysis.
a) Acceptance sampling
b) Design of experiment
c) p-chart
d) Specifications on the quality characteristic
Answer: d
Explanation: In case we don’t have any information about the mean, spread and the shape of the probability distribution of one CTQ characteristic of a process, we can use specifications on the CTQ characteristic.
10. Which of these is not a major use of the PCA ?
a) Prediction of how well the process will hold the tolerances
b) Reduction of variability in process
c) Establishing an interval between sampling
d) Stating the need of the Acceptance sampling
Answer: d
Explanation: PCA or the Process Capability Analysis is used to predict how well the process will hold the tolerance and to reduce the variability. It also gives us an interval between sampling.
11. When in PCA, we have only the sample unit of product without any direct observation of the process or the time history of the production, the PCA is also called ____________
a) Product inspection
b) Product characterization
c) Product sampling
d) Process design
Answer: b
Explanation: In the studies of PCA, when we have availability of only the sample units of product without any direct observation of the process or the time history of the production, PCA study is now called Product Characterization.
12. In product characterization, we cannot say anything about _____________
a) Dynamic behavior of the process
b) Distribution of quality characteristic
c) Process yield
d) Fraction conforming to specifications
Answer: a
Explanation: In product characterization, we can only state the process yield information or the information about the distribution of the product quality characteristic or the fraction conforming to specifications data.
13. Which of these is not one of the primary techniques used to find out the process capability?
a) Histogram
b) Probability plots
c) Control charts
d) Acceptance sampling
Answer: d
Explanation: Histogram or probability plots, control charts and designed experiments are the 3 primary techniques used to find out process capability.
14. Product characterization cannot be used for the determination of the state of the statistical control of the process by which, the products are made.
a) True
b) False
Answer: a
Explanation: During product characterization, we do not have the historical data of production and the direct observation on the process. So we cannot say anything about the state of the process making the products to be characterized.
15. Process capability can be expressed as a percentage outside of the specifications.
a) True
b) False
Answer: a
Explanation: We use the process mean, spread, and the shape of the probability plot of the quality characteristic for the estimation of Process Capability. Alternatively, Process capability can be expressed as a percentage outside of the specifications.
This set of Statistical Quality Control Questions and Answers for Aptitude test focuses on “Attribute Control Charts – Process Capability Analysis using a Histogram or a Probability Plot – 2”.
1. PCA can’t be used _____________
a) To determine the process state of statistical control
b) To reduce the variability in the process
c) For specifying the location of defect
d) For modifying or selecting a process
Answer: c
Explanation: PCA can’t be used to specify the location of defect. It can be used to determine the process state, to reduce the variability present in it and to select or modify any process.
2. How many primary techniques are there which are used in PCA?
a) 3
b) 2
c) 4
d) 5
Answer: a
Explanation: There are 3 primary techniques, which are used in process capability analysis. These are; histogram or probability plots, control charts, and designed experiments.
3. Which can be used as an alternative for histogram for PCA?
a) Check sheet
b) Defect concentration diagram
c) Stem-and-Leaf Plot
d) Scatter Diagram
Answer: c
Explanation: Stem-and-leaf plots can be used alternatively for the process capability analysis, instead of using the histogram plots. It needs almost more than 100 observations.
4. How many observations are needed at least, to plot a histogram which gives proper information about the process capability?
a) 100
b) 120
c) 50
d) 10
Answer: a
Explanation: There must be at least 100 observations or more, to plot a histogram or a stem-and-leaf plot to be stable enough, which gives efficient information about the process capability.
5. Which of these is needed with the histogram to estimate the process capability?
a) Acceptance sampling
b) Scatter diagram
c) Sample average
d) Check sheet
Answer: c
Explanation: The histogram, when used along with the sample average x and sample standard deviation s, provides the sufficient information about the process capability.
6. Which is not an advantage of histogram to check the process capability?
a) Immediate impression of process
b) Visual impression of process
c) Fast interpretation
d) Easy to find the location of defects
Answer: d
Explanation: By using the histogram, to determine the process capability, we can get an immediate and visual impression of process. We can also interpret the data very quickly. But it cannot tell the location of defects.
7. When the number of product units outside the LSL and the number of product units outside USL are same, the process is called?
a) Centered process with poor process capability
b) Centered process with great process capability
c) Poorly centered process with poor process capability
d) Poorly centered process with great process capability
Answer: a
Explanation: When the numbers of product units outside the LSL, and the USL, are same. This means the process is centered, but there are some product units lying outside the specification limits. This gives poor process capability.
8. Probability plotting is an alternative of _________ to find process capability.
a) Control charts
b) Histogram
c) Designed experiments
d) Acceptance sampling
Answer: b
Explanation: Probability plotting is an alternative to the histogram to find out the capability of the process which produces the product, which is to be used in the industry.
9. Probability plotting is not used to find __________
a) The shape of distribution
b) The center of distribution
c) The spread of distribution
d) The origin of defects
Answer: d
Explanation: By using the probability plots instead of the histogram, find out the process capability, we can determine the shape, center, and the spread of the distribution.
10. Which is more efficient in the case of small samples to find out the shape and spread of the distribution?
a) Histogram
b) Probability plots
c) Designed experiments
d) PDCA
Answer: b
Explanation: The advantage of probability plotting is that, it is unnecessary to divide the range of the variable into class intervals, and it often produces reasonable results for moderately small samples .
11. Histogram generally is a __________
a) Graphical representation of intervals of data plotted against frequency
b) Ranked data versus the sample cumulative frequency
c) Ranked data versus the frequency
d) Diagram of unit giving the location of defects
Answer: a
Explanation: A histogram is generally described as a graphical representation of range of data divided into intervals plotted against the frequency of data.
12. Probability plots are graph of ranked data plotted against the cumulative frequency, with a vertical scale chosen so that the cumulative distribution of assumed type is a _________
a) Parabola
b) Straight line
c) Hyperbola
d) Pulse
Answer: b
Explanation: Probability plots are the graphs of the ranked data versus the sample cumulative frequency on special paper. These plots have the vertical scale so chosen, that the cumulative distribution of the assumed type is a straight line.
13. Which of these is a disadvantage of probability plot?
a) They are not an objective procedure
b) They don’t tell about the defect
c) They are hard to interpret
d) They can’t tell about mean of the data
Answer: a
Explanation: Probability plots are quite useful but, they have a major disadvantage that they are not an objective procedure. There can be many conclusions for same set of data.
14. Probability plots are straight line.
a) True
b) False
Answer: a
Explanation: Probability plots are the plot of ranked data against the cumulative frequency. The cumulative probability plots are mostly straight lines.
15. When there is an interactive effect of processes on the tolerances, process capability can’t be used for the planning of production processes.
a) True
b) False
Answer: b
Explanation: It is the main use of process capability to plan the production processes when there is an interactive effect of processes on the tolerances.
16. It is often desirable to supplement probability plots with more formal statistically based goodness-of-fit tests.
a) True
b) False
Answer: a
Explanation: There is a disadvantage of probability plots that they are not objective procedures, i.e. they give more than one conclusions of a single data. That’s why; it’s often desirable to supplement probability plots with more formal statistically based goodness-of-fit tests.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Process Capability Ratios – 1”.
1. Process capability ratio is expressed as ____________
a) \
\
\
\(C_p=\frac{USL+LSL}{6σ}\)
Answer: a
Explanation: Process capability ratio is denoted as C p and expressed as,
\(C_p=\frac{USL-LSL}{6σ}\)
2. USL in the formula of PCR is ____________
a) Upper safe limit
b) Ultimate safe limit
c) Upper specification limit
d) Ultimate specification limit
Answer: c
Explanation: USL in the formula of PCR or process capability ratio is the Upper Specification Limit. This denotes the maximum tolerable value for a quality characteristic.
3. The LSL in the C p is called ____________
a) Lower safe limit
b) Largest safe limit
c) Largest specification limit
d) Lower specification limit
Answer: d
Explanation: The LSL, in the formula of the Process Capability Ratio, C p is the Lower Specification Limit value of the quality characteristic. It specifies the lowest possible value for the quality characteristic for the process to be in statistical control.
4. When the process standard deviation σ is unknown for a process, to calculate the PCR we use _____________
a) The mean
b) The variance
c) The sample average
d) The estimate of σ
Answer: d
Explanation: It is common that the process standard deviation is not known for the process. In this case, to calculate the PCR, we use the estimate of the process standard deviation.
5. We use _____________ to calculate the estimate of the process standard deviation.
a) Sample standard deviation
b) Sample average
c) Sample variance
d) Sample mean
Answer: a
Explanation: When the process standard deviation is not known, we use the estimate of the process standard deviation. This estimate is calculated by using the sample standard deviation, s.
6. How is s calculated?
a) s = R /2
b) s = R /2d 2
c) s = R /d 2
d) s = 2 R /d 2
Answer: c
Explanation: The sample standard deviation, s, used for the estimation of the process standard deviation, is calculated by the formula:
s = R /d 2 .
7. The estimate of the C p is calculated by ______________
a) \
\
\
\(\frac{USL-LSL}{12\hat{\sigma}}\)
Answer: a
Explanation: The estimate of the process capability ratio C p is calculated by,
\(\widehat{C_P} = \frac{USL-LSL}{6\hat{\sigma}}\)
8. If the estimate of the process standard deviation is 0.1368, and the USL for the quality characteristic is 2.00, and the LSL=2.00, what will be the value of the estimate of PCR C p is _____________
a) 1.21
b) 1.19
c) 1.13
d) 1.31
Answer: b
Explanation: We know that,
\(\widehat{C_P} = \frac{USL-LSL}{6\hat{\sigma}}\)
So, when we put values, we get, \(\widehat{C_P}\)=1.192.
9. The percentage of the specification band used by the process is calculated by formula _______________
a) P=C p
b) P=\
P=C p *100
d) P=\(\frac{1}{C_p}\)
Answer: b
Explanation: The percentages of the specification band used by any process is calculated by the following formula,
P=\(\frac{1}{C_p}\)*100
10. If the process capability ratio C p is 1.532, what percentages of the specification band will be used by the process?
a) 65.27%
b) 75.11%
c) 44.21%
d) 21.42%
Answer: a
Explanation: The percentages of specification band used by the process is calculated by,
P=\(\frac{1}{C_p}\)*100
When we put values, we get, P=65.27%.
11. The one sided process capability ratio that uses the USL of the quality characteristic, is written as ____
a) \
\
\
\(C_{pu}=\frac{USL-μ}{3σ}\)
Answer: d
Explanation: The one sided process capability ratio that uses the USL of the quality characteristic is written as,
\(C_{pu}=\frac{USL-μ}{3σ}\)
12. For a process, the process standard deviation is 32. If the mean of the process is 364, and the upper specification limit is 400, what will be the one side process capability ratio which uses USL of the quality characteristic?
a) 1.242
b) 1.135
c) 1.125
d) 0.991
Answer: c
Explanation: We know that,
\(C_{pu}=\frac{USL-μ}{3σ}\)
Putting values, we get, C pu =1.125.
13. What is the formula to calculate the one sided process capability which uses lower specification limit?
a) \
\
\
\(C_{pl}=\frac{μ-LSL}{6σ}\)
Answer: c
Explanation: The formula to calculate the one sided process capability which uses lower specification limit of the quality characteristic,
\(C_{pl}=\frac{μ-LSL}{3σ}\)
14. Process capability ratio C p cannot be zero.
a) True
b) False
Answer: a
Explanation: Process capability ratio C p cannot be zero as the USL and LSL difference can’t be zero and the standard deviation of the process also can’t be infinite. So we can say that that C p can’t be zero.
15. Increase in process capability ratio C p states that there is an increase in the percentage of the specification band used.
a) True
b) False
Answer: b
Explanation: We know that,
P=\(\frac{1}{C_p}\)*100
So an increase in the PCR C p shows a decrease in the percentage of the specification band used.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Process Capability Ratios – 2”.
1. If the Lower specification limit for one quality characteristic of one product, is 200, and the mean of the process is estimated to be 264 with an estimate of one sided process capability ratio \
43
b) 59
c) 32
d) 18
Answer: c
Explanation: We know that,
We know that,
\(\widehat{C_{pl}} = \frac{\hat{μ}-LSL}{3\hat{σ}}\)
Putting values, we get, The process standard deviation estimate, \(\hat{σ}\)=32.
2. If the LSL of the quality characteristic increases, the process capability ratio C p will ____________
a) Increase
b) Decrease
c) Will remain same
d) Can’t say
Answer: b
Explanation: We know that,
\(C_p=\frac{USL-LSL}{6σ}\)
So an increase in the LSL value will decrease the PCR.
3. PCR is a measure of __________
a) The ability of the process to manufacture the products that meet the specifications
b) The ability of the operator to remove the variability
c) The probability of the mean to be equal to the USL
d) The probability of the mean to be equal to the LSL
Answer: a
Explanation: Process capability ratio is used as the measure of the ability of the process to manufacture the products, which meet the specifications set for it.
4. The process fallout ___________ with increase in the process capability ratio C p , for a normally distributed process.
a) Increases
b) Decreases
c) Remains same
d) Can’t be described an increase or decrease
Answer: b
Explanation: The process fallout in ppm decreases with increase in the Process Capability Ratio C p . This relation can only be applied to the normally distributed processes.
5. A PCR C p =1.00 indicates the fallout rate of ___________ ppm for two-sided specifications.
a) 2
b) 7
c) 967
d) 2700
Answer: d
Explanation: A process capability ratio is a measure of the ability of a process to produce products, which meet their specifications. A PCR C p =1.00 indicates a fallout rate of 2700 ppm for two-sided specifications.
6. A process capability ratio C p =____________ indicates the fallout rate of 1 ppm for one-sided specifications for a normally distributed process.
a) 1.2
b) 1.8
c) 1.9
d) 1.6
Answer: d
Explanation: A process capability ratio C p =1.6 indicates the fallout rate of 1 ppm for the one – sided specifications. It also indicates the fallout rate of 2 ppm for two-sided specifications.
7. A correct interpretation of the PCR C p does not need the consideration of ___________ of process.
a) Operator
b) Mean
c) Variance
d) Stability
Answer: a
Explanation: The correct interpretation of the C p does not need the consideration of the operator of the process, as this factor is already included in the earlier steps taken before this interpretation.
8. What is generally the minimum value of two sided specification PCR C p for a existing process?
a) 1.11
b) 1.33
c) 1.89
d) 1.60
Answer: b
Explanation: The minimum value of the two sided specifications PCR C p for an existing process is generally taken as 1.33. For one sided specifications, PCR C p is taken as 1.25.
9. What is the minimum value of Process Capability Ratio C p for one sided specifications, for a new process which is normally distributed?
a) 1.45
b) 1.19
c) 1.72
d) 1.5
Answer: a
Explanation: The minimum value of the Process Capability Ratio C p for one sided specifications, is taken as 1.45 for any new process which is assumed to be normally distributed.
10. If the quality characteristic in consideration is a strength parameter of the existing process, what will be the minimum value of C p for two sided specifications for this?
a) 1.22
b) 1.52
c) 1.50
d) 1.36
Answer: c
Explanation: If the quality characteristic in consideration is a strength parameter of the existing process, the minimum value of Process Capability Ratio C p for two sided specifications 1.50.
11. Which of these is a necessary assumption made for the calculation of the PCR C p ?
a) The quality characteristic has a normal distribution
b) The quality characteristic has a lognormal distribution
c) The quality characteristic has an exponential distribution
d) The quality characteristic has a Poisson distribution
Answer: a
Explanation: For the calculation of the Process capability ratio, it is assumed that the quality characteristic, for which the PCR is to be calculated, should have a normal distribution.
12. What is the minimum two – sided specification PCR for the safety parameter for a new process?
a) 1.48
b) 1.45
c) 1.67
d) 1.32
Answer: c
Explanation: For a new process, a two – sided specification Process Capability Ratio for the safety parameter should have a minimum value of 1.67.
13. What is the value of process fallout in defectives ppm for one sided specifications of a PCR C p = 1.10?
a) 484
b) 201
c) 2700
d) 100
Answer: a
Explanation: The value of process fallout in defectives ppm for the one – sided specifications of a PCR C p =1.10 is 484. This value for two sided specifications of the same PCR C p is 967.
14. For calculating the PCR C p , the process should be in statistical control.
a) True
b) False
Answer: a
Explanation: Before calculating Process Capability ratio for a process, it is a necessary assumption that a process should be in the state of statistical control.
15. For the calculation of the PCR C p the process mean must be centered in the case of two-sided specifications.
a) True
b) False
Answer: a
Explanation: It is assumed that the process mean is centered between the lower and upper specification in the case of two sided specifications, before the calculation of PCR C p .
This set of Statistical Quality Control Assessment Questions and Answers focuses on ” Process Capability Ratios – 3″.
1. What should be the minimum value of PCR C p for a six-sigma company?
a) 2.0
b) 1.0
c) 3.0
d) 1.3
Answer: a
Explanation: A six-sigma company would require the process mean, when in control, will not be closer to than the six-sigma deviations from the nearest specification limit. For this, C p ≥2.0
2. What should be the position of the process mean for the calculation of the PCR C p ?
a) Near to the USL
b) Near to LSL
c) Centered between LSL and USL
d) C p Does not take this into account
Answer: d
Explanation: The process capability ratio PCR C p does not take into account where the process mean is located relative to the specifications.
3. What is the function of C p ?
a) To measure the value of process mean
b) To measure the value of the process standard deviation
c) To measure the USL and LSL of the quality characteristic
d) To measure the spread of the specifications relative to the six sigma spread
Answer: d
Explanation: The process capability ratio C p does not take into account the process centering. It simply measures the spread of the specifications relative to the six sigma spread of the process.
4. Which one of the processes performs the best?
a) The one which operates near the USL
b) The one which operates near the LSL
c) The one which operates at the midpoint of the interval between the specifications
d) The one which operates outside the specifications
Answer: c
Explanation: The process which operates at the midpoint of the interval between the specifications, has the best performance among all the processes.
5. 4 processes have same value of C p and the USL and LSL values equal to 62 and 38 respectively. Which one of those performs the best?
a) Centered at 50
b) Centered at 44
c) Centered at 54
d) Centered at 63
Answer: a
Explanation: The process, which is centered at the midpoint of the interval between the specifications limits, has the maximum probability of performing the best.
midpoint=\(\frac{USL+LSL}{2}\)
6. If C p =C pk which of these is true?
a) Process is centered at the LSL of the specifications
b) Process is centered at the USL of the specifications
c) Process is centered at the midpoint of the specification
d) Process is centered at the outside the specification limits
Answer: c
Explanation: Generally, when C p =C pk , the process is centered at the midpoint of the specifications. This means the process performance is a lot better.
7. Which of these gives the best performance of the process?
a) C p = C pk
b) C p – C pk = C pk
c) C p > C pk
d) C p < C pk
Answer: a
Explanation: When the process capability ratio C p is equal to the process capability ratio C pk , it is generally stating that the process is centered at the midpoint of the specifications. This gives the best performance of the process at a particular value of C p .
8. Which of these is correct?
a) C pk =min(C pu , C p )
b) C pk =min(C pu , C pl )
c) C pk =max(C pu , C p )
d) C pk =max(C p , C pl )
Answer: b
Explanation: The process capability ratio C pk is the minimum of the two, C pu and C pl . So,
C pk =min(C pu , C pl )
9. Which of these is a correct definition for the C pk ?
a) C pk is the one sided PCR for the specification limit nearest to the standard deviation of the process
b) C pk is the one sided PCR for the specification limit farest to the standard deviation of the process
c) C pk is the one sided PCR for the specification limit farest to the average of the process
d) C pk is the one sided PCR for the specification limit nearest to the average of the process
Answer: a
Explanation: C pk is the one sided PCR for the specification limit nearest to the standard deviation of the process.
10. If C p >C pk what conclusion can be carried out about the process?
a) The process has its operation at the midpoint of the interval between the specifications
b) The process operates off-center
c) The process performance is the best possible for that C p
d) The process can’t perform any better for same C p
Answer: b
Explanation: Generally, if C p > C pk , we can say that the process is not centered at the midpoint of the specifications, i.e. the process operates off-center.
11. For a process, the upper specification limit and lower specification limits are 62 and 38 respectively. If the process has its mean equal to 53, and its standard deviation 2, what is the value of C pk for this process?
a) 1.7
b) 2.5
c) 1.0
d) 1.5
Answer: d
Explanation: We know that,
C pk =min(C pu , C pl )
Calculating the values of C pu and C pl , we get C pu =1.5 and C pl =2.5. So we get C pk =1.5.
12. If C p =2.0,C pu =1.5,C pl =2.5 what will be the value of C pk ?
a) 2.5
b) 2.0
c) 1.5
d) 0
Answer: c
Explanation: The process capability ratio C pk is expressed by the following equation,
C pk =min(C pu , C pl )
Here C pu =1.5, so C pk =1.5.
13. If C pk =1.0 and C p =2.0, what will be the actual fallout for the process in ppm?
a) 200
b) 2700
c) 1
d) 1350
Answer: d
Explanation: If C pk < C p , C pk is considered to evaluate the process fallout as the process is off center and the one sided process capability must be used.
14. C pk measures the maximum of the two, C pu and C pl .
a) True
b) False
Answer: b
Explanation: As we write,
C pk =min(C pu , C pl )
We consider that the value of C pk will be equal to the one lower than the other between C pk and C pl .
15. For the calculation of C pk the quality characteristic should have a lognormal distribution.
a) True
b) False
Answer: b
Explanation: PCR C pk is evaluated using the assumptions made for C p . So as for C p , it is assumed that the quality characteristic has a normal distribution, it is also valid for C pk .
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Process Capability Ratios – 4”.
1. To remove the errors in the estimation of the PCR, the _________ is used.
a) Acceptance Sampling
b) Sample mean
c) Sample variance
d) Confidence intervals
Answer: d
Explanation: In actual practice, we only observe the estimate of PCR, not the real PCR. As the estimate can have a large error probability, it is always a good idea to use confidence intervals.
2. Which of these measures the actual capability of the process?
a) C p
b) C pu
c) C pk
d) C pl
Answer: c
Explanation: C pk always shows the actual capability as by interpretation of it, we can sense the potential improvement possible in the process by centering it.
3. Which of these measures the potential capability of the process?
a) C p
b) C pu
c) C pk
d) C pl
Answer: a
Explanation: Actual process capability can only be measured using the process capability ratio C p as it does not take process centering into account. It ignores the potential improvement possible.
4. If a process actual fallout is 1350 ppm, and its C p =2.0, what will be its fallout after centering?
a) 1 ppm
b) 4 ppm
c) 0.009 ppm
d) 0.0018 ppm
Answer: d
Explanation: Process fallout is 1350, so its C pk =1.0. So after potential improvement, when its C pk = C p , we get the process fallout = 0.0018 ppm.
5. Which of these can be used to analyze the process capability of the process which has the quality characteristic distributed non-normally?
a) C p
b) C pk
c) C pl
d) C pc
Answer: d
Explanation: There have been many attempts to extend the definition of the standard capability indices to the case of non-normal distribution. One of them is C pc .
6. How is C pc expressed?
a) \
\
\
\(C_{pc}=\frac{USL-LSL}{\sqrt{\frac{π}{2} E|X-T|}}\)
Answer: b
Explanation: C pc is expressed as,
\(C_{pc}=\frac{USL-LSL}{6 \sqrt{\frac{π}{2} E|X-T|}}\)
7. What is the value of T in the expression of C pc ?
a) T=\
b) T=
c) T=\
d) T=\
Answer: a
Explanation: The value of T in the expression of C pc is as follows.
T=\
8. What does the second subscript in C pc ?
a) Course
b) Capability
c) Confidence
d) Clarity
Answer: c
Explanation: In the process capability index C pc , designed by Lucin ̃o, which takes process non-normality into account, the second subscript denotes confidence as in Confidence intervals.
9. Which of these is not desirable to be worked alone with?
a) C p
b) C pc
c) C pk
d) C pu
Answer: c
Explanation: Two processes having different centering can have a same C pk . So we need to evaluate C p to check the equality of PCR C pk with it. This way we get perfect centering between the specifications.
10. For any fixed value between LSL and USL, C pk depends _________ on σ.
a) Inversely
b) Directly
c) Negatively
d) Positively
Answer: a
Explanation: As we know,
C pc =min(C pu ,C pl )
Where, C pu and C pl , both vary inversely according to σ. So C pc is inversely dependent on σ.
11. What will be the value of C pc as σ approaches to zero?
a) Zero
b) One
c) Infinity
d) Minus Infinity
Answer: c
Explanation: As Process Capability Ratio C pc varies inversely with σ, when the value of σ approaches zero, the value of the PCR C pc becomes infinity.
12. Which of these besides the PCR C pc can be used to know more about the process centering?
a) C p
b) C pm
c) C pu
d) C pl
Answer: b
Explanation: As C pc also needs some complementary process capability ratio such as C p , to know more about process centering, another PCR C pm is developed to know more about process centering.
13. What is the value of C pm ?
a) \
\
\
\(C_{pm}=\frac{USL-LSl}{6τ}\)
Answer: d
Explanation: The value of C pm Process capability ratio is given as,
\(C_{pm}=\frac{USL-LSl}{6τ}\)
14. ‘C pk ’ can also be used to determine the process capability of non-normal data.
a) True
b) False
Answer: b
Explanation: C pk and C p are based on an assumption that the quality characteristic varies on a normal distribution. So we cannot use the process capability ratio C pk to determine the process capability of the non-normal data.
15. ‘C p ’ and ‘C pk ’ both can’t be used alone to make decisions about the process centering.
a) True
b) False
Answer: a
Explanation: To predict the process centering, we must have information about C p and C pk . This way we can check the equality between C p and C pk and say if the process is centered.
16. Kotz and Lovelace were in strong opposition of _______________
a) P p
b) C p
c) C pk
d) C pm
Answer: a
Explanation: Kotz and Lovelace were in strong opposition of P p because they were used when the process was not in control. They said no index can give useful predictive information about process capability.
This set of Statistical Quality Control Problems focuses on “Process Capability Ratios – 5”.
1. What is the value of τ in the expression of C pm process capability ratio?
a) \
\
\
\(τ=\sqrt{E[^2]}\)
Answer: a
Explanation: The value of τ in the expression of C pm is the square root of expected squared deviation from target T, i.e.
\(τ=\sqrt{[E\big\{x-T)^2]}\)
2. Which of these is equal to the squared value of τ?
a) σ 2 - 2
b) σ 2 + 2
c) σ 2 - 2
d) σ 2 + 2
Answer: b
Explanation: The value of squared value of τ,
τ 2 =E[ 2 ]=E[ 2 ]+ 2 =σ 2 + 2 .
3. Which of these is having same value as C pm ?
a) \
\
\
\(\frac{USL+LSL}{6\sqrt{σ^2+^2}}\)
Answer: c
Explanation: As we know,
τ 2 =σ 2 + 2
We may write,
C pm = \(\frac{USL-LSL}{6\sqrt{σ^2+^2}}\)
4. What is value of ξ in the expression for C pm Process Capability Ratio?
a) \
\
\
\(\frac{μ-T}{σ}\)
Answer: d
Explanation: The value of ξ in the expression of C pm is given by,
ξ = \(\frac{μ-T}{σ}\)
5. If standard values are not given, the value of C pm is estimated by expression _____
a) \
\
\
\(\widehat{C_{pm}}=\frac{C_p}{\sqrt{1-V^2}}\)
Answer: a
Explanation: The estimate of C pm , when the standard values of σ, μ are not given, is calculated by expression,
\(\widehat{C_{pm}}=\frac{C_p}{\sqrt{1+V^2}}\)
6. What is the value of V in the expression of the estimate of C pm Process Capability Ratio?
a) \
\
\
\(\frac{\bar{x}-T}{σ}\)
Answer: a
Explanation: The value of V is used at the place of ξ in the case when the standard values of σ and μ are not available. The value of V is equal to,
V = \(\frac{\bar{x}-T}{s}\)
7. What is the value of C pm when the value |μ-T|→∞ ?
a) 1
b) 0
c) ∞
d) -∞
Answer: b
Explanation: The value of C pm approaches to zero asymptotically as the value |μ-T|→∞. This is because,
C pm = \(\frac{USL-LSL}{6\sqrt{σ^2+^2}}\)
8. What is the value of C pm when the value μ=T?
a) C pm =1
b) C pm =0
c) C pm =C p
d) C pm <C p
Answer: c
Explanation: As we know,
C pm = \(\frac{USL-LSL}{6\sqrt{σ^2+^2}}\)
When, μ=T we get C pm =C p .
9. What is the value of C pkm process capability ratio?
a) \
\
\
\(C_{pkm}=\frac{C_{pk}}{\sqrt{1-ξ^2}}\)
Answer: b
Explanation: The value of PCR C pkm is given by
\(C_{pkm}=\frac{C_{pk}}{\sqrt{1+ξ^2}}\)
10. What is the value of ξ in the expression for C pkm PCR?
a) \
\
\
\(\frac{μ+T}{6σ}\)
Answer: b
Explanation: The value of ξ in the expression of C pkm is given by,
ξ = \(\frac{μ-T}{σ}\)
11. C pkm is also called _________ generation process capability ratio.
a) Third
b) Fourth
c) Second
d) First
Answer: a
Explanation: C pkm is sometimes called third generation process capability ratio since it is constructed from the second generation PCR C pm and C pk .
12. C p is called the _________ generation PCR.
a) First
b) Second
c) Third
d) Fourth
Answer: a
Explanation: C p is called the first generation process capability ratio as it was designed before any other process capability ratio like C pk or C pm .
13. Which of these is one of the first generation process capability ratios?
a) C pk
b) C pm
c) C pc
d) C pu
Answer: d
Explanation: As we know, the first Process capability ratios, that were designed, were C p , C pu , C pl . So they are called the first generation process capability ratios.
14. C pm is the second generation process capability ratio.
a) True
b) False
Answer: a
Explanation: C pk and C pm were designed after the first generation capability ratios, which were not able to consider the process centering fact into their evaluation. So they are called the second gen. process capability ratios.
15. Confidence intervals are not needed for \
True
b) False
Answer: b
Explanation: All the estimates of the process capability ratios are point estimates. This means there is a strong possibility of error due to statistical fluctuation. So confidence interval is needed also for \(\widehat{C_{pkm}}\).
16. What does usage of s instead of R /d 2 in the confidence intervals indicate?
a) s=1
b) Process must be in control
c) Process may be in the out-of-control state
d) d 2 <1
Answer: b
Explanation: When process is not in control, the value of s has a big difference with the value of R /d 2 . So for the calculation of confidence interval s is used. This means, for a PCR to have a meaning, the process must be in control.
17. P p is a ___
a) Process capability ratio
b) Process stability ratio
c) Process prediction index
d) Process performance index
Answer: d
Explanation: AIAG has recommended using the process performance indices to be used instead of the process capability ratios when process is not in control. P p is one of them.
18. What is the value of process performance index estimate (P p )?
a) \
\
\
\(\frac{USL-LSL}{6s}\)
Answer: d
Explanation: The estimate of process performance index P p has the value,
P p = \(\frac{USL-LSL}{6s}\)
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Process Capability Analysis Using Control Charts”.
1. Which one of these tells explicitly about the potential capability of the process?
a) Histogram
b) Probability plots
c) PCRs
d) Control charts
Answer: d
Explanation: Histogram, process capability ratios and the probability plots all can only summarize the performance of process. They don’t necessarily display the process potential capability.
2. Which one of these addresses the issue of statistical control?
a) Design of process experiments
b) Probability plots
c) Control charts
d) Histogram
Answer: c
Explanation: The probability plots, histogram and PCRs are just tools to summarize the process performance. They don’t address the issue of statistical control. Control charts only, are capable of doing that.
3. Which one these show that, if the systematic pattern in process output is eliminated, it would produce a lesser variability in the quality characteristic?
a) Probability plots
b) Control Charts
c) Histogram
d) PCRs
Answer: b
Explanation: As control charts address the issue of statistical control of the process, they ensure that there must be random patterns in the process output, so that variability is reduced.
4. Which one of these is primary technique of the process capability analysis?
a) Histogram
b) Process capability ratios
c) Probability plots
d) Control charts
Answer: d
Explanation: As control charts are main technique to reduce variability, by taking the statistical control into consideration, control charts are regarded as the primary technique of the process capability analysis.
5. Which chart should be used, if possible?
a) x & R Charts
b) u Charts
c) p Charts
d) c Charts
Answer: a
Explanation: As x & R charts are the variable control charts and the others are attribute control charts, taking the fact “variable charts give better information than the others” into consideration, x & R charts are favored over the others.
6. Which one of these charts doesn’t need the specification of the quality characteristic?
a) x & R Charts
b) p-charts
c) c-charts
d) u-charts
Answer: a
Explanation: The variable control charts don’t need the specification of the quality characteristic, so x & R charts don’t need the specifications of the quality characteristic.
7. Which one of these helps us to find both instantaneous and the variability across time variability?
a) p-charts
b) np-charts
c) x & R Charts
d) c-charts
Answer: c
Explanation: The charts for x & R are quite useful in determination of the instantaneous and the variable across time variability of a quality characteristic.
8. What is the second name of instantaneous variability?
a) Long term variability
b) Short term variability
c) Long distance variability
d) Variability across time
Answer: b
Explanation: Instantaneous variability is the variability for a small tenure of process operation. It is also known as the short term variability.
9. Which one of these is similar to the long-term variability?
a) Instantaneous variability
b) Variability across time
c) Large variability
d) Large distance variability
Answer: b
Explanation: The variability across time phrase is also known as the long term variability, as this phrase is used for the variability calculated for a big interval of time.
10. Which control charts will be useful if the process output data for the process capability study are collected in different time periods?
a) x & R Charts
b) p-charts
c) c-charts
d) u-charts
Answer: a
Explanation: As variable control charts are capable of determining the process capability of both types, i.e. the instantaneous variability, and the variability across time, they are used, when the data for process capability analysis is collected in different shifts.
11. If the C pl of a safety-related parameter, is 0.69, what can we say about the process capability?
a) Process is capable
b) Process is not capable
c) Process is capable but will become not-capable in sooner time
d) Process is not-capable but will become capable in sooner time
Answer: b
Explanation: For a safety-related parameter, we know that the one-sided process capability ratio must be greater than, 1.50 for an existing process, or 1.67 for a new process. So the above mentioned process capability is inadequate.
12. Which one of these can be used as the monitoring device to show the effect of changes in the process on process performance?
a) Control charts
b) Designed experiments
c) Histogram
d) Probability plots
Answer: a
Explanation: Control charts give the time to time data about the process output in a visual display. So we can use them as a monitoring device to show the effect of changes in the process, on process performance.
13. If C pu =1.82 for strength quality characteristic for a new process, can we use the process?
a) Process is capable
b) Process is not capable
c) Process is capable but will improve in sooner time
d) Process is not capable but will improve in sooner time
Answer: a
Explanation: As strength is a safety parameter, the minimum value for C pu for it for a new process is 1.67. So we can use the above mentioned new process.
14. p-charts can be used rather than using the x & R charts, in the case of variable time spans of data collection.
a) True
b) False
Answer: b
Explanation: As variable charts are more capable of finding the process capability to analyze both instantaneous variability and the variability across time, x & R charts are more used instead of p-charts.
15. Histograms explain the process statistical control better than the control charts.
a) True
b) False
Answer: b
Explanation: As control charts are real time representation of the quality characteristic of process output according to time, they explain the process statistical control better than any other method.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Process Capability Analysis using Designed Experiments and Attribute Data”.
1. In the design of experiments, the ____________ variables are adjusted first.
a) Output controllable
b) Output uncontrollable
c) Input uncontrollable
d) Input controllable
Answer: d
Explanation: In the design of experiments, the value of input controllable variables is varied to find the desired output quality of a process.
2. Which of these can be varied?
a) Input Controllable variables
b) Input uncontrollable variables
c) Output controllable variables
d) Output uncontrollable variables
Answer: a
Explanation: Only the input controllable variables can be varied. This is because all the other variables are uncontrollable like input uncontrollable variables etc.
3. Which of the following is not done in the designing of experiments?
a) Varying the input variables
b) Observation of the output
c) Making a relationship between input and output
d) Process control
Answer: d
Explanation: In the design of experiments, the controllable inputs of a process are varied, and the output of the process is observed simultaneously, then a relationship between the inputs and the outputs, is made.
4. Which of these is not a function of designing of experiments?
a) To find the set of process variables influential on output
b) To determine the optimum level of the variables
c) To estimate maximum output possible with optimum value of variables
d) To eliminate some uncontrollable inputs
Answer: d
Explanation: Designed experiments are helpful in finding, which set of process variables is influential on the output, and at what levels these variables should be held to optimize the process performance.
5. Which of these is the easiest method for finding the sources of variability?
a) Designed Experiments
b) p-charts
c) n-charts
d) Acceptance sampling
Answer: a
Explanation: Designed experiments are useful in determining the optimum level of the process variables for a better process performance, so we can find the sources of variability simultaneously.
6. Which of these is not a method to judge the process capability?
a) Control charts
b) Attributes data
c) Acceptance sampling
d) Designed experiments
Answer: b
Explanation: Process capability can be determined by the usage of either of control charts, or attributes data or designed experiments. The process capability ratios, histogram and the probability plots also can be used.
7. DPU is ________
a) Decrement per unit
b) Defects per unit
c) Design per unit
d) Decimals per unit
Answer: b
Explanation: The term “DPU” is used in the determination of process capability by using the attribute data. Here, DPU stands for Defects per unit.
8. DPMO stands for ________
a) Defects per million opportunities
b) Decrement per minute opportunity
c) Defects per minute opportunity
d) Designation per million opportunities
Answer: a
Explanation: DPMO represents the defects per million opportunities. It takes the complexity of an unit into account when calculating the process capability.
9. Which of these is the correct formula to calculate DPMO which uses the number of opportunities?
a) DPMO=\
DPMO=\
DPMO=\
DPMO=\(\frac{Total\, number \,of \,units}{Number \,of \,units \,* \,Number \,of \,Opportunities}\)
Answer: d
Explanation: DPMO is calculated by using the following formula,
DPMO=\(\frac{Total\, number \,of \,units}{Number \,of \,units \,* \,Number \,of \,Opportunities}\)
As this formula contains the number of opportunities, it takes the number of times that an error can be occurred. So it is much accurate.
10. If there are 85 defects in the lot containing 150 units, what is the value of DPU?
a) 0.45
b) 0.57
c) 0.89
d) 1.52
Answer: b
Explanation: DPU is calculated by using the following formula,
DPU = \(\frac{Total \,number \,of \,defects}{Total \,number \,of \,units}\)
By calculating the DPU, we get, DPU=0.57.
11. If the DPMO for a sample containing 45 units is 0.237. If the total number of defects is 32, what will be the number of opportunities?
a) 3
b) 1
c) 6
d) 7
Answer: a
Explanation: DPMO is calculated using the formula,
DPMO=\(\frac{Total\, number \,of \,units}{Number \,of \,units \,* \,Number \,of \,Opportunities}\)
Putting the values, we get, the number of opportunities= 3.
12. PPM defectives can be a measure of the process capability. Although an equivalent sigma level can also be used.
a) True
b) False
Answer: a
Explanation: As a particular sigma level shows a particular amount of defectives, e.g. 2700 ppm is equivalent to a 3-sigma process; an equivalent sigma level can be used instead of using the ppm defectives as a measure of the process capability.
13. DPU can be evaluated only when the variable data is available.
a) True
b) False
Answer: b
Explanation: As DPU suggests the defects found per unit of the product, which is an attribute data, we do not need the variable data to calculate the DPU.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Gauge and Measurement System Capability Studies – 1”.
1. Which of these is not an objective of most measurement systems capability studies?
a) Determining how much of variability is due to the gauge or instrument
b) Isolating the components of variability in measurement system
c) Assessing whether the instrument is capable
d) Checking whether the instrument can make defective pieces
Answer: d
Explanation: Most measurement system capability studies are having objectives like determination of the amount of variability due to the gauge or instrument, to isolate the components of variability in the measurement system etc.
2. Which of these is one of the two R’s of measurement systems capability?
a) Reducibility
b) Recyclability
c) Reproducibility
d) Reusability
Answer: c
Explanation: The two R’s of the measurement systems capability are “Reproducibility” and “Repeatability”. They play very important role in the measurement system capability studies.
3. What is repeatability of a measurement instrument?
a) Getting the maximum value of the measured dimension
b) Getting the minimum value of the measured dimension
c) Getting the exact value of the measured dimension
d) Same observation value for same measured dimension for several times
Answer: d
Explanation: Repeatability of a measurement instrument is defined as getting the same observed value if we measure the same unit several times under identical conditions.
4. What is reproducibility of a measurement system?
a) Difference between the observed values when the measuring conditions are same
b) Difference between the observed values when the measuring conditions are different
c) Ease in measurement when the measuring conditions are same
d) Ease in measurement when the measuring conditions are different
Answer: b
Explanation: Reproducibility is how much difference we get in observed values, when the units are measured under different conditions like different operators, time periods etc.
5. Measurement system capability ____________ when the good unit is judged to be defective.
a) Increases
b) Decreases
c) Does not change
d) Can’t be evaluated
Answer: b
Explanation: The capability of a measurement system decreases substantially when the good unit in the output of a process is judged to be defective. This is a fault of the measurement system.
6. Measurement system capability increases if the bad unit is judged to be _____________
a) Conforming
b) Nonconforming
c) Not defective
d) Satisfying the specifications
Answer: b
Explanation: When a measurement system identifies a bad unit by classifying it as nonconforming, the measurement system capability is increased.
7. If a good unit is judged to be conforming, what will be the effect on the measurement system capability?
a) Increase
b) Decrease
c) Remain same
d) Can’t be calculated in this case
Answer: a
Explanation: If a good unit is judged to be conforming to the standards and specifications, it indicates the efficiency of the system. So the capability of the measurement system is increased in this case.
8. What is the linearity of a measurement system?
a) Differences in observed accuracies and/or precision over a range of measurements made
b) Differences in theoretical accuracies and/or precision over a range of measurements made
c) Differences in products dimensions of products from a same process
d) Differences in the product quality from a same process
Answer: a
Explanation: Linearity of a measurement system is defined as the differences in observed accuracy and/or precision, experienced over the range of the measurements made by the system.
9. When the measurement system is not linear ____________ is needed.
a) Process to be changed by which the units are made
b) Calibration of the measurement system
c) Change in the process parameters
d) No change
Answer: b
Explanation: If there are problems with linearity of the measurement system, they can be due to calibration or maintenance issues, so proper calibration or the replacement of the measurement system can solve the problems.
10. Which of these is not a cause of the stability defects in the measurement systems?
a) Warm up effects
b) Environmental factors
c) Inconsistent operator performance
d) Maintenance errors
Answer: d
Explanation: Stability or different levels of variability in different operating regimes, can result from warm-up effects, environmental factors, inconsistent operator performance, and inadequate standard operating procedures.
11. Bias of a measurement instrument is defined as ___________
a) The difference between the observed value and the true value
b) The difference between the observed value and the lowest possible value
c) The difference between the observed value and the estimated value
d) The difference between the observed value and the highest possible value
Answer: a
Explanation: The bias of a measurement system reflects the difference between the observed value and the true value obtained from a master or gold standard, or from a different measurement technique reliable to produce accurate values.
12. It becomes very difficult to monitor, control, improve, or effectively manage a process with _____________ measurement system.
a) A passable
b) A satisfactory
c) An adequate
d) An inadequate
Answer: d
Explanation: If the measurement system is not capable to produce good results, it becomes very difficult to monitor, control, improve, or effectively manage a process with an inadequate measurement system.
13. MSA stands for ________________
a) Million standards analysis
b) Major system analysis
c) Measurement system analysis
d) Master system analysis
Answer: c
Explanation: MSA is an acronym used for Measurement System Analysis, which is done in the measurement system capability analysis.
14. An ineffective measurement system can dramatically impact business performance.
a) True
b) False
Answer: a
Explanation: It is true that an ineffective measurement system can dramatically impact business performance because it leads to uninformed and usually bad decision making.
15. Measurement system can also cause variability in the products.
a) True
b) False
Answer: a
Explanation: Generally, in an activity that involves measurements, some of the observed variability will be inherent in the units or items that are being measured, and some of the variability will result from the measurement system used.
This set of Basic Statistical Quality Control Questions and Answers focuses on “Gauge and Measurement System Capability Studies – 2”.
1. What of these can be used as a reasonable model for measurement system capability studies? Here y,x and ε denote the observed measurement, true measurement, and the measurement error respectively.
a) y=x-2ε
b) y=x+ε
c) y=x-ε
d) y=x+2ε
Answer: b
Explanation: The reasonable model, which can be used for measurement system capability is expressed as,
y=x+ε
2. The variance of the total observed measurement is expressed by __________
a) \
\
\
\(\sigma_{Total}^2=\sigma_P^2+\sigma_{Gauge}^2\)
Answer: d
Explanation: The variance of the total observed measurement uses the value of the sum of the variances of product errors, and the gauge errors. It is calculated by,
\(\sigma_{Total}^2=\sigma_P^2+\sigma_{Gauge}^2\)
3. The P/T ratio stands for ___________
a) Probability to tolerance ratio
b) Precision to time ratio
c) Probability to total ratio
d) Precision to tolerance ratio
Answer: d
Explanation: The P/T ratio is used in the measurement system capability analysis. In P/T ratio, P/T refers to Precision to tolerance ratio.
4. What is the value of the P/T ratio?
a) \
\
\
\(\frac{P}{T}=\frac{k\hat{σ}_{gauge}}{USL+LSL}\)
Answer: b
Explanation: The P/T ratio is calculated for the evaluation of the gauge capability. It uses the value of the \(\hat{σ_{gauge}}\) The value of P/T ratio is given by,
\(\frac{P}{T}=\frac{k\hat{σ}_{gauge}}{USL-LSL}\)
5. If the number of standard deviations between the usual natural tolerance limits of a normal distribution, what is the value used for k in the P/T ratio?
a) 5.15
b) 8
c) 6
d) 5.60
Answer: c
Explanation: The value k=6 in the P/T ratio corresponds to the number of standard deviations, between the usual natural tolerance limits for a normal distribution.
6. For a process, which has, USL and LSL equal to 60, and 5 respectively, and the value of \
0.087
b) 0.077
c) 0.067
d) 0.097
Answer: d
Explanation: We know that,
\(\frac{P}{T}=\frac{k\hat{σ}_{gauge}}{USL-LSL}\)
Putting the values in the question, we get P/T=0.097.
7. Which of these indicate an adequate measurement system?
a) \
\
\
\(\frac{P}{T}=0.3\)
Answer: a
Explanation: If the value of P/T ratio is lesser than or equal to 0.1, the measurement system used is predicted to be adequate for the selected process.
8. The options are the P/T ratios for different measurement systems. Which of these shows an adequate measurement system?
a) 0.21
b) 0.13
c) 0.18
d) 0.06
Answer: d
Explanation: A P/T ratio lesser than 0.1 indicates an appropriate measurement system for any process. So here, 0.06<0.1, so it is an example of acceptable measurement systems.
9. The cause of calling a measurement system adequate because it has P/T ratio lesser than 0.1, is ____________
a) A measurement device should be calibrated in units one-tenth large as the accuracy required in final measurement
b) A measurement device should be calibrated in units one-third large as the accuracy required in final measurement
c) A measurement device should be calibrated in units one-fourth large as the accuracy required in final measurement
d) A measurement device should be calibrated in units three-tenth large as the accuracy required in final measurement
Answer: a
Explanation: Values of estimated P/T ratio of 0.1 or less indicate adequate measurement system. It’s based upon the general rule, which requires the measurement device to be calibrated in units one-tenth large, as the accuracy required in the final measurement.
10. Which of these can be used as the estimate of standard deviation of total variability which is including both product variability, and the gauge variability?
a) The sample mean
b) The sample variance
c) The sample standard deviation
d) No of defects in the sample
Answer: b
Explanation: The sample variance can be used as the estimate of the standard deviation of the total variability, which includes both, the product variability, and the gauge variability.
11. If the sample variance of a process is, 10.05, and the gauge capability standard deviation is estimated to be 0.79. What will be the value of the estimate of the standard deviation of the product variability?
a) 9.26
b) 3.04
c) 2.03
d) 8.91
Answer: b
Explanation: As we know that,
\(\sigma_{Total}^2=\sigma_P^2+\sigma_{Gauge}^2\)
If the estimates are to be used, the same equation can be written for the corresponding estimate values. Putting the values as mentioned, we get, \(\hat{\sigma}_p^2\)=3.04.
12. Which of these show a correct expression for the ρ p ?
a) \
\
\
\(ρ_p=\frac{\sigma_{gauge}^2}{2σ_{total}^2}\)
Answer: c
Explanation: The gauge capability ratio ρ p is the ratio of the variances of the product error and the gauge errors. It is expressed as,
\(ρ_p=\frac{\sigma_P^2}{σ_{Gauge}^2}\)
13. The gauge capability ratio ρ M is expressed as ____________
a) \
\
\
\(ρ_p=\frac{\sigma_{gauge}^2}{2σ_{total}^2}\)
Answer: b
Explanation: The value of gauge capability ratio ρ M is a ratio of the variances of the gauge errors and the total observed errors. It is expressed as,
\(ρ_p=\frac{\sigma_{gauge}^2}{σ_{Total}^2}\)
14. The general rule, that is used to define a measurement system adequate by using P/T ratio equal to or less than 0.1, can be used without caution.
a) True
b) False
Answer: b
Explanation: The caution should be used in accepting this general rule of thumb in all cases. A gauge must be capable to measure product accurately enough and precisely enough, for the analyst to make a correct decision. This may not necessarily require P/T <=0.1.
15. ρ P =1-ρ M .
a) True
b) False
Answer: a
Explanation: It is known that,
\(ρ_p=\frac{\sigma_{gauge}^2}{σ_{Total}^2}; ρ_p=\frac{\sigma_P^2}{σ_{Gauge}^2}; σ_{Total}^2=σ_P^2+σ_{Gauge}^2\)
Combining all three equations, we get,
ρ P =1-ρ M .
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Gauge and Measurement System Capability Studies – 3”.
1. SNR represents __________
a) Sign to noise ratio
b) Sample to nonconforming ratio
c) Size to noise ratio
d) Signal to noise ratio
Answer: d
Explanation: A measure of measurement system adequacy is defined by AIAG, which is called SNR or signal to noise ratio. It is used to check the satisfactoriness of the measurement system.
2. SNR is equal to _________
a) \
\
\
\(\frac{2ρ_p}{1-ρ_p}\)
Answer: a
Explanation: SNR is the measure of measurement system adequacy. It uses the gauge capability ratio ρ p . It is expressed as,
SNR=\(\sqrt{\frac{2ρ_p}{1-ρ_p}}\)
3. What is the correct definition of SNR?
a) The number of signals for 30 noise indications on control chart
b) The number of signals for single noise indication on control chart
c) The number of signals with no noise from the measurements
d) The number of distinct levels or categories that can be reliably obtained from the measurements
Answer: d
Explanation: SNR is defined by AIAG as the number of distinct levels or categories that can be reliably obtained from the measurement.
4. If the estimate of p p is 0.9214, what will be the estimate of SNR?
a) 4.84
b) 6.87
c) 9.12
d) 2.13
Answer: a
Explanation: We know that,
SNR=\(\sqrt{\frac{2ρ_p}{1-ρ_p}}\)
By putting the value of the estimate of ρ p we get SNR estimate = 4.84.
5. Which of these is not one of the gauge capability ratios?
a) SNR
b) C p
c) ρ M
d) ρ p
Answer: d
Explanation: SNR, ρ p and ρ M are some examples of the gauge capability ratios, which are used to check the adequacy of the measurement system used.
6. What is the full-form of DR in the context of gauge capability?
a) Decrease rate
b) Direction ratios
c) Discrimination ratios
d) Dimension rates
Answer: c
Explanation: Like the other gauge capability ratios like SNR, ρ p , and ρ M ; DR is also a measure of gauge capability. Its full-form is known as discrimination ratio.
7. What is the value of DR?
a) DR=\
DR=\
DR=\
DR=\(\frac{1-ρ_p}{1+ρ_p}\)
Answer: d
Explanation: The value of the discrimination ratio, which is used as a measure of adequacy of a measurement system, and which uses the ratio ρ p , is used to evaluate the gauge capability. It is expressed as,
DR=\(\frac{1-ρ_p}{1+ρ_p}\)
8. For a measurement system ρ p =0.8914. What will be the value of DR for that?
a) 17.416
b) 0.0571
c) 24.46
d) 0.0652
Answer: a
Explanation: We know that,
DR=\(\frac{1-ρ_p}{1+ρ_p}\)
Putting the values, we get DR=17.416.
9. For a gauge to be capable, generally DR must be greater than ___________
a) 1
b) 0.4
c) 0.1
d) 4
Answer: d
Explanation: It is generally stated that the DR for a capable gauge is, generally having a value higher than 4. This means for a gauge to be capable, the DR must exceed 4.
10. Which of these tells about the inherent variability of the gauge?
a) Accuracy
b) Precision
c) DR ratio
d) SNR
Answer: b
Explanation: Precision of a measurement system can be a measure of the inherent variability of the gauge. This is because the inherent variability moves the output of a gauge strongly to one direction or to a point.
11. The picture indicates gauge measurements where the dark circle is true value and the dots as observed value. This picture shows _____
statistical-quality-control-questions-answers-gauge-measurement-system-capability-studies-3-q11
a) More accuracy, less precision
b) Neither accuracy nor precision
c) More precision, less accuracy
d) High accuracy and precision
Answer: a
Explanation: The average of the observed values is almost equal to the true value. So from the definition of accuracy, we can say this gauge is more accurate than precise.
12. The standard model for a gauge R & R study is also called ____________
a) Random effects model analysis of variance
b) Continuous effects model analysis of variance
c) Discrete effects model analysis of variance
d) Attribute effects model analysis of variance
Answer: a
Explanation: The standard model for a gauge R & R study is also called a random effects model analysis of variance. It is generally used while calculating the gauge capability.
13. The more the gauge capability, the more are the ____________
a) Tolerances
b) Specifications
c) Chances of rejection
d) Chances of no error in measurements
Answer: d
Explanation: The gauge capability represents the ability of gauge to measure the dimension or unit correctly. The more is the gauge capability, the more are the chances of no error in measurements.
14. Even if the instantaneous value measured for a dimension by a gauge is not equal to the true value, but the gauge can still be accurate.
a) True
b) False
Answer: a
Explanation: The accuracy of a gauge refers to the ability of the instrument to measure the true value correctly, on average. So even if the instantaneous value by a gauge is not equal to the true value, but average of it is almost equal to the true value, the gauge is accurate.
15. An equipment can be more precise but not accurate at the same time.
a) True
b) False
Answer: a
Explanation: Accuracy refers to the ability of the instrument to measure the true value correctly on average, but precision is a measure of inherent variability in the measurement system, so both can be present, even in the absence of either one of them.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Gauge and Measurement System Capability Studies – 4”.
1. What is reproducibility?
a) Variability due to different operators using the gauge or different time periods, or environments
b) Variability due to error in process
c) Variability reflecting the basic inherent precision of the gauge itself
d) Variability reflecting the basic inherent precision of the process
Answer: a
Explanation: Reproducibility is defined as the variability due to different operators using the gauge, or different time periods, or different environments, or in general, different conditions.
2. What is repeatability?
a) Variability due to different operators using the gauge or different time periods, or environments
b) Variability due to error in process
c) Variability reflecting the basic inherent precision of the gauge itself
d) Variability reflecting the basic inherent precision of the process
Answer: c
Explanation: Repeatability is defined as reflecting the basic inherent precision of the gauge itself, through which we are measuring the unit.
3. What can we write the variance in measurement error or gauge in the terms of variance in repeatability, and variance of reproducibility?
a) \
\
\
\(\sigma_{gauge}^2=\sigma_{repeatability}-\sigma_{reproducibility}^2\)
Answer: b
Explanation: We can write the measurement error or gauge error variance as follows,
\(\sigma_{gauge}^2=\sigma_{repeatability}^2+\sigma_{reproducibility}^2\).
4. Which of these is called an R & R study?
a) Experiment used to measure components of \
Experiment used to measure components of \
Experiment used to measure components of \
Experiment used to measure components of \(\sigma_{repeatability}^2\)
Answer: a
Explanation: The experiment used to measure the components of \(\sigma_{gauge}^2\) is usually called a gauge R & R study, for two components of \(\sigma_{gauge}^2\).
5. What is the other name of random effects model analysis of variance?
a) ANOVA
b) RENOVA
c) FNOVA
d) MENOVA
Answer: a
Explanation: The random effects model analysis of variance is also called the ANOVA model analysis. This is a part of analysis of variance method.
6. If the value of the variance of the product variability increases, what will be the effect on the variance of the gauge?
a) Increase
b) Decrease
c) No change
d) Can’t be predicted
Answer: d
Explanation: As there is no dependency of the variance of the gauge on the product variability, even if the product variability increases, we cannot predict the changes in gauge variability variance.
7. If the variance of the total observed measurement is constant and the gauge capability ratio ρ_M is increased, what will be change in the gauge measurement variance?
a) Increase
b) Decrease
c) No change
d) Can’t be predicted
Answer: a
Explanation: As we know,
\Missing or unrecognized delimiter for \right\)
So if the variance of the total observed measurement is kept constant with the gauge capability ratio ρ M increasing, the variance of gauge measurement will be increased.
8. If the variance of product variability is increased while keeping the total observed measurement variance constant, what will be effect on the gauge capability ratio ρ P ?
a) Increase
b) Decrease
c) No change
d) Can’t be predicted
Answer: a
Explanation: As we know,
ρ P ∝ σ p 2
So, the gauge capability ratio ρ P will increase with increase with increase in the variance in product variability σ p 2 .
9. If the value of product variance is increased with the variance of the total observed measurement kept constant, the value of the gauge capability ratio ρ M increases.
a) True
b) False
Answer: b
Explanation: As we know,
\(\sigma_{Total}^2 = \sigma_{gauge}^2 + \sigma_P^2; ρ_M=\frac{\sigma_{gauge}^2}{\sigma_{Total}^2}\)
Putting the equations together, we get to know that the value of ρ M increases with decrease in the product variance.
10. A DR of value 5.89 indicates an adequate measurement system.
a) True
b) False
Answer: a
Explanation: It is known that an Discrimination ratio of value more than 4 indicates an adequate measurement system. So a DR of value 5.89 indicates an adequate measurement system.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – Cumulative Sum Control Chart – 1”.
1. Cusum control charts were originated in ____________
a) 1950s
b) 1960s
c) 1920s
d) 1980s
Answer: a
Explanation: The emphasis on more and more variability reduction, yield enhancement and process improvement, forced people to create new process monitoring techniques. Cusum charts were originated due to the same reason in 1950s.
2. Which of the control charts are used in phase I application of SPC?
a) Shewhart control charts
b) Cusum control charts
c) Both, Shewhart control charts and Cusum control charts
d) Neither one of the Shewhart control charts or the Cusum control charts
Answer: a
Explanation: Shewhart control charts are used for preliminary process monitoring, so the Shewhart control charts are used in the phase I implementation of SPC.
3. Which of these is not one of the phenomenons likely to occur while the phase I application of SPC?
a) Process likely to be out of control
b) Process changing its state continuously between in-control and out-of-control
c) Experiencing assignable causes
d) Large shifts in the monitored parameters
Answer: b
Explanation: Shewhart control charts are used in phase I implementation of the SPC, where process is likely to be out-of-control, and experiencing assignable causes, which may result in large shifts in the monitored parameters.
4. Which of these is a disadvantage of Shewhart control charts?
a) Can be for both attributes and variables
b) Gives process information
c) Using only the information about the last sample observations
d) Uses every information about the process
Answer: c
Explanation: The major disadvantage of a Shewhart control chart is that, it uses only the information about the process contained in the last sample observation, and it ignores the information given by the entire sequence of points.
5. Shewhart control charts are insensitive to ________ process shifts.
a) Small
b) Medium
c) Big
d) Very Big
Answer: a
Explanation: As the Shewhart control charts ignore the information about the entire sequence of the points, it ignores the small process shifts, say on the order of 1.5σ or less.
6. Which of these is a reason why the Shewhart control charts are not used in phase II application of SPC?
a) Because they can count larger shifts in process
b) Because they can analyze information about the last sample observation
c) Because they become insensitive to lower order process shifts
d) Because they are impractical
Answer: c
Explanation: Shewhart control charts are insensitive towards small process shifts. As the process shifts in the phase II application are of very less order, Shewhart control charts are not generally used for phase II application of SPC.
7. Which of these can be applied to Shewhart control charts to enhance their performance in the phase II application of the SPC?
a) Control limits
b) Mean of the process
c) Standard deviation of the process
d) Warning limits
Answer: d
Explanation: To enhance the performance of the Shewhart control charts in the phase II application of the SPC, we can use the warning limits and other sensitizing rules with the Shewhart control charts.
8. What is the effect of warning limits and other sensitizing rules over the average run length of the process?
a) Increase
b) Decrease
c) Remains same
d) Does not change the ARL
Answer: a
Explanation: The warning limits and the other sensitizing rules are used to improve the performance of Shewhart control charts, to get better results in the Phase II applications of the SPC. But it also decreases the ARL of the process.
9. Which of the error types are likely to occur when the warning limits are used with Shewhart control charts the phase II of SPC?
a) Type I
b) Type II
c) Type I and Type II
d) Type III
Answer: a
Explanation: As ARL of the process reduces, due to the use of warning limits with the Shewhart control charts, we conclude a process out-of-control when it is actually in-control, which is actually type I error.
10. What is the full form of the Cusum control charts?
a) Curetted sum control charts
b) Corrected sum control charts
c) Compressive sum control charts
d) Cumulative sum control charts
Answer: d
Explanation: The Cusum chart is an alternative of the Shewhart control charts. Its full name is cumulative sum control chart, which is used in the place of the Shewhart control charts with warning limits.
11. In which phase of SPC are the Cusum control charts are used?
a) Phase I
b) Phase II
c) Both, phase I and II
d) Neither in phase I nor in phase II
Answer: b
Explanation: The Cusum control charts, or the cumulative control charts were originated due to the degraded performance of Shewhart control charts in the phase II of SPC. So they are used in the phase II only.
12. What is the value of the quantity plotter on the Cusum charts?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: a
Explanation: Cusum control charts are plotted instead of Shewhart control charts in phase II applications of SPC. The quantity plotted on the charts is written as,
\
\)
13. What is the name of the quantity plotted C i on Cusum charts?
a) Cumulative sum from ith sample
b) Cumulative sum to ith sample
c) Cumulative sum to and ith sample
d) Cumulative sum from and ith sample
Answer: c
Explanation: The quantity plotted on the Cusum charts or the cumulative sum control charts is C i . It is called, Cumulative sum to and ith sample.
14. Shewhart control charts are better for phase II applications than the Cusum charts.
a) True
b) False
Answer: b
Explanation: As Shewhart control charts fail to detect small shifts in process in phase II of SPC, Cusum charts are better in the detection of small shifts in process.
15. Type I possibility in the phase II application of the Shewhart control charts with warning limits.
a) True
b) False
Answer: a
Explanation: The application of Shewhart control charts with warning limits, in the phase II application of the SPC, reduces the ARL of process. So we get an error signal frequently even if the process in-control, increasing the possibility of type I error.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – Cumulative Sum Control Chart – 2”.
1. Which of these is a reason, why the Cusum charts are better than the Shewhart control charts?
a) Because they are having information about only one sample
b) Because the quantity plotted on the Shewhart control charts is variable
c) Because the quantity plotted on the Cusum chart contains information about more than one sample
d) Because the quantity plotted on the Cusum control charts is containing information about a single sample
Answer: c
Explanation: The quantity plotted in the Cusum charts, contains more information as it contains information about more than one samples (C i cumulative sum to and including an ith sample).
2. Which charts are particularly more effective for sample size one?
a) p-charts
b) c-charts
c) X bar and s charts
d) Cusum charts
Answer: d
Explanation: Because of the reason that Cusum control charts can detect small process shifts easily, and they contain information about more than one sample, they are better used for sample size=1.
3. Which charts are more effective for the chemical and process industries?
a) p-charts
b) c-charts
c) X bar and s charts
d) Cusum charts
Answer: d
Explanation: Cusum charts are more effective when the rational subgrouping with sample size 1 concept is used. As chemical and process industries use the concept of rational subgrouping with sample size 1, Cusum charts are more productive in them.
4. The processes where discrete part manufacturing is done, which charts are better?
a) p-charts
b) c-charts
c) Cusum charts
d) X bar and R charts
Answer: c
Explanation: The process where discrete part manufacturing is done, the sample size is usually 1. As the Cusum charts are better and more effective than Shewhart control charts when it comes to sample size 1, they are used in the mentioned situations.
5. Which of these control charts will have a better performance in the discrete part manufacturing assembly, where automatic measurement of each part is done?
a) X bar charts
b) Cusum charts
c) u-charts
d) c-charts
Answer: b
Explanation: As automatic measurement of each part is indicating a sample size of 1 unit, the Cusum charts are used for the processes of such type. The reason behind it being that, they are effective for the industries having a sample size, mostly 1.
6. Cumulative control charts were first presented by ____________
a) Shewhart
b) ASQC
c) ASQ
d) Page
Answer: d
Explanation: The initial proposal for the use of the cumulative sum control charts, was first introduced by Page in the year of 1954. Then the improvements and the use of the charts started in the industries.
7. If the mean changes to a higher value, what will be the effect on the cumulative sum?
a) Decrease
b) Increase
c) Can decrease or increase
d) Can’t be predicted.
Answer: b
Explanation: If the value of the mean of the process shifts to a higher value i.e. a positive drift, we note form the formula of Cusum,
\
\)
That it will introduce a positive drift in the cumulative sum too.
8. If the value of the cumulative sum shifts to a lower value what is likely to be the reason for it?
a) The decrease in the value of the mean
b) The increase in the value of the mean
c) No change in the value of mean
d) Decrement in the value of the standard deviation
Answer: a
Explanation: It is noted while operating on the Cusum chart that, whenever there is a downward drop in the process mean, there is a negative drop in the value of the cumulative sum.
9. How are the changes in the conditions of the process, known using the Cusum charts?
a) By the change in Cusum value
b) By the change in the sample standard deviation
c) By plotting a s chart first
d) By plotting an R chart first
Answer: a
Explanation: As a positive drift, and a negative drift in the Cusum or cumulative sum value indicates a shifting of mean in positive or negative direction respectively, we can sense the shifting of mean continuously by checking the Cusum value, and hence the process state.
10. A significant trend upwards in the process cumulative sum chart indicates ____________
a) Shifting of mean in the positive direction
b) Shifting of mean in the negative direction
c) No shift in the process condition
d) No shift in the mean
Answer: a
Explanation: A significant trend developed in the Cusum chart due to the positive drift of the Cusum value, points toward the positive shift of the mean of the process. An assignable cause is present in this case.
11. Which of these conditions don’t describe an assignable cause in the process?
a) A positive shift in the Cusum value
b) A negative shift in the Cusum value
c) Random change in Cusum value
d) Continuous upward drift of Cusum value after a continuous downward drift in the Cusum value
Answer: c
Explanation: A random change in the Cusum value indicates that the mean is shifting randomly. This indicates an in-control process condition, of the process for which the Cusum chart is made for.
12. Which of these is another form of the cumulative sum value plotted on the cumulative sum chart?
a) \
+\sum_{j=1}^{i-1}
\)
b) \
+C_{i-1}\)
c) \
+\sum_{j=1}^{i-1}
\)
d) \
-C_{i-1}\)
Answer: b
Explanation: As we know,
\
\)
This can be written as,
\
+\sum_{j=1}^{i-1}
=
+C_{i-1}\)
13. If the value of x i =9.29, and C i-1 =-2.56, what will be the value of the cumulative sum for this sample, if the value of μ 0 =10?
a) -3.27
b) -5.13
c) 3.27
d) 5.13
Answer: a
Explanation: We know that,
C i =(x i -μ)+C i-1
Putting the values from the question, we get,
C i =-3.27
14. Upward or downward shifts in the Cusum value, directly indicate changes in process condition.
a) True
b) False
Answer: a
Explanation: As we know, the upward and downward shifts in the Cusum values are caused due to process mean shift, and the process mean shift indicates a shift in the process condition, hence, the shifts in the Cusum values directly indicates changes in process condition.
15. A change of process mean changes the value of C i >C i-1 , so we can say that the process mean has shifted upwards.
a) True
b) False
Answer: a
Explanation: As the positive shift in the process mean indicates a positive drift in the Cusum chart, we can say that, the inequality C i >C i-1 indicates the upward shift of the process mean too.
This set of Advanced Statistical Quality Control Questions and Answers focuses on “Time-Weighted – Cumulative Sum Control Chart – 3”.
1. How many ways are there to represent the Cusum on the Cusum charts?
a) 3
b) 4
c) 2
d) 5
Answer: c
Explanation: There are two ways invented for the representation of the Cusums on the Cusum charts; namely Tabular Cusum, and V-mask form of the Cusum.
2. Which of these is a name of any way to represent the Cusum charts?
a) Logarithmic
b) Algorithmic
c) Exponential
d) Normal
Answer: b
Explanation: There are two ways of representing the Cusums. The first one is the tabular Cusum, and the second one is the V mask form of Cusum.
3. The tabular Cusums are made to monitor the ____________ of the process.
a) Variance
b) Mean
c) Standard deviation
d) Capability
Answer: b
Explanation: The tabular Cusum is generally made to monitor the changes occurring in the mean of the process with respect to the time. This is preferred over the V-mask type.
4. Which of these is not one of the assumptions made for the construction of Tabular Cusum charts for the individual measurements?
a) The selected variable varies according to a lognormal distribution
b) The selected variable varies according to a normal distribution
c) The selected variable varies according to a binomial distribution
d) The selected variable varies according to an exponential distribution
Answer: b
Explanation: For the construction of the tabular cusum charts, it is assumed that the variable for which the Cusum charts are to be plotted, varies on a normal distribution.
5. Which of these is not an assumption made for the construction of the tabular Cusum charts?
a) The selected variable varies on a normal distribution
b) The mean of the normal distribution is available
c) The estimate or the exact value of the standard deviation of the variable
d) The assumptions are very inconsistent with phase II application of SPC
Answer: d
Explanation: Tabular Cusum charts are made with the assumptions that, the selected variable varies on normal distribution with a known mean and standard deviation. The standard deviation estimate may also be used.
6. If the process standard deviation increases, how will the cusum chart for monitoring process variability indicate the out-of-control state?
a) The value of S i + will increase
b) The value of S i + will decrease
c) The value of S i – will decrease with decrease in the value of S i +
d) The value of S i – will increase
Answer: a
Explanation: If the process standard deviation increases, S i + value will be increased consequently. It will eventually exceed the decision interval h and we will get an out-of-control signal.
7. Which of these is an indication of out-of-control process with low standard deviation?
a) Increase in the value of S i +
b) Decrease in the value of S i + with increase in the value of S i –
c) Increase in the value of S i –
d) Decrease in the value of S i – with decrease in the value of S i –
Answer: c
Explanation: If the process becomes out-of-control with low value of the process standard deviation, the value of S i – will increase and eventually exceed the decision interval, h.
8. Who was the first person to recommend plotting the cusum charts for mean and standard deviation on same graph?
a) Hawkins
b) Roberts
c) Atkinson
d) Crowder
Answer: a
Explanation: Although it is customary to plot the cusum charts for mean and cusum charts for standard deviation on different graphs, Hawkins was the first person who recommended to plot them on same graph.
9. If the deployment of the cusum is extended to the case of averages of the rational subgroups where sample size n>1, what will be done?
a) The value of x i be replaced by x i
b) The value of x i be replaced by x i
c) The value of C i + be replaced by value of C i –
d) The value of C i – be replaced by the value of C i +
Answer: a
Explanation: When the rational subgrouping procedure is adopted with sample size n>1, the cusum charts are built by simply replacing x i by x i and σ by σ/√n.
10. A cusum for normal variance is having the value of upper cusum equal to __________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: b
Explanation: The value of the upper cusum for the cusum for normal variance has the value of σ 2 replaced by the sample variance S i 2 . It is expressed as,
\
\)
11. A cusum for a normal variance has the value of lower cusum equal to ___________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: The value of the lower cusum for the normal variance is denoted by \
\)
12. If the value of C – i-1 +S 2 +k is lesser than , what will be the value of C i – equal to?
a) Lesser than -1
b) Higher than 1
c) Lower than 0
d) 0
Answer: d
Explanation: We know that
\
\)
Also the cusum values, whether it is the upper cusum, or lower cusum, are either zero or positive. So even if the value of C – i-1 +S 2 +k is lesser than , the value of C i – will be zero.
13. The V-mask procedure was proposed by ___________
a) Crowder
b) Roberts
c) Brinson
d) Barnard
Answer: d
Explanation: There are two procedures to plot the cusum charts. One is called the tabular cusum, and the other is called the V-mask procedure. The V-mask procedure was first introduced by Barnard .
14. Only the value of lower cusum can be negative.
a) True
b) False
Answer: b
Explanation: We know that the value of lower cusum is denoted by,
\
\)
So we get to know that the value of the lower cusum is always greater than or equal to zero.
15. Tabular Cusums can be constructed for both, individual measurements, and the averages of the rational subgroups.
a) True
b) False
Answer: a
Explanation: It is a big advantage of the tabular Cusum charts that they can be constructed for both cases; for individual measurements, and the averages of the rational subgroups.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – Cumulative Sum Control Chart – 4”.
1. Which of these is a correct expression for the one sided upper cusum?
a) \
+C_{i-1}^+]\)
b) \
+C_{i-1}^+]\)
c) \
-x_i+C_{i-1}^+ ]\)
d) \
-x_i+C_{i-1}^+]\)
Answer: b
Explanation: The tabular cusum works by accumulating derivations from mean that are above target with one statistic C i + , which is called one sided upper cusum. It is expressed as,
\
+C_{i-1}^+]\)
2. What is the value of one sided lower cusum?
a) \
+C_{i-1}^-]\)
b) \
+C_{i-1}^-]\)
c) \
-x_i+C_{i-1}^-]\)
d) \
-x_i+C_{i-1}^-]\)
Answer: d
Explanation: The tabular cusum works also by accumulating derivations from the mean that are below the target with one statistic C i – . The value of this value is called the one sided lower cusum and its expressed as,
\
-x_i+C_{i-1}^-]\)
3. What is K called in the expressions of the one-sided Cusums?
a) Regarded value
b) Related value
c) Resultant value
d) Reference value
Answer: d
Explanation: The values of the statistics used in the construction of a tabular cusum are called one sided upper and lower Cusums. K is used in their expressions which is called Reference value.
4. Which of these is another name of the reference value?
a) Regarding value
b) Stoppage value
c) Stack value
d) Assignable value
Answer: c
Explanation: The constant K in the expression of the one-sided Cusums, is called the reference value. It is also called slack value, or allowance.
5. The value of the reference value is chosen ____________
a) 3/4 ways between mean and the out-of-control mean towards the mean
b) 1/2 ways between mean and the out-of-control mean
c) 3/4 ways between mean and the out-of-control mean towards the out-of-control mean
d) 1/4 ways between mean and the out-of-control mean towards the mean
Answer: b
Explanation: The value of the reference value or K is chosen such that, it stay halfway between the target mean and the out-of-control value of the mean, that we are interested in finding out quickly.
6. What is the value δ is called, when used in the expression of K?
a) The shift of mean in standard deviation units
b) The shift of variance in standard deviation units
c) The shift of standard deviation in mean units
d) The shift of mean in variance units
Answer: a
Explanation: The value δ is used in the expression of K or the reference value.
K=\(\frac{\delta\sigma}{2}\)
Here this value δ is called the shift of mean in standard deviation units.
7. What is the value of μ 1 in the terms of the actual target mean μ 0 , and the shift?
a) μ 1 =μ 0 +δσ
b) μ 1 =μ 0 +2δσ
c) μ 1 =μ 0 +\
μ 1 =μ 0 -δσ
Answer: a
Explanation: The value μ 1 is called the out-of-control mean. It is the value of mean when process changes its state to out-of-control. Its value is,
μ 1 =μ 0 +δσ.
8. What is the value of K in the terms of out-of-control mean and the target mean?
a) \
\
\
\(K=3\frac{|μ_1-μ_0|}{4}\)
Answer: b
Explanation: The value K is called the allowance. This is used in the one-sided upper and lower Cusums expressions. It is expressed as,
\(K=\frac{|μ_1-μ_0|}{2}\)
9. What is the starting value of one-sided upper cusum?
a) 1
b) 6
c) 5
d) 0
Answer: d
Explanation: The one-sided upper cusum is expressed as follows,
\
+C_{i-1}^+]\)
The starting value of this would be C 0 + which is always taken as 0.
10. What is the starting value of the one-sided lower cusum?
a) 1.5
b) -1.5
c) 0
d) .3
Answer: c
Explanation: The lower one sided cusum is expressed as follows,
\
-x_i+C_{i-1}^-]\)
This value is called the starting value of lower one-sided cusum when the value of i=0. The value of C 0 – =0.
11. If the value of the quantity ”\
-x_i+C_{i-1}^-\)” becomes negative, what will be the value of the value C i – ?
a) Negative
b) Positive
c) Zero
d) Can be both, ±1
Answer: c
Explanation: We know that,
\
-x_i+C_{i-1}^-]\)
So, as the value of quantity \
-x_i+C_{i-1}^-\) becomes negative, it becomes lesser than 0. So we take the maximum value between zero and it, i.e. C i – =0
12. Which of these is always correct?
a) \
\
\
\(C_i^-<-1.323\)
Answer: a
Explanation: As the value of C i – is chosen as the maximum between zero and quantity \
-x_i+C_{i-1}^-\), we always have C i – ≥0.
13. After the value of C i – increasing than the value of _____ the process is said to be out-of-control.
a) Control interval
b) Decision interval
c) Distribution interval
d) Calculation interval
Answer: b
Explanation: There is a certain value decided for both C i – and C i + , after increasing from which, the process is said to be out-of-control. It is called decision interval.
14. If the value of μ 0 > μ 1 , K will have a negative value.
a) True
b) False
Answer: b
Explanation: We know that,
\(K=\frac{|μ_1-μ_0|}{2}\)
So even if μ 0 > μ 1 , K will have its value greater than 0.
15. The generally used value of K is the only value, which substantially impact the performance of the cusum.
a) True
b) False
Answer: b
Explanation: The cusum performance not only depends on the allowance or reference value K, but it also depends on the decision interval H. It is because H is related to the value of C i + or C i – above which the process becomes out-of-control.
This set of Statistical Quality Control Questions & Answers for Exams focuses on “Time-Weighted – Cumulative Sum Control Chart – 5”.
1. What are the graphical displays of tabular cusum called?
a) Cusum regular charts
b) Cusum regulated display
c) Cusum level display
d) Cusum status charts
Answer: d
Explanation: It is useful to present a graphical display for the tabular cusum. These charts are sometimes called the cusum status charts.
2. Which of these is a correct statement for cusum status charts?
a) A plot between C i + or C i – and the sample number
b) A plot between the C i + or C i – and the sample mean
c) A plot between the C i + or C i – and the sample variance
d) A plot between the C i + or C i – and the sample standard deviation
Answer: a
Explanation: Cusum status charts are constructed by plotting C i + or C i – versus the sample number. These are used to get an quick information about the process from the displayed chart.
3. Each vertical bar in cusum status chart represents __________
a) The value of C i + and C i –
b) The value of C i –
c) The value of C i +
d) Neither the value of C i + nor C i –
Answer: a
Explanation: Cusum status charts are the charts made to represent the cusum data on a graphical display, where each vertical bar in the cusum status charts, represents the value of C i + and C i – .
4. What is analogous to the control limits as in Shewhart control charts, in the cusum status charts?
a) The reference value
b) The allowance value
c) The value of the decision intervals
d) The value of C i + and C i –
Answer: c
Explanation: The value of the decision interval H can be used as the control limits in the cusum charts as in the Shewhart charts. This gives us an indication about when the process is out-of-control.
5. In the Minitab version of the cusum control charts, which of these is used?
a) \
\
\
\(C_i^-=min\left\{0,x_i-μ_0-k-C_{i-1}^-\right\}\)
Answer: b
Explanation: The Minitab version of the tabular cusum charts uses the negative of C i – i.e. it uses the following expression for C i – ,
\(C_i^-=min\left\{0,x_i-μ_0+k+C_{i-1}^-\right\}\).
6. Which of these is true for the Minitab version of tabular cusum charts?
a) \
\
\
\(C_i^-≤5\)
Answer: d
Explanation: The Minitab version of the tabular cusum uses the negative expression for the C i – , so the value of C i – always remains less than or equal to zero.
7. Which of these steps, is not carried out when the process becomes out-of-control when using the cusum control charts?
a) Search for an assignable cause
b) Taking corrective action
c) Restarting the control chart from zero
d) Continuing the control chart
Answer: d
Explanation: When an out-of-control situation is encountered while using the cusum control charts, we use the same procedure, used as in the Shewhart control charts. First, identifying the assignable cause, then second, taking corrective action, and then restarting the control chart from zero.
8. Which of these control charts may accurately tell when the assignable cause has occurred?
a) Cusum control charts
b) p-charts
c) c-charts
d) x bar and s charts
Answer: a
Explanation: As the pattern changes in the cusums due to an assignable cause, the changed pattern gets recorded in all the next cusums. So, we can identify from which cusum, the pattern changed, and hence the assignable cause location can be known.
9. Which of these can be used to calculate the estimate of mean of the process in the case of the tabular cusums when C i + > H?
a) \
\
\
\(\hat{μ}=\left\{μ_0+K+\frac{C_i^-}{N^-}\right\}\)
Answer: b
Explanation: In cases when a manipulative variable is required to take an out-of-control process back to target valueμ 0 , we use the new process mean estimate, which is written as,
\(\hat{μ}=\left\{μ_0+K+\frac{C_i^+}{N^+}\right\}\)
10. Which of these following expressions can be used to calculate the estimate of mean of the process in the case of the tabular cusums when C i – > H?
a) \
\
\
\(\hat{μ}=\left\{μ_0+K-\frac{C_i^-}{N^-}\right\}\)
Answer: b
Explanation: The cases when there is a need to estimate the changed process mean, the expression of new process mean estimate is written as,
\(\hat{μ}=\left\{μ_0-K-\frac{C_i^-}{N^-}\right\}\)
11. What of these can be used as the decision interval for the tabular cusum charts?
a) 2σ
b) 1σ
c) 5σ
d) 4σ
Answer: c
Explanation: It is a generalized rule that, the decision intervals should always be five times of the standard deviations of the process.
12. What is the expression for the two sided cusum ARL?
a) \
\
\
\(\frac{1}{ARL}=\frac{1}{ARL^-}+\frac{1}{ARL^+}\)
Answer: d
Explanation: The expression for the ARL of the two sided cusum from the ARLs of the two one sided statistics, is written as,
\(\frac{1}{ARL}=\frac{1}{ARL^-}+\frac{1}{ARL^+}\)
It is used to estimate the average time till an out-of-control signal comes from the cusum chart.
13. Which of these is an expression for Siegmund’s approximation?
a) \
\
\
\(ARL=\frac{exp-2∆b-1}{2∆^2}\)
Answer: a
Explanation: The Siegmund’s approximation is used to evaluate the average run length for the cusum charts. It is expressed as,
\(ARL=\frac{exp+2∆b-1}{2∆^2}\)
14. The value of K and H should be determined according to the ARL required for the corresponding cusum chart.
a) True
b) False
Answer: a
Explanation: ARL value tells about the time until the next out-of-control signal. So it is also an estimate to predict the type I error possibility. Hence, K and H values should be determined according to the ARL required.
15. The desired ARL is obtained by using the Siegmund’s approximation.
a) True
b) False
Answer: a
Explanation: A desired ARL predicts the probability of occurrence of a type I error. This ARL is obtained by using the Siegmund’s approximation.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – Cumulative Sum Control Chart – 6”.
1. What is the standardized value used for x i in the standardized cusum chart?
a) \
\
\
\(y_i=\frac{x_i-μ_0}{6σ}\)
Answer: b
Explanation: Many users of the cusum prefer to standardize the variable x i , before performing the calculations. The standardized value of the variable is,
\(y_i=\frac{x_i-μ_0}{σ}\)
2. What is the value of one sided upper cusum of the standardized cusum chart?
a) \
\
\
\(C_i^+=min\left\{0,y_i-k+C_{i-1}^+\right\}\)
Answer: a
Explanation: The standardized cusum chart uses the standardized value of variable x i , i.e. y i , so the value of the one-sided upper cusum of the standardized cusum chart will be,
\(C_i^+=max\left\{0,y_i-k+C_{i-1}^+\right\}\)
3. What is the value of the one-sided lower cusum of the standardized cusum chart?
a) \
\
\
\(C_i^+=max\left\{0,-y_i-k+C_{i-1}^-\right\}\)
Answer: c
Explanation: The standardized cusum uses the different variable to calculate the upper and lower cusums. The lower cusum of the standardized cusum chart is expressed as,
\(C_i^-=max\left\{0,-y_i-k+C_{i-1}^-\right\}\)
4. Which of these is an advantage of the standardized cusum chart?
a) There can be same means chosen for different processes
b) There can be same standard deviations chosen for different processes
c) The choices of k and h parameters are not scale dependent
d) No variability at all
Answer: c
Explanation: As in the standardized cusum charts, many charts can now have the same values of k and h, and because of the fact that the choice of k and h is dependent over the value of process standard deviation in normal cusum, the standardized cusum has k and h not scale dependent or σ dependent.
5. Combined Cusum-Shewhart procedure is applied _____________
a) On-line control
b) On-line measure
c) Off-line control
d) On-line measure
Answer: a
Explanation: As Combined Cusum-Shewhart procedure is used while using the cusum charts to detect the large process shifts while keeping the process continued, this is called an on-line control.
6. To apply Shewhart-cusum combined procedure, the Shewhart control limits should be applied almost _________ standard deviation from the center.
a) 2
b) 1
c) 1.5
d) 3.5
Answer: d
Explanation: Combined Cusum-Shewhart procedure is an on-line control method. It is used to detect larger process shifts. The Shewhart control limits in the procedure are put approx. 3.5 standard deviations away from center.
7. What is the full form of FIR feature in the cusum charts?
a) First initial response
b) Fast initial response
c) First initiation response
d) Free initial response
Answer: b
Explanation: The cusum control charts sensitivity at the process start-up is improved by the means of the FIR feature of the cusum charts. It means the Fast Initial Response.
8. What is the meaning of the 50% headstart?
a) The value of C 0 – equal to H/2
b) The value of C 0 + equal to H/2
c) Both the values of C 0 + and C 0 – equal to H/2
d) Both the values of C 0 + and C 0 – lesser than H/2
Answer: c
Explanation: The FIR feature of the cusum charts essentially sets the starting values of both, the values of C 0 + and C 0 – equal to typically, H/2. This is called 50% headstart.
9. What is the standardized variable value for the cusum charts from Hawkins?
a) \
\
\
\(v_i=\frac{2\sqrt{|y_i|}-0.822}{0.349}\)
Answer: a
Explanation: Hawkins presented a new standardized variable v i to be used in the standardized cusum chart. It had a value equal to,
\(v_i=\frac{\sqrt{|y_i|}-0.822}{0.349}\)
10. The standardized variable v i was subjected to vary more with respect to ____________ than process mean.
a) Sample mean
b) Sample variance
c) Process variance
d) Process standard deviation
Answer: c
Explanation: Hawkins had presented the std. variable v i to construct a cusum chart, which could possibly monitor process variability, by the variation of the variance of the process.
11. The two-sided standardized scale, i.e. standard deviation cusums will have its upper cusum value equal to ___________
a) \
\
\
\(S_i^+=max\left\{0,v_i-k+S_{i-1}^-\right\}\)
Answer: a
Explanation: The two-sided standardized scale cusums was first presented by Hawkins. This has the upper cusum value equal to,
\(S_i^+=max\left\{0,v_i-k+S_{i-1}^+\right\}\)
12. What is the value of lower cusum in the standardized scale cusum chart for process variability?
a) \
\
\
\(S_i^+=max\left\{0,-v_i-k+S_{i-1}^+\right\}\)
Answer: c
Explanation: The value of the lower cusum used in the standardized scale cusum chart, used for monitoring the process variability, is having the negative value of what was used in the normal cusum. This is written as,
\(S_i^-=max\left\{0,-v_i-k+S_{i-1}^-\right\}\)
13. The values of S i + or S i – at the starting are ____ if the FIR feature is not used.
a) 1
b) H
c) H/2
d) 0
Answer: d
Explanation: The values of S i + or S i – are H/2 mostly when the fast initial response feature is used. When it is not used the values of them become zero.
14. Only two-sided cusums are useful all over the industries.
a) True
b) False
Answer: b
Explanation: There are situations in which only a single one-sided cusum procedure is useful. For example, in the chemical process industry, where viscosity of a liquid can be allowed to drop below one value but should not increase rapidly. So this increase is monitored in the upper cusum.
15. Some cusums can have different sensitivity of the lower cusum than the upper cusum.
a) True
b) False
Answer: a
Explanation: Although cusum charts are designed to have same sensitivity in both the directions, upper and lower, but some cusums can have different sensitivity in the both directions.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – EWMA Control Chart – 1”.
1. What is the full form of E in the EWMA chart?
a) Exponentially
b) Experimentally
c) Exactly
d) Estimated
Answer: a
Explanation: The EWMA charts are a better alternative to the Shewhart control charts. The full form of E in the EWMA chart is exponentially.
2. What is the full form of EWMA?
a) Exponentially weighted moving average
b) Exponentially weighted measured approximate
c) Exponentially weighted moving approximate
d) Exponentially weighted measured average
Answer: a
Explanation: The EWMA charts were developed after the cusum charts. They have a full form of Exponentially Weighted Moving Average charts.
3. EWMA charts are better than Shewhart control charts in detecting the ___________ shifts.
a) Large process
b) Medium process
c) Small process
d) Every process
Answer: c
Explanation: EWMA charts and Cusum charts are used instead of the Shewhart control charts as the capability of these charts to find small process shifts, is greater than the Shewhart control charts’ capability.
4. Which of these is not an advantage of EWMA control charts?
a) Almost equivalent performance to the cusum charts
b) Easier set up than cusum charts
c) Easier to operate than cusum charts
d) Intricacy is a little higher than cusum charts
Answer: d
Explanation: The EWMA control charts are the control charts, which have almost equivalent performance to the cusum charts, and in some ways, they are easier to set up and operate.
5. Who did first introduce the EWMA charts?
a) Lucas
b) Saccucci
c) Roberts
d) Crowder
Answer: c
Explanation: The EWMA charts, which are also called the Exponentially Weighted Moving Average charts, were first introduced in the year of 1990 by Roberts. It was a revolutionary step towards process control.
6. What is the correct expression for the EWMA?
a) z i = λx i + z i-1
b) z i = λx i – z i-1
c) z i = λx i + z i+1
d) z i = λx i + z i-1
Answer: a
Explanation: The EWMA charts use the exponentially weighted moving average as the quantity plotted. The EWMA used is expressed as,
z i = λx i + z i-1 .
7. Which of these is correct for λ in EWMA expression?
a) 0 > λ
b) 1 < λ
c) 1 ≤ λ
d) 0 < λ ≤ 1
Answer: d
Explanation: The EWMA expression is written as,
z i = λx i + z i-1
Where λ is a constant, and it is having a value, 0 < λ ≤ 1.
8. What is the starting value of the EWMA?
a) Zero
b) Process target mean
c) Process target variance
d) Process target standard deviation
Answer: b
Explanation: The starting value of the exponentially weighted moving averages is z 0 and its starting value is equal to the process target . So,
z 0 = μ 0 .
9. Which of these is another name of EWMA charts?
a) GMA charts
b) AMA charts
c) EMA charts
d) RMA charts
Answer: a
Explanation: Because of the fact, that the weights decline geometrically when connected by a smooth curve, EWMA charts are also called the GMA charts.
10. GMA stands for ___________
a) Geosum moved average
b) Geometric Moving average
c) Geometrically moved average
d) Geocentric moving average
Answer: b
Explanation: GMA stands for Geometric moving average. It is another name of the EWMA charts because the weights decrease geometrically when connected by a smooth curve.
11. Which of these is the use of the EWMA charts?
a) In time series modeling
b) In Real-time processing
c) In acceptance sampling
d) In designing of experiments
Answer: a
Explanation: The EWMA charts are better at forecasting future results and predicting future trends in the control chart. So they are used in time series modeling and in forecasting future process behavior.
12. Which of these is an ideal chart for individual measurements among these all?
a) p-chart
b) Cusum charts
c) EWMA charts
d) x bar and R chart
Answer: c
Explanation: Since EWMA can be viewed as a weighted average of all past and current observations, it is very sensitive to the normality assumption. It is therefore an ideal control chart to use with individual observations.
13. If the observations x i are independent random variables with variance σ 2 then according to the EWMA charts, the variance of z i will be ________________
a) \
[1+^{2i}]\)
b) \
[1+^{2i}]\)
c) \
[1-^{2i}]\)
d) \
[1-^{2i}]\)
Answer: c
Explanation: The variance of the variable z i when used with EWMA, we get to know that, weights decline geometrically. So the variance becomes,
\
[1-^{2i}]\)
14. The weights of EWMA charts may also increase.
a) True
b) False
Answer: b
Explanation: We know that weights of EWMA charts decline with a geometric pattern. So they are called GMA charts. It makes clear that weights of EWMA charts can never increase.
15. The starting value of the variable z i is always equal to μ 0 .
a) True
b) False
Answer: b
Explanation: The starting value of the variable z i is mostly taken as μ 0 or the target value for mean but, sometimes the average of preliminary data is used as the starting value of EWMA. So that,
z 0 = x .
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – EWMA Control Chart – 2”.
1. The EWMA charts are constructed by plotting _______________ versus the sample number i or the time.
a) The sample observation, x i
b) The sample variance, σ 2
c) The variance of variable z i i.e. \
The variable z i
Answer: d
Explanation: The EWMA charts are the plots for knowing the trends in data due to assignable causes. They are constructed by plotting the variable z i versus the sample number i or time.
2. The UCL of the EWMA charts is given by ______________
a) UCL=\
UCL=\
UCL=\
UCL=\(μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-^{2i}\big\}]}\)
Answer: a
Explanation: The upper control limit of the EWMA charts is L times the standard deviation of the variable z i added, to the variable z i starting value μ 0 . It is expressed by,
UCL=\(μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-^{2i}\big\}]}\)
3. What is the center line value of the EWMA charts?
a) 0
b) 1
c) x
d) μ 0
Answer: d
Explanation: The center line value of the EWMA charts is the target value, i.e. the mean without shift. It is expressed as,
CL=μ 0
4. What is the value of L=2.27 upper control limit for a EWMA chart which has value of λ=0.10 and the value of the standard deviation σ=1 and the value of μ 0 =10?
a) 10.41
b) 9.59
c) 10.27
d) 10.10
Answer: c
Explanation: We know that,
UCL=\(μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-^{2i}\big\}]}\)
Putting the values, we get UCL=10.27.
5. What is the value of LCL of EWMA control charts?
a) LCL=\
LCL=\
LCL=\
LCL=\(μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-^{2i}\big\}]}\)
Answer: b
Explanation: We know that the LCL of the EWMA charts are just L times the standard deviation. So the LCL of the EWMA charts is given by,
LCL=\(μ_0-Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-^{2i}\big\}]}\)
6. When the EWMA charts are in operation for quite a long period, what will be the value of the UCL of the chart?
a) UCL=\
UCL=\
UCL=\
UCL=\(μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}\)
Answer: a
Explanation: EWMA charts, when are in operation for a long period, the term, 1- 2i becomes unity as i becomes large. So the UCL can be written as,
UCL=\(μ_0+Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}\)
7. What will be the value of the UCL for a control chart, which has the value of λ=0.10, L=2.7, σ=1 and μ 0 =10? The value of i can be assumed quite large.
a) UCL=10.67
b) UCL=10.62
c) UCL=9.38
d) UCL=9.37
Answer: b
Explanation: We know that when the value of i is high, we calculate UCL by the following expression.
UCL=\(μ_0+Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}\)
Putting the values, we get UCL=10.62.
8. What is the LCL value when the value of i becomes high in the EWMA charts?
a) LCL=\
LCL=\
LCL=\
LCL=\(μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}\)
Answer: d
Explanation: We know that the difference between the UCL and the CL is the same as the difference between the CL and the LCL. So we can write the expression for LCL when the process is running for quite a long period as,
LCL=\(μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}\)
9. After some time, i in the equation of LCL becomes high. What will be the value of λ when this occurs? You may use, LCL=9.38, the standard deviation= 1 and the target mean = 10.
a) 0.1
b) 2.8
c) 3.0
d) 6.0
Answer: a
Explanation: We know that,
LCL=\(μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}\)
Here, we consider the high value of i. When we put the given values, we get, λ=0.1.
10. Which of the parameters are chosen wisely in the case of EWMA charts to get a desired ARL?
a) Only L
b) Only σ
c) Both, L and λ
d) Only λ
Answer: c
Explanation: We know that in the parameters used for constructing EWMA charts, only L and λ are the design parameters. So we choose them wisely to obtain desired ARL.
11. When the value of λ is small, and if the value of EWMA is on one side of the CL and the shift of mean occurs in the opposite direction, EWMA takes several periods to react to that shift. This is called _____
a) Weight effect
b) Stationary effect
c) Inertia effect
d) Tendency effect
Answer: c
Explanation: The staying of EWMA on a side till a very heavy shift occurs in the process mean; is called the inertia effect of the EWMA charts. This occurs only when the value of λ is very small.
12. How is the effectiveness of EWMA charts by inertia effect?
a) Increase
b) Decrease
c) No change
d) Can either decrease or increase
Answer: b
Explanation: As inertial effects increase the time for EWMA to react to a process mean shift, it ultimately is influencing the effectiveness of EWMA charts. This decreases the effectiveness of EWMA.
13. Which of these values of λ gives the weights given to current and previous observations matching as closely as possible the weights when these observations give by a Shewhart chart with the WE rules?
a) 0.7
b) 0.9
c) 0.2
d) 0.4
Answer: d
Explanation: The value of λ is recommended by Hunter to be equal to 0.4 to get the weights given to current and previous observations, matching closely to the weights when these observations give by a Shewhart chart, with Western Electric rules.
14. The control limits of EWMA charts do not depend on the sample number when the sample number becomes quite large.
a) True
b) False
Answer: a
Explanation: We know that the sample number is denoted by i in the control limit expression of EWMA charts. As i becomes large, it eliminates its term from the control limit expressions and they don’t depend on i. They become,
Control limits of EWMA = \(μ_0±Lσ\sqrt{\frac{λ}{2-λ}}\)
15. ARL of the process, when EWMA is used, does not depend on σ.
a) True
b) False
Answer: b
Explanation: The design parameters of the EWMA charts are the multiple of sigma and the value of λ. It is possible to choose only these, to give desired ARL performance for the EWMA charts.
This set of Tricky Statistical Quality Control Questions and Answers focuses on “Time-Weighted – EWMA Control Chart – 3”.
1. The largest standard deviation of the sample mean from the target, not leading to an immediate out-of-control signal, is called _______________ of a control chart.
a) ARL
b) UCL
c) Signal resolution
d) Signal resistance
Answer: d
Explanation: Signal Resistance is a very important property of any control chart. It is defined as the maximum standard deviation of the sample mean from the target or in-control value, not leading to an immediate out-of-control signal.
2. For a Shewhart control chart, what is the value of SR?
a) Z α/2
b) 2
c) L
d) 2L
Answer: c
Explanation: The value of SR is the maximum standard deviation of mean off-target without any out-of-control signal. This is L for the Shewhart control chart, i.e.
SR( x )=L
Where L is the multiplier used to obtain the control limits.
3. Which of these is a having a constant SR?
a) A x chart
b) A Cusum chart
c) An EWMA chart
d) Every control chart
Answer: a
Explanation: The Shewhart control charts have the signal resistance value equal to the multiplier used to obtain the control limit, which is constant. As a x chart is also a Shewhart control chart, so for this also, “ SR=constant”.
4. Which of these is not having a constant SR value?
a) c-chart
b) R-chart
c) p-chart
d) EWMA chart
Answer: d
Explanation: We know that only Shewhart control chart have the SR value equal to constant. As EWMA chart is not a Shewhart chart, it does not have a constant SR value.
5. What is the value of SR for an EWMA chart?
a) SR=\
SR=\
SR=\
SR=\(\frac{L\sqrt{\frac{λ}{2-λ}}+w}{λ}\)
Answer: b
Explanation: We have derived an equation of an EWMA chart SR which includes the value of the EWMA statistic. The SR is given by,
SR=\(\frac{L\sqrt{\frac{λ}{2-λ}}-w}{λ}\)
6. What is the value of the term “w” called in the SR value of a EWMA chart?
a) First EWMA value
b) Last EWMA value
c) Value of EWMA statistic
d) Weight average
Answer: c
Explanation: The SR value of EWMA chart is given by,
SR=\(\frac{L\sqrt{\frac{λ}{2-λ}}-w}{λ}\)
Here, w is called the value of EWMA statistic.
7. Which of these is the maximum value of the SR averaged over all values of EWMA statistic?
a) \
=L \frac{}{λ}\)
b) \
=L\sqrt{\frac{}{λ}}\)
c) \
=L+\sqrt{\frac{}{λ}}\)
d) \
=L-\sqrt{\frac{}{λ}}\)
Answer: b
Explanation: If the chart has asymptotic limits, then the maximum value of the signal resistance averaged over all values of EWMA statistic is,
\
=L\sqrt{\frac{}{λ}}\)
8. Small values of λ gives ______ values of the SR .
a) Larger
b) Lower
c) Equal
d) Both, larger and smaller
Answer: a
Explanation: As we know that,
SR=\
∝1/λ, so SR will be larger for smaller values of λ.
9. Which of these methods can be used to counteract the signal resistance in case of small values of λ with EWMA charts?
a) To use a cusum chart paralleled
b) To use a Shewhart control chart paralleled
c) To just wait for EWMA chart to be plotted till 50 observations
d) To increase the value of process mean target
Answer: b
Explanation: To counteract the signal resistance’s higher value in the case of low λ value, we use the Shewhart control charts in conjunction with an EWMA chart.
10. Who were the first persons to recommend the use of Shewhart control charts with the EWMA charts with low values of λ?
a) Woodall and Mahmoud
b) Woodall and Roberts
c) Roberts and Mahmoud
d) Roberts and Crowder
Answer: a
Explanation: Woodall and Mahmoud recommended to always use a Shewhart control chart with a EWMA chart , as one way to counteract the signal resistance’s high value.
11. Which of these is not an advantage of using a Shewhart control chart with the EWMA chart?
a) Detection of larger process shifts
b) ARL adjustment
c) Counteraction to the high signal resistance value
d) No defects
Answer: d
Explanation: We know that Shewhart control charts are used with the EWMA control charts. They provide advantages like easy ARL adjustment, counteraction to high signal resistance value, and detection of larger process shifts, without lowering the ability to find lower process shifts.
12. Which of these is not an advantage of low value of λ in the case of EWMA charts?
a) Good ARL performance
b) Insensitivity to the normality of the data
c) Good ability to detect the larger process shifts
d) Easy detection of small process shifts
Answer: d
Explanation: Low values of λ are quite helpful in finding small process shifts, good AR performance, and in increasing the insensitivity of the EWMA charts to the normality of the data. They don’t give the charts ability to detect the larger process shifts.
13. In the case of Shewhart charts, if the data is not according to the normal distribution, the actual in-control ARL (ARL 0 ) will be ____________ the expected value.
a) Considerably higher than
b) Considerably lower than
c) Equal to
d) Almost equal to
Answer: b
Explanation: It is experimentally determined that if the process data does not follow the normal distribution, then the Shewhart control charts ARL value will be quite less than the value expected based on the normality assumption.
14. The signal resistance value depends on the value chosen for λ.
a) True
b) False
Answer: a
Explanation: We know that,
SR=\
depends only on the value of the λ.
15. The EWMA charts are recommended for phase II applications as it is totally a nonparametric procedure.
a) True
b) False
Answer: a
Explanation: We know that the variation in ARL, which is caused by the non-normality of data, is very low in the case of low λ with the EWMA charts. So they are recommended for phase II applications as it is a perfectly nonparametric procedure.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Time-Weighted – EWMA Control Chart – 4”.
1. What is the initial S in EWMS stand for?
a) Severity error
b) Signal error
c) Square error
d) Simple error
Answer: c
Explanation: The EWMS charts were used to monitor the variability in the process. They are variants of EWMA chart. EWMS stands for Exponentially Weighted Mean Square error.
2. Who were the first people to introduce the EWMA charts to monitor process standard deviation?
a) McGregor and Harris
b) Harris and Roberts
c) Crowder and Roberts
d) Harris and Roberts
Answer: a
Explanation: It was Macgregor, and Harris , who were the first people to identify the importance of EWMA charts to monitor process std deviation. They recommended using EWMS charts to do so.
3. What is the value of EWMS?
a) \
^2- S_{i-1}^2\)
b) \
^2- S_{i-1}^2\)
c) \
^2+ S_{i-1}^2\)
d) \
^2+ S_{i-1}^2\)
Answer: c
Explanation: EWMS is defined as the exponentially weighted mean square error. It is expressed as,\
^2+ S_{i-1}^2\)
4. “S i 2 /σ 2 ” has an approximate __________ distribution.
a) Normal
b) Lognormal
c) Exponential
d) Chi-square
Answer: d
Explanation: If the observations in EWMS are independent and normally distributed, then S i 2 /σ 2 will have an approximate chi-square distribution, with v=/λ degrees of freedom.
5. EWRMS chart plots __________ on the control chart.
a) Exponentially weighted root moving square error
b) Exponentially weighted root mean square error
c) Exponentially weighted root mean signal error
d) Exponentially weighted root moving signal error
Answer: b
Explanation: If we have a target value for the mean, and if we want to monitor process standard deviation, we use the EWRMS charts which use Exponentially Weighted Root Mean Square error.
6. EWRMS charts have the upper limit of ____________
a) UCL=\
UCL=\
UCL=\
UCL=\(\sqrt{\frac{χ_{v,\frac{α}{2}}^2}{v}}\)
Answer: a
Explanation: The EWRMS chart is plotted by obtaining the value of Exponentially Weighted Root Mean Square error, which has an approximate chi-square distribution. It has an upper limit of,
UCL=\(σ_0 \sqrt{\frac{χ_{v,\frac{α}{2}}^2}{v}}\)
7. What is the lower limit of the EWRMS chart?
a) LCL=\
LCL=\
LCL=\
LCL=\(\frac{σ_0}{2} \sqrt{\frac{χ_{v,1-\frac{α}{2}}^2}{v}}\)
Answer: c
Explanation: We know that the EWRMS chart has an approximate chi-square distribution when the observations are independent and normally distributed. So LCL of the EWRMS chart will be,
LCL=\(σ_0 \sqrt{\frac{χ_{v,1-\frac{α}{2}}^2}{v}}\)
8. EWRMS chart is sensitive to _____________
a) Process mean only
b) Process standard deviation only
c) Neither process mean nor standard deviation
d) Both, process mean and standard deviation
Answer: d
Explanation: EWRMS chart is constructed to monitor the process standard deviation. But this also, is considered that the process mean is also a factor affecting the EWRMS chart. So EWRMS chart is sensitive to both, the process mean and the standard deviation.
9. EWMV is ____________
a) Exponentially weighted mean variability
b) Exponentially weighted moving variance
c) Exponentially weighted mean variance
d) Exponentially weighted moving variability
Answer: b
Explanation: EWMV is the exponentially weighted moving variance. It is a term derived from the EWRMS and it is used in measuring variability of the process with EWMA charts.
10. What is the value of EWMV?
a) \
^2- S_{i-1}^2\)
b) \
^2± S_{i-1}^2\)
c) \
^2- S_{i-1}^2\)
d) \
^2+ S_{i-1}^2\)
Answer: d
Explanation: The EWMV is said to be the exponentially weighted moving average. It is expressed by the following expression,
\
^2+ S_{i-1}^2\)
11. What is the upper limit for the EWMA for Poisson data?
a) UCL=\
UCL=\
UCL=\
UCL=\(μ_0+A_L \sqrt{\frac{λμ_0}{2-λ}\big\{1-^{2i}\big\}}\)
Answer: a
Explanation: The EWMA is generally plotted for normal data. It is also said that the EWMA chart with low value of λ is nonparametric. When EWMA is plotted for Poisson data, the UCL is,
UCL=\(μ_0+A_U \sqrt{\frac{λμ_0}{2-λ}\big\{1-^{2i}\big\}}\)
12. LCL for EWMA chart for Poisson distribution is written as ____________
a) LCL=\
LCL=\
LCL=\
LCL=\(μ_0-A_L \sqrt{\frac{λμ_0}{2-λ}\big\{1-^{2i}\big\}}\)
Answer: d
Explanation: EWMA charts are made to monitor the process mean shift. The EWMA chart for Poisson distribution is plotted with an assumption that, the data from the process follows Poisson distribution, It has LCL of,
LCL=\(μ_0-A_L \sqrt{\frac{λμ_0}{2-λ}\big\{1-^{2i}\big\}}\)
13. “A L ” in the expression of the LCL of EWMA charts for Poisson distribution, is ______
a) Lower control limit factor
b) Lower allowance factor
c) Life Allowance factor
d) Last Allowance factor
Answer: a
Explanation: The factors A U and A L are included in the expressions of UCL and LCL of the EWMA charts plotted for Poisson data, respectively. They are called upper and lower control limit factors respectively.
14. EWMA recursion is different in the case of the EWMA charts for normal data and EWMA charts for Poisson data.
a) True
b) False
Answer: b
Explanation: The recursion of EWMA is the property of the EWMA statistic. That is why, the EWMA charts show trend due to an assignable cause. This recursion is always same, whether the EWMA charts are plotted for normal data or for the Poisson data.
15. The Poisson EWMA has considerably better ability to detect assignable causes than Shewhart c-chart.
a) True
b) False
Answer: a
Explanation: As Poisson EWMA has a better ability of detecting small process shifts in the phase II applications of SPC, they are also having a better ability to detect assignable causes than Shewhart c-chart.
16. Which of these can be used as a forecast of where the process mean will be at the next time period?
a) p-chart
b) c-chart
c) EWMA chart
d) R-chart
Answer: c
Explanation: The EWMA charts have both monitoring, and forecasting abilities. They can actually predict the position of process mean at the next time period. The statistic, z i is actually a forecast of the value of the process mean at time i+1.
17. The EWMA chart can be used as a basis for a dynamic process-control algorithm.
a) True
b) False
Answer: a
Explanation: Due to the fact that the EWMA charts can both monitor the process, and forecast the process mean in future time, they can be useful in process control. They can also detect small process shifts. So they may be used as the basis of a dynamic process control algorithm.
18. The moving average span w at a time I is defined as ____________
a) \
\
\
\(M_i=\frac{x_i+x_{i-1}+⋯x_{i-w+1}}{i}\)
Answer: c
Explanation: The moving average is also a time weighted chart as EWMA charts and the Cusum charts. The moving average chart uses Moving Average span w at time i, which is expressed as,
\(M_i=\frac{x_i+x_{i-1}+⋯x_{i-w+1}}{w}\)
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 1”.
1. Which of these is the correct expansion of DNOM?
a) Deviation from Lognormal
b) Derived from normal
c) Deviation from normal
d) Deviation from normality
Answer: c
Explanation: The DNOM in DNOM control chart denote the deviation from normal. This technique is used for the process quality control in the case of short process run or production runs.
2. Which of these is used as a technique for process control for short production runs?
a) Using deviation from nominal to plot the control chart
b) Using the variable value to plot the control chart
c) Using the variable change to plot the control chart
d) Using the variable decrease or increase to plot the chart
Answer: a
Explanation: The short production runs are having very wide applications in modern industries. The statistical process control technique used for the short production runs is generally measuring deviation from nominal value to plot the control chart, instead of the variable control chart.
3. If the M i is the actual sample measurement, and T A is a nominal value, what will be the DNOM?
a) x i = M i – 2T A
b) x i = M i – T A
c) x i = 2T A – M i
d) x i = T A + M i
Answer: b
Explanation: The deviation from the normal or nominal value is used for the plotting of DNOM control chart. The deviation from the normal is evaluated as,
x i = M i – T A .
4. The mean of the variable measured is replaced by the mean of ________ value in the case of the DNOM chart.
a) Normal value
b) Deviation from normal
c) Range
d) Deviation from normal range
Answer: b
Explanation: The DNOM chart is used for short production runs, which used DNOM or deviation from normal as the x as in the case of x chart. SO the mean of variable measured is replaced by the mean of the DNOM values.
5. If in a manufacturing process, the measurements for a certain dimension for a part are, 50, 51 and 52; where the nominal value of the dimension measured is 50, what will be the DNOM mean?
a) 1.0
b) 50
c) 51
d) 0.33
Answer: a
Explanation: We know that the deviations from nominal are,
x i = M i – T A
By putting values, we get DNOM’s 0, 1 and 2. The mean DNOM will be 1.0.
6. How many samples are generally taken before calculating the control limits for the DNOM x bar and R charts?
a) 10
b) 40
c) 35
d) 20
Answer: d
Explanation: We use the DNOM mean for the plot of the x bar and R charts using the deviation from the normal. It is recommended that we should take almost 20 samples before calculating the control limits for DNOM x bar and R chart.
7. Which of these assumptions are made while using the DNOM approach?
a) The process standard deviation is same for all parts
b) The process standard deviation is different for all parts
c) The samples size is variable
d) The sample size is 1
Answer: a
Explanation: DNOM approach is defined as the method to plot control charts by calculating the deviation from the nominal value. While using this approach, it is assumed that, the process standard deviation is same for all parts.
8. Which of these is true for DNOM approach?
a) The sample size is variable
b) The process standard deviation is different for all parts
c) The sample size varies from one part to another
d) The sample size remains same for all parts
Answer: d
Explanation: It is desirable that, the sample sizes, which are used to find out the observations data, should be constant, when the DNOM approach is to be used for the control charting.
9. What is used instead of Nominal values when the nominal value is not given in the case of DNOM approach?
a) Historical process standard deviation
b) New standard deviation
c) Historical process average
d) Historical process variance
Answer: c
Explanation: The DNOM approach is based on using the deviation from normal data instead of the variable itself. When the nominal value is not given for this, the historical process average is used instead of the nominal value.
10. Which of these does not need to have the nominal value specified frequently?
a) When part has two-sided specifications
b) When part has one-sided specifications
c) When the part can’t have one sided specifications
d) When the part has either one-sided or two sided specifications
Answer: b
Explanation: It is generally very obvious that the DNOM approach needs the nominal value quite frequently but, when the part has one-sided specifications, the nominal is not specified frequently.
11. Standard hypothesis testing procedures are used to compare __________ process average to a desired process target to determine whether the process mean is different from the target.
a) New
b) Historical
c) Correct
d) Disturbed
Answer: b
Explanation: The DNOM uses the historical process average data when the nominal is not available. Standard hypothesis test procedures are used to compare this with the process target, to check their difference.
12. Which of these will work correctly and efficiently, when the process standard deviations are different for different part numbers?
a) Standardized x bar and R charts
b) Standardized attribute charts
c) DNOM charts
d) Both, the cusum and EWMA charts
Answer: a
Explanation: While creating the DNOM based x bar and R charts, it’s assumed that the different parts have same standard deviations. When they are not same, the standardized charts are used.
13. Standardized R chart which uses control limits D3 and D4 has the value of \Missing open brace for subscript \
\
\
\(R_i^s=\frac{R_i}{R_j}\)
Answer: d
Explanation: The standardized R chart is used when the parts have different standard deviations. It has the quantity plotted as,
\(R_i^s=\frac{R_i}{R_j}\)
14. Deviation from the process target charts work efficiently when the process standard deviation is not same.
a) True
b) False
Answer: b
Explanation: While constructing the DNOM charts, it’s believed that the process standard deviations are same. But if they are not, they don’t perform well. The same occurs for the Deviation from the Process target charts.
15. Sample size is allowed to vary for the construction of the DNOM based control charts.
a) True
b) False
Answer: b
Explanation: It is a necessary assumption for the DNOM based x bar and R charts that, the Sample size must be constant for all part numbers. Without this the control charts don’t give effective results.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 2”.
1. What is the UCL of the standardized R chart for short production runs?
a) D 4
b) D 5
c) D 1
d) D 2
Answer: a
Explanation: The standardized R chart is the best way to study the short production run data when the part standard deviations are not same. This chart has UCL value of D 4 .
2. What is the value of LCL of the standardized R chart for short runs?
a) D 4
b) D 2
c) D 1
d) D 3
Answer: d
Explanation: The standardized R chart is used to study the short production runs. The effectiveness of this chart depends on the control limits chosen. It has UCL and LCL of values, D 4 and D 3 respectively.
3. What is the value of standard variable plotted in the standardized x bar chart for short run production?
a) \
\
\
\(\bar{x}_i^s=\frac{2\bar{M}_i-T_j}{\bar{R}_j}\)
Answer: c
Explanation: The standardized charts are plotted for both R and x bar. The x bar chart has the value of the standard variable as,
\(\bar{x}_i^s=\frac{\bar{M}_i-T_j}{\bar{R}_j}\)
4. The standardized x bar chart has the value of LCL when used for short production runs is ____________
a) 3
b) –A 2
c) 1
d) A 2
Answer: d
Explanation: The standardized x bar chart is used to find the process performance of short production runs. This chart has the LCL value equal to –A 2 .
5. Which of these has the same difference between the LCL and Center line, and the UCL and the center line?
a) Standardized R chart for short production runs
b) Standardized x bar charts for short production runs
c) Both, standardized x bar and R charts for short production runs
d) Neither one of the standardized R chart and standardized x bar charts for short production runs
Answer: b
Explanation: The x bar chart for short production runs, has the LCL and UCL values equal to ∓A 2 respectively. So the difference from the centerline for them is equal.
6. The term M i , which is used in the expression of the standard variable used in standardized x bar charts for short production runs, is defined as ___________
a) Average of averages of standard deviations
b) Average of original mean measurements
c) Moving average
d) Exponentially weighted moving average
Answer: b
Explanation: The term M i is the average of original mean measurements. It is used in the expression of the standard variable used in the standardized x bar charts for short production runs.
7. R j is the ____________
a) Target value for range
b) Target value for ratios
c) Result of jth part
d) Ratio of jth part
Answer: a
Explanation: R j is used in the expression of the standard value variable in the standardized R chart. It is defined as the target value for range for each part number.
8. T j stands for __________
a) Target value for relative range
b) Target value for relative mean
c) target value for x bar for each part number
d) Target value for tested value of the variable
Answer: c
Explanation: The term T j is used in the expression of the standardized x bar variable, which is used in the standardized x bar chart for the short production runs. It is the target value for the x bar for each part number.
9. Which of these is correct?
a) \
\
\
\(\bar{R}_j=\frac{d_2}{c_4}\)
Answer: a
Explanation: The value of target for the range and the x bar, i.e. R j and T j are specified using the specifications for T j , and using the history for R j . Here we can say that,
\(\bar{R}_j=\frac{Sd_2}{c_4}\)
10. Standardized control chart approach for short production runs was presented by ____________
a) Rodriguez
b) Roberts
c) Crowder
d) Farnum
Answer: d
Explanation: Farnum has presented a generalized approach to the DNOM procedure. The standardized control chart approach is a special case of his method.
11. According to Farnum, the coefficient of variation is _____________
a) \
\
\
\(\frac{2μ}{σ}\)
Answer: a
Explanation: Farnum was the first person to present the generalized approach to the DNOM procedure. He defined the coefficient of variation as σ/μ.
12. Which of these is true according to Farnum?
a) The coefficient of the variation should be randomly varying
b) The coefficient of the variation should be systematically varying
c) The coefficient of the variation should be varying, maybe randomly or systematically
d) The coefficient of the variation must be constant
Answer: d
Explanation: Farnum defined the coefficient of variation as σ/μ, which he used to give his generalized approach to DNOM procedure. According to him, the coefficient of variation should be constant, which probably occurs fairly often in practice.
13. Which of these is correct target value for standardized control chart for fraction nonconforming for shorter production runs?
a) p
b) n p
c) c
d) R̅
Answer: a
Explanation: The short production runs concept can also be used for attributes control chart. The standardized p chart for short production runs uses the target value equal to p .
14. The standardized control chart approach for short production runs can only be applied on variable control charts.
a) True
b) False
Answer: b
Explanation: The standardized control chart approach for short production runs is generally applied to the x bar and R charts but, it can also be applied on the attributes charts such as p-chart, c-chart, and np-chart.
15. The DNOM approach does not need the measurements to be taken at all.
a) True
b) False
Answer: b
Explanation: The DNOM is calculated by,
x i =M i -T A
Where M i stands for the measurements of the dimension. So to calculate DNOM value we need to have the value of the dimension measured.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 3”.
1. What is the statistic used to plot on control chart for a standardized p-chart for short production runs?
a) \
\
\
\(Z_i=\frac{\hat{p}_i+\bar{p}}{\frac{\sqrt{\bar{p}
}}{n}}\)
Answer: b
Explanation: The p-charts are plotted on the data of no of samples not conforming. The statistic plotted on the standardized version of it, is expressed as,
\(Z_i=\frac{\hat{p}_i-\bar{p}}{\frac{\sqrt{\bar{p}
}}{n}}\)
2. What is the value of standard deviation for the standardized p-chart for short production runs?
a) \
\
\
\(\sqrt{\frac{\bar{p}
}{2n}}\)
Answer: a
Explanation: The standardized control charts are used to monitor processes for short production runs. The standardized p-chart which is used in short production runs is having the standard deviation value equal to,
\(\sqrt{\frac{\bar{p}
}{n}}\)
3. What is the standard deviation value for the standardized c chart used for the short production runs?
a) \ \
\
\(\hat{c}\)
Answer: b
Explanation: The standard deviation value for the standardized c chart remains constant if it is used for the short production runs. It is expressed as,
\(\sqrt{\bar{c}}\)
4. What is the value of the statistic to be plotted on the standardized c-chart which is designed to run in the short production?
a) \
\
\
\(Z_i=\frac{c_i+\bar{c}}{\sqrt{\bar{c}}}\)
Answer: b
Explanation: The c chart is plotted keeping the nonconformity number data as the observations. The statistic which is to be standardized version of it, when the case is of the short production runs, is
\(Z_i=\frac{c_i-2\bar{c}}{\sqrt{\bar{c}}}\)
5. What is the value of the target value for the number nonconforming chart?
a) n p
b) 2 p
c) 3 p
d) 2n p
Answer: a
Explanation: The number nonconforming chart is called the np chart too. It plots the number of nonconforming samples in total samples. The target value for the number nonconforming chart is n p .
6. The attribute u i when plotted on standardized control chart for short production runs, the statistic plotted on the chart has the value equal to __________
a) \
\
\
\(Z_i=\frac{u_i-2\bar{u}}{\sqrt{\frac{\bar{u}}{n}}}\)
Answer: a
Explanation: The attribute u is called the average number of nonconformities per unit. The standardized u-chart is plotted for statistic,
\(Z_i=\frac{u_i-\bar{u}}{\sqrt{\frac{\bar{u}}{n}}}\)
7. What is the standard deviation of the number of nonconformities per unit, when the standardized u-chart is used for monitoring the process?
a) \
\
\
\(\sqrt{\bar{u}}\)
Answer: b
Explanation: The deviation of the number of nonconformities per unit is plotted on Shewhart u-chart. Its standard deviation is expressed as,
standard deviation=\(\sqrt{\frac{\bar{u}}{n}}\)
8. What is the value of the statistic plotted on the standardized np-chart?
a) \
\
\
\(z_i=\frac{n\hat{p}_i-n\bar{p}}{\sqrt{n\bar{p}
}}\)
Answer: d
Explanation: The np chart monitors the process by using the data of the nonconforming samples. The statistic plotted on the standardized version of it for short production run is,
\(z_i=\frac{n\hat{p}_i-n\bar{p}}{\sqrt{n\bar{p}
}}\)
9. The upper limit of the standardized c-chart for short production runs is ________
a) 1
b) 2
c) -3
d) +3
Answer: d
Explanation: The c chart is used to plot the number of nonconformities on the control chart. The standardized version of it has the UCL of +3 and LCL of -3.
10. The LCL of the standardized p-chart is ________
a) -2
b) +3
c) -1
d) -3
Answer: d
Explanation: The LCL and UCL of the standardized p-chart are at the same units away from the center line which is at zero. The UCL and the LCL of the standardized p-chart are ±3.
11. The center line of the np-chart is at _______
a) -2
b) +2
c) 0
d) 3
Answer: c
Explanation: The control chart, which plots the number of nonconforming samples, is called np-chart. This charts when standardized, have their center line at the zero.
12. Which of these correctly shows the correct values for standardized u-chart in the order of CL, UCL and LCL?
a) 0, 3 and -3
b) 3, 0 and -3
c) 0, -3 and 3
d) -3, 0 and 3
Answer: a
Explanation: THE u-chart, when standardized, has the center line at zero value. The values of the UCL and LCL of the standardized version of u-chart, plotted for short production runs, are ±3.
13. Which of these does not have the LCL and UCL at ∓3?
a) Standardized u-chart
b) Standardized R-chart
c) Standardized c-chart
d) Standardized p-chart
Answer: b
Explanation: The standardized attribute charts are the only charts which have their upper and lower control limits at ±3. Variable control charts do not have their control limits at ±3.
14. The values of control limits for the standardized c-chart are different from the control limits of the standardized p-chart.
a) True
b) False
Answer: b
Explanation: Both, the p-chart and the c-chart when standardized, have their control limits at ±3 value. So the values of the control limits for both, standardized c-chart, and standardized p-chart, are same.
15. The control limits for the p-chart and the standardized p-chart are different.
a) True
b) False
Answer: a
Explanation: The p-chart has the control limits at,
\(p \pm 3\sqrt{\frac{p}{n}}\)
Here p is the fraction nonconforming value. Whereas, the standardized p-chart has its control limits at ±3 values.
This set of Tough Statistical Quality Control Questions and Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 4”.
1. EWMA and cusum charts can be used for the short production runs because ____________
a) They are good in phase II
b) They have shorter ARL
c) They don’t detect process shift
d) They detect larger process shift
Answer: b
Explanation: EWMA charts and the cusum charts both, find applications in the short production run cases. This is due to the fact that the EWMA charts and the Cusum charts both have shorter ARL than the Shewhart charts.
2. Which of these is suitable for the short run environment?
a) x bar chart
b) R chart
c) p-chart
d) Self starting cusum chart
Answer: d
Explanation: The Cusum charts are effective for subgroups size of one. The self starting version of the Cusum charts is particularly suitable for the short run environment due to the same fact.
3. Which of these is not one of the areas for which, the self starting cusum approach uses regular process measurements for?
a) For establishing the cusum
b) For calibrating the cusum
c) For acceptance sampling
d) For process monitoring
Answer: c
Explanation: The self starting approach of cusum charts uses regular process measurements for both establishing the cusum, and calibrating of cusum. These measurements are also used for process monitoring.
4. Which of these is eliminated by the self starting approach?
a) Phase I parameter estimation phase
b) Phase I parameter measurement phase
c) Phase II parameter estimation phase
d) Phase II parameter measurement phase
Answer: a
Explanation: The self starting approach uses regular process measurements for establishing and calibrating the cusum, and for process monitoring. Thus, this eliminates the phase I parameter estimation stage.
5. The false alarm rate is _______ when small number of subgroups is used for the Shewhart control charts.
a) Decreased
b) Increased
c) Remains same
d) Changed randomly
Answer: b
Explanation: The number of subgroups used in calculating the trial control limits for Shewhart control charts, impacts the false alarm rate of the chart. This rate is inflated when small number if subgroups is used.
6. Who was the first person to study the effect of the number of subgroups used for x bar and R chart control limit calculation on the false alarm rate?
a) Hillier
b) Robertson
c) Wang
d) Ricardo
Answer: a
Explanation: The false alarm rate of the Shewhart control charts also depends over the number of subgroups used to construct the trial control limits. This was first studied by Hillier .
7. Which of these charts can be used instead of Shewhart control charts, in the case of small number of subgroups used to construct the charts, which were recommended by Quesenberry?
a) R-charts
b) C-charts
c) P-charts
d) Q-charts
Answer: d
Explanation: Quesenberry recommended the Q charts to be used in the place of the Shewhart control charts, in the case of small subgroup number used to calculate the control limits. This was because small number of subgroups used with Shewhart control charts, increased their FAR.
8. Which method was suggested by Del Castillo and Montgomery, to be used instead of the Q-chart to get better ARL performance?
a) Kalman Filter
b) Acceptance Sampling
c) Design of experiments
d) Deleting some samples completely
Answer: a
Explanation: Del Castillo and Montgomery found the ARL performance of the Q-charts very inadequate. So, they recommended the Kalman filter that has a better ARL performance than Q-chart.
9. Which of these is not a purpose for which control charts are produced for?
a) Control of the process
b) Reduction of variability
c) Increment of variability
d) Continuous process improvement
Answer: c
Explanation: The control charts are generally plotted for the main reasons like control of the process or statistical monitoring, reduction of variability, and continuous process improvement.
10. Which of these is used when the process has achieved high capability level?
a) Trial control limits
b) Actual control limits
c) Modified control limits
d) Easy control limits
Answer: c
Explanation: At the stage, when the process has achieved a high level of capability, there is some relaxation provided to the level of the surveillance provided. This is generally done with modified control limits in the case of x charts.
11. Which of these is not a name of Modified control limits for x̅ charts?
a) Acceptance limits
b) Reject limits
c) Reserve limits
d) Liquid limits
Answer: b
Explanation: The relaxation in the case of x charts in the case of high process capability is provided with the use of the modified limits. These are also called reject limits.
12. Which of these is a method used when the process capability is quite high?
a) Acceptance sampling
b) Design of experiments
c) Acceptance control charts
d) Shewhart control charts
Answer: c
Explanation: When the process capability becomes very high, the relaxation of the level of surveillance is provided by using two methods; first is the reject limits, and the second is the acceptance control chart.
13. Which of these does not say that the modified control limits must be applied?
a) C pk ≫1
b) Process spread is considerably small
c) Natural variability becomes high
d) Process capability becomes quite good
Answer: c
Explanation: The modified control limits are used when the process capability is very high. The PCR C pk ≫1, very less process spread, and low natural variability indicate the high process capability. So, modified control charts are used in these conditions.
14. The six-sigma approach when implemented for a long time, the process may need modified control charts.
a) True
b) False
Answer: a
Explanation: As six-sigma approach to variability reduction focuses on improving process until the minimum value of C pk =2. So this means high process capability, which encourages modified control charts usage.
15. If the C pk =0.9899, the usage of modified control charts is very necessary.
a) True
b) False
Answer: b
Explanation: The modified control charts have the use of the modified control limits, which are used when the process capability is high or C pk ≫1. As C pk =0.9899, the usage of modified control charts is not necessary.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 5”.
1. When the type I error is specified, which of these is correct expression for the UCL of the modified control charts?
a) \Missing or unrecognized delimiter for \right \sigma \)
b) \Missing or unrecognized delimiter for \right \sigma \)
c) \Missing or unrecognized delimiter for \right \sigma \)
d) \Missing or unrecognized delimiter for \right \sigma \)
Answer: b
Explanation: The modified control charts are the alternatives of ordinary Shewhart control charts, when the process capability is high. These have UCL as,
\Missing or unrecognized delimiter for \right \sigma \)
2. Which of these is a correct expression for the UCL of the modified control charts when the type I error is not specified?
a) \Missing or unrecognized delimiter for \right \sigma \)
b) \Missing or unrecognized delimiter for \right \sigma \)
c) \Missing or unrecognized delimiter for \right \sigma \)
d) \Missing or unrecognized delimiter for \right \sigma \)
Answer: c
Explanation: The UCL of the modified control charts can be done by both, specifying the type I error or by not specifying the type I error. When the type I error is not specified, the UCL is,
\Missing or unrecognized delimiter for \right \sigma \)
3. When the type I error is not specified, the value of LCL of the modified control charts is ____________
a) \Missing or unrecognized delimiter for \right \sigma \)
b) \Missing or unrecognized delimiter for \right \sigma \)
c) \Missing or unrecognized delimiter for \right \sigma \)
d) \Missing or unrecognized delimiter for \right \sigma \)
Answer: a
Explanation: The LCL of the modified control chart also uses the type I error. If the type I error is not to be specified. Its value is written by the expression,
\Missing or unrecognized delimiter for \right \sigma \)
4. ____________ sigma limits are recommended for modified control charts.
a) 4
b) 2
c) 3
d) 6
Answer: b
Explanation: The modified control charts are effective when the spread of process is quite less than the limits used. So 2-sigma limits are recommended for the modified control charts.
5. Which of the hypothesis can be tested using the modified control charts?
a) μ L ≤ μ ≤ μ U
b) μ L ≤ μ
c) μ ≤ μ U
d) μ L = μ = μ U
Answer: a
Explanation: The modified control charts allow the mean shift between two particular mean values. This means that the hypothesis μ L ≤ μ ≤ μ U can be tested using the modified control charts.
6. To design a modified control chart, we must have a good estimate of _________ available.
a) μ
b) σ
c) μ 2
d) process variance
Answer: b
Explanation: The modified control chart limits are totally based upon the estimate of the process standard deviation. So, to design a modified control chart, we must have a good estimate of σ available.
7. If process variability shifts, the modified control charts are ____________
a) Not appropriate
b) Totally appropriate
c) Good to use
d) Perfect for the mapping the shifts
Answer: a
Explanation: A modified control chart needs a good estimate of the process standard deviation. As the process standard deviation is depended upon the process variability, the modified control charts are not appropriate to use.
8. If there is a chance of shifting of the process variability, which chart may be used with the modified control chart?
a) c-chart
b) p-chart
c) R-chart
d) x̅ -chart
Answer: c
Explanation: As the process variability shifts can be monitored using a Shewhart R chart or an s-chart, if there is a chance of shifting of the process variability, they can be used in conjunction with modified control charts.
9. From which chart the initial estimate of the process variability is determined?
a) R-chart
b) c-chart
c) p-chart
d) Cusum charts
Answer: a
Explanation: As R charts and s-charts are good ways to estimate the process standard deviation, hence the process variability too, they are used to determine the initial estimate of the process variability.
10. When the type I error is specified, the LCL of the modified control chart is ___________
a) \Missing or unrecognized delimiter for \right\)
b) \Missing or unrecognized delimiter for \right\)
c) \Missing or unrecognized delimiter for \right\)
d) \Missing or unrecognized delimiter for \right\)
Answer: d
Explanation: As the type I error is not specified, the value of the 100 percentage point of the normal distribution Z δ is used. The LCL is,
\Missing or unrecognized delimiter for \right\)
11. The approach of using a x chart to monitor the fraction of nonconforming units or the fraction of the units exceeding the specifications, is called the ___________
a) Shewhart control charts
b) Cusum charts
c) EWMA charts
d) Acceptance control charts
Answer: d
Explanation: There is an approach to using an x chart to monitor the fraction of nonconforming or defective units, or the fraction of the units exceeding the specifications, which is called Acceptance control chart.
12. Who was the first person to develop the acceptance control charts?
a) Astern
b) Roy
c) Freund
d) Crowder
Answer: c
Explanation: The first person, to develop a method to use the x chart to monitor the fraction of defective units, was Freund . He first developed the technique of the acceptance control charts.
13. A modified control charts limit expression does not contain ___________
a) δ
b) σ
c) μ
d) α
Answer: c
Explanation: The modified control charts was based on a specific sample size n, a process nonconforming δ, and type I error probability α. It does not depend on μ.
14. The p-chart is the only chart to monitor the fraction of nonconforming units.
a) True
b) False
Answer: b
Explanation: The acceptance charts is based on the approach of using the x̅ chart to monitor the fraction of nonconforming units. So p-chart is not the only chart to monitor the fraction nonconforming units.
15. If the process mean increases very much, the modified control charts are not appropriate.
a) True
b) False
Answer: a
Explanation: As modified control charts are based on the assumption that the process has a very high capability, and the mean does not shift very significantly, if the mean shifts very much, the modified control chart becomes inappropriate.
This set of Statistical Quality Control Objective Questions & Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 6”.
1. Which of the error probability does the acceptance control chart take into account?
a) Type I
b) Type II
c) Neither of the type I not the type II
d) Both, type I and type II
Answer: d
Explanation: The acceptance control charts were first designed by Freund. These charts take both, the probability of type I error and the probability of type II error into account.
2. How many approaches can be used to develop the acceptance control charts?
a) 2
b) 1
c) 4
d) 5
Answer: a
Explanation: The acceptance control charts can be designed by using two approaches. First uses the specified n, and a process fraction nonconforming γ that we would like to reject with probability 1-β. Another uses the values of n, δ and α.
3. What is the UCL of the acceptance control chart when the sample size n, the fraction nonconforming γ are specified?
a) \Missing or unrecognized delimiter for \right\)
b) \Missing or unrecognized delimiter for \right\)
c) \Missing or unrecognized delimiter for \right\)
d) \Missing or unrecognized delimiter for \right\)
Answer: a
Explanation: The Upper control limit of the acceptance control chart, when sample size n, and the fraction nonconforming for 1-β probability rejection are specified, is expressed as,
\Missing or unrecognized delimiter for \right\)
4. What is the value of the LCL of the acceptance control chart when sample size n and the fraction nonconforming for 1-β probability of rejection are specified?
a) \Missing or unrecognized delimiter for \right\)
b) \Missing or unrecognized delimiter for \right\)
c) \Missing or unrecognized delimiter for \right\)
d) \Missing or unrecognized delimiter for \right\)
Answer: d
Explanation: The LCL value for the acceptance control chart uses the value of lower specification limit instead of the USL. The LCL value, when n, and γ are specified to reject with 1-β probability are specified, is
\Missing or unrecognized delimiter for \right\)
5. When n, γ, and β are specified, the values of control limits are between ________
a) μ L And μ U
b) LSL And μ U
c) μ L And USL
d) μ L And LSL
Answer: a
Explanation: We know that the control limits of an acceptance control chart are,
\Missing or unrecognized delimiter for \right\); \Missing or unrecognized delimiter for \right\)
By the above equations, we get the control limit values between μ L and μ U .
6. When the values of n, δ, and α, are specified, the LCL of the acceptance control charts, lie in between ________
a) μ L And USL
b) μ L And LSL
c) USL And μ U
d) LSL And μ U
Answer: b
Explanation: When the values of n, β, and γ are specified, the control limits for an acceptance control chart lie between μ L and μ U . But, when the values of n, δ, and α, are specified, the LCL lies in between μ L and LSL.
7. When the values of δ and α are specified instead of the β and γ, what is the position of UCL of the acceptance control charts?
a) Between USL and μ U
b) Between LSL and μ U
c) Between μ L and USL
d) Between μ L and LSL
Answer: a
Explanation: The value of UCL lies between μ L and μ U when the sample size n, β and γ are specified, but when the values of δ and α are specified instead of β and γ, the UCL of acceptance control charts lies between the USL and μ v .
8. What is the expression for n which gives desired values of α, β, γ, and δ for any acceptance control chart?
a) \Missing or unrecognized delimiter for \right^2\)
b) \Missing or unrecognized delimiter for \right^2\)
c) \Missing or unrecognized delimiter for \right^2\)
d) \Missing or unrecognized delimiter for \right^2\)
Answer: b
Explanation: It is possible to choose a sample size for an acceptance control chart so that the specified values of α, β, γ, and δ are obtained. The required sample size will be,
\Missing or unrecognized delimiter for \right^2\)
9. if δ= 0.01, α=0.00135, γ=0.05, and β=0.2, what will be the sample size for the desired values of these?
a) n=38
b) n=2
c) n=32
d) n=45
Answer: c
Explanation: We know that,
\Missing or unrecognized delimiter for \right^2\)
Putting the required values, we get,
\Missing or unrecognized delimiter for \right^2\)=31.43≅32.
10. Modified control chart design is not based upon ____________
a) Type I error probability
b) Sample size n
c) Type II error probability
d) Process fraction nonconforming
Answer: c
Explanation: The modified control chart design includes the development of the control limits for it, which includes only the sample size n, the process fraction nonconforming δ, and the type I error probability.
11. 2-sigma limits are recommended for the modified control charts. Which of these is not one of the reasons for this?
a) The spread must be low
b) The limits must be greater than the spread
c) Gives smaller β-risk
d) Gives higher β-risk
Answer: d
Explanation: The 2-sigma limits are recommended for the modified control charts, because the spread of the process is quite low than the limits, plus it also gives smaller β-risk.
12. Low type I error probability is obtained when 2-sigma limits are used for modified control chart.
a) True
b) False
Answer: b
Explanation: The use of 2-sigma limits gives the smaller β-risk, which is an indicator of low type II error probability. Thus the α-risk or the type I error probability is increased.
13. It is impossible to get desired values of α, β, γ, and δ in the acceptance control charts.
a) True
b) False
Answer: b
Explanation: It is possible to choose such a value for the sample size n for getting desired values of α, β, γ, and δ in the acceptance control charts. This may be done by using the following expression,
\Missing or unrecognized delimiter for \right^2\)
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Control Charting Techniques – Control Charts for Multiple-Stream Processes – 1”.
1. Which of these is a correct definition for an MSP?
a) Process in which same value of data comes from several sources
b) Process in which data at a point in time consisting of measurement from several sources
c) Process in which many values of data come from single source
d) Process in which data at a point in time consisting of measurement from single sources
Answer: b
Explanation: A MSP is defined as the process with data at a point in time, consisting of measurements from several different sources or streams.
2. What does P stand for in MSP?
a) Point
b) Pressurization
c) Prices
d) Process
Answer: d
Explanation: The MSP is a process in which data at a particular instant of time, comes from several individual streams. It has a full form of Multiple Stream Process.
3. When are the sources, or streams, in an MSP, considered to be identical?
a) When the process is out-of-control
b) When the process is in-control
c) When the process is running
d) The sources are never identical in any of the states of process
Answer: b
Explanation: In a Multiple Stream Process, when the process is in the state of statistical control, the sources or the streams, from which the data are obtained, are considered to be identical.
4. Each of the stream can _________ adjusted __________ in the case of an MSP.
a) Never, either individually or in small groups
b) Always, individually
c) Always, either individually or in small groups
d) Always, only in small groups
Answer: c
Explanation: In every multiple stream process, it is always a characteristic of it that, each of the stream from which the data is obtained, can be adjusted either individually, or in small groups.
5. What conclusion can we get about the process from the following statement?
“A machine has several heads, with each head producing identical units of product.”
a) Process is a MSP
b) Process is a SSP
c) Process is in-control
d) Process is out-of-control
Answer: a
Explanation: The above process can have different set of data from different streams at any instant of time, so it is an example of the Multiple Stream Process.
6. If the data of the streams of the MSP are all correlated , what control procedure may be adopted?
a) Different control chart for different stream
b) One control chart for some streams and one for a group
c) Different control charts for different groups of streams
d) One control chart for the any stream of all streams
Answer: d
Explanation: If the data of the MSP are all correlated perfectly, the control procedure, which may be adopted, can be to design one control chart for any one stream of all streams.
7. Which of these is a type of situation occurring when an assignable cause has come up in the case of MSP?
a) The whole process is in control
b) All streams are on-target
c) All streams are off target
d) Every stream is on-target
Answer: c
Explanation: There are at least two cases of situations involving the occurrence of assignable causes in MSP: first when the output of one or a few streams is off-target, second, when the output of all streams has shifted off-target.
8. When only the output of only one stream is off target we need to control _________
a) The whole MSP
b) Only that Process stream
c) Every process steam
d) Every process stream except the steam which has off target output
Answer: b
Explanation: When the output of only one stream of the MSP is off-target, that means there is an assignable cause in it. So we need to control only that stream.
9. Which technique is used to monitor MSPs?
a) Cusum control charts
b) EWMA charts
c) Shewhart charts
d) Group control charts
Answer: d
Explanation: The MSP processes have different data obtained from many individual streams. The data from the group of streams are controlled using the technique of Group control charts.
10. Who was the first person to discover the technique of GCC ?
a) Freund
b) Maxwell
c) Boyd
d) Montgomery
Answer: c
Explanation: The MSP or the multiple stream processes are controlled using the technique of GCC. These charts were first introduced by Boyd in 1950.
11. Which of these is an assumed standard distribution for the quality characteristic data from different streams of MSP for construction of GCC?
a) Lognormal
b) Normal
c) Poisson
d) Binomial
Answer: b
Explanation: While the constructing GCC for the MSPs, it is assumed that the data obtained from the streams of the MSP, is normally distributed. It is a standard assumed distribution for the data.
12. Which of these is not an assumption made for the GCC?
a) Same target value for all the streams
b) Same inherent variability for all the streams
c) All streams data are perfectly correlated
d) Each stream data is based on the normal distribution
Answer: c
Explanation: For the construction of GCC for the MSP, it is assumed that all the streams of the MSPs have same target value, and same inherent variability, and all the data is normally distributed.
13. If there are 6 streams of a process and sample size is 4, what is true?
a) 6 units individually from any 4 streams
b) 4 units individually from all the 6 streams
c) 10 units from any stream
d) 4 units from any one stream
Answer: b
Explanation: The sampling for the GCC construction for any MSP, the sampling is done as if separate control charts were to be set up for each stream. So there will be 4 units taken from each of the six streams.
14. Assignable cause may be present even when any one or a few streams have their output off-target.
a) True
b) False
Answer: a
Explanation: There are two possible conditions in an MSP when the assignable cause is present in the streams. One, when the output of any one stream or a group of stream is off target, and second when the whole set of streams have off target output.
15. The practical situation of MSP says that all the data from the streams are perfectly correlated.
a) True
b) False
Answer: b
Explanation: The assignable cause shifts all the process stream outputs off-target if the data is perfectly correlated. But in practical situations, the data is moderately correlated.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Control Charting Techniques – Control Charts for Multiple-Stream Processes – 2”.
1. The upper control limits for the GCC will be ____________ for the x̅ chart.
a) UCL=\
UCL=\
UCL=\
UCL=\(\bar{\bar{x}}+A_2 R\)
Answer: a
Explanation: The GCC are used to control the multiple stream processes. These control charts are having the upper control limits as,
UCL=\(\bar{\bar{x}}+A_2 \overline{R}\)
2. What is the value of the difference between the UCL and the LCL of x̅ chart of the GCC?
a) 1
b) 0
c) 3A 2 R
d) 2A 2 R
Answer: d
Explanation: As we know that the control limits for the GCC are,
UCL=\(\bar{\bar{x}}+A_2 \overline{R}\); LCL=\(\bar{\bar{x}}-A_2 \overline{R}\)
So the difference between the control limits of GCC will be 2A 2 R
3. What is the value of the UCL of the R chart constructed for the GCC?
a) UCL = D 4 R
b) UCL = D 3 R
c) UCL = R ̅+ D 3 R
d) UCL = R ̅+ D 4 R
Answer: a
Explanation: The GCC is prepared with the same way as the simple x and R charts are constructed for individual process streams. The value of the UCL of the R chart of the GCC is,
UCL=D 4 R
4. What is the ratio of the UCL to LCL of the same GCC R chart?
a) UCL/LCL=D 2 /D 3
b) UCL/LCL=D 3 /D 4
c) UCL/LCL=D 4 /D 3
d) UCL/LCL=D 4 /D 2
Answer: c
Explanation: We know that the values of the control limits of the R chart for the GCC are,
UCL=D 4 R ;LCL=D 3 R
So their ratio will be,
UCL/LCL=D 4 /D 3 .
5. If the sample size is 4, what will be the value of the upper control limit of the GCC R chart?
a) UCL=2.672R̅
b) UCL=2.282R̅
c) UCL=2.114R̅
d) UCL=2.574R̅
Answer: b
Explanation: We know that, for sample size= 4, the value of D 4 =2.282. So putting it into the equation of the UCL of R chart of the GCC gives,
UCL=2.282 R
6. What will be the value of the LCL of the x chart of the GCC when the value of x =12, and the value of R̅= 0.1 and sample size = 4?
a) 12.0729
b) 12.9271
c) 11.9271
d) 11.0729
Answer: a
Explanation: We know that, for n=4, A 2 =0.0729 Putting the above data into the equation of the UCL of the x̅ chart of GCC, we get,
UCL=12.0729.
7. When the subgroup size is 4 for any process samples, what will be the value of the R chart LCL of the GCC?
a) 5
b) 1
c) 0
d) 2
Answer: c
Explanation: We know that for n=4, D 3 has the value of zero. Putting it into the equation of the LCL of the R chart for the GCC, gives us, LCL=0.
8. In GCC we plot ___________ means observed at any time period.
a) Only the maximum
b) Only the minimum
c) Mean of the
d) Both, the maximum and the minimum
Answer: d
Explanation: When the GCC are used to control the process, we plot only the largest and the smallest of the s means observed at any time period on the x̅ chart. If they lie inside the control limits, so will others do.
9. ___________ range will be plotted on the R chart of the GCC.
a) Maximum
b) Minimum
c) Both the maximum and the minimum
d) Neither the maximum nor the minimum
Answer: a
Explanation: Only the largest range is plotted on the range chart because if it will exceed the control limits, there is a possibility that other ranges will also do the same. The lower limit of the Range chart is mostly zero.
10. Each plotted point on the GCC is identified by __________
a) The mean value of the stream
b) The stream that produced it
c) The standard deviation of the stream
d) The variance of the stream
Answer: b
Explanation: The maximum and minimum means are plotted on the x̅ chart, and only the maximum range is plotted on the range chart while using the GCC. Each point on the GCC is identified by the stream that produced it.
11. A MSP is out of control if a point on the GCC exceeds __________
a) 3-sigma limit
b) 2-sigma limit
c) 1.5-sigma limit
d) 1-sigma limit
Answer: a
Explanation: The GCC is used to control the MSP or the multiple stream processes. The data from the different streams is plotted on it, if any point on the GCC exceeds a 3-sigma limit, the process is out of control.
12. What is the value of the one sided in-control ARL for the GCC?
a) \_0=\frac{s^r}{s-1}\)
b) \_0=\frac{s^r-1}{s-1}\)
c) \_0=\frac{s^{2r}-1}{s+1}\)
d) \_0=\frac{s^r+1}{s+1}\)
Answer: b
Explanation: If a process has s streams and r is the consecutive times a particular stream has given largest value, the one sided in-control ARL will be,
\_0=\frac{s^r-1}{s-1}\)
13. If there are 6 streams of a MSP, and one of the streams has given 4 consecutive times the largest value, what will be the value of the one sided in-control ARL for this event?
a) 200
b) 400
c) 259
d) 421
Answer: c
Explanation: Given: s=6, r=4. We know that,
\_0=\frac{s^r-1}{s-1}\)
By putting the values, we get, ARL 0 =259.
14. Runs tests can be applied on the GCC.
a) True
b) False
Answer: b
Explanation: Runs tests cannot be applied on the GCC for any MSP, as the conventional run tests were not developed to test averages or ranges that are the extremes of a group of averages or ranges.
15. If the number of times a particular stream gives the maximum value of average or range for the GCC, increases, the value of one sided in-control ARL for that event will increase.
a) True
b) False
Answer: a
Explanation: As we know that,
\_0=\frac{s^r-1}{s-1}\)
As the value of r increases, the value of ARL 0 also increases.
This set of Statistical Quality Control Question Paper focuses on “Control Charting Techniques – Control Charts for Multiple-Stream Processes – 3”.
1. Which of these is a suitable pair of s, r, while using the GCC?
a) 2, 6
b) 3, 7
c) 9, 2
d) 6, 3
Answer: b
Explanation: The ARL of the MSP depends on the value of the s and r of the process. So to choose r is in our hand. and are the most used pairs which give desired results. Others will give too many false alarms.
2. The expected number of trials until r consecutive largest or smallest means come from the same stream, is called _________ of the MSP.
a) ARL
b) ARL 0
c) ARL 0
d) Predicted ARL
Answer: c
Explanation: The two sided in-control ARL of any process or ARL 0 is defined as the expected number of trials until r consecutive largest, or r consecutive smallest means come from a particular stream of any MSP, while MSP is in control.
3. Who was the first person to find out the one sided in-control ARL for any event of consecutive largest values being obtained from a single process stream?
a) Bond
b) Nelson
c) Robertson
d) Clarkes
Answer: b
Explanation: The one sided in-control ARL for any GCC of any MSP was first explained by Nelson . He expressed it as,
\_0=\frac{s^r-1}{s-1}\)
4. Who gave the Markov chain approach to compute the ARL 0 ?
a) Brook and Evans
b) Mortell
c) Runger
d) Nelson
Answer: a
Explanation: The Markov Chain Approach was given by the pair of persons, Brook and Evans, in the year of 1972. This was used by Nelson and Stephenson to calculate the ARL 0 .
5. Who was not one of the scientists who used the Markov chain approach to compute the two sided in-control ARL for any GCC of any MSP?
a) Mortell
b) Robertson
c) Runger
d) Stephenson
Answer: a
Explanation: Mortell and Runger , and Nelson and Stephenson used the Markov chain approach of Brook and Evans to compute ARL 0 .
6. Which of these is a lower bound on the ARL 0 ?
a) \_0=2\frac{s^r-1}{s-1}\)
b) \_0=\frac{2}{s^r-1}\)
c) \_0=\frac{s^r-1}{s-1}\)
d) \_0=\frac{s^r-1}{2}\)
Answer: d
Explanation: The ARL 0 close form expression could not be evaluated. So Nelson and Stephenson give a lower bound on ARL 0 which is written as,
\_0=\frac{s^r-1}{2}\)
7. Which of these is one of the drawbacks of GCC?
a) Each stream does not need to be sampled at a time
b) There is no information about nonextreme streams at each trial. Thus no past value related EWMA can be constructed
c) It can’t be used for MSPs
d) It is very easy to construct
Answer: b
Explanation: In GCC, there is no information about nonextreme streams at each trial. So, we can’t utilize past values to form an E.W.M.A. or a cusum statistic to improve on GCC performance.
8. Dependent streams of an MSP are also called ____________
a) Accuracy streams
b) Cross Acceptance
c) Cross-correlated
d) Uncorrelated
Answer: c
Explanation: The Dependent streams of a multiple stream process are the streams that have their data related to each other. So these streams are also called cross-correlated streams.
9. Which of these is the correct model proposed for MSP by Mortell and Runger to accommodate the practical case of the dependent streams?
a) x tjk = μ – A t + ϵ tjk
b) x tjk = μ + A t + ϵ tjk
c) x tjk = μ – A t – ϵ tjk
d) x tjk = μ + A t – ϵ tjk
Answer: b
Explanation: Mortell and Runger proposed the following model for the MSP to accommodate the practical case of cross-correlated streams:
x tjk = μ + A t + ϵ tjk
Here cross-correlated streams have their data related to each other.
10. What does the term x tjk correspond to in the model proposed by Mortell and Runger?
a) k Th measurement on the jTh stream at time t
b) j Th measurement on the kTh stream at time t
c) t Th measurement on the jTh stream at time k
d) k Th measurement on the tTh stream at time j
Answer: a
Explanation: The model proposed by Mortell and Runger, to accommodate the case of cross-correlated stream data of MSP, is expressed as,
x tjk = μ + A t + ϵ tjk
Here the term x tjk corresponds to k Th measurement on the j Th stream at time t.
11. The model given by Mortell and Runger represents two types of variability, σ a 2 accounting for the variance over time ___________ and σ 2 accounting for the variation between the streams at specific time t.
a) Common to all streams
b) Specific to one stream
c) Specific to 2 streams
d) Specific to a group of 3 streams
Answer: a
Explanation: The MSP representation given by Mortell and Runger, has the total variation allocated into two sources, σ a 2 accounting for the variation over time common to all streams, and σ 2 accounting for the variation between the streams at specific time t.
12. What is the correct value of cross correlation given by the Mortell and Runger model?
a) \
\
\
\(ρ=\frac{2σ_a^2}{
}\)
Answer: c
Explanation: The MSP model was given by Mortell and Runger in the year of 1995. According to it, the value of the cross-correlation given by the model was expressed as,
\(ρ=\frac{σ_a^2}{
}\)
13. Mortell and Runger proposed monitoring the average at time t of the means across all the streams with _________ control chart to detect an overall assignable cause.
a) A cusum
b) A EWMA
c) A p-chart
d) An individuals
Answer: d
Explanation: Mortell and Runger proposed the MSP model. They recommended using an Individuals Chart to compare the average at time t of the means across all streams, to find an assignable cause.
14. The Mortell and Runger model did not propose to monitor the range of stream’s means at time t.
a) True
b) False
Answer: b
Explanation: According to Mortell and Runger, it was quite necessary to monitor the range of the stream’s means at time t. This range was denoted by,
R t = max( x tj )-min( x tj ).
15. The proposed control charts on residuals are quite good than the GCC.
a) True
b) False
Answer: a
Explanation: The Mortell and Runger model given in 1995, recommended to monitor the range of stream’s means at time t or the maximum residual at time t,
max( x ij )- x i
This chart is better than the GCC, especially when the variation in the process means over time is greater than the between-stream variability.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “What is Experimental Design – 1”.
1. A designed experiment is a test or series of tests in which ____________ changes are made to the input variables so that we may observe and identify corresponding changes in the output response.
a) Systematic
b) Random
c) Purposeful
d) Non-purposeful
Answer: c
Explanation: A designed experiment is a series of test in which we change the input variables according to our choice to obtain a desired change in the output. These changes are perfectly purposeful.
2. Which of these does not come into the general model of a process?
a) Input
b) Controllable input factors
c) Uncontrollable inputs factors
d) Acceptance sampling
Answer: d
Explanation: The general model of process describes a process to which the input is provided in the presence of controllable factors and uncontrollable input factors, to get an output.
3. The uncontrollable factors are the factor ___________
a) That varies according to a normal distribution
b) That does not vary at all
c) That can be controlled by the user
d) That cannot be changed according to the wish of the user
Answer: d
Explanation: There are two types of factors present with a process input while running period of a process, the controllable, and the uncontrollable input factors. The factors, which can be changed according to user’s wish, are called the uncontrollable factors.
4. The uncontrollable factors are also called ____________
a) Designed factors
b) Noise factors
c) Acceptance factors
d) Sound factors
Answer: b
Explanation: The uncontrollable input factors are present when the process is running, and these factors can’t be varied according to the user’s wish. So these factors are also called the noise factors.
5. Which of these steps are not conducted when the design of experiment procedure is adopted?
a) Determining which variable is most influential to output
b) Determining where to set the influential controllable factors so that output is near the nominal requirement
c) Deleting the uncontrollable factors
d) Determining where to set the influential controllable inputs so that the variability in the output is smallest
Answer: c
Explanation: While conducting the design of experiment procedure, we determine the influential factors which affect the output, vary the determined variables to get a required output value.
6. Experimental design methods are not used ____________
a) Evaluating the process capability
b) In process development
c) In process troubleshooting to improve process performance
d) To obtain a process that is robust and insensitive to external sources of variability
Answer: a
Explanation: The experimental design methods may be used either in process development or process troubleshooting to improve process performance, or to obtain a process that is robust or insensitive to external variability causes.
7. The designed experiments are the part of ___________ step of DMAIC process.
a) Define
b) Measure
c) Analyze
d) Control
Answer: c
Explanation: The major procedure parts, which are used in the designing of the experiments, are including the analysis of data and their improvement. So they are an integral part of the Analyze and Improve step of DMAIC process.
8. The design of the experiment is used to determine the variables which are ___________ affecting the state of the process.
a) The most
b) The least
c) Not
d) Not changing or
Answer: a
Explanation: The design of the experiment procedure is, determining the variables which most affect the process state and the output. We vary these variables to get a relationship between them and the output.
9. Which of these can be used to develop a new process?
a) Design of experiments
b) Acceptance sampling
c) Control charts
d) Histogram
Answer: a
Explanation: The design of the experiment procedure is an active statistical method, which can be used to control the values of the output, by obtaining the relationship between the input factors and output. SO this can be used to design a new process with great process capability.
This set of Statistical Quality Control written test Questions & Answers focuses on “What is Experimental Design – 2”.
1. Which of these can be obtained by using the Experimental design?
a) Reduced process capability
b) Increased variability
c) Increased cost
d) Reduced distance from the nominal value
Answer: d
Explanation: With the use of the designed experiments, it is possible to increase the conformance of the product to the nominal value. So it can be used in designing a new product.
2. Which of these can be used in the DFSS activities?
a) Design of experiments
b) Histogram
c) Acceptance sampling
d) Stem and Leaf plot
Answer: a
Explanation: The DFSS activities are said to be the Design for Six-sigma activities. The process of design of experiments is very useful in these activities.
3. The DFSS stands for _______________
a) Deleting for smaller standards
b) Degradation for Six-sigma
c) Design for six-sigma
d) Development for six-sigma
Answer: c
Explanation: The DFSS stands for the Design for Six – Sigma activities. These activities are used to setup a process environment to get the six-sigma output.
4. SPC and Design of experiments are very closely interrelated.
a) True
b) False
Answer: a
Explanation: SPC and experimental design of experiments are two very interrelated tools for process improvement and optimization. For example, if a low C p process is in-control; Designed experiments could be used to increase process capability.
5. SPC is a passive statistical method.
a) True
b) False
Answer: a
Explanation: SPC is a passive statistical method as, in this procedure; we have to wait for the process information to come, to decide a step to control the process. However, if the process is in-control, we can’t do very much to reduce the variability.
6. Which of these can be used in the process of engineering design?
a) Design of experiment
b) Control charting
c) Acceptance sampling
d) Cusum charts
Answer: a
Explanation: The design of experiments can also play a major role in engineering design activities, where new products are developed and existing ones are improved. So this is also used in DFSS activities.
7. Which of these is not an application of experimental design in the field of the Engineering design?
a) Evaluation and comparison of basic design configurations
b) Evaluation of material alternatives
c) Evaluation of key product design parameters that impact performance
d) Evaluation of correct process to develop the product
Answer: d
Explanation: The Design of the experiment procedure helps in engineering design by helping in evaluation of material alternatives, key product design parameters, and basic design configurations.
8. Use of designed experiments does not give _________
a) Improved yield
b) Reduced variability
c) Closer conformance to the nominal
d) Increased development time
Answer: d
Explanation: The use of the design of the experiments procedure is quite useful in process development. This gives an improved yield, reduced variability and closer conformance to nominal, and reduced development time.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Acceptance Sampling Introduction – 1”.
1. Which of these is not a correct statement for Acceptance Sampling?
a) Concerned with inspection of products
b) Concerned with decision making regarding products
c) One of the oldest aspects of quality assurance
d) One of the oldest aspects of quality control
Answer: d
Explanation: Acceptance sampling is a procedure used in quality assurance, which is concerned with inspection of manufactured products and the decision making regarding their state.
2. Which technique was used majorly in 1930s and 1940s for incoming or receiving inspection?
a) SPC
b) Histogram
c) c-chart
d) Acceptance sampling
Answer: d
Explanation: Acceptance sampling was one of the major components of the field of statistical quality control, and was used primarily for incoming and receiving inspection, in the 1930s and 1940s.
3. Decision making regarding the lot disposition is sometimes called _____________
a) Lot rejection
b) Lot acceptation
c) Lot sentencing
d) Lot wording
Answer: c
Explanation: Based on the acceptance sampling information, there is decision making regarding the lot information. This decision is either to accept or reject the lot. So it is called, the Lot Sentencing.
4. Acceptance sampling can be used as _____________
a) Incoming inspection activity
b) Outgoing inspection activity
c) Both, incoming and outgoing inspection activity
d) Neither incoming nor outgoing inspection activity
Answer: c
Explanation: The Acceptance sampling is customarily thought to be the receiving or incoming inspection activity, but there are a few cases where it can be used as the outgoing inspection activity.
5. The purpose of Acceptance sampling is to _____________
a) Sentence lots
b) Estimate lot quality
c) Estimate lot defectives
d) Estimate lot conformity
Answer: a
Explanation: The Acceptance sampling procedure is necessarily a lot sentencing procedure. It cannot be used to estimate the lot quality or lot conformity to the standard specifications.
6. Which of these procedures doesn’t provide a direct form of quality control?
a) Control charts
b) Acceptance sampling
c) Design of experiments
d) Cusum charts
Answer: b
Explanation: The Acceptance sampling procedure doesn’t provide a direct form of quality control. Acceptance sampling simply accepts and rejects lots; even if all lots are of same quality, some are accepted and some are not.
7. Which of these can be used as an audit tool to ensure the output of a process conforms to requirements?
a) Cusum charts
b) EWMA charts
c) Acceptance sampling
d) np-charts
Answer: c
Explanation: The most effective use of acceptance sampling is not to “inspect quality into the product”, but rather as an audit tool to ensure that the output is similar to the required specifications.
8. Which of these is not used in sampling?
a) 0% inspection
b) 100% inspection
c) Acceptance sampling
d) 5% inspection
Answer: d
Explanation: There are generally three techniques used in the inspection of a lot; first is 0% inspection, second is 100% inspection, and third is acceptance sampling.
9. The no-inspection alternative of sampling is used when ______________
a) The supplier’s process is so good that defective units are never encountered
b) The supplier’s process is so bad that almost every unit is defective
c) The component is extremely critical
d) The component is moderately critical
Answer: a
Explanation: When the supplier’s process is so good that the defective units are never encountered. This shows that the supplier process capability is quite high.
10. When is the 100% inspection done?
a) The supplier’s process is so good that defective units are never encountered
b) The supplier’s process is so bad that almost every unit is defective
c) The component is extremely critical
d) The component is moderately critical
Answer: c
Explanation: The option of full inspection is chosen only when the product component is extremely critical, and can affect the whole product performance when used defective, e.g. aircraft nuts and bolts.
11. Acceptance sampling is not used when _____________
a) The test is destructive
b) The cost of 100% inspection is quite high
c) The supplier’s process capability is very high
d) Although the supplier process is satisfactory but a program is needed for continuous monitoring
Answer: c
Explanation: The acceptance sampling is used when the test is destructive, or the cost of 100% inspection is quite high, or when we need a continuous monitoring program.
12. When the inspection error rate is sufficiently high, which of these is used as the sampling technique?
a) 0% inspection
b) 100% inspection
c) 50% inspection
d) Acceptance sampling
Answer: d
Explanation: When there are many items to be inspected and the inspection error rate is sufficiently high, that 100% inspection might cause a high % of defective units to be passed, the procedure of acceptance sampling is adopted.
13. Which of these is not used for a lot quality inspection purposes?
a) EWMA Control chart
b) Cusum chart
c) Shewhart control charts
d) Acceptance Sampling
Answer: d
Explanation: The Acceptance sampling procedure is used for decision making of either acceptance or rejection of a lot. It can’t be used as a lot quality estimators.
14. Designed experiments may benefit the lot quality improvement process more than the Acceptance sampling.
a) True
b) False
Answer: a
Explanation: As Acceptance sampling is just a lot sentencing process, it can’t estimate the quality of products in the lot. But designed experiments ensure good quality of the process output before even production.
15. A good lot can be rejected through the use of acceptance sampling.
a) True
b) False
Answer: a
Explanation: As acceptance sampling considers only a particular number of sample units, and does not make use of the data of other units, there is a slight possibility that it will reject the good lot.
This set of Statistical Quality Control Puzzles focuses on “Acceptance Sampling Introduction – 2”.
1. Which of these is not an advantage of acceptance sampling over the 100% sampling plan?
a) Less expensive
b) Highly costly
c) Applicable to destructive testing
d) Lesser manpower is needed
Answer: b
Explanation: As lesser number of units are checked while using the acceptance sampling plan instead of the 100% sampling plan, it is less expensive, applicable to destructive testing, and needs lesser manpower.
2. Which of these is used when the test of the component is non-destructive, cheap and fast?
a) 0% inspection
b) Acceptance sampling
c) 100% sampling
d) 50% sampling
Answer: d
Explanation: As it will give better results in the case of 100% inspection when the test procedure of a component is non-destructive, cheap, and fast, the full lot sampling procedure is adopted.
3. Which has the lowest number of manpower required?
a) Acceptance sampling
b) 0% inspection
c) 100% inspection
d) 50% inspection
Answer: b
Explanation: As 0% inspection does not need any equipment or personnel to check the product component supplied by the supplier, it has the lowest number of manpower required.
4. Which has lesser probability of handling damage in between 100% inspection, and the acceptance sampling procedure?
a) 100% inspection
b) Acceptance sampling
c) Both have equal handling damage
d) Can’t be predicted
Answer: b
Explanation: As there is a lesser number of units used as samples in the acceptance sampling, there is less handling of the product, and hence reduced damage is there in the case of acceptance sampling.
5. Which of these quite successfully fulfills the following sentence?
“The rejection of entire lots as opposed to the simple return of defectives often provides stronger motivation to the supplier for quality improvements.”
a) 0% sampling
b) 100% sampling
c) Random % of the lot sampling
d) Acceptance sampling
Answer: d
Explanation: As in the case of acceptance sampling, there are only a few units selected to be returned to the supplier, instead of returning the whole lot, it motivates the supplier for quality improvements.
6. Which is the most expensive for the same testing process and product component to be tested?
a) Acceptance sampling
b) 100% sampling
c) 0% sampling
d) 50% sampling
Answer: b
Explanation: As the 100% sampling procedure requires the checking of whole lot, it requires more resources to be devoted towards the sampling process. This makes it the most expensive procedure of all.
7. Which generates lesser information about the products and the manufacturing process, 100% sampling or Acceptance sampling?
a) 100% sampling
b) Acceptance sampling
c) Both generate equal amount of information
d) Can’t be predicted
Answer: b
Explanation: As we there are lesser number of units as samples in the case of acceptance sampling rather than the 100% sampling procedure, it generates lesser information about the process and the product.
8. Which has the most probability of rejecting the good lot?
a) Acceptance sampling
b) 100% sampling
c) 0% sampling
d) Can’t be predicted
Answer: a
Explanation: As the 100% inspection accepts the lot after the inspection of whole lot, and there are rarely any defectives in the case of no-inspection, acceptance sampling is having the highest probability of rejecting the good lot.
9. Which of these requires planning and documentation of the sampling procedure?
a) Acceptance sampling
b) 100% sampling
c) 0% sampling
d) 50% sampling
Answer: a
Explanation: The 100% sampling checks the whole lot, so it does not require any planning or documentation, whereas the acceptance sampling requires it.
10. Which of these does not require sampling documentation at all?
a) 0% sampling
b) 100% inspection
c) Acceptance inspection
d) 50% inspection
Answer: a
Explanation: As there is no sampling done in the case of 0% inspection or sampling, there is no need to do the documentation for the same. So it doesn’t require any sampling documentation at all.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Acceptance Sampling Introduction – 3”.
1. Why are larger lots preferred over smaller lots in the case of acceptance sampling?
a) Because it’s economical
b) Because it is costly
c) Because it is time consuming
d) Because it is complicated to sample larger lots
Answer: a
Explanation: The larger lots are preferred over smaller lots as it saves time, and it is very easy to sample larger lots. The larger lots are also mostly homogeneous.
2. Sequential sampling is an extension of ____________
a) Single sampling plan
b) Double-sampling plan
c) Multiple-sampling plan
d) 0% sampling
Answer: c
Explanation: As the multiple-sampling plan is associated with the case when N numbers of samples are taken of n units, it has an intimate extension as Sequential sampling.
3. Which is most economical of these?
a) Single sampling plan
b) Double-sampling plan
c) Multiple-sampling plan
d) 100% sampling
Answer: a
Explanation: As single sampling plan necessarily requires only one sample of n units, we may say that it is the most economical of all the acceptance sampling plans.
4. How many sampling plans are there in the case of acceptance sampling?
a) 1
b) 2
c) 3
d) 5
Answer: c
Explanation: There are only 3 sampling plans available in the case acceptance sampling, namely: single-sampling plan, double-sampling plan, and multiple-sampling plan.
5. What is done in single sampling plan?
a) Only one unit is checked
b) Only the first lot is checked 100%
c) Only n samples of 1 unit are checked
d) Only one sample of n units is checked
Answer: d
Explanation: Single-sampling plan is a lot sentencing procedure in which one sample of n units is selected and checked, at random from the lot. Lot is sentenced based upon only this sample.
6. Double-sampling plan is __________
a) Only 2 units are checked
b) Only the first and last lot is checked 100%
c) Only two samples of n units are checked
d) Only two samples of n units are checked
Answer: d
Explanation: In the double-sampling plan, there are only two samples of n units are checked. The lot disposition is totally based upon the information from the first and second sample. There may not be two samples necessary to make decision.
7. There are necessarily 2 samples of n units taken and checked in the case of double sampling plan.
a) True
b) False
Answer: b
Explanation: In the Double sampling plan, it is not necessary to take two samples. Sometimes we can accept or reject the lot based upon the information from the first lot.
8. Lots in the case of acceptance-sampling plan may not be homogeneous.
a) True
b) False
Answer: b
Explanation: It is necessary to have the lots to be homogenous in the case of acceptance sampling. This is done by producing them on same machines by same operators, and keeping the other environmental factors to be the same.
This set of Statistical Quality Control Multiple Choice Questions & Answers focuses on “Acceptance Sampling Procedures – Military Standard 105E”.
1. MIL STD 105E was first issued in ________
a) 1949
b) 1937
c) 1945
d) 1950
Answer: d
Explanation: The standard sampling procedures for inspection by attributes were developed during World War II. So the MIL STD 105E was first issued in 1950.
2. How many revisions have been done in the case of MIL STD 105E?
a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: After the original version of MIL STD 105E been issued in 1950, there have been 4 revisions in it since then, and the latest version was issued in 1989.
3. MIL STD 105E is a _________
a) Acceptance sampling plan
b) Acceptance sampling system
c) Both Acceptance sampling plan and Acceptance sampling system
d) Neither acceptance sampling plan nor system
Answer: b
Explanation: The acceptance sampling scheme is a an overall strategy specifying the way in which the sampling plans are to be used. MIL STD 105E is a collection of acceptance sampling schemes. So it’s an acceptance sampling system.
4. AQL is the __________
a) Acceptable Quality Level
b) Acceptance Quality Level
c) Associated Quality Level
d) Assigned Quality level
Answer: a
Explanation: The main focal point of MIL STD 105E is AQL. The AQL stands for Acceptable Quality Level. This is the lowest quality of a product to be accepted.
5. Which of these is a derivative civilian standard which is quite similar to MIL STD 105E?
a) ANSI/ASQC Z1.4
b) ANSI/ASQC Z1.1
c) ANSI/ASQC Z1.8
d) ANSI/ASQC Z1.7
Answer: a
Explanation: There is a derivative civilian standard originated from MIL STD 105E, ANSI/ASQC Z1.4. It is quite similar to the MIL STD 105E standard.
6. Which of these ISO standards were adopted from MIL STD 105E?
a) ISO 14000
b) ISO 9000
c) ISO 2859
d) ISO 6000
Answer: c
Explanation: MIL STD 105E was also adopted by the international organization for standards as ISO 2859. This is quite similar to ANSI/ASQC Z1.4.
7. How many sample inspection levels are used in MIL STD 105E?
a) 1
b) 2
c) 5
d) 3
Answer: d
Explanation: The sample size used in MIL STD 105E is determined by lot size and by the choice of inspection levels. Three general levels are provided.
8. Level I of inspection of MIL STD 105E requires about __________ amount inspection as Level II.
a) Zero
b) Half
c) 100%
d) 3/4th
Answer: b
Explanation: There are generally 3 levels of inspection in the case of MIL STD 105E. The level I requires about one-half the amount of inspection as Level II.
9. Tightened inspection is instituted when at least ______ out of 5 consecutive lots have been rejected on original submission.
a) 1
b) 4
c) 2
d) 5
Answer: c
Explanation: When the normal inspection is in effect, tightened inspection is instituted when at least two of five consecutive lots have been rejected on original submission.
10. When is the normal inspection to be instituted instead of tightened inspection?
a) When 3 lots are rejected consecutively
b) When 2 lots are accepted consecutively
c) When 5 lots are rejected consecutively
d) When 5 lots are accepted consecutively
Answer: d
Explanation: When tightened inspection is in effect, normal inspection is instituted when five consecutive lots or batches are accepted on original inspection.
11. When should the inspection with the provision of MIL STD 105E be terminated?
a) When at least 10 consecutive lots remain on tightened inspection
b) When at least 7 consecutive lots remain on normal inspection
c) When at least 5 consecutive lots remain on tightened inspection
d) When at least 10 consecutive lots remain on normal inspection
Answer: a
Explanation: In the event that ten consecutive lots remain on tightened inspection, inspection under the provision of MIL STD 105E should be terminated, and action should be taken at the supplier level to improve quality of submitted lots.
