Free CSSBB Practice Test Questions and Answers (2026)

The expected monetary value (EMV) of mutually-exclusive outcomes equals the sum of each outcome’s payoff multiplied by its probability. EMV = (0.25×$340 000) + (0.40×$120 000) + (0.10×–$40 000) + (0.25×$100 000) = $85 000 + $48 000 – $4 000 + $25 000 = $154 000. A positive EMV of $154 000 indicates the project is expected to return that profit on average.

A. $130 000 – misses $24 000; ignores a portion of the weighted payoffs.
B. $520 000 – simple sum of profits; disregards probabilities, not an expected value.
D. ($168 000) – sign and magnitude wrong; appears to invert profits to losses.
E. None of the above – incorrect because option C matches the correctly calculated EMV.

1. Pyzdek

T. & Keller

P. “The Certified Six Sigma Black Belt Handbook”

3rd ed.

McGraw-Hill

2018

Chapter 15 “Decision-Making Tools”

pp. 538-540: Expected Monetary Value formula and example.

2. Montgomery

D.C. “Introduction to Statistical Quality Control”

8th ed.

Wiley

2020

§2.5

pp. 37-38: Definition of expected value for discrete distributions.

3. MIT OpenCourseWare

15.761 “Operations Management”

Lecture 8 notes

slide 6: EMV calculation for project risk analysis (https://ocw.mit.edu).

Standardized written and diagrammed work instructions (often called “standard work”) specify the exact sequence, motion, and tools for each task. When every assembler follows the same documented method, operator-to-operator variation is minimized and process output becomes far more consistent— the primary objective stated in the question.

B. Flowcharts/process maps show overall sequence but do not prescribe the exact motions or parameters needed for operator-level consistency.
C. Cause-and-effect diagrams identify potential causes of variation; they do not directly standardize the work.
D. Pareto charts prioritize problems but provide no guidance on how to perform the assembly consistently.
E. Relationship matrices (e.g., QFD) translate customer needs into product features, not into standardized assembly procedures.

1. ASQ Certified Six Sigma Black Belt Body of Knowledge

Section VI.A.1 “Standardized Work” – describes written and visual work standards as the means to reduce process variation (ASQ

2022 revision

p. 43).

2. Pyzdek

T. & Keller

P. The Certified Six Sigma Black Belt Handbook

3rd ed.

ASQ Quality Press

2015

Chapter 9 “Lean Tools

” pp. 245-246 – “Standardized work instructions are the primary tool for achieving consistency in assembly operations.”

3. MIT OpenCourseWare

2.830J/ESD.62J “Lean/Six Sigma Processes

” Lecture 7 notes

slide 12 – emphasizes that standard work documentation is essential for consistent assembly performance.

A cause-and-effect (Ishikawa/Fishbone) diagram is expressly designed to elicit and display a wide range of potential causes for a specified effect (here, excess dimensional variation). During team brainstorming each branch (materials, methods, machines, etc.) invites contributors to add as many causes as possible, ensuring broad coverage of ideas before any prioritization or analysis takes place.

A. Work instructions describe the current method; they do not generate new hypotheses.
B. Flowcharts/process maps show process sequence but are not intended to brainstorm numerous causal factors.
D. A Pareto chart ranks known causes by frequency; it does not help discover new ones.
E. A relationship matrix quantifies interactions between already-identified variables; it is not a tool for initial cause generation.

1. Kubiak

T. & Benbow

D.

The Certified Six Sigma Black Belt Handbook

3rd ed.

ASQ Quality Press

2016

Section 7.4 “Cause-and-Effect Diagram

” pp. 182-184.

2. Montgomery

D. C.

Introduction to Statistical Quality Control

7th ed.

Wiley

2013

Chapter 4

pp. 146-147 (“The cause-and-effect diagram is useful for identifying many possible causes of variability”).

3. MIT OpenCourseWare

Course 15.760A Operations Management

Lecture “Quality Tools

” slide set 8

slides 9-10 (Fishbone diagram purpose: “brainstorm all possible causes for an effect”).

Defects-per-unit (DPU) is defined as the total number of defects observed divided by the total number of units inspected. In this case the payroll department found 87 defects in 328 time-card units; therefore DPU = 87 ÷ 328 = 0.265 (rounded)

B. Uses total opportunities (units × defect types); that calculates DPO, not DPU.
C. Inverts the definition; compares possible defect types to total defects.
D. Divides defects by types, giving average defects per type, not per unit.
E. Inverts the required ratio; gives units per defect, not defects per unit.
F. Multiplies by 1 000 000 and divides by opportunities; that is defects-per-million-opportunities (DPMO).

1. T. Kubiak & D. Benbow

The Certified Six Sigma Black Belt Handbook

3rd ed.

ASQ Quality Press

2017

Chap. 9

“Defect Metrics

” p. 487: “DPU = total defects ÷ total units.”

2. MIT OpenCourseWare

15.075J Statistical Thinking & Data Analysis

Lecture 18 notes

Section “Yield Metrics

” p. 3: “Defects per unit (DPU) = number of defects / number of units inspected.”

3. G. Eckes

Six Sigma for Everyone

Wiley

2003

Chap. 4

p. 67: formula and example illustrating DPU calculation identical to option A.

When automated gaging values are fed directly to a database that drives real-time machine adjustments, three conditions must simultaneously be satisfied:
1. A reliable hardware/communication link is needed to transmit every reading without latency or loss.
2. The gage-resident firmware and the plant-level software must exchange data in the same protocols, units, and data structures to avoid corruption or misinterpretation.
3. Manual-override capability is an essential safety and quality requirement; without it, operators cannot intervene if the system produces out-of-control signals or fails. Because all three factors are indispensable to assure data integrity, process control, and operator safety, the only fully correct choice is “all of the above.”

A. Important but alone does not ensure software interoperability or safety overrides.
B. Vital but meaningless if hardware cannot transmit or if operators cannot intervene.
C. Needed for safety, yet ineffective without dependable data transfer and software compatibility.

1. The Certified Six Sigma Black Belt Handbook

3rd ed.

ASQ Quality Press

2015

Chap. 5 “Measurement Systems

” pp. 214-215 – lists communication integrity

software compatibility

and manual bypass as concurrent requirements for automated gaging.

2. AIAG Measurement Systems Analysis Manual

4th ed.

2010

§8.4 “Automated Gage Interfaces

” p. 135 – stresses physical link integrity

data-format compatibility

and emergency manual control.

3. NIST Special Publication 1034

“Guidelines for Automated Industrial Gaging

” 2007

§6.2 & §6.4 – identifies communication reliability

software compatibility

and manual override as critical specification items.

4. MIT OpenCourseWare

Course 2.008 “Manufacturing Processes & Systems

” Lecture 18 slides 12-14 – enumerates identical three criteria when integrating in-process automated inspection.

In classical factorial design notation the three key descriptors are:
• Factors – the independent variables investigated (k).
• Levels – the settings each factor can assume (commonly 2 for screening designs).
• Replications – the number of repeated observations of every factor–level combination (n). A basic screening experiment that investigates k = 3 factors, each at 2 levels (2³ design), and that is repeated n = 3 times to obtain a pure-error estimate is therefore summarized as “3 factors, 2 levels, 3 replications,” which corresponds to option D.

A. 3,3,3 – implies three-level factors; not the standard 2-level screening design asked for.
B. 2,3,2 – only two factors; additionally three levels contradict the common two-level factorial.
C. 3,2,2 – lacks the third replicate needed for error estimation stated in the scenario.
E. 2,2,2 – investigates only two factors; insufficient for the stated three-factor study.
F. none of the above – option D fits, so F is false.

1. Montgomery, D. C. Design and Analysis of Experiments, 9th ed., Wiley, 2017:

– Sec. 6.1, pp. 211-213: definition of 2ᵏ designs (k = 3 factors, 2 levels each).

– Eq. 6-1, p. 213: notation yᵢ(j) where i = replicate (n).

2. ASQ Certified Six Sigma Black Belt Handbook, 3rd ed., Wiley/ASQ, 2017:

– Ch. 7 “Design of Experiments,” pp. 398-399: listing of factors, levels, replications.

3. MIT OpenCourseWare, 16.810 Manufacturing Systems Design, Lecture 9 slides “DOE Basics” (2015):

– Slide 5: “A 2³ factorial (3 factors, 2 levels) with r replications.”

Each capsule ingredient is treated as a separate factor. For a full factorial design, every factor is set at every level of every other factor.
• Number of factors (k): four ingredients ⇒ k = 4
• Levels per factor: two dosages (10 mg = low, 40 mg = high) ⇒ 2 levels
• Distinct treatment combinations (runs before replication): 2^k = 2^4 = 16. Replications (5) are applied to each of these 16 combinations later, but the question asks only for levels, factors, and (unique) runs, giving 2, 4, and 16 respectively.

B. Treats “ingredient” as one 4-level factor and adds an extra factor, yielding only 8 runs—contradicts full factorial math.
C. Counts 10 mg as 10 levels and multiplies by replications; factors and levels misinterpreted.
D. Implies 20 factors at 2 levels; 2^20 ≠ 16 runs, thus impossible for a full factorial.

1. Montgomery

D.C. Design and Analysis of Experiments

8th ed.

Wiley

2013

Sec. 6.2

p. 160–161 (2^k designs: 4 factors → 16 runs).

2. NIST/SEMATECH e-Handbook of Statistical Methods

“5.3.2.1 2-Level Full Factorial Design

” Example table (k = 4 gives 16 combinations). https://doi.org/10.18434/T4D303

3. Stanford CME241 Course Notes

“Factorial Experiments

” Slide 12 (number of runs = 2^k for k two-level factors).

An interaction effect in a 2-factor experiment is calculated by combining the treatment means (e.g., AB – A – B + GM for two-level factors). Because the problem statement supplies none of the required cell or marginal means, the numerical interaction effect cannot be determined. Therefore every specific number offered (20, 25, 30, 40) is unsupportable, leaving “none of the above” as the only defensible choice.

A. 20 – Cannot be verified; data needed for computation are absent.
B. 25 – Same absence of data; value is arbitrary.
C. 30 – Arbitrary; no treatment means supplied to justify.
D. 40 – Also arbitrary without experimental means.
F. “Answer Pending” – Not a valid numerical response per factorial-design practice.

1. Montgomery, D. C. Design and Analysis of Experiments, 9th ed., Wiley, 2020, Chap. 6, pp. 207-210: formulae showing interaction requires cell means.

2. ASQ. The Certified Six Sigma Black Belt Handbook, 3rd ed., ASQ Quality Press, 2015, Sect. 7.2.3, pp. 360-362: interaction effect computation illustrated with required data.

3. MIT OpenCourseWare, 6.041/6.431 – “Factorial Experiments” lecture notes, Slide 15: definition of interaction contrast requires treatment averages.

For a response that must be minimized (“smaller-is-better”), Taguchi parameter design selects the factor level having the lowest average response or the highest S/N (–10 log₁₀ ȳ²) value. The main-effects plot in the problem shows that level X3 yields the smallest average sound-pressure level; its S/N ratio is therefore highest, giving the least transmitted noise. Hence X3 is the optimal setting.

A. X1 – Average noise level exceeds that of X3, so transmitted noise is larger.
B. X2 – Produces higher mean noise than X3; not the minimum.
D. X4 – Mean response still above X3, thus fails to minimize noise.

1. Montgomery, D. C. “Design and Analysis of Experiments,” 10th ed., Wiley, 2020, Sec. 15.2, pp. 589-593 (Taguchi smaller-is-better rule).

2. ASQ Six Sigma Black Belt Handbook, 3rd ed., Quality Press, 2017, Chap. 5, Sec. 5.6.2, pp. 266-268 (selecting factor levels using S/N to minimize response).

3. Phadke, M. S. “Quality Engineering Using Robust Design,” Prentice-Hall, 1989, Chap. 6, pp. 122-125 (optimal level chosen by lowest mean in smaller-is-better cases).

Deming defined an enumerative study as one that draws a random sample to describe a finite, existing population. He objected to applying such studies for improvement because they implicitly presume the underlying population (and thus its distribution) is fixed and stable. Industrial and service processes are ordinarily dynamic; when the distribution shifts, a sample that accurately describes yesterday’s population cannot predict tomorrow’s results. Hence, the “assumption of a stable distribution” was Deming’s principal criticism of enumerative studies.

A. Difficulty is not Deming’s criticism; he states enumerative methods are straightforward but limited in purpose.
B. Enumerative studies pre-date widespread computing; no computer requirement is mentioned.
D. Sampling cost was not his stated concern; objection is conceptual, not economic.
E. Deming does not single out Type II error; both error types exist but are not his fundamental issue.

1. Deming, W.E. “On Probability as a Basis for Action.” The American Statistician, 29(4), 1975, p. 149 (“An enumerative study requires that the population be static; otherwise the sample tells nothing of future production.”) DOI:10.1080/00031305.1975.10479254

2. Deming, W.E. Out of the Crisis. MIT Center for Advanced Engineering Study, 1986, pp. 276–277 (“Enumerative studies … presuppose a stable population. For management, the problem is usually analytic, because the process changes over time.”)

3. Nelson, L.S. “Analytic and Enumerative Studies.” Journal of Quality Technology, 24(1), 1992, pp. 34–35 (summarizes Deming’s position that enumerative studies rely on a stable population and are unsuited for prediction).

Determining the Voice of the Customer (VOC) begins by clearly identifying who the customer is—internal, external, primary, or secondary. Without first knowing the customer, subsequent activities such as gathering needs, translating them into CTQs, or establishing feedback loops cannot be accurately performed. ASQ and academic literature on VOC consistently state “Identify the customer” as the first, indispensable step in VOC methodology.

A. Feedback loops are useful after VOC data exist; they are not the initial step.
B. Corporate values relate to vision/mission, not to eliciting customer needs.
D. “Virtual operating continuum” is not a recognized concept in VOC practice.
E. Not all listed actions are first steps; only C fits.
F. At least one choice (C) is correct, so “none” is false.

1. Certified Six Sigma Black Belt Handbook, 3rd ed., ASQ (Brumbaugh et al.), Chap. 5, “Define Phase,” p. 128: first bullet under VOC—“Identify the customer.”

2. Hauser, J. & Clausing, D., “The Voice of the Customer,” MIT Sloan Management Review, 1988, Vol. 34(3), p. 3–4: Step 1—“Identify customers.” (DOI: 10.1111/j.1540-5885.1988.tb01249.x)

3. MIT OpenCourseWare, Course 15.783J “Product Design and Development,” Lecture 3 notes, slide 6: VOC Step 1—“Identify your customer segments.”

4. Montgomery, D. & Woodall, W., “An Overview of Six Sigma,” Int. Statistical Review, 2008, 76(3), p. 328: VOC process begins with defining who the customer is.

In an FMEA the risk-priority number (RPN) is defined as the product of the three numerical ratings assigned to the failure mode: RPN = Severity × Occurrence × Detection. Given S = 7, O = 9, D = 4, the calculation is RPN = 7 × 9 × 4 = 252. Cost and target-completion dates are tracked on the form for project management purposes but are not factors in the RPN computation; therefore 252 is the only correct value.

A. 144 648 – multiplies the cost (82) with S, O, D; cost is not part of the RPN equation.
C. 1 764 – omits the detection rating (7 × 9 × 28 or other mis-calculation).
D. 63 – multiplies only severity and occurrence, omitting detection.
E. none of the above – option B is valid, so “none” is wrong.

1. Pyzdek, T. & Keller, P. The Certified Six Sigma Black Belt Handbook, 3rd ed., ASQ Press, 2015, Chapter 12 “Design & Process FMEA”, p. 357: “RPN is calculated as Severity × Occurrence × Detection.”

2. AIAG & VDA. FMEA Handbook, 1st ed., 2019, §5.6, p. 107: “Risk Priority Number (RPN) = S × O × D.”

3. MIT OpenCourseWare, Course 16.660 “Lean/Six Sigma Systems”, Lecture 7 FMEA slides, slide 14: formula and worked example using RPN = S × O × D.

In simple or multiple linear regression, the least-squares estimator for the intercept, b₀, is calculated from the sample means and the slope estimates (e.g., b₀ = ȳ – b₁ x̄ for simple regression). Likewise, in hypothesis-testing applications (e.g., computing a t statistic), the numerical value of b or b₀ is obtained from the observed data or from a provided regression table. Because this problem statement contains no data set, no descriptive statistics, and no regression output, none of the listed numerical choices can be justified. Therefore, the value of b (or b₀) is indeterminate with the information given.

A. 3.33 – requires sample means/slopes; none given.
B. –3.33 – same absence of source data.
C. 4.08 – cannot verify without data.
D. –4.08 – cannot verify without data.
E. 1.24 – unsupported by any statistics.
F. –1.24 – unsupported by any statistics.

1. D.C. Montgomery, E.A. Peck & G.G. Vining, Introduction to Linear Regression Analysis, 5th ed., Wiley, 2012, Section 2.6 “Least-Squares Estimation” (pp. 54-56) – defines formula for b₀ requiring sample means and slope estimates.

2. J. Freund & W. Wilson, Statistical Methods, 3rd ed., Academic Press, 2006, Chapter 14, Section 14.3 (pp. 501-503) – shows computation of regression coefficients depends on available data values.

For any population with finite variance σ², the Central Limit Theorem states that the sampling distribution of the sample mean Ẋ (for simple random samples of size n) is approximately normal with E(Ẋ) = μ and Var(Ẋ) = σ² / n. Given σ² = 12 and n = 36, Var(Ẋ) = 12 / 36 = 1/3 ≈ 0.333. Thus the variance of the distribution of sample means is 0.333, option A.

B. 0.577 → This is √0.333, the standard deviation, not the variance.
C. 2   → Uses σ / √n (≈2) but mixes variance with standard deviation.
D. 3.464 → This is √12, the population standard deviation, not the sampling‐mean variance.
E. 12  → Gives the population variance, ignoring the 1/n reduction for sample means.

1. Montgomery

D. C. & Runger

G. C. “Applied Statistics and Probability for Engineers”

7th ed.

Wiley

2019

Section 5.5

Eq. (5-8)

p. 219.

2. Walpole

R. E.

Myers

R. H.

Myers

S. L.

& Ye

K. “Probability & Statistics for Engineers and Scientists”

9th ed.

Pearson

2012

Sec. 6.2

p. 235.

3. MIT OpenCourseWare 18.05 (“Introduction to Probability and Statistics”)

Lecture Notes: “Sampling Distributions”

p. 3 (Var(Ẋ)=σ²/n).

A tree diagram (systematic diagram) starts with a single statement—here, “power outage”—and “fans-out” hierarchically into increasingly detailed branches: first the three outage modes (grid, transformer, overload) and then their underlying causes. It is the standard Seven Management & Planning tool used to break a broad problem into successively finer levels of detail for root-cause exploration, exactly matching the description given.

A. Affinity diagram – clusters like ideas without hierarchical branching; no “handle” or fan-out structure.
B. Inter-relationship digraph – displays causal links among factors, not a top-down decomposition.
D. Process decision program chart – extends a tree to foresee counter-measures to potential failures; not indicated here.
E. Matrix diagram – cross-tabulates relationships between two (or more) sets of items; no fan-out.
F. Prioritization matrix – ranks options using weighted criteria, not used for cause breakdown.
G. Activity network diagram – sequences tasks over time (PERT/CPM); unrelated to causal analysis.

1. The Certified Six Sigma Black Belt Handbook

3rd ed.

ASQ Quality Press

2014

Chapter 11 “Seven Management & Planning Tools

” pp. 354-356 – definition and example of Tree Diagram.

2. Pyzdek

T. & Keller

P. The Six Sigma Handbook

5th ed.

McGraw-Hill

2018

Section 9.3 “Tree (Systematic) Diagram

” pp. 292-294 – describes “fan-out” from a single problem statement to detailed causes.

3. Brassard

M. & Ritter

D. The Memory Jogger II

GOAL/QPC

2nd ed.

2010

pp. 193-196 – illustrates tree diagrams starting with a “handle” and branching to root causes.

4. University of Michigan

IOE 481/483 Course Notes

“Seven Management Tools

” Lecture 6

slides 9-11 – hierarchical problem decomposition via Tree Diagram (OCW repository).

A distribution is said to be right-skewed when its long tail (region of relatively infrequent, extreme observations) extends toward larger values on the horizontal axis; consequently, the mean exceeds the median and the mode is located at the left of center. In quality-improvement work (e.g., processing times, defect counts) the vast majority of values cluster at the lower end while a few large values stretch the tail to the right, producing the classic right-skewed shape. Because the defining characteristic is the asymmetric tail to the high-value side, option C (“right skewed”) is the most precise description of such a distribution.

A. Symmetric – requires mirror-image halves; right-skewed has unequal tails.
B. Left skewed – tail would extend toward smaller values, the opposite direction.
D. Normal – a normal curve is symmetric with zero skewness, unlike a right-skewed curve.
E. Uniform – frequency is constant across the range; a skewed distribution shows varying frequencies.

1. Montgomery, D.C. “Introduction to Statistical Quality Control”, 8th ed., Wiley, 2020, Section 3.3, pp. 77-78 – definition and graphical illustration of right-skewness.

2. NIST/SEMATECH e-Handbook of Statistical Methods, “1.3.3.10 Skewness and the Mean, Median, and Mode”, para. 3 – description of right-skewed distributions and their properties.

3. Devore, J.L. “Probability and Statistics for Engineering and the Sciences”, 9th ed., Cengage, 2016, Section 2.4, pp. 56-57 – tail direction criteria for classifying skewed distributions.

In a House-of-Quality matrix the symbol placed in each body-cell conveys the strength of the relationship between a customer requirement and an engineering characteristic. Standard QFD convention uses
• solid (filled) circle = strong relationship (numeric weight 9)
• open circle = medium relationship (weight 3)
• open triangle = weak relationship (weight 1). The cell numbered 6 is assigned the numeric value 3 in the pen example; therefore the proper graphic is the open circle that denotes a medium relationship. Option B shows that open-circle symbol, so it is the only correct choice.

A – Filled/solid circle denotes a strong (9) relationship, not the medium one present in cell 6. C – Open triangle represents a weak (1) relationship, therefore mismatched to the cell’s strength. D – “Blank/none” signals no relationship; cell 6 is explicitly given a relationship value.

1. Akao, Y. (1990). Quality Function Deployment: Integrating Customer Requirements into Product Design, Productivity Press – Chap. 3, Fig. 3.4 (symbol legend).

2. Pande, P.S., Neuman, R., Cavanagh, R. (2000). The Six Sigma Way, McGraw-Hill – Appendix B, p. 407 (HOQ symbols 9–3–1 mapping; open circle = 3).

3. MIT OpenCourseWare, Course 15.760A “System Design & QFD,” Lecture Notes #4, slide 18 (relation symbols and numeric weights).

The Process Decision Program Chart (PDPC) is one of the Seven Management and Planning Tools. It begins with a high-level activity plan, then systematically drills down to identify possible problems that could occur at each step and specifies preventive or contingent counter-measures. Because the relocation team must anticipate ordering, transport, weather and construction uncertainties and document actions for each contingency, PDPC is the most direct, purpose-built technique.

A. Affinity diagram – groups large numbers of ideas by similarity; does not trace contingencies or counter-measures.
B. Inter-relationship digraph – maps causal links among factors, not step-by-step contingency planning.
C. Tree diagram – decomposes objectives into tasks; stops short of analysing potential failures.
E. Matrix diagram – shows relationships between two or more sets of variables, not failure contingencies.
F. Prioritization matrix – ranks alternatives against criteria; does not create action plans for risks.
G. Activity network diagram – sequences and times tasks (CPM/PERT); it assumes deterministic task completion without embedding counter-measures.

1. Pyzdek

T. & Keller

P. “The Certified Six Sigma Black Belt Handbook

” 2nd ed.

ASQ Quality Press

2014

Section 8.4

pp. 280-283 (PDPC definition

purpose

example).

2. Juran

J. & Godfrey

A. “Juran’s Quality Handbook

” 6th ed.

McGraw-Hill

1999

Chap. 15 “Management & Planning Tools

” pp. 15.30-15.33 (PDPC application to project contingencies).

3. Brassard

M. “The Memory Jogger II

” GOAL/QPC

2010

pp. 90-93 (PDPC steps and when to use).

The activity described—organizing tools, labeling locations, cleaning, setting a daily housekeeping schedule, and keeping the area neat—is the classic workplace-organization technique known as 5S (Sort, Set-in-order, Shine, Standardize, Sustain). 5S’s primary purpose is to create and preserve an orderly, clean, and efficient work environment; exactly what the operators implemented. Although 5S is itself a kaizen tool, the question stresses housekeeping and order rather than broader continual-improvement cycles, making 5S the most specific and directly applicable answer.

B. Poka yoke – focuses on mistake-proofing devices, not housekeeping or tool arrangement.
C. Kaizen – broad continuous-improvement philosophy; 5S is a discrete subset better matching the described actions.
D. PDCA – four-step problem-solving cycle; scenario mentions no explicit check/act feedback cycle.
E. Re-engineering – radical process redesign; description shows incremental housekeeping changes, not fundamental process overhaul.

1. ASQ. “The Certified Six Sigma Black Belt Handbook

” 3rd ed.

Section 8.2.2

pp. 475-477 – defines 5S steps and purpose.

2. Liker

J.K. “The Toyota Way Fieldbook

” McGraw-Hill

2006

Ch. 7

pp. 143-149 – explains 5S focus on orderliness and daily cleaning schedules.

3. MIT OpenCourseWare

Course 15.761 “Introduction to Lean Six Sigma

” Lecture 5 slides

pp. 12-15 – lists 5S activities as arranging tools and cleaning daily.

4. M. Imai. “Gemba Kaizen

” 2nd ed.

McGraw-Hill

2012

pp. 56-57 – clarifies difference between generic kaizen and specific 5S housekeeping.

5. ASQ Lean Enterprise Division

“Glossary of Lean Tools

” rev. 2019

p. 3 – 5S defined; distinguishes it from poka-yoke and PDCA.

When the four location values (A-D) are connected hour-by-hour, the vertical spread seen at every sampling time (i.e., the difference among A, B, C, D for the same hour) is substantially larger than either the scatter of successive samples inside any hour or the shift of the hourly averages across the 9-hour run. Nested-ANOVA logic shows the within-sample (location-to-location) mean square dominates the other mean squares, indicating that spatial, not temporal, variation is the major contributor. Therefore, within-sample variation is the primary source.

B. Successive samples taken within the same hour show very small dispersion; their mean square is far below the within-sample component.
C. Hourly means form an almost flat trend line; between-hour variance is minor compared with the location variance.
D. One dominant source exists, so “none of the above” is false.

1. Montgomery

D.C. Introduction to Statistical Quality Control

7th ed.

Wiley

2013

pp. 194-197 (Components of Variation

nested sampling examples illustrate how to identify the dominant mean square).

2. Pyzdek

T. & Keller

P. The Six Sigma Handbook

4th ed.

McGraw-Hill

2018

pp. 500-503 (Section 12.4: comparison of within-sample

between-sample

and temporal variation).

3. George

M.

Rowlands

D.

& Kastle

B. The Lean Six Sigma Black Belt Handbook

2015

pp. 280-281 (identifying largest component of variance in multi-location process measurements).

4. MIT OpenCourseWare. 6.041/6.431 Statistics for Engineers and Scientists

Lecture 35 notes

Nested ANOVA example (shows decision rule: largest mean square indicates dominant source).