CSSBB Exam Dumps - Certified Six Sigma Black Belt

The correct answer is B. 2ⁿ⁻¹ . In DOE, a full factorial experiment for n factors at two levels requires 2ⁿ runs . A half-fraction design uses exactly one-half of those runs, so the number of runs becomes:

(1/2)(2ⁿ) = 2ⁿ⁻¹

The CSSBB material identifies these as two-level fractional factorial designs and explains that fractional designs are used to reduce the number of experimental trials while still learning about the most important factor effects. This is a standard Six Sigma Improve-phase concept because DOE is often used to optimize a process efficiently without running every possible treatment combination.

The other options represent different design sizes. 2ⁿ is the full factorial. 2ⁿ⁻² would be a quarter-fraction, and 2ⁿ⁻³ would be an eighth-fraction. Therefore, when the question specifically asks for a half-fraction run for a two-level design with n factors , the correct mathematical expression is 2ⁿ⁻¹ . That makes option B the verified answer.

The correct answer is C. 7 minutes . The CSSBB material defines takt time as the available production time divided by the rate of customer demand and also gives the formula as net operating time per period / customer requirements per period . Here, the available work time is the 8-hour day minus 1 hour for meetings and breaks, leaving 7 hours of net operating time. Converting to minutes gives 420 minutes per day . The customer requires 60 invoices per day , so:

Takt time = 420 / 60 = 7 minutes per invoice

That matches option C . The listed “total work time per invoice = 5 minutes” is useful for comparing actual process cycle time to takt time, but it is not used to calculate takt time itself. In Lean Six Sigma, takt time reflects the pace required to meet demand , not the actual current processing time. Since the process must complete one invoice every 7 minutes to satisfy customer demand, the verified answer is 7 minutes per invoice .

An andon board is a visual control device used to alert leaders and team members that there is a problem requiring attention in the work area. In Lean Six Sigma environments, andon systems provide immediate visibility of abnormal conditions such as machine stoppages, quality issues, material shortages, or workflow interruptions. This supports rapid response, containment, and restoration of normal process conditions. The value of the andon board in the Control Phase is that it makes process status visible in real time, helping teams sustain improvements and react quickly when the process begins to drift or fail. Heijunka refers to production leveling, kaizen refers to continuous improvement, and jidoka refers to built-in quality and autonomation, often involving stopping the process when an abnormality occurs. While jidoka is related conceptually, the specific technique that visibly alerts the leader or team leader is the andon board. Therefore, the correct answer is C, because it is the standard visual signaling method used to notify leadership that a problem exists somewhere in the area.

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The correct answer is C. Work breakdown structure . The CSSBB source states that the work breakdown structure (WBS) is a detailed plan that expands the project statement of work into a detailed listing of activities required to complete the project . It further explains that each project task is divided into smaller activities until responsibility can be clearly assigned.

This makes WBS the best tool for identifying the specific tasks needed to complete a project. The same CSSBB section also notes that the WBS helps identify detailed activities for the plan, while tools such as PERT , Gantt charts , and critical path method are used more for evaluating project timelines and sequencing once the activities are known. A PERT chart and an activity network diagram focus on ordering tasks and estimating durations, while CPM identifies the critical path through the network. None of those is as directly focused on decomposing the project into detailed task elements as the WBS. Therefore, the verified CSSBB answer is C. Work breakdown structure .

The correct answer is C. Process control measurements . In the CSSBB material, appraisal costs are defined as the costs associated with measuring, evaluating, or auditing products or services to ensure conformance to standards and requirements. The detailed appraisal-cost checklist includes items such as inspection and test , in-process inspection , laboratory testing , equipment calibration , and test reporting , all of which are measurement- and evaluation-related activities. Process control measurements fit this category because they are used to assess current process output or performance against requirements.

By contrast, quality planning is a prevention cost, because planning is intended to prevent defects before they occur. The CSSBB prevention-cost list explicitly includes planning and design reviews . New product review and design evaluation testing are more closely aligned with early design/prevention activities than with routine appraisal of conformance. Since appraisal costs focus on checking, measuring, and verifying quality, process control measurements is the best CSSBB-based answer.

The correct test is the paired t-test because the experiment uses matched conditions on the same trucks. Each truck receives both tire brands, which means the wear results are naturally paired within the same operating environment. That pairing controls for truck-to-truck variation such as driver behavior, route conditions, load characteristics, suspension differences, and maintenance history. In Six Sigma Analyze Phase work, when two treatments are applied in a matched or dependent way and the goal is to compare mean differences within each matched unit, the paired t-test is the appropriate method. A two-sample t test would be used if the two tire groups were independent, but they are not independent here because both brands are evaluated on the same trucks. Z tests are not appropriate in this situation because population variances are not known and the sample size is relatively small. The paired t-test provides a more sensitive and statistically correct comparison because it isolates the treatment effect from much of the background variation. Therefore, the correct answer is D, a paired t-test.

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The learning and growth perspective is the correct answer because employee satisfaction, absenteeism, and turnover are workforce-related indicators tied directly to organizational capability, culture, development, and long-term performance capacity. In the balanced scorecard framework, the learning and growth perspective captures the infrastructure that enables future improvement, including employee skills, training, morale, engagement, leadership development, and knowledge systems. These are considered leading or predictive indicators because they influence how well internal processes will function and, ultimately, how customers and financial results will perform. If employees are dissatisfied or turnover is high, the organization may experience lower productivity, weaker service quality, reduced innovation, and instability in process performance. Financial and customer measures are typically more lagging in nature, while internal business process measures track how operations perform. The learning and growth category addresses the people and capability base that drives those outcomes. Therefore, the correct answer is D, since the stated KPIs are predictors associated with the learning and growth perspective of the balanced scorecard.

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In a one-way ANOVA, the F statistic is calculated by dividing the mean square for treatment by the mean square for error. From the table, the treatment sum of squares is 333 with 4 degrees of freedom, and the error sum of squares is 310 with 20 degrees of freedom.

First calculate the treatment mean square:

MS treatment = 333 / 4 = 83.25

Then calculate the error mean square:

MS error = 310 / 20 = 15.5

Now compute the F statistic:

F = 83.25 / 15.5 = 5.37

That value corresponds to choice D.

ANOVA is an Analyze Phase tool used by Black Belts to test whether there are statistically significant differences among treatment means. The F ratio compares variation between groups to variation within groups. A larger F value suggests that treatment effects may be real rather than due to random error. In this case, the calculations from the given sums of squares and degrees of freedom clearly produce 5.37. Therefore, the correct answer is D, 5.37.