IASSC Lean Six Sigma Green Belt Exam Prep FREE Sample Exam

Reworking mislabelled cartons discovered in the warehouse is which type of cost?

Reworking cartons found before dispatch is an external failure cost for the shipper.

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Carton rework counts as appraisal because warehouse staff noticed the wrong labels.

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Rework discovered internally is classified as prevention of future labelling errors.

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Reworking cartons found before dispatch is an internal failure cost for the shipper.

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A billing team at a regional clinic network handles 9,400 claims a month, and 11 percent bounce back for missing referral codes, forcing rework and delaying payment by weeks. Which draft line works best as the problem statement?

The team will roll out a new referral checklist software package next quarter to stop claim denials at the source.

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Claims lacking referral codes trigger costly rework, so leadership should immediately fund extra billing staff.

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Eleven percent of monthly claims return for missing referral codes, creating rework and postponing reimbursement for weeks.

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The department aims to cut returned claims from eleven percent down to three percent within six months of starting this spring.

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A failure mode has severity 6, occurrence 3, and detection 7. What risk priority number does that produce?

The risk priority number is 16, found by adding 6 plus 3 plus 7 together.

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The risk priority number is 42, found by multiplying 6 times 7 while ignoring occurrence.

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The risk priority number equals 63, found by multiplying 3 times 7 and adding 42 more.

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The risk priority number equals 126, found by multiplying 6 times 3 times 7 during this review.

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Why do teams divide by n minus 1 when computing a sample variance?

Dividing by n minus 1 removes the largest observation from the variance calculation.

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Dividing by n minus 1 converts the variance into standard deviation units automatically.

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Using n minus 1 as the divisor forces the sample variance to equal the population variance exactly in effect.

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Using n minus 1 as the divisor corrects the tendency of a sample to understate population variation.

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For loaf weight on bread loaves, the Pioneer Cove Manufacturing R&R report lists percent study variation of 25%. What verdict does that figure support?

At 25%, the measurement system is marginal: the figure sits inside the 10% to 30% band, so the system is questionable and its use depends on how critical the measurement is at the Ivory Street plant.

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A 25% reading means the system is comfortably acceptable for bread loaves, judged by how much of the loaf weight tolerance the gage consumes at Pioneer Cove Manufacturing in the bread loaves operation.

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A 25% reading means the system is unacceptable, because R&R percentages are graded on field performance of bread loaves after shipment.

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No verdict is possible from 25% alone, since study variation percentages carry guideposts only for pass/fail inspection of bread loaves.

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At the Foundry Street plant, a report on fill height shows Ppk noticeably below Cpk for the same characteristic and period. How should Bruno Holm read the indices involved?

The gap suggests the process moved around over the longer window, because the overall variation behind Ppk includes that extra movement.

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The choice is cosmetic for paint cans, since Pp and Ppk at Maple Meadow Industries come from the limits alone, just like their short-term cousins.

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Centering vanishes inside Ppk for paint cans, guaranteeing the index never falls below the Pp figure for fill height.

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The convention runs backward for paint cans: Cpk takes the overall spread, Ppk the subgroup spread, at plants like Maple Meadow Industries.

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A report lists a standard deviation of 9 minutes for individual waits and a standard error of 1.5 minutes. What does each describe?

The 9 minutes describes spread among individual waits; the 1.5 describes spread among average waits from samples.

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The 9 minutes describes spread among sample averages; the 1.5 describes spread among the individual waits seen.

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Both figures describe individual waits, but the 1.5 was computed after removing outliers from the waiting data.

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Both figures describe sampling precision, but the 9 was computed using a smaller set of repeated samples drawn.

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A trial halves rework from 12 percent to 6 percent, a gain managers call valuable, yet p equals 0.09 from a modest sample. What should happen next?

Rework gains of this size are automatically significant, so the reported p-value must be a typo.

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The signal deserves a larger confirmatory study, because the observed gain matters if it holds.

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The change should be reversed immediately, because p above 0.05 proves the old method superior.

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The team should lower standards until p complies, because worthwhile ideas deserve significance.

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An Anderson-Darling test on 40 wait times returns p equal to 0.003 when normality is the null. What does that signal before a planned t-test?

Normality looks excellent, because tiny p-values always flatter the assumption being examined.

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Wait times are proven fraudulent, because genuine service data never depart from bell shapes.

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The data depart significantly from normality, prompting a closer look at skew and outliers.

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The t-test becomes mandatory, because assumption failures must be punished with parametric tests.

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A scatter of house prices against distance from a city centre slopes downward and thins out: expensive homes vary wildly in price, cheaper ones cluster tightly. Which regression assumption is visibly strained?

The independence of errors, because neighbouring houses clearly share a postcode and therefore cannot enter one regression.

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The existence of an intercept, because a thinning scatter implies the line should start at zero price at zero distance.

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The normality of the predictor, because distance values thinning out shows the predictor itself must follow a bell shape.

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Constant error variance, because the vertical spread of prices changes systematically along the distance axis of the plot.

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A coefficient table shows humidity slope 0.22 with a 95 percent interval from -0.10 to 0.54 for paint shop rework. What should the team take from that interval?

Humidity's effect is exactly 0.22 in 95 percent of shops, so the interval mainly confirms the slope for audiences elsewhere.

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The reported interval proves humidity helps slightly, because most of its length lies above zero and majorities decide statistical practical questions.

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Rework should be forecast by ignoring humidity, because an interval touching negative values erases the variable from the model.

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The data stay compatible with a small negative effect, no effect, or a moderate positive one, so humidity's partial role remains unresolved.

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An insurer samples 200 claims daily; the average defective share is 0.10. Using three-sigma limits, which pair describes the proportion chart for a 200-claim day?

The band is narrower, running from about 0.064 at the bottom to about 0.136 at the top for that day.

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Both boundaries collapse onto 0.100, the average rate itself.

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For a 200-claim day the band runs from about 0.036 at the bottom to about 0.164 at the top.

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A top boundary near 0.310 with no positive bottom boundary applies.

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IASSC Lean Six Sigma Green Belt Exam Prep FREE Sample Exam — Fixed Set

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