CSSGB Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

1. 1. RPN = S × O × D.
2. 2. No table constants are needed; ratings are on 1–10 scales.
3. 3. A: 10 × 3 × 4 = 120; B: 4 × 5 × 6 = 120.
4. 4. The RPNs are equal (120 each).
5. 5. Because A has severity 10 (consumer safety), it is addressed first; the RPN tie does not mean equal risk.

Act on A first, because its severity rating is 10: Correct. Both RPNs are 120 (10 × 3 × 4 and 4 × 5 × 6), but A can harm a consumer with an allergy; severity 9–10 modes get priority regardless of RPN.

Act on B first, because its O and D ratings are higher: Higher O and D do not outweigh a potential safety effect; B's worst effect is a cosmetic or seal-quality issue.

Act on either first, because the two RPNs are equal: Equal products can hide very different risk profiles; RPN is an ordinal ranking aid, not a measure of risk.

Act on neither, because an RPN of 120 is low: No action threshold was set, and fixed RPN cut-offs ignore severity; a severity-10 mode needs review at any RPN.

Always review severity alongside RPN: a high-severity failure mode warrants action even when its RPN equals or falls below that of a minor mode.

Answer: B

1. 1. Takt = available time / demand; effective step time = work time per file / parallel staff.
2. 2. No table constants are needed; available time = 432 min, demand = 90 files.
3. 3. Takt = 432 / 90 = 4.8 min. Effective times: intake 3.5 / 1 = 3.5; appraisal ordering 5.4 / 1 = 5.4; credit review 9.0 / 2 = 4.5; final approval 12.0 / 3 = 4.0 min per file.
4. 4. Only appraisal ordering (5.4 min) exceeds takt (4.8 min).
5. 5. Appraisal ordering is the constraint; under TOC the team exploits it first (no idle time, no incomplete files), then subordinates the other steps, then elevates if needed.

Intake: Intake releases a file every 3.5 min, well under the 4.8-min takt; it starts the flow but does not limit it.

Appraisal ordering: Correct. Takt = 432 / 90 = 4.8 min per file; appraisal ordering releases a file every 5.4 min, the only step slower than takt, so it is the constraint to exploit first.

Credit review: Two analysts give 9.0 / 2 = 4.5 min per file, which meets the 4.8-min takt.

Final approval: Has the longest work time, but three underwriters give 12.0 / 3 = 4.0 min per file, faster than takt.

The constraint is the step whose effective time per unit (work time ÷ parallel resources) is longest, not the step with the longest raw work time.

Answer: B

1. 1. RTY = FPY₁ × FPY₂ × FPY₃ × FPY₄, where FPY = units passed first time / units entering the step.
2. 2. From the table: 1,940/2,000; 1,920/1,980; 1,850/1,970; 1,930/1,960 (rework and scrap are both excluded from first-pass good).
3. 3. RTY = 0.9700 × 0.9697 × 0.9391 × 0.9847.
4. 4. RTY = 0.8698 ≈ 0.870.
5. 5. Only about 87% of jars pass all four steps right the first time, although the final yield looks like 97.8%; labeling is the weakest step.

0.863: Subtracts the summed step defect rates from 1 (1 − 0.1365); yields of serial steps multiply, they do not subtract additively.

0.870: Correct. RTY = Π(first-pass yields) = (1,940/2,000)(1,920/1,980)(1,850/1,970)(1,930/1,960) = 0.9700 × 0.9697 × 0.9391 × 0.9847 = 0.870.

0.966: Averages the four first-pass yields instead of multiplying them.

0.978: Final (throughput) yield, 1,955/2,000: it counts reworked jars as good and hides the hidden factory that RTY exposes.

RTY multiplies the first-pass yields of serial steps, so rework counts as a loss even when the unit is eventually shipped.

Answer: B

1. 1. Project duration = length of the longest path through the network (critical path); slack = critical length − path length.
2. 2. Durations (weeks) from the table: A 4, B 6, C 3, D 5, E 4, F 2, G 1; no table constants needed.
3. 3. A–B–E–G = 4 + 6 + 4 + 1 = 15; A–D–E–G = 4 + 5 + 4 + 1 = 14; A–C–F–G = 4 + 3 + 2 + 1 = 10.
4. 4. Critical path A–B–E–G, duration 15 weeks.
5. 5. Any delay to A, B, E or G delays go-live; D has 1 week of slack and C–F have 5 weeks.

A–C–F–G, 10 weeks: This is the shortest path; the project cannot finish until the longest path is complete.

A–B–E–G, 15 weeks: Correct. Paths: A–B–E–G = 4 + 6 + 4 + 1 = 15; A–D–E–G = 14; A–C–F–G = 10. The longest path is critical, so the project takes 15 weeks.

A–D–E–G, 14 weeks: A near-critical path with 1 week of slack; E cannot start until B (6 weeks) is also done.

all seven activities in series, 25 weeks: Adds every duration as if nothing ran in parallel, ignoring the precedence network.

The critical path is the longest path through the activity network; its length is the shortest possible project duration.

Answer: A

"From January to June, 18% of billing calls led to a repeat call within 7 days, versus a 10% target.": Correct. It states the measurable symptom, the time frame, the baseline (18%) and the gap to target, without assuming a cause or solution.

"Customer satisfaction with billing support is too low and must be improved as quickly as possible.": Vague: no metric, baseline, time frame or magnitude, so success cannot be judged.

"Agents need more training on the billing system so that customers will stop calling back.": Embeds a solution (training) in the problem statement before any data have been analyzed.

"Confusing invoices cause the repeat calls, which must be cut in half by the end of the year.": Assumes the root cause (confusing invoices) in the Define phase; causes are confirmed in Analyze.

A charter problem statement describes what is wrong, where and when, how big it is (baseline) and the gap to target, with no assumed cause or solution.

Answer: B

1. 1. σ̂_within = R̄/d₂; Cpk = min[(USL − X̄̄)/3σ̂, (X̄̄ − LSL)/3σ̂].
2. 2. d₂ = 2.059 for n = 4; R̄ = 0.206 mm.
3. 3. σ̂ = 0.206/2.059 = 0.100 mm; Cpu = 0.24/0.300 = 0.80; Cpl = 0.36/0.300 = 1.20.
4. 4. Cpk = 0.80.
5. 5. Cpk < 1, so the process is not capable at the upper limit; centering at 18.30 mm would raise Cpk to Cp = 1.00.

0.39: Uses R̄ itself as σ without dividing by d₂.

0.80: Correct. σ̂ = R̄/d₂ = 0.206/2.059 = 0.100 mm; Cpu = (18.60 − 18.36)/(3 × 0.100) = 0.80 and Cpl = 1.20, so Cpk = min = 0.80.

1.00: This is Cp = 0.60/(6 × 0.100), which ignores the off-center mean.

1.20: This is Cpl, the distance to the farther limit; Cpk takes the smaller of Cpu and Cpl.

Cpk uses the within-subgroup σ (R̄/d₂) and the distance from the mean to the nearer specification limit; Cp ignores centering.

Answer: B

the natural spread is wider than the tolerance, so variation must be cut first: The data span about 494–502 g with 6s ≈ 9.3 g, slightly less than the 10 g tolerance (Cp ≈ 1.07); the spread is not wider than the tolerance, and location is the bigger problem.

the spread is about equal to the tolerance, and the process is centered low: Correct. 6s ≈ 9.3 g is about equal to the 10 g tolerance (Cp ≈ 1.07), but the mean is about 498.1 g, so the lower tail crosses 495 g; re-centering on 500 g would put the natural limits (about 495.3–504.7 g) inside both specifications.

the process is capable, because most of the histogram lies inside the specification: "Most inside" is not capability; 2 bottles are already below the LSL and the process is off target.

the specification limits should be widened to match the natural process limits: Specifications come from the customer (voice of the customer), not from the process; widening them to fit the process is not a valid fix.

Compare the natural process limits (voice of the process) with the specification limits (voice of the customer) for both spread and location: an off-center process produces defects even when its spread is no wider than the tolerance.

Answer: D

1. 1. σ²_GRR = σ²_repeatability + σ²_reproducibility; σ²_total = σ²_GRR + σ²_part; %GRR = σ_GRR/σ_total × 100.
2. 2. Variance components given: 20.25, 4.00 and 100.00 µm²; guideline: > 30% unacceptable.
3. 3. σ²_GRR = 24.25 → σ_GRR = 4.92 µm; σ²_total = 124.25 → σ_total = 11.15 µm.
4. 4. %GRR = 4.92/11.15 × 100 = 44.2%.
5. 5. Unacceptable; repeatability (equipment variation) is about five times the reproducibility variance, so improve the gage or fixturing before retraining operators.

19.5%; may be acceptable, with repeatability the larger component: This is the percent contribution (variance ratio 24.25/124.25); %GRR compares standard deviations, so take square roots first.

44.2%; unacceptable, with reproducibility (operators) the larger component: The %GRR is right, but reproducibility (variance 4.00) is far smaller than repeatability (20.25).

49.2%; unacceptable, with repeatability the larger component: Divides σ_GRR by the part-to-part σ (4.92/10.00) instead of the total σ.

44.2%; unacceptable, with repeatability (equipment) the larger component: Correct. σ_GRR = √(20.25 + 4.00) = 4.92 µm; σ_total = √(24.25 + 100) = 11.15 µm; %GRR = 44.2% (> 30%, unacceptable), and repeatability dominates, so the gage or fixture needs attention first.

Add variances, not standard deviations, then take the square root; a large repeatability share points to the instrument, a large reproducibility share to operators or method.

Answer: B

1. 1. OR gate: P = 1 − Π(1 − pᵢ); AND gate: P = Π pᵢ (independent events).
2. 2. Given: P(utility) = 0.05 per year; P(UPS fails on demand) = 0.20; P(generator fails on demand) = 0.30.
3. 3. P(backup fails) = 1 − (1 − 0.20)(1 − 0.30) = 1 − 0.56 = 0.44; P(top) = 0.05 × 0.44.
4. 4. P(top) = 0.0220 (2.2% per year).
5. 5. The AND gate gives redundancy: improving either the utility or the backup chain cuts the top-event probability.

0.0030: Treats every gate as AND, multiplying all three probabilities.

0.0220: Correct. OR gate: P(backup fails) = 1 − (0.80)(0.70) = 0.44; AND gate: P(top) = 0.05 × 0.44 = 0.0220.

0.0250: Adds the OR-gate probabilities (0.20 + 0.30 = 0.50), which double-counts the case where both fail; the sum is only a rare-event approximation and overstates the exact value.

0.4680: Treats every gate as OR, so any single failure would cause the top event.

In fault tree analysis, multiply probabilities through an AND gate and use 1 − Π(1 − p) through an OR gate (independent events); work from the basic events up to the top event.

Answer: D

1. 1. Paired t = d̄ / (s_d / √n), with df = n − 1.
2. 2. n = 8 pairs, df = 7; two-tailed α = 0.05 gives t₀.₀₂₅,₇ = 2.365.
3. 3. s_d/√n = 3.0/√8 = 3.0/2.828 = 1.061; t = 4.5/1.061.
4. 4. t = 4.24.
5. 5. 4.24 > 2.365, so reject H0: the new checklist changes processing time (it saves about 4.5 minutes per file).

t = 1.50; fail to reject H0: Divides d̄ by s_d without the √n in the standard error.

t = 12.00; reject H0: Multiplies by n = 8 instead of √8 when forming the standard error.

t = 4.24; fail to reject H0: The statistic is right, but |t| = 4.24 exceeds the critical value 2.365, so H0 must be rejected.

t = 4.24; reject H0: Correct. t = d̄ / (s_d/√n) = 4.5 / (3.0/√8) = 4.5 / 1.061 = 4.24; |t| > 2.365, so H0 is rejected.

The paired t-test treats the differences as one sample: t = d̄/(s_d/√n) with n − 1 degrees of freedom.

Answer: B

positional (within-part) variation: Gate-end and far-end readings differ by only about 0.02 mm on every part, a small share of the total spread.

temporal (time-to-time) variation: Correct. The time-group averages shift from about 2.51 mm to 2.60 mm to 2.43 mm, a range of about 0.16 mm, far larger than the within-part (≈0.02 mm) or part-to-part (≈0.01 mm) spread.

cyclical (part-to-part) variation: Consecutive parts taken at the same time differ by about 0.01 mm, the smallest family on the chart.

all three families about equally: The families are clearly unequal: the shift between sampling times is several times larger than the other two.

Read a multi-vari chart by comparing the spread within a unit, between consecutive units and between time periods; investigate the inputs that change on the scale of the largest family (here, things that drift over hours, such as material lot or mold temperature).

Answer: B

1. Internal time after conversion = total changeover − elements converted to external.
2. Total = 80 min; "Yes" elements = 12 + 15 + 6 = 33 min.
3. Internal = 80 − 33 (or 3 + 10 + 14 + 20).
4. Internal = 47 min.
5. The press downtime falls about 41% (33/80) before any streamlining of internal elements.

33 min: This is the time converted to external setup (12 + 15 + 6), not the time the press is still stopped.

47 min: Correct. Internal time = elements that need the press stopped: 3 + 10 + 14 + 20 = 47 min (80 − 33).

59 min: Moves only preheating and tool gathering (80 − 15 − 6) and leaves mold fetching internal.

65 min: Moves only the preheat (80 − 15), the largest "Yes" element, and leaves the other external work internal.

SMED first separates and then converts internal work to external, so only elements that truly require the machine stopped remain in downtime.

Answer: D

1. Main effect = mean(+) − mean(−); with an interaction, compare cell means.
2. Lighting (task = +): (4.0 + 7.0)/2 − (8.0 + 3.0)/2 = 5.5 − 5.5 = 0.
3. Labels (new = +): (7.0 + 3.0)/2 − (8.0 + 4.0)/2 = 5.0 − 6.0 = −1.0; AB = (8.0 + 7.0)/2 − (4.0 + 3.0)/2 = 4.0.
4. The interaction (4.0) dwarfs both main effects; the lowest cell is standard lighting/new labels at 3.0.
5. Recommend standard lighting with new labels, then confirm with verification runs.

either lighting with new labels, since the lighting main effect is zero: The lighting main effect is zero only on average; because of the interaction, task lighting with new labels gives 7.0 versus 3.0 with standard lighting.

task lighting with old labels: Gives 4.0 errors per 1,000 lines, worse than the 3.0 achieved with standard lighting and new labels.

task lighting with new labels: Picking each factor's "improved" level gives 7.0, one of the worst cells, because the factors interact.

standard lighting with new labels: Correct. This cell has the lowest mean (3.0). With a strong interaction, the best setting is read from the cell means (the interaction plot), not from the main effects alone.

When the interaction is large, main effects can mislead; choose the operating point from the cell means shown on the interaction plot.

Answer: B

1. Effect of speed at each preheat level = mean(high speed) − mean(low speed).
2. Preheat low: 28 − 12 = 16; preheat high: 15 − 14 = 1.
3. AB interaction = (16 − 1)/2 = 7.5 defects per board.
4. The slopes differ, so the lines are non-parallel.
5. Conclusion: a strong speed × preheat interaction; running high speed is acceptable only with high preheat.

the factors act independently, because both lines rise from low to high speed: Lines sloping the same direction does not mean independence; the slopes differ greatly (16 vs 1), which is an interaction.

an interaction: higher speed adds far more defects at low preheat: Correct. The lines are clearly non-parallel: raising speed adds 16 defects at low preheat but only 1 at high preheat, so the effect of speed depends on the preheat level.

conveyor speed has no effect, because the lines would cross if extended: The lines do not cross on the plot, and speed clearly changes defects at low preheat; an interaction does not cancel a factor's effect.

preheat can be dropped, because it changes defects by only 2 at low speed: True only at low speed; at high speed preheat changes defects from 28 to 15, so dropping it would ignore the interaction.

Non-parallel lines on an interaction plot mean the effect of one factor depends on the level of the other; interpret main effects only in light of the interaction.

Answer: D

Row 1: Continuous monitoring with an alarm and a reaction plan that stops, contains and corrects is complete for this characteristic.

Row 2: This row has a measurement method, frequency, chart and a reaction plan that contains, corrects and escalates.

Row 3: 100% automatic scanning with a reject gate is a strong control, and the reaction plan says what happens to rejects.

Row 4: Correct. The safety-critical torque row has no real reaction plan: recording readings does not say who contains suspect parts or how the process is corrected when the chart signals.

Every control-plan row needs a reaction plan that contains suspect product, corrects the process and escalates, and the most critical characteristics need the strongest controls.

Answer: C

1. 1. p chart: UCL = p̄ + 3√[p̄(1 − p̄)/n].
2. 2. p̄ = 240/4,000 = 0.06; n = 200 claims per subgroup (day).
3. 3. √(0.06 × 0.94/200) = √0.000282 = 0.0168; 3 × 0.0168 = 0.0504.
4. 4. UCL = 0.06 + 0.0504 = 0.1104 (LCL = 0.06 − 0.0504 = 0.0096).
5. 5. A day with more than about 11% defective claims would signal a special cause.

0.0608: This omits the square root: 0.06 + 3 × (0.06 × 0.94/200).

0.0768: This is p̄ plus one standard error (0.06 + 0.0168): a 1σ limit, not 3σ.

0.1104: Correct. p̄ = 240/4,000 = 0.06; UCL = 0.06 + 3√(0.06 × 0.94/200) = 0.06 + 3 × 0.0168 = 0.1104.

0.2193: This uses n = 20 (the number of days) instead of the 200 claims in each daily sample.

For a p chart, n is the number of units inspected in each subgroup, not the number of subgroups; the limits are p̄ ± 3√[p̄(1 − p̄)/n].

Answer: A

a special cause that has shifted the process mean upward from subgroup 11: Correct. Subgroups 11–20 are 10 consecutive points above the 12.000 mm center line, which meets the stated 8-point rule: the process mean has shifted upward (for example, tool wear or an offset change) even though every point is inside the limits.

only common-cause variation, because no point falls beyond either control limit: A point beyond a limit is not the only signal; the stated run rule is met, so the chart is not showing common cause alone.

a special cause that has increased the spread within subgroups: Within-subgroup spread is monitored on the R chart, which the stem says is in control; the X̄ chart shows a change in level.

stratification, with the points hugging the center line: Stratification shows as points crowded unusually close to the center line; here the later points sit consistently above it.

Run rules detect sustained shifts that stay inside the limits; a run on one side of the X̄ center line means the mean has moved, while within-subgroup spread is judged on the R chart.