CSSBB Free Practice Questions (Review Mode)

Free sample questions

6 questions from this set, shown in full with the correct answer and the reasoning behind it. Read them as a study resource.

1. Q1.Which team role is responsible for ensuring that a quieter member's contribution reaches the group?

   • A.The team leader, who is accountable for the team's output
   • B.The champion, who sponsors the project and its resources
   • C.The member concerned, who must learn to assert their own contribution
   • D.The facilitator, who owns participation as part of the process✓ Correct answer

   Explanation

   Balanced participation is a process outcome, and the facilitator owns the process: agenda, airtime, decision method and conflict. Placing the burden on the quiet member is the primary foil, and it is the default that produces teams where the same three people decide everything.

   Reference: CSSBB BoK III

2. Q2.A CTQ tree for an online order process is partially built.
   
   $$\begin{array}{l|l|l} \textbf{Need} & \textbf{Driver} & \textbf{CTQ} \\ \hline \text{Easy to order} & \text{Few steps} & \text{Checkout in} \le 3 \text{ clicks} \\ \text{Easy to order} & \text{Clear pricing} & \text{Total shown before payment} \\ \text{Arrives on time} & \text{Fast dispatch} & \text{Reliable delivery} \\ \text{Arrives intact} & \text{Good packing} & \text{Damage rate} < 0.5\% \end{array}$$
   
   Which CTQ entry is inadequate, and why?

   • A.Reliable delivery, because it restates the need without a measure✓ Correct answer
   • B.Checkout in three clicks, because click count is not a customer concern
   • C.Total shown before payment, because it is a feature and not a measure
   • D.Damage rate below 0.5%, because no target date has been attached to it

   Explanation

   The bottom level of a CTQ tree must be specific and measurable, and \"reliable delivery\" is a restatement of the need one column to its left. Something like \"delivered by the promised date in 98% of orders\" would complete the branch. Restating the need in different words is by far the most common failure when building the tree.

   Reference: CSSBB BoK IV

3. Q3.A measurement system has a standard deviation of 0.08 mm against a tolerance of $\pm 0.60$ mm.
   
   What is the precision-to-tolerance ratio, and is the system adequate?

   • A.6.7%, so the system is excellent for this characteristic
   • B.80%, so the system is inadequate for this characteristic
   • C.40%, so the system is inadequate for this characteristic✓ Correct answer
   • D.13%, so the system is acceptable for this characteristic

   Explanation

   The tolerance is the full width, which is 1.20 mm, not the plus-or-minus figure:
   
   $$P/T = \frac{6(0.08)}{1.20} = 0.40 = 40\%$$
   
   Anything above 30% is normally unacceptable. Using 0.60 as the tolerance would give 80%, and omitting the factor of six would give 6.7%, so both common errors are represented among the options.

   Reference: CSSBB BoK V

4. Q4.A box plot compares cycle time across three shifts.
   
   What does the display indicate, and what should the team investigate first?

   

   • A.Day shift is the benchmark, so its method should be copied across the others
   • B.Night shift variation is far greater, so investigate consistency✓ Correct answer
   • C.Night shift is slower on average, so investigate why its median has risen
   • D.The three shifts perform equivalently, so no investigation is warranted here

   Explanation

   The medians differ by about three minutes while the night shift interquartile range is roughly three times the others, with a long upper tail and outliers. Comparing only averages would report the shifts as broadly similar and miss the real signal, which is that something on nights sometimes takes very much longer.

   Reference: CSSBB BoK V

5. Q5.A contingency table records on-time delivery by carrier.
   
   $$\begin{array}{l|c|c|c} & \textbf{On time} & \textbf{Late} & \textbf{Total} \\ \hline \text{Carrier X} & 168 & 32 & 200 \\ \text{Carrier Y} & 111 & 39 & 150 \\ \hline \text{Total} & 279 & 71 & 350 \end{array}$$
   
   What is the expected number of late deliveries for Carrier X if carrier and outcome are independent?

   • A.45.7
   • B.40.6✓ Correct answer
   • C.32
   • D.35.5

   Explanation

   Expected counts come from the marginal totals:
   
   $$E = \frac{200 \times 71}{350} = 40.6$$
   
   32 is the observed count, and the gap between 32 and 40.6 is precisely what the chi-square test evaluates. 35.5 would be the result of splitting the 71 late deliveries evenly between the two carriers, which ignores their different volumes.

   Reference: CSSBB BoK VI

6. Q6.A Pareto analysis of downtime causes gives the following.
   
   $$\begin{array}{l|c|c} \textbf{Cause} & \textbf{Minutes} & \textbf{Cumulative \%} \\ \hline \text{Changeover} & 480 & 40.0 \\ \text{Jams} & 300 & 65.0 \\ \text{Material wait} & 216 & 83.0 \\ \text{Breakdowns} & 120 & 93.0 \\ \text{Other} & 84 & 100.0 \end{array}$$
   
   The team proposes a breakdown reduction project. What is the strongest objection?

   • A.Breakdowns are only 10% of downtime, while the top two are 65%✓ Correct answer
   • B.Breakdowns cannot be reduced without capital investment in equipment
   • C.The Pareto principle requires that exactly 80% of causes be addressed
   • D.Downtime in minutes is the wrong unit and should be converted to cost

   Explanation

   Even eliminating breakdowns entirely would recover a tenth of the downtime, while changeover and jams together hold nearly two thirds. Converting to cost is a legitimate refinement and is the primary foil: it could in principle reorder the chart, and nothing in the stem suggests breakdown minutes are worth six times changeover minutes.

   Reference: CSSBB BoK VI

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