CSSYB - 3. Measure Phase

Free sample questions

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

1. Q1.A plant reports mean cycle time separately for three shifts. The shifts did not produce the same number of units.
   
   $$\begin{array}{l|c|c} \textbf{Shift} & \textbf{Units produced} & \textbf{Mean cycle time (min)} \\ \hline \text{A} & 120 & 8.0 \\ \text{B} & 60 & 11.0 \\ \text{C} & 20 & 14.0 \end{array}$$
   
   What is the mean cycle time across all units produced that day?

   • A.9.5 minutes✓ Correct answer
   • B.11.0 minutes
   • C.8.0 minutes
   • D.12.5 minutes

   Explanation

   Three shift means cannot be averaged directly when the shifts made different numbers of units. Each mean has to be weighted by its own unit count:
   
   $$\bar{X} = \frac{\sum n_i \bar{X}_i}{\sum n_i}$$
   
   Substituting the shift figures:
   
   $$\bar{X} = \frac{120(8.0) + 60(11.0) + 20(14.0)}{200} = \frac{1900}{200} = 9.5$$
   
   11.0 minutes is the trap. It is the plain average of the three shift means, which treats the 20-unit shift as equally important as the 120-unit shift. 12.5 minutes applies the same weights to the wrong shifts.

   Reference: CSSYB BoK III

2. Q2.The histogram below shows order fulfillment time for 200 orders.
   
   Which ordering of the three measures of central tendency is correct for this distribution?

   

   • A.Mean is lowest, median is in the middle, mode is highest
   • B.Mean, median and mode all fall at the same value
   • C.Mode is lowest, median is in the middle, mean is highest✓ Correct answer
   • D.Median is lowest, mode is in the middle, mean is highest

   Explanation

   The long tail runs to the right, so this distribution is right skewed. The mode is the tallest bar, so it sits at the peak, which is on the left. The mean uses every value, so the few very long fulfillment times in the tail pull it to the right, away from the peak. The median lands between the two. The order is mode, then median, then mean. Reverse the picture and the order reverses too: in a left skewed distribution the mean is lowest. Only a symmetric distribution puts all three at the same value.

   Reference: CSSYB BoK III

3. Q3.A gauge is checked against three certified reference standards. Each standard is measured ten times and the average reading is recorded.
   
   $$\begin{array}{c|c|c} \textbf{Reference (mm)} & \textbf{Average reading (mm)} & \textbf{Bias (mm)} \\ \hline 2.00 & 2.40 & +0.40 \\ 5.00 & 5.40 & +0.40 \\ 8.00 & 8.40 & +0.40 \end{array}$$
   
   What does this result show?

   • A.A linearity problem, because bias appears at every reference value
   • B.A constant bias across the range, with acceptable linearity✓ Correct answer
   • C.A stability problem, because the readings differ from the references
   • D.A repeatability problem, because ten readings were needed at each point

   Explanation

   Bias is the difference between the average reading and the true value. Here it is $+0.40$ mm at every reference, so the gauge reads high by the same amount everywhere. Linearity asks a different question: does the bias change as you move across the operating range? It does not change here, so linearity is fine. A gauge with good linearity and a constant offset is the easy case, because a calibration adjustment removes the whole $0.40$ mm. Stability would require readings taken at different times, and repeatability would require looking at the spread of the ten readings rather than their average.

   Reference: CSSYB BoK III

4. Q4.The data collection form below was used by inspectors on a painted door panel line for one week.
   
   What does this form show the team that a plain defect count would not?

   

   • A.Which defect type occurs most often, ranked from largest to smallest
   • B.Where on the part the defects occur, which points toward a physical cause✓ Correct answer
   • C.How the defect rate changed from one day to the next during the week
   • D.Whether the difference between the inspectors is statistically significant

   Explanation

   This is a defect location check sheet, sometimes called a concentration diagram. The marks are made on a picture of the part, so the pattern of where defects land is the data. Fifteen of the eighteen defects in one corner is a strong physical clue: something contacts, drips onto or masks that corner. A running total of 18 defects would never reveal it. Ranking defect types is what a Pareto chart does, and changes over time need a time axis, which this form does not have.
   

   Reference: CSSYB BoK III

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