CQPA Free Practice Questions (Review Mode)

Free sample questions

3 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 check sheet recorded defects by shift over one week, with equal production on each shift.
   
   $$\begin{array}{l|c|c|c|c} \hline \textbf{Shift} & \textbf{Scratch} & \textbf{Dent} & \textbf{Short fill} & \textbf{Total} \\ \hline \text{Day} & 12 & 4 & 9 & 25 \\ \text{Evening} & 14 & 5 & 8 & 27 \\ \text{Night} & 13 & 31 & 10 & 54 \\ \hline \end{array}$$
   
   What does this stratification reveal?

   • A.Scratches are the dominant problem across all shifts
   • B.The three shifts perform equivalently once totals are compared
   • C.One defect type on one shift accounts for the difference✓ Correct answer
   • D.Night shift performs worse across every defect type

   Explanation

   Night shift's scratch and short fill counts sit alongside the other shifts; the entire gap comes from dents, at 31 against 4 and 5. Concluding that night shift is worse overall is the primary foil, and it is exactly the wrong conclusion to act on, because it would send the investigation to the shift as a whole instead of to the one dent-producing condition.

   Reference: CQPA BoK II

2. Q2.Two machines were sampled and summarized below.
   
   $$\begin{array}{l|c|c} \hline \textbf{Machine} & \textbf{Mean} & \textbf{Std dev} \\ \hline \text{M1} & 50.0 & 0.8 \\ \text{M2} & 50.0 & 2.4 \\ \hline \end{array}$$
   
   The specification is $50.0 \pm 2.0$. What does this comparison indicate?

   • A.Only M1 can meet the specification consistently✓ Correct answer
   • B.Both machines will produce equivalent output, since the means match
   • C.M2 is preferable because its wider spread covers more of the tolerance
   • D.Neither machine can meet the specification at these settings

   Explanation

   Both are centered, so the tolerance is met or missed on spread alone: M1's natural spread of $\pm 3\sigma$ is $\pm 2.4$ against a tolerance of $\pm 2.0$, while M2's is $\pm 7.2$, three times the tolerance width. Equal means is the primary foil, and the item exists to make the point that centering without adequate spread control does not deliver conforming product.

   Reference: CQPA BoK III

3. Q3.A hypothesis test comparing two fixtures produced the output below.
   
   $$\begin{array}{l|c} \hline \textbf{Quantity} & \textbf{Value} \\ \hline \text{Significance level } \alpha & 0.05 \\ p\text{-value} & 0.18 \\ \text{Sample size per group} & 12 \\ \hline \end{array}$$
   
   What is the correct conclusion?

   • A.Reject the null hypothesis, because the p-value is below alpha
   • B.Do not reject the null hypothesis, because the p-value exceeds alpha✓ Correct answer
   • C.Reject the null hypothesis, because the p-value exceeds alpha
   • D.Do not reject the null hypothesis, because the p-value is below alpha

   Explanation

   The null hypothesis is rejected when the p-value falls below the significance level, and $0.18$ is well above $0.05$, so it is not rejected. Both halves of the answer must be right: two options pair a correct decision with a false statement about the comparison, which is what makes this item discriminate.

   Reference: CQPA BoK III

Practise CQPA the way the exam actually asks it

• 25 exam-style CQPA questions
• Full explanation for every question
• Mobile friendly, no install required
• Instant feedback as you answer

Questions about this practice test

Related CQPA practice tests

• CQPA Full 100 Randomly Drawn Questions Mock Exam (240 Minutes)100 practice questionsExam · 240 minView quiz→
• CQPA Free Mock Exam (60 Minutes)25 practice questionsExam · 60 minFreeView quiz→
• CQPA - 1. Team Dynamics and Processes25 practice questionsReviewView quiz→
• CQPA - 2. Basic and Advanced Quality Tools25 practice questionsReviewView quiz→