Final Mock Exam-Solution

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1 ISEN 350: 2023 Fall Semester Final Mock Exam (1-4) Multiple Choice Questions: Select the correct answer: 1. Within variability is always larger than overall variability. A) True B) False 2. When the process variation increases, how will the potential capability change? A) Increases B) Decreases C) Remain the same D) There is no specific pattern in the capability index change. 3. When the process variation increases, how will the actual capability change? A) Increases B) Decreases C) Remain the same D) There is no specific pattern in the capability index change. 4. When a control chart has a larger probability of false alarm, 𝐴𝐴𝐴𝐴𝐿𝐿 0 is: A) Smaller B) Larger C) The same D) There is no specific pattern in the ARL0
2 (5-15) In a precision machining facility, a team is tasked with manufacturing cylindrical metal rods used in aerospace applications. To meet the stringent requirements of the aerospace industry, the rods must have a diameter within the specification limits of 49.4 mm to 50.6 mm, and the target diameter is 50 mm. The manufacturer plans to construct 𝑋𝑋 � chart and S chart. (5-7) The quality control team collected two samples at time T=1 and T=2, each of sample size three. The diameters (in mm) of these rods are recorded in the table below: Sample X1 X2 X3 𝑋𝑋 � 𝑠𝑠 T=1 50.1 50.5 49.8 50.13 0.3512 T=2 49.9 50.7 50.4 50.33 0.4041 Mean 50.23 0.3777 5. What is the estimated mean of 𝑋𝑋 � (i.e., : 𝐸𝐸 [ 𝑋𝑋 � ] )? 50.23 6. What is the estimated mean of 𝑠𝑠 : E[s]? 0.3777 7. What is the estimated standard deviation of 𝑋𝑋 , 𝜎𝜎 � ? 𝜎𝜎 � = 𝑠𝑠̅ 𝑐𝑐 4 = 0.3777 0.8862 = 0.4262 (8-15) Suppose the quality control team has collected more data over a month using samples of size 3 and estimated the individual measurement X’s mean ( 𝑋𝑋 � ̅ ) as 50.1 and standard deviation ( 𝜎𝜎 � ) as 0.3. Use three sigma control limits. 8. Calculate the Lower Control Limit (LCL) of 𝑋𝑋 � chart. 𝑋𝑋 � � − 3 𝜎𝜎 � √𝑛𝑛 = 50.1 − 3 × 0.3 √ 3 = 49.5804 9. Calculate the Lower Control Limit (LCL) of 𝑠𝑠 chart. LCL=0 Because 𝑐𝑐 4 𝜎𝜎 � − 3 � 1 − 𝑐𝑐 4 2 𝜎𝜎 � = − 0.5037 × 0.4262 < 0 10. All points on both control charts fall between the control limits (In other words, the process is in control). What is the lower natural tolerance limit of the process? 𝑋𝑋 � � − 3 𝜎𝜎 � = 50.1 − 0.9 = 49.2
3 11. What is the probability that a rod is nonconforming (fallout probability)? 1 − 𝑃𝑃 (49.4 < 𝑋𝑋 < 50.6) where 𝑋𝑋 ~ 𝑁𝑁 ( 𝜇𝜇 = 50.1, 𝜎𝜎 = 0.3) 0.0576 (12-13) Obtain Potential Capability: 12. What formula do you need to use? A) min � 𝑈𝑈𝑈𝑈𝑈𝑈−𝜇𝜇 3𝜎𝜎 , 𝜇𝜇−𝑈𝑈𝑈𝑈𝑈𝑈 3𝜎𝜎 � B) 𝑼𝑼𝑼𝑼𝑼𝑼−𝑼𝑼𝑼𝑼𝑼𝑼 𝟔𝟔𝟔𝟔 C) Φ −1 (1 − 𝑝𝑝 ) D) 𝑈𝑈𝑈𝑈𝑈𝑈−𝑈𝑈𝑈𝑈𝑈𝑈 6�𝜎𝜎 2 + ( 𝜇𝜇−𝑇𝑇 ) 2 E) 𝜎𝜎 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝜎𝜎 𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑤𝑤𝑖𝑖 13. Calculate the value: 50.6 − 49.4 6 × 0.3 (14-15)Obtain Taguchi Index: 14. What formula do you need to use? A) min � 𝑈𝑈𝑈𝑈𝑈𝑈−𝜇𝜇 3𝜎𝜎 , 𝜇𝜇−𝑈𝑈𝑈𝑈𝑈𝑈 3𝜎𝜎 � B) 𝑈𝑈𝑈𝑈𝑈𝑈−𝑈𝑈𝑈𝑈𝑈𝑈 6𝜎𝜎 C) Φ −1 (1 − 𝑝𝑝 ) D) 𝑼𝑼𝑼𝑼𝑼𝑼−𝑼𝑼𝑼𝑼𝑼𝑼 𝟔𝟔�𝟔𝟔 𝟐𝟐 + ( 𝝁𝝁−𝑻𝑻 ) 𝟐𝟐 E) 𝜎𝜎 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝜎𝜎 𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑤𝑤𝑖𝑖 15. Calculate the value: 50.6 − 49.4 6 × √ 0.3 2 + (50.1 − 50) 2 = 0.6325
4 (16-18) A control chart for the number of nonconforming is to be established, based on samples of size 40. To start the control chart, 30 samples were selected and the number nonconforming in each sample was determined, yielding ∑ 𝐷𝐷 𝑖𝑖 30 𝑖𝑖=1 = 150 . Use three sigma control limits. 16. If X is defined as the number of nonconforming in each sample, what distribution does X follow? A) Poisson Distribution (Parameter 𝜆𝜆 ) B) Binomial Distribution (Parameter 𝒏𝒏 , 𝒑𝒑 ) C) Geometric Distribution (Parameter 𝑝𝑝 ) D) Normal Distribution (Parameter 𝜇𝜇 , 𝜎𝜎 ) 17. What is the estimated p? 𝑝𝑝 = 150 30 × 40 = 0.125 18. What is the Lower Control Limit (LCL) of the nonconforming chart (np-chart)? LCL=0 Because 𝑛𝑛 = 40 , 𝑛𝑛𝑝𝑝 − 3 �𝑛𝑛𝑝𝑝 (1 − 𝑝𝑝 ) = − 1.2750 (19-20). Suppose after a month of data collection (using a sample size of 40) from in control process, you estimated the fraction of the nonconforming as 0.051. However, the process fraction nonconforming now shifted to 0.15 (out-of-control). 19. What is the probability that the shift would be detected on the first subsequent sample? Because 𝑛𝑛 = 40 , 𝑝𝑝 = 0.051, 𝑈𝑈𝑈𝑈𝐿𝐿 = 𝑛𝑛𝑝𝑝 + 3 �𝑛𝑛𝑝𝑝 (1 − 𝑝𝑝 ) = 6.2142 𝐿𝐿𝑈𝑈𝐿𝐿 = 0 because 𝑛𝑛𝑝𝑝 − 3 �𝑛𝑛𝑝𝑝 (1 − 𝑝𝑝 ) < 0 𝑃𝑃 ( 𝑋𝑋 > 𝑈𝑈𝑈𝑈𝐿𝐿 ) = 𝑃𝑃 ( 𝑋𝑋 > 6.2142) = 1 − 𝑃𝑃 ( 𝑋𝑋 ≤ 6.2142) where 𝑋𝑋 ~ 𝐵𝐵𝐵𝐵𝑛𝑛𝐵𝐵 ( 𝑛𝑛 = 40, 𝑝𝑝 = 0.15) 𝑃𝑃 ( 𝑋𝑋 > 𝑈𝑈𝑈𝑈𝐿𝐿 ) = 0.3933 20. Average Run Length? 1/0.3933
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