Question

1. ABC company is producing printed circuit boards (PCBs) used in laptop computers. The associated manufacturing process which is in 3s-control seems to fluctuate in quality recently. The following table presents the number of nonconformities observed in 26 successive samples of 100 PCBs over time. One nonconformity represents one defect on a PCB. But one PCB may have more than one nonconformity.   

Sample number	Number of conformities	Sample Number	Number of conformities
1	21	14	19
2	24	15	10
3	16	16	17
4	12	17	13
5	15	18	22
6	5	19	18
7	28	20	39
8	20	21	30
9	31	22	24
10	25	23	16
11	20	24	19
12	24	25	17
13	16	26	15

Questions

a. What control chart should ABC use to monitor the PCB manufacturing process if the company wants to monitor the number of nonconformity over time.             

b. What are the upper and lower control limits? (MUST show the control chart you choose and demonstrate the calculations of the limits)                                 

c. Which sample(s) shows the process is out of control?                           

d. What are the new control limits after removing the samples identified in #c ? (MUST show the new control chart)                                                              

Answer 1

Answer:

a. p chart is the right chart to monitor the manufacturing process for non confromity.

b. mean norn conforming units =516/(100*26) =0.1984

UCLp = pbar + 3 * [ pbar*(1-pbar)/n]^0.5 = 0.1984+0.1196 = 0.318

LCLp = pbar - 3 * [ pbar*(1-pbar)/n]^0.5 =0.0787

c. It can be seen that there are only two samples out of control limits, which are sample number 6 and sample number 20. No. 6 is not a non conformity as there is no harm with defects less than required.

d. Removing the two samples, the remaining 24 samples will show the values

pbar = 472/24*100 =0.1966

UCLp = 0.1966 + 3*[ 0.1966*(1-0.1966)/100]^0.5 =0.1966+0.1122 =0.3088

LCLp = 0.1966 - 3*[ 0.1966*(1-0.1966)/100]^0.5 =0.1966-0.1122 =0.0773

It can be seen that now, one more sample i.e. sample 9 is out of control.

If this sample is also removed, the new limits will be

pbar =441 /23*100 =0.1917

UCLp =0.1917+3*[ 0.1917*(1-0.1917)/100}^0.5 = 0.3097

LCLP = 0.1917-3*[ 0.1917*(1-0.1917)/100}^0.5 = 0.0736

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