10. The quality manager of a steel sheet production company wants to construct a
ID: 361367 • Letter: 1
Question
10. The quality manager of a steel sheet production company wants to construct a chart for determining if the 3 leveling machines are under control with regard to the defect of visible surface scratches. What is the estimate of the standard deviation of the sampling distribution of sample proportions for whenever this process is under control considering below results if he inspects 600 sheets from each machine?
11- What are the control chart upper and lower control limits for an alpha risk of .02?
Leveliing Machine Total Defectives A 25 B 17 C 30Explanation / Answer
Since alpha risk = 0.02 , it leaves 0.01 in each tail
That corresponds to Z value for probability of = 1 – 0.01 = 0.99
Thus Z value for probability of 0.99 = NORMSINV ( 0.99) = 2.3263
Given are :
Sample size = n = 600 ( sheets from each machine)
Total number of defective sheets = 25 + 17 + 30 = 72
Therefore,
Proportion of defective sheets, Pbar
= Total number of defective sheets / ( Number of machines x Number of defective sheets/ machine)
= 72/ ( 3 x 600)
= 72/1800
= 0.04
Therefore ,
Estimate of Standard deviation
= Sd
= Square root ( pbar x ( 1 – pbar)/ n)
= Square root ( 0.04 x 0.96 / 600)
= 0.008
ESTIMATE OF STANDARD DEVIATION = 0.008 DEFECTIVES
Control Limits for Control charts :
Upper Control Limit = UCL = Pbar + Z x Sd = 0.04 + 2.3263 x 0.008 = 0.04 + 0.0186 = 0.0586
Lower Control Limit = LCL = Pbar – Zx Sd = 0.04 – 2.3263 x 0.008 = 0.04 – 0.0186 = 0.0214
Thus Control Limits for proportion of defectives will be : 0.0214 – 0.0586
Following are the proportion of defectives for each machine :
Machine A = 25/600 = 0.0416
Machine B = 17/ 600 = 0.0283
Machine C = 30/600 = 0.05
Note: Alternately, an np chart also can be drawn which would set control limits for number of defectives instead of proportion of defectives as done in above. The centreline in case of np chart would be “n.Pbar” instead of “pbar” as in P chart. Similarly, Standard deviation would be calculated as “Square root (n x pbar x ( 1 – pbar)) “ instead of “Square root ( pbar x ( 1 – pbar)/n) as done in p chart
ESTIMATE OF STANDARD DEVIATION = 0.008 DEFECTIVES
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.