Charts QP-104 [PDF]

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Title:




 – S Chart



Problem:



The net weight (in oz) of a dry bleach product is to be monitored by
 &



s- chart using a sample size of n = 5. Data of 20 preliminary samples are as shown. Construct s- chart. Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Total



x1



x2



x3



x4



x5



X-bar



15.8 16.3 16.1 16.3 16.1 16.1 16.1 16.2 16.3 16.6 16.2 15.9 16.4 16.5 16.4 16 16.4 16 16.4 16.4



16.3 15.9 16.2 16.2 16.1 15.8 16.3 16.1 16.2 16.3 16.4 16.6 16.1 16.3 16.1 16.2 16.2 16.2 16 16.4



16.2 15.9 16.5 15.9 16.4 16.7 16.5 16.2 16.4 16.4 15.9 16.7 16.6 16.2 16.3 16.3 16.4 16.4 16.3 16.5



16.1 16.2 16.4 16.4 16.5 16.6 16.1 16.1 16.3 16.1 16.3 16.2 16.4 16.3 16.2 16.3 16.3 16.5 16.4 16



16.6 16.4 16.3 16.2 16 16.4 16.5 16.3 16.5 16.5 16.4 16.5 16.1 16.4 16.2 16.2 16.2 16.1 16.4 15.8



16.20 16.14 16.30 16.20 16.22 16.32 16.30 16.18 16.34 16.38 16.24 16.38 16.32 16.34 16.24 16.20 16.30 16.24 16.30 16.22 325.36



 = 3.93 / 20 = 0.1965 and



Control limits for s -chart  UCL = B4* =2.09*0.196 =0.419 CL = =0.196 LCL = B3  =0*0.196 =0




 = 325.36 / 20 = 16.268



S.D. 0.29 0.23 0.16 0.19 0.22 0.37 0.20 0.08 0.11 0.19 0.21 0.33 0.22 0.11 0.11 0.12 0.10 0.21 0.17 0.30 3.93



B3 =0, B4 =2.089 Sample Sample Number UCL CL LCL S.D. Number UCL CL LCL S.D. 1 0.410 0.196 0 0.292 11 0.410 0.196 0 0.207 2 0.410 0.196 0 0.230 12 0.410 0.196 0 0.327 3 0.410 0.196 0 0.158 13 0.410 0.196 0 0.217 4 0.410 0.196 0 0.187 14 0.410 0.196 0 0.114 5 0.410 0.196 0 0.217 15 0.410 0.196 0 0.114 6 0.410 0.196 0 0.370 16 0.410 0.196 0 0.122 7 0.410 0.196 0 0.200 17 0.410 0.196 0 0.100 8 0.410 0.196 0 0.084 18 0.410 0.196 0 0.207 9 0.410 0.196 0 0.114 19 0.410 0.196 0 0.173 10 0.410 0.196 0 0.192 20 0.410 0.196 0 0.303



0.45



s-chart



0.4 0.35 0.3



s . d



S.D.



0.25



UCL 0.2 CL 0.15



LCL



0.1 0.05 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20



sample no.



Conclusion: From the above chart, it can be concluded that all the observations are inside the control limits. Hence, the process variability is in control.



-chart: Control limits for  chart:



A3 1.427



X-BAR CHART UCL = + A3* LCL = - A3* CL = 16.55



16.27



15.99



Sample Sample Number UCL CL LCL X-bar Number UCL CL LCL X-bar 1 16.55 16.27 15.99 16.20 11 16.55 16.27 15.99 16.24 2 16.55 16.27 15.99 16.14 12 16.55 16.27 15.99 16.38 3 16.55 16.27 15.99 16.30 13 16.55 16.27 15.99 16.32 4 16.55 16.27 15.99 16.20 14 16.55 16.27 15.99 16.34 5 16.55 16.27 15.99 16.22 15 16.55 16.27 15.99 16.24 6 16.55 16.27 15.99 16.32 16 16.55 16.27 15.99 16.20 7 16.55 16.27 15.99 16.30 17 16.55 16.27 15.99 16.30 8 16.55 16.27 15.99 16.18 18 16.55 16.27 15.99 16.24 9 16.55 16.27 15.99 16.34 19 16.55 16.27 15.99 16.30 10 16.55 16.27 15.99 16.38 20 16.55 16.27 15.99 16.22



x bar-Chart 16.6 16.5 16.4 M E A N



16.3



mean



16.2



UCL



16.1 CL



16



LCL



15.9 15.8 15.7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20



sample no.



Conclusion: From the X-bar, S-chart it can be concluded that all the points of the data are inside the control limits. And all the observations are very close to the Central Limit. There are no assignable causes present in the data.



Practical Title:
 – Chart and MR Chart Problem -1: The viscosity of aircraft quality characteristic the product is produced in batches, & because each batch takes several hours to product, the production rate is to slow to allow sample sizes greater than one. The viscosity of the previous is batches are shown is bellow table. Batch no 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15



viscosity (x) Mr. (moving range) 33.75 33.05 0.7 34 0.95 33.81 0.19 33.46 0.35 34.02 0.56 33.68 0.34 33.27 0.41 33.49 0.22 33.2 0.29 33.62 0.42 33 0.62 33.54 0.54 33.12 0.42 33.84 0.72



Solution: We use the moving range chart to estimate the process variability when the production rate is very slow & sample size (n) > 1. The moving range is defined as; MR i = | xi – xi – 1|



x-bar MR (avg) S.D



33.52333 0.480714 0.335552



Control limit for MR chart: chart:



MR – CHART CL= 0.480714



UCL= D4* 1.57482 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15



M o v i n g



UCL 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494 1.570494



A v e r a g e



n=2, D3=0, D4=3.276



CL 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714 0.480714



LCL 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0



LCL= D3* 0 MR 0.7 0.95 0.19 0.35 0.56 0.34 0.41 0.22 0.29 0.42 0.62 0.54 0.42 0.72



MR-chart



1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0



MR UCL CL LCL



1



2



3



4



5



6



7



8



9



10 11 12 13 14 15



Batch Number



Conclusion:



Since none of the sample observation lies beyond the control limits of the



MR – chart, the process is in statistical control.



 – Chart Chart: Control limit for  D2=1.128,  =33.52,  =0.48, S=0.336 n=2 UCLMR= UCLS= CL= LCLS= LCLMR=



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15



UCL MR 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018 34.8018



X BAR- CHART /D2  + 3*  + 3*  XBAR  + 3*  /D2  - 3*



UCL s 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300 34.5300



CL 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233 33.5233



LCL s 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167 32.5167



34.80 34.53 33.52 32.52 32.24



LCL MR Viscosity 32.2448 33.75 32.2448 33.05 32.2448 34 32.2448 33.81 32.2448 33.46 32.2448 34.02 32.2448 33.68 32.2448 33.27 32.2448 33.49 32.2448 33.2 32.2448 33.62 32.2448 33 32.2448 33.54 32.2448 33.12 32.2448 33.84



 – Chart:  XBAR - CHART 35 34.5 34



R a n g e



33.5



UCLMR



33



UCLS



32.5



CL



32



LCLS



31.5



LCLMR



31



x



30.5 1



2



3



4



5



6



7



8



9



10



11



12



13



14



15



Sample Numbers



Conclusion:  – chart, the Since none of the sample observation lies beyond the control limits of the  process is in statistical control. There are no assignable causes present in the data.



Another Method:  – Chart: Control limit for  Chart: D2=1.128,  =33.52,  =0.48, S=0.336 n=2 UCLMR= CL= LCLMR=



X BAR- CHART /D2  + 3* XBAR /D2  - 3*



You can take only upper limit and lower limit for MR chart.



34.80 33.52 32.24



x



UCL CL LCL 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24 34.80 33.52 32.24



33.75 33.05 34.00 33.81 33.46 34.02 33.68 33.27 33.49 33.20 33.62 33.00 33.54 33.12 33.84



 – Chart: Draw  X-bar MR chart 35 34.5 34 UCL



33.5



CL



33 32.5



LCL



32



X-bar



31.5 31 30.5 1



2



3



4



5



6



7 8 9 10 11 12 13 14 15 Sample Number



Conclusion:  –MR chart, Since none of the sample observation lies beyond the control limits of the  the process is in statistical control. There are no assignable causes present in the data.



The variation exists due to the chance causes, there is no existence of assignable causes.



Practical Title: P Chart Problem -1: A sample of 50 TV voltage stabilizer checks the quality for 15 days & finds the fraction of non conforming units & number of defectives as follows; Construct the 3sigma trial control limit for p – Chart. Days No. of defect fraction defective (pi) 1 5 0.10 2 10 0.20 3 3 0.06 4 2 0.04 5 8 0.16 6 1 0.02 7 4 0.08 8 3 0.06 9 1 0.02 10 8 0.16 11 6 0.12 12 7 0.14 13 4 0.08 14 5 0.10 15 3 0.06



Solution:



Here, Total no. of. Defects are 70 & pi = 1.40  n m  = (1 - )



0.093 50 15 0.907



Control limit for P – Chart: P - chart  



UCL CL LCL



  3           3    



0.2168 0.0933 -0.0301



Sr.no 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15



UCL 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751 0.216751



CL 0.09333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333 0.093333



LCL 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0



p 0.1 0.2 0.06 0.04 0.16 0.02 0.08 0.06 0.02 0.16 0.12 0.14 0.08 0.1 0.06



Draw P chart: p-chart fraction defective



0.25 0.2 0.15



UCL



0.1



CL LCL



0.05



p 0 1



2



3



4



5



6



7



8



9 10 11 12 13 14 15



sample no.



Conclusion: Since none of the sample points lies beyond the control limits of the P chart, the process is in statistical control.



Practical Title: np - chart Problem:



Frozen orange juice concentrate is packed in 6-oz cardboard cans. These



cans are formed on a machine by spinning them from cardboard stock and attaching a metal bottom panel. By inspection of a can, we may determine whether, when filled, it could possibly leak either on the side seam or around the bottom joint. Such a nonconforming can has an improper seal on either the side seam or the bottom panel. We wish to setup a control chart to improve the fraction of nonconforming cans produced by this machine. To establish the control chart, 30 samples of n= 50 cans each were selected at half-hour intervals over a three-shift period in which the machine was in continuous operation. The data are follows: Number of Number of Number of Sample sample sample Nonconforming Nonconforming Nonconforming number number number cans, (Di) cans, (Di) cans, (Di) 1 2 3 4 5 6 7 8 9 10



12 15 8 10 4 7 16 9 14 10



n= m=  =  = n* = n* * =



11 12 13 14 15 16 17 18 19 20



5 6 17 12 22 8 10 5 13 11



21 22 23 24 25 26 27 28 29 30



50 30 0.2313 0.76867 11.5667 8.89091



20 18 24 15 9 12 7 13 9 6



Control limits for np -chart np -chart     3        n*     3       



UCL = CL = LCL = Sample No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15



UCL 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196



CL 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667



LCL Di 2.621377 12 2.621377 15 2.621377 8 2.621377 10 2.621377 4 2.621377 7 2.621377 16 2.621377 9 2.621377 14 2.621377 10 2.621377 5 2.621377 6 2.621377 17 2.621377 12 2.621377 22



Sample No. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30



UCL 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196 20.51196



20.51 11.57 2.621 CL 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667 11.56667



LCL 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377 2.621377



Di 8 10 5 13 11 20 18 24 15 9 12 7 13 9 6



30



np-chart



25 20 No of nonconforming 15 cans(Di)



UCL



10



LCL



CL



Di



5 0 1



3



5



7



9 11 13 15 17 19 21 23 25 27 29 Sample number



Conclusion: Since, corresponding observation no.15 & 23 are outside the control limits, the process is out of control. This suggests the presence of assignable causes of variation which should be traced & corrected as outliers are present. We will recalculate the control limits neglecting the Outliers.



ReRe-calculated control limits of npnp-chart:



UCL = CL = LCL = Sample No. 1 2 3 4 5 6 7



np -chart   3        n*     3       



UCL CL 19.46486 10.75 19.46486 10.75 19.46486 10.75 19.46486 10.75 19.46486 10.75 19.46486 10.75 19.46486 10.75



LCL Di 2.035142 12 2.035142 15 2.035142 8 2.035142 10 2.035142 4 2.035142 7 2.035142 16



Sample No. 15 16 17 18 19 20 21



18.53 10.03 1.537



UCL 19.46486 19.46486 19.46486 19.46486 19.46486 19.46486 19.46486



CL 10.75 10.75 10.75 10.75 10.75 10.75 10.75



LCL 2.035142 2.035142 2.035142 2.035142 2.035142 2.035142 2.035142



Di 8 10 5 13 11 20 18



8 9 10 11 12 13 14



19.46486 19.46486 19.46486 19.46486 19.46486 19.46486 19.46486



10.75 10.75 10.75 10.75 10.75 10.75 10.75



2.035142 2.035142 2.035142 2.035142 2.035142 2.035142 2.035142



9 14 10 5 6 17 12



22 23 24 25 26 27 28



19.46486 19.46486 19.46486 19.46486 19.46486 19.46486 19.46486



10.75 10.75 10.75 10.75 10.75 10.75 10.75



2.035142 15 2.035142 9 2.035142 12 2.035142 7 2.035142 13 2.035142 9 2.035142 6



Revised np-chart



25 20



UCL



15



CL



Di 10



LCL Di



5 0 1 2 3 4 5 6 7 8 9 10111213141516171819202122232425262728 Sample Numbers



Conclusion: Since, corresponding observation no. 21 is outside the control limits, the process is out of control. This suggests the presence of assignable causes of variation which should be traced & corrected. As outlier is present we will recalculate the control limits neglecting the Outlier. Re-calculated control limits of np-chart: UCL = CL = LCL = Sample No. UCL 1 19.01962 2 19.01962 3 19.01962 4 19.01962 5 19.01962



np -chart   3        n*     3        CL 10.40741 10.40741 10.40741 10.40741 10.40741



LCL Di 1.7952 12 1.7952 15 1.7952 8 1.7952 10 1.7952 4



Sample No. 14 15 16 17 18



17.64 9.367 1.09 UCL 19.01962 19.01962 19.01962 19.01962 19.01962



CL 10.40741 10.40741 10.40741 10.40741 10.40741



LCL 1.7952 1.7952 1.7952 1.7952 1.7952



Di 12 8 10 5 13



6 7 8 9 10 11 12 13



19.01962 19.01962 19.01962 19.01962 19.01962 19.01962 19.01962 19.01962



D i



20 18 16 14 12 10 8 6 4 2 0



10.40741 10.40741 10.40741 10.40741 10.40741 10.40741 10.40741 10.40741



1.7952 1.7952 1.7952 1.7952 1.7952 1.7952 1.7952 1.7952



7 16 9 14 10 5 6 17



19 20 21 22 23 24 25 26 27



19.01962 19.01962 19.01962 19.01962 19.01962 19.01962 19.01962 19.01962 19.01962



10.40741 10.40741 10.40741 10.40741 10.40741 10.40741 10.40741 10.40741 10.40741



1.7952 1.7952 1.7952 1.7952 1.7952 1.7952 1.7952 1.7952 1.7952



Revised np-chart



UCL CL LCL Di



1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627 Sample Numbers



Conclusion: From the above np-chart, all the points are inside the statistical control limits after re-revising the np-chart. Hence the process is in control. There are no assignable causes present in the data.



11 18 15 9 12 7 13 9 6



Practical Title: C- Chart Problem:



The following data represent the no. of non – conformities per 1000 Meter



in telephone cable from analysis of these data would you conclude that the process is in statistical control? What control procedure would you recommend for future production. Sample no.



Number of non-conformities



Sample no.



Number of non-conformities



1



1



12



6



2



1



13



9



3



3



14



11



4



7



15



15



5



8



16



8



6



10



17



3



7



5



18



6



8



13



19



7



9



0



20



4



10



19



21



9



11



24



22



20



Total non-conformities = 189  = 189 / 22 = 8.59091



Control limits for c - chart  c - chart UCL =



 + 3*(sqrt())



CL =







8.59091



LCL =



 - 3*(sqrt())



-0.2022



17.384



NOTE: When LCL value is in negative take LCL = 0.



1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22



UCL 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398 17.38398



CL LCL 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0 8.590909 0



No. of no conformities 1 1 3 7 8 10 5 13 0 19 24 6 9 11 15 8 3 6 7 4 9 20



C - CHART 30



n o n -



c o n f o r m i t i e s



25 20



UCL CL



15



LCL 10



NC



5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 sample no.



Conclusion: Since, corresponding observation no. 10, 11, 22 are outside to the control limits, the process is out of control. This suggests the presence of assignable causes of variation which should be traced & corrected. As outlier is present, we will recalculate the control limits neglecting the Outlier.



ReRe-calculated control limits of cc-chart: c - chart UCL =



 + 3*(sqrt())



CL =







LCL =



 - 3*(sqrt())



Sample no. 1 2 3 4 5 6 7 8 9



UCL 14.36 14.36 14.36 14.36 14.36 14.36 14.36 14.36 14.36



14.357 6.63158



CL 6.63 6.63 6.63 6.63 6.63 6.63 6.63 6.63 6.63



-1.094



LCL



c-bar 0 0 0 0 0 0 0 0 0



1 1 3 7 8 10 5 13 0



Sample no. 12 13 14 15 16 17 18 19 20 21



UCL 14.36 14.36 14.36 14.36 14.36 14.36 14.36 14.36 14.36 14.36



CL 6.63 6.63 6.63 6.63 6.63 6.63 6.63 6.63 6.63 6.63



C-chart



16 14 12 C 10 b 8



UCL CL



a 6 r 4



LCL c-bar



2 0 1



2



3



4



5



6



7



8 9 12 13 14 15 16 17 18 19 20 21 Sample Numbers



LCL c-bar 0 6 0 9 0 11 0 15 0 8 0 3 0 6 0 7 0 4 0 9



Conclusion: Since, corresponding observation no. 15 is outside the control limits, the process is out of control. This suggests the presence of assignable causes of variation which should be traced & corrected. As outlier is present, we will recalculate the control limits neglecting the Outlier.



ReRe-calculated control limits of cc-chart: c – chart



Sample no. 1 2 3 4 5 6 7 8 9



UCL =



 + 3*(sqrt())



CL =







6.16667



LCL =



 - 3*(sqrt())



-1.2832



UCL 13.62 13.62 13.62 13.62 13.62 13.62 13.62 13.62 13.62



CL 6.17 6.17 6.17 6.17 6.17 6.17 6.17 6.17 6.17



LCL



13.616



c-bar 0 0 0 0 0 0 0 0 0



1 1 3 7 8 10 5 13 0



Sample no. 12 13 14 16 17 18 19 20 21



UCL 13.62 13.62 13.62 13.62 13.62 13.62 13.62 13.62 13.62



CL 6.17 6.17 6.17 6.17 6.17 6.17 6.17 6.17 6.17



c-chart



16 14 No.of Nonconformities



LCL c-bar 0 6 0 9 0 11 0 8 0 3 0 6 0 7 0 4 0 9



UCL



12 10



CL



8 6



LCL



4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Sample No.



No.of Nonconformit ies



Conclusion: Since none of the sample points lies beyond the control limits of the C- chart, the process is in statistical control.



Practical Title: u - chart Problem:



A personal computer manufacturer wishes to establish a control chart for



nonconformities per unit on the final assembly line. The sample size is selected as five computers. Data on the no. of non- conformities in 20 samples of 5 computers each are shown in table.



Sample Number, i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20



Sample size n 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5



Average Number of Total Number of Nonconformities Nonconformities, xi per Unit, ui=xi/n 10 2 12 2.4 8 1.6 14 2.8 10 2 16 3.2 11 2.2 7 1.4 10 2 15 3 9 1.8 5 1 7 1.4 11 2.2 12 2.4 6 1.2 8 1.6 10 2 7 1.4 5 1 193 38.6



 = 38.6 / 20 = 1.93



Control limits for u - chart  u -bar chart UCL =



 + 3*(sqrt( /))



3.79







1.93



 + 3*(sqrt( /))



0.07



CL = LCL = Sample No. (i) 1 2 3 4 5 6 7 8 9 10



UCL 3.79 3.79 3.79 3.79 3.79 3.79 3.79 3.79 3.79 3.79



CL 1.93 1.93 1.93 1.93 1.93 1.93 1.93 1.93 1.93 1.93



o f u



ui



0.07 0.07 0.07 0.07 0.07 0.07 0.07 0.07 0.07 0.07



2 2.4 1.6 2.8 2 3.2 2.2 1.4 2 3



Sample UCL CL LCL No. (i) 11 3.79 1.93 0.07 12 3.79 1.93 0.07 13 3.79 1.93 0.07 14 3.79 1.93 0.07 15 3.79 1.93 0.07 16 3.79 1.93 0.07 17 3.79 1.93 0.07 18 3.79 1.93 0.07 19 3.79 1.93 0.07 20 3.79 1.93 0.07



ui 1.8 1 1.4 2.2 2.4 1.2 1.6 2 1.4 1



u-chart



4.00 M e a n



LCL



3.50 3.00 2.50



UCL



2.00



CL



1.50



LCL



1.00



ui



0.50 0.00 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sample number



Conclusion: Since none of the sample no. lies beyond control limits of the u –chart, the process is in statistical control.