A textbook of Computer Based Numerical and Statiscal Techniques part 50 pptx

476 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. Mean Chart Mean of 10 sample mean 442 44.2 10 10 x X === ∑ Mean Range of 10 sample ranges 58 5.8 10 10 R R === ∑ As we have, for 5,n = 2 0.58 A = , 3 0, D = 4 2.115 D = 3-σ control limits for x – chart are: 2 2 UCL 44.2 0.58 5.8 47.567 LCL 44.2 0.58 5.8 40.836 CL 44.2 x x x XAR XAR X =+ =+×= =− =−×= == Range Chart: 3-σ Control Limits for R chart are: 4 3 UCL 2.115 5.8 12.267 LCL 0 5.8 0 CL 5.8 R R R DR DR R == ×= ==×= == 60 50 40 30 20 10 0 Sample Mean Mean Chart 0 2 4 6 10 128 Sample Number FIG. 11.17 From mean chart we see that 2nd, 3rd, 6th and 7th samples lies outside the control limits. Hence the process is out of control. This shows that some assignable causes of variation are operating which should be detected and removed. STATISTICAL QUALITY CONTROL 477 9 8 7 6 5 4 3 2 1 0 Sample Range 0 2 4 6 8 10 12 Sample Number FIG. 11.18 Since all the points with in the control limits. Hence the process is in statistical control. Example 2. The following are the mean lengths and ranges of lengths of a finished product from 10 samples each of size 5. The specification limits for length are ±200 5 cm . Construct x – and R-chart and examine whether the process is under control and state your recommendation. Sample No. 12345678910 x 201 198 202 200 203 204 199 196 199 201 R 5073472856 Assume for n = 5, A 2 = 0.577, D 3 = 0, D 4 = 2.115. Sol. In given problem specification limits for length are given 200 ± 5 cm. Hence standard deviation is unknown. (1) Control Limits for x – -chart are: Central limit, CL x = µ = 200 2 UCL 200 0.577 4.7 x AR =µ+ = + × 202.712= ; 47 10 10 R R == ∑ 2 LCL 200 0.577 4.7 x AR =µ− = − × = 197.288 4.7R = (2) Control limits for R-Chart are: 4 3 UCL 9.941 2.115 4.7 CL 4.7 LCL 0 0 4.7 R R R DR R DR ===× == == =× from control charts for mean and range, the process is in statistical control in R — -Chart because all points lies with in the control limits where as in x — -chart, process is out of control because sample 5, 6 and 8 lies outside the control limits. The process therefore should be halted to check 478 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES whether there are any assignable causes. If assignable causes found, the process should be re-adjusted to remove assignable cause. 205 204 203 202 201 200 199 198 197 196 195 Sample Mean 0 2 4 6 8 10 12 Sample Number Mean Chart FIG. 11.19 Range Chart 10 8 6 4 2 0 Sample Range 0 5 10 15 Sample Number FIG. 11.20 Example 3. In a glass factory, the task of quality control was done with the help of mean (x — ) and standard deviation σ charts. 18 samples of 10 items each were chosen and then values ∑X and ∑S were found to be 595.8 and 8.28 respectively. Determine the 3-σ limits for mean and standard deviation chart. Given that n = 10, A 1 = 1.03, B 3 = 0.28, B 4 = 1.72, ∑S = 8.28. Sol. No. of samples 18 S — = ∑ 18 S = 8.28 18 = 0.46 hence, 3-σ control limits for standard deviation chart are: UCL – S = B 4 .S — = 1.72 × 0.48 = 0.7912 LCL S – = B 3 .S — = 0.28 – 0.46 = 0.1288 CL S – = 0.46 3-σ control limits for mean chart (x — ) are: X — = ∑ 18 x = 595.8 18 = 33.1 STATISTICAL QUALITY CONTROL 479 UCL X — = x — + A 1 σ = 33.1 + 1.03 × 0.46 UCL X — = 33.57 LCL X — = x — – A 1 σ = 33.1 – 1.03 × 0.46 LCL X — = 32.63 CL X — = 33.1. Example 4. If the average fraction defective of a large sample of a product is 0.1537, calculate the control limits when subgroup size is 2,000. Sol. Here, Sample size n = 2,000 for each sample Average fraction defective = 0.1537 i.e., P = 0.1537 ⇒ Q =1 – P = 1 – 0.1537 Q = 0.8463 Hence, 3–σ control limits for P-Chart are : ± 3 PQ P n UCL P = × + 0.1537 0.8463 0.1537 3 2, 000 UCL P = 0.1537 + 0.02418 = 0.17788 LCL P = × − 0.1537 0.8463 0.1537 3 2, 000 LCL 0.1537 0.02418 0.12952 P =− = CL 0.5137 P = . Example 5. The following data gives the number of defectives in 10 independent samples of varying sizes from a production process. Sample no. 12345678910 Sample size 2000 1500 1400 1350 1250 1760 1875 1955 3125 1575 No. of defectives 425 430 216 341 225 322 280 306 337 305 Draw the control chart for fraction defective. Sol. (In problem 4 sample size is fixed whereas in this problem sample size is variable) Since it is a problem of variable sample size so control chart for fraction defective can be drawn in two ways. (1) By first way, we set up two sets of control limits, one based on the maximum sample size, 3125n = and the second based on minimum sample size 1, 250.n = (a) For 3, 125;n = UCL 0.200,= LCL 0.159= (b) For 1, 250;n = UCL 0.212,= LCL 0.147= 480 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 12 10 8 6 4 2 0 Number of Defectives 0 100 200 300 400 500 Sample Number FIG. 11.21 Control Chart for Fraction Defective Since there are 4 points lies outside (based on minimum sample size) of control limits, so process is of out of control. (2) By second way, 3- σ limit for each sample separately obtained by using formula 3 PQ P n ± where Total no. of defectives Total sample size d P n == ∑ ∑ and n is corresponding sample size. 3187 0.1791 1 0.8209 17790 d PQP n ∑ == = ⇒=−= ∑ ∴ () PQ = 0.1791 × 0.8209 = 0.1470231 nn nn n dd dd d P = d/nP = d/n P = d/nP = d/n P = d/n 1/n1/n 1/n1/n 1/n P Q n P Q n 3 × 3 × 3 × 3 × 3 × P Q n UCLUCL UCLUCL UCL LCLLCL LCLLCL LCL 2000 425 0.2125 0.0005 0.000735 0.008573 0.025719 0.205 0.153 1500 430 0.2867 0.00066 0.000098 0.009899 0.029698 0.209 0.149 1400 216 0.1543 0.00071 0.000105 0.010247 0.030741 0.210 0.148 1350 341 0.2526 0.00074 0.000109 0.010440 0.031321 0.210 0.148 1250 225 0.1800 0.00080 0.000118 0.010863 0.032588 0.212 0.147 1760 322 0.1829 0.00057 0.000084 0.009138 0.027413 0.207 0.152 1875 280 0.1495 0.00053 0.000078 0.008854 0.026562 0.206 0.153 1995 306 0.1565 0.00051 0.000075 0.008672 0.026015 0.205 0.153 3125 337 0.1078 0.00032 0.000047 0.006856 0.020567 0.200 0.159 1575 305 0.1937 0.00063 0.000093 0.009659 0.028977 0.0208 0.150 17790 3187 STATISTICAL QUALITY CONTROL 481 500 400 300 200 100 0 Number of Defectives 0 1000 2000 3000 4000 Sample Size Sample points corresponding to sample no. 1, 2, 4, 7 and 9 lie outside the control limits. Hence, process is out of control. FIG. 11.22 Example 6. A daily sample of 30 items was taken over a period of 14 days in order to establish attributes control limits. If 21 defectives were found, what should be upper and lower control limits of the proportion of defectives? Sol. Since a sample of 30 items is taken daily over a period of 14 days. Total No. of items inspected = 30 × 14 = 420 No. of defective found = 21 n = 30 ∴Average fraction defective P — = 21 420 = 0.05 ∴ UCL P = 3 PQ P n + where 1QP=− = () () 1 0.05 0.95 30.053 30 PP P n − × +=+ UCL P = 0.05 + 3 × 0.0398 UCL P = 0.1694 LCL P = () 1 3 PP P n − − = 0.05 – 0.1194 < 0 (negative) ∴ LCL P = 0. Example 7. The past record of a factory using quality control melthods show that on the average 4 articles produced are defective out of a batch of 100. What is the maximum number of defective articles likely to be encountered in the batch of 100, when the production process is in a state of control? Sol. n = Sample size = 400 P = Process fraction defective = 4 100 = 0.04 Q = 1 – P = 0.96 482 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Let d be the number of defectives in a sample size of n. i.e., np. The 3–σ limit for number of defectives are given by () () 3.Ed sEd ± or 3np nPQ± 400 0.04 3 400 0.04 0.96=× ± ×× 16 3 15.36 16 3 3.9192=± =±× 16 11.7576=± () 4.2424, 27.7576 = Therefore if the production process is in a statistical control, the number of defective items to be encountered in a batch of 400 should lie within the control limits, viz. (4.2424, 27.7576), i.e., (4, 28). Hence the maximum number of defective items in this batch is 28. Example 8. In a blade manufacturing factory, 1000 blades are examined daily. Following information shows number of defective blades obtained there. Draw the np-chart and give your comment? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . 910128715101210 8 7 13141516 Date No of Defective Sol. Here 10000,n = 15k = (sample no.) If P denotes the fraction defectives produced by the entire process then 166 0.011 15 1000 P P kn ∑ == = × ∴ 1000 0.011 11np =× = Hence control limits are () () () CL 11 UCL 3 (1 ) 11 3 11 1 0.011 UCL 20.894 LCL 3 1 11 3 11 1 0.011 LCL 1.106 np np np p np np p == == −− =+ −− = =− −− =− −− = Since all the 15 points lies within the control limits, the process is under control. STATISTICAL QUALITY CONTROL 483 20 15 10 5 0 No. of defectives 0 5 10 15 20 Date FIG. 11.23 Example 9. The number of mistakes made by an accounts clerk is given below: Week 1234567891011121314151617181920 No.ofMistakes10201010123310071010 Establish a suitable control chart and state how it should be used in future in order to control the mistakes of the clerk. Sol. The control chart to be used for the given problem is the number of defects chart i.e., C-chart. Average no. of mistakes. c — = 24 1.2 20 20 C∑ == Thus the control limits for c — -chart are; (i) UCL = 3cc+ = 1.2 3 1.2+ = 4.49 (ii) CL = c — = 1.2 (iii) LCL = 3cc− = 1.2 3 1.2− = 2.09 ≈ 0 3 The number of mistakes during the 16th week lies outside the UCL the process is not under control. Now to establish the suitable control chart for future, we homogenize the data for future control by eliminating the data corresponding to the 16th week. 17 0.895. 19 new C == Hence the revised control limits for c chart are: UCL 3 0.895 3 0.895 3.73 LCL 3 0.895 3 0.895 1.94 0 17 CL 0.895. 19 cc cc C =+ = + = =− = − =− ≅ == = 484 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES So the revised C-chart for revised control limit is in statistical control, i.e., all the points lies within the control limits. 8 6 4 2 0 No. of Mistakes Week FIG. 11.24 Example 10. During the examination of equal length of cloth, the following are the number of defects observed. 2340567432 Draw a control chart for the number of defects and comment whether the process is under control or not? Sol. Let the no. of defects per unit (equal length) be denoted by c. The average no. of defects in 10 samples 36 3.6 20 10 c c ∑ === Hence 3–σ limit for c-chart are: 3cc± 3.6 3 3.6=± 3.6 3 1.8974=±× 3.6 5.6922=± UCL C – = 3.6 + 5.6922 = 9.2922 LCL C – = 3.6 – 5.6922 = – 2.0922 ≈ 0 CL C – = 3.6 (LCL C – = 0 because no. of defects per unit cannot be negative) STATISTICAL QUALITY CONTROL 485 8 6 4 2 0 Number of Defect 0 2 4 6 8 10 12 Sample Number FIG. 11.25 Since all the points are within the control limits therefore the process is in statistical control. Example 11. An automobile producer wishes to control the number of defects per automobile. The data for 16 such automobiles is shown below: Sample No. 12345678910111213141516 No. of defects243218105 2 313412 1. Set up the control lmits for c-charts. 2. Do these data come from a controlled process ? If not, calculated the revised control charts limits. Sol. Here k = 16 Average no. of defects in 16 units 142 2.625 16 cC k =∑= = Thus, the control limits for c-chart are: UCL = 3 2.625 3 2.625 2.625 4.861 7.486cc+=+ =+= CL = 2.625c = LCL = 3 2.625 3 2.625 2.625 4.861 2.236 0cc−= − = − =− ≈ 10 8 6 4 2 0 No. of Defect Sample Number FIG. 11.26 . 11.20 Example 3. In a glass factory, the task of quality control was done with the help of mean (x — ) and standard deviation σ charts. 18 samples of 10 items each were chosen and then values ∑X and ∑S were found. 476 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. Mean Chart Mean of 10 sample mean 442 44.2 10 10 x X === ∑ Mean Range of 10 sample ranges 58 5.8 10 10 R R === ∑ As we have, for. halted to check 478 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES whether there are any assignable causes. If assignable causes found, the process should be re-adjusted to remove assignable