Question

4. (10 points) Samples of three (3) resistors are taken periodically, and the resistances in ohms, are measured. The following table presents the sample means (R), and sample standard deviations (s) for 25 consecutive samples. (X) ), ranges Sample R 5.114 0.146 0.077 0.085 0.031 1 0.158 2 5.144 3 5.220 0.057 0.081 0.099 5.196 0.158 5 5.176 0.172 0.030 0.017 0.059 0.099 0.053 6 5.222 7 5.209 5.212 0.118 0.157 0.085 9 5.238 10 0.104 0.054 5.152 11 5.163 0.051 0.026 12 5.221 0.105 0.055 0.132 0.071 13 5.144 14 5.098 0.123 0.062 0.083 0.042 15 5.070 0.038 16 5.029 0.073 17 5.045 0.161 0.087 0.071 5.008 5.029 0.138 18 19 0.082 0.042 20 5,038 0.109 0,055 21 0.066 0,034 4.962 22 5.033 0.078 0.041 23 4.993 0.085 0.044 24 0.126 0.066 4.961 25 4.976 0.094 0.047 = 5.1061, R = 0.1082, and 3 0.05688. The means are (a) Use an R chart to determine if the process variation is under control and delete any samples necessary to bring the process variation under control. Explain. (b) Use R to estimate oy (see p. 773). Also, 3 to estimate Oy (see p. 782). use (c) Construct the Western Electric X chart. (d) Is the process in control? If not, for each Western Electric rule, identify when the process is first detected to be out of control?

Answer 1

Page No:1 Given ; R = 0.1082 x S = 5.1061 = 0.05688 a) We know that control limits of R chart Las follow UCL = D. R | CL = Ř LCL = Dz R where R = 0.1082 ... Given K = Total number of sample = 25 De= 1.54 1 D z = 0,549 We get UCLA = 0.167 Cha0.1082 LOLA = 0.059 :: We get Idea about process variation under control or not Ru : 0.051 € (LCL, UCLA) Roz = 0.057 € (0.059, 01162) Ros= 0.172 € (0.059 0.167) RoG = 0.030 4 20.059, 0:167) .: We conclude that process is not under the control limits

Page No. 2 b) Use Ŕ to estimate of E R = 0.1082 = 0.026 de 3.931 where dz value calculated in table. statistical Use Š → 6 to estimate 68 = Š ca where 12 z This niz vh-I' net C4 =1212512 where n=25 24 1368 433 65.8496 zgg16800 164 = 0.991 1. 6 f = 0.0 5688 = 0.057 0.99 |

c) Page NO:3 Construct the Western Electric T chart: T = 5.1061 , = 0.1082, A2 = 0.153 UCLE = 8 +AzR che = Ă LOLE = T - Az R control limits as follow - We get ñ chart's UCLA = 5.123 CLE = 5.1061 LCLE = 5.089 d) is the » No process in control ? process is out of control detected | Identify when the process is first to be out of control ? → at Xz = 5.14¢ ¢ ( 5.089, 5.123

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