TQM &amp; SPC Problem Solutions

# B cl p 04 ucl 04 3 08157 lcl 04 3 00157 round to zero

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b. CL = p = .04 UCL = .04 + (3) 200 / ) 96 (. 04 . = .08157 LCL = .04 - (3) 200 / ) 96 (. 04 . = .-.00157 round to zero c. One of the 6 sample points from the second shift (.04, .06, .10 , .02, .05, .03) fall outside the control limits, we conclude that the process is out of statistical control. 6. a. X-Bar chart R Chart CL = x = 30 CL = 5 UCL = 30 + A 2 R UCL = D 4 (5) = 30 + .419(5) = 1.924(5) = 32.095 = 9.62 LCL = 30 - .419(5) LCL = D 3 (5) = 27.905 = 0.076(5) = .38 b. The sample average is 31.9 and the sample range is 11. The sample is in control on average measurement (27.905 < 31.9 < 32.095) and out of control on range (11 > 9.62). 7. a. x = 3.1 R = 1.2 X Chart: UCL = x + A 2 R = 3.1 + 0.577(1.2) = 3.8 LCL = x - A 2 R = 3.1 - 0.577(1.2) = 2.4 R Chart: UCL = D 4 R = 2.115(1.2) = 2.5 LCL = D 3 R = 0(1.2) = 0 9 - 4

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b. Sample 4 is out of control based on the x-bar chart, and sample 2 is out of control based on the R chart. In both cases, the out-of-control reading is border-line. Nevertheless, the producer should look for assignable causes in the two samples in question. c. The variables chart is much easier to use because the sample size is only five. However, the producer should be certain that the circuit voltage measurement will serve as a good indicator of defective boards. 8. a. x = 3.0 R = .0015 X Chart: UCL = 3.0 + 0.729(.0015) = 3.00109 LCL = 3.0 - 0.729(.0015) = 2.99891 R Chart: UCL = 2.282(.0015) = .0034 LCL = 0(.0015) = 0 X Chart: UCL: 3.00109 x : 3.0 LCL: 2.99891 R Chart: UCL: .0034 R : .0015 LCL: 0.0 b. Samples 1, 3, and 5 are within the control limits of the x-bar chart, but samples 2 and 4 are outside the control limits. Sample 4 is also outside the control limits of the R chart. Based on this, we would conclude that the process is out of control. c. Tolerances are based on measurements of individual parts, not on averages of samples of 4. Since the sample averages are out of tolerances in some of the samples, it is likely that some of the individual parts are out of tolerance too. But, we can’t be sure unless we have the measurement of each individual part. 9. a. x = 106 R = 10 X Chart: UCL = 106 + 0.483(10) = 110.83 LCL = 106 - 0.483(10) = 101.17 9 - 5
R Chart: UCL = 2.004(10) = 20.04 LCL = 0(10) = 0 b. For the past six days the average is 102.67, and the range is 28. Although the average is lower than the grand average it is not below the LCL of the average. However, the range for the process for the last six days is above the UCL of the range. Thus, we conclude that the process is outside its control limits on range.

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