Figure 3: sample size = 10000 0.9 0.8 lacf Opacf 0.7 0.6 0.5 0.4 0.3 acf and pact
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- Auto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: a) What is the value of X? x= mm (round your response to two decimal places). Day 1 2 3 4 5 Mean x (mm) 158.9 155.2 155.6 157.5 156.6 Range R (mm) 4.2 4.4 4.3 4.8 4.3Auto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: a) What is the value of x? x= 156.76 mm (round your response to two decimal places). b) What is the value of R? Day 1 2 3 4 5 Mean x (mm) 158.9 155.2 155.6 157.5 156.6 R = 4.40 mm (round your response to two decimal places). c) What are the UCL and LCL using 3-sigma? Upper Control Limit (UCL) = mm (round your response to two decimal places). Range R (mm) 4.2 4.4 4.3 4.8 4.3 ÇAuto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: a) What is the value of x? = x= 156.76 mm (round your response to two decimal places). b) What is the value of R? R= mm (round your response to two decimal places). Day 1 2 3 4 5 Mean x (mm) 158.9 155.2 155.6 157.5 156.6 Range R (mm) 4.2 4.4 4.3 4.8 4.3
- Auto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: a) What is the value of x? = x= 155.56 mm (round your response to two decimal places). b) What is the value of R? R 4.48 mm (round your response to two decimal places). c) What are the UCL; and LCL; using 3-sigma? Day 1 2 3 4 5 Upper Control Limit (UCL) = 156.94 mm (round your response to two decimal places). Lower Control Limit (LCL-) = 154.18 mm (round your response to two decimal places). d) What are the UCLR and LCLR using 3-sigma? Upper Control Limit (UCL) = 7.96 mm (round your response to two decimal places). Mean x (mm) 154.9 153.2 155.6 155.5 158.6 Range R (mm) 4.0 4.8 3.9 5.0 4.7 Lower Control Limit (LCL) = 1.00 mm (round your response to two decimal places). e) If the true diameter mean should be 155 mm and you want…At Webster Chemical Company, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Even when the process is in control, an average of four lumps per tube of caulk will remain. Testing for the presence of lumps destroys the product, so an analyst takes random samples. The following results are obtained: Tube No. Lumps Tube No. Lumps Tube No. Lumps 1 6 5 6 9 5 2 5 6 4 10 0 3 0 7 1 11 9 4 4 8 6 12 2 Determine the c-chart two-sigma upper and lower control limits for this process. Is the process in statistical control?A process is in statistical control with = 202,5 y s = 2,0. Specifications are atLSL = 196 and USL = 206.a) Estimate the process capability with an appropriate process capability ratio.b) What is the potential capability of this process?c) Items that are produced below the lower specification limit must bescrapped, while items that are above the upper specification limit can bereworked. What proportion of the process output is scrap and whatproportion is rework?d) Because scrap is more expensive than rework, the process has been centered closer to the upper specification limit. If scrap is twice as expensive as rework, is the process mean at the best possible location? What value of the process target would you recommend?
- For a particular process, the mean value, Xbar, is 1.011 and the mean range value, Rbar, is 0.102. If the product specification is 1.000 +/- 0.050, calculate the process capability for a sample size is 4. Question options: 0.194 0.237 0.297 0.397 0.549 0.624 0.734 0.929Using samples of 200 credit card statements, an auditor found the following:Sample 1 2 3 4Number with errors 4 2 5 9a. Determine the fraction defective in each sample.b. If the true fraction defective for this process is unknown, what is your estimate of it?c. What is your estimate of the mean and standard deviation of the sampling distribution of fractions defective for samples of this size?d. What control limits would give an alpha risk of .03 for this process?e. What alpha risk would control limits of .047 and .003 provide?f. Using control limits of .047 and .003, is the process in control?g. Suppose that the long-term fraction defective of the process is known to be 2 percent. What arethe values of the mean and standard deviation of the sampling distribution?h. Construct a control chart for the process, assuming a fraction defective of 2 percent, using twosigma control limits. Is the process in control?4. For a certain drilling machine of the Putnam Manufacturing Company, control limits foran x−Chart have been determined as: UCLx = 86.8 and LCLx = 83.3 (a) Sample averages for the most recent six samples, in sequential order, are 83.2, 83.9,84.7, 85.4, 85.9, and 86.7. What should be concluded about the process average?Should some action be taken? Why or why not? (b) What would you conclude (and why) about the process average if the sample averagesfor the most recent six samples, in sequential order, were 85.6, 84.9, 86.2, 85.2, 85.5,and 85.1?
- Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped. Management is concerned about whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes were taken, each tube was weighed, and the weights in Table were obtained. Ounces of Caulking Per Tube Tube Number Sample 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7.98 8.33 7.89 8.24 7.87 8.13 8.34 8.22 7.77 8.18 8.13 8.14 8.02 8.08 7.91 7.83 7.92 8.11 7.94 8.51 8.04 8.05 7.99 8.13 8.44 8.41 8.00 7.90 8.10 8.14 7.68 8.28 7.89 8.16 7.81 8.12 7.81 8.09 7.93 7.97 8.14 8.13 8.11 8.16 8.09 8.07 7.88 8.14 a. Assume that only six samples are sufficient and develop the control charts for the mean and the range.b. Plot the observations on the control chart and comment on your findings.Cartons of Plaster are supposed to weigh exactly 32 ounces. Inspectors want to develop process control charts. They take five samples of six (4) boxes each and weigh them. Sample means (X-bar) are: 33.8, 34.6, 34.7, 34.1, and 34.2 respectively. Also, the corresponding ranges are: 1.1,0.3,0.4,0.7, and O.3, respectively. The lower and upper control limits of the R- chart are respectively O a. O and 0.15 Ob. 0.03 and 1.28 Oc. O and 1.28 O d. 0.01 and 1.13 O e. None is correctYou work for Raider Data Systems where thousands of insurance records are entered by clerks each day for a variety of client firms. You are in charge of setting control limits to include 99.73% of the random variation in the data entry process when it is in control. Samples that you collected from 20 employees are shown below. You carefully examine 100 records entered by each employee and count the number of errors entered by each clerk. You also compute the proportion defective in each sample. Using a p-chart, what are the upper and lower control limits? Sample Errors Made Proportion Defective 1 4 0.04 2 5 0.05 3 6 0.06 4 3 0.03 5 8 0.08