For small samples, R chart is relatively sensitive to changes in process standard deviation. False True
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For small samples, R chart is relatively sensitive to changes in process standard deviation.
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- Need help on calculating control limits for x-bar and r-chart based on the samples below: Sample No X-BAR (Sample Mean) R (Sample Range) X-BAR R 1 28.2151942 9.5450186 2 30.5767201 11.1032318 3 29.0253789 5.52221327 4 29.9431791 6.27087883 5 29.947177 12.7568255 6 29.6848334 11.4927456 7 29.2539019 4.72514932 8 29.6694257 13.9330532 9 30.1590205 6.33461804 10 28.6867307 6.49522527 11 30.5960109 8.13717125 12 29.4390565 5.80286433 13 31.863426 3.40227918 14 32.3716448 8.62217828 15 29.7052597 10.0026916 16 30.0060069 4.25003967 17 32.8491753 4.6515662 18 33.5023103 9.62764088 19 28.1876404 9.6980397 20 30.5043768 6.46724539 21 29.6856417 7.65071069 22 32.5728395 9.2483604 23 28.2363719 7.54975234 24 29.6579999 11.3224787 25 28.1319897 13.213391 26 28.8811004 8.08405181 27 31.6730765 12.9865678 28 28.2078142 5.96315947 29 27.9430689 7.39518992 30 29.6269815 9.48802129 31 34.3599108 7.66563648 32 28.2423234 7.34122645…3.2. The Road King Tire Company in Birmingham wants to moni- tor the quality of the tires it manufactures. Each day the company quality-control manager takes a sample of 100 tires, tests them, and determines the number of defective tires. The results of 20 samples have been recorded as follows: Number of Number of Sample Defectives Sample Defectives 1 14 11 18 12 12 10 9. 13 19 10 14 20 11 15 17 7 16 18 7 8 17 18 8 14 18 22 9. 16 19 24 10 17 20 23 3. 4,At Quick Car Wash, the wash process is advertised to take less than 8 minutes. Consequently, management has set a target average of 440 seconds for the wash process. Suppose the average range for a sample of 9 cars is 10 seconds. Use the accompanying table to establish control limits for sample means and ranges for the car wash process. Factors for calculating three-sigma limits for the x-chart and R-chart Size of Sample (n) Factor for UCL and LCL for x-chart (A2) Factor for LCL for R-Chart (D3) Factor for UCL for R-Chart (D4) 2 1.880 0 3.267 3 1.023 0 2.575 4 0.729 0 2.282 5 0.577 0 2.115 6 0.483 0 2.004 7 0.419 0.076 1.924 8 0.373 0.136 1.864 9 0.337 0.184 1.816 10 0.308 0.223 Part 2 The UCLR equals enter your response here seconds and the LCLR equals enter your response here seconds. (Enter your responses rounded to two decimal places.)
- Find the control limits for a 3-sigma u chart with process average number of nonconformities per inspection unit equaling to 3 and sample size n=3. a. UCL=18 LCL=0 b. UCL=9 LCL=0 c. UCL=36 LCL=18 d. UCL=6 LCL=0Consider a p-control chart with 3-sigma limits at 0.02 and 0.08: (A) What is the sample size used? (B) If the process average shifts to p1=0.10, the probability of detecting the shift on the first subsequent sample =Twenty samples (K=20) of 200 observations (n = 200) were taken by an operator at a workstation in a production process. The number of defective items in each sample was recorded as follows. Sample Number of Defects 1. 12 18 10 4. 15 16 6. 19 7. 17 8 12 11 10 14 11 16 12 15 13 13 14 16 15 18 16 17 17 18 18 20 19 21 20 22 Management wants to develop ap-chart using 3-sigma limits. What is the Lower Control Limit (LCL)? (calculation 2 sig fig)
- 1. The data shown in Table 1 are x and R values for 20 samples of size n= 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals recorded (i.e., 31.6 should be 0.50316). Please show all your work for full credit. (a) Set up x and R charts on this process. Does the process seem to be in statistical control? If necessary, revise the trial control limits. (b) Assume that diameter is normally distributed. Estimate the process standard deviation. Sample R Sample R 1 31.6 4 11 29.8 4 33.0 3 12 34.0 4 35.0 4 13 33.0 10 4 32.2 4 14 34.8 4 5 33.8 38.4 31.6 15 35.6 7 3 16 30.8 7 4 17 33.0 5 8 36.8 10 18 31.6 3 9. 35.0 15 19 28.2 9 10 34.0 6 20 33.8 Table 1: Bearing Diameter DataFind the control limits for a 3-sigma u chart with process average number of nonconformities per inspection unit equaling to 3 and sample size n=3. Select one: a. UCL=36 LCL=18 b. UCL=9 LCL=0 c. UCL=6 LCL=0 d. UCL=18 LCL=0Explain how a person using 2-sigma control charts will more easily find samples “out of bounds” than 3-sigma control charts. What are some possible consequences of this fact?
- In a fabric manufacturing factory, the quality control process using control charts from SPC. In an hour there are a total of 5 samples are taken each having 5 observations regarding the thickness of fabric in measured in millimeters. In a particular hour, the sample means (X-bar) are noted to be: 172.11, 219.51 . 208.24, 112.44 and 123.30 respectively. In the same sample, the corresponding ranges are: 13.1713.3815.34, 13.04, and 13.02 respectively What are the lower and upper control limits for the R chart?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 5 pistons produced each day, the mean and the range of this diameter have been as follows: Day Mean (mm) Range R (mm) 158 4.3 151.2 4.4 155.7 4.2 153.5 4.8 156.6 4.5 What is the UCL using 3-sigma?(round your response to two decimal places). 1. 2. 4.An X-bar control chart designed with L=3 and constant sample size "n" has UCL=130.013 and LCL=125.987, if the process is under control and its standard deviation is sigma=1.5, what is the sample size "n"? not enough information 4 6 10 8.