Make an np chart for the following data: n non-conforming 400 12 400 10 400 8 400 14 400 21 400 16 400 18 400 15 400 12 400 17
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- Make an np chart for the following data:
n |
non-conforming |
400 |
12 |
400 |
10 |
400 |
8 |
400 |
14 |
400 |
21 |
400 |
16 |
400 |
18 |
400 |
15 |
400 |
12 |
400 |
17 |
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- 4. Refer to Table S6.1 - Factors for Computing Control Chart Limits (3 sigma) LOADING... for this problem. 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: Day Mean x (mm) Range R (mm) 1 156.9 4.4 2 155.2 4.4 3 155.6 4.3 4 153.5 4.8 5 156.6 4.7 Part 2 a) What is the value of x? x= _______ mm (round your response to two decimal places). b) What is the value of Upper R overbarR ? Upper R overbarRequals= _______ mm (round your response to two decimal places). c) What are the UCL Subscript x overbarUCLx and LCL Subscript x overbarLCLx using 3-sigma ? Upper Control Limit (UCL Subscript x overbarUCLx ) = _________ mm (round your response to two…Refer to Table S6.1 - Factors for Computing Control Chart Limits (3 sigma) for this problem. Sample Size, n Mean Factor, A2 Upper Range, D4 Lower Range, D3 2 1.880 3.268 0 3 1.023 2.574 0 4 0.729 2.282 0 5 0.577 2.115 0 6 0.483 2.004 0 7 0.419 1.924 0.076 8 0.373 1.864 0.136 9 0.337 1.816 0.184 10 0.308 1.777 0.223 12 0.266 1.716 0.284 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: Day Mean Bold x overbarx (mm) Range R (mm) 1 154.9154.9 4.04.0 2 151.2151.2 4.44.4 3 153.6153.6 3.93.9 4 155.5155.5 5.05.0 5 158.6158.6 4.3 a) What is the value of x? X = ____ mm b) What is the value of R? R =…Consider the ANOVA table that follows. Analysis of Variance Source DF SS MS F p Regression 2 77.907 38.954 4.14 0.021 Residual Error 62 583.693 9.414 Total 64 661.600 a-1. Determine the standard error of estimate. (Round your answer to 3 decimal places.) a-2. About 95% of the residuals will be between what two values? (Round your answers to 3 decimal places.) b-1. Determine the coefficient of multiple determination. (Round your answer to 3 decimal places.) b-2. Interpret the coefficient of multiple determination. (Round your answer to 1 decimal place.) Determine the coefficient of multiple determination, adjusted for the degrees of freedom. (Round your answer to 3 decimal places.) Prev Question 5 of 30 Total5 of 3
- A medical facility does MRIs for sports injuries. Occasionally a test yields inconclusive results andmust be repeated. Using the following sample data and n = 200, determine the upper and lowercontrol limits for the fraction of retests using two-sigma limits. Is the process in control?SAMPLE1 2 3 4 5 6 7 8 9 10 11 12 13Number of retests 1 2 2 0 2 1 2 0 2 7 3 2 1Refer to Table S6.1 - Factors for Computing Control Chart Limits (3 sigma) LOADING... for this problem. Sampling 4 pieces of precision-cut wire (to be used in computer assembly) every hour for the past 24 hours has produced the following results: Hour x R Hour x R Hour x R Hour x R 1 3.15" 0.65" 7 3.15" 0.58" 13 3.11" 0.85" 19 4.51" 1.66" 2 3.00 1.23 8 2.65 1.13 14 2.93 1.31 20 2.89 1.04 3 3.22 1.43 9 3.02 0.66 15 3.22 1.11 21 2.55 1.08 4 3.39 1.21 10 2.85 1.28 16 2.84 0.55 22 3.28 0.41 5 3.07 1.12 11 2.73 1.12 17 2.96 1.43 23 2.84 1.63 6 2.76 0.42 12 3.07 0.45 18 2.84 1.34 24 2.64 0.92 Based on the sampling done,…5-3 Sample data:Mean (Microns) (N=4) Minimum Maximum3.821 3.764 3.9993.946 3.804 4.2514.049 4.039 4.4543.793 3.779 3.9233.635 3.534 4.0684.046 3.578 4.5043.569 2.974 3.9084.084 3.779 5.6673.414 2.806 4.2565.319 5.168 6.3464.096 4.092 4.2234.108 3.957 4.3084.037 3.982 4.3933.789 3.691 3.8534.066 4.023 4.269 Constants: Sample Size N A2 D3 D4 2 1.88 0 3.27 3 1.02 0 2.57 4 0.73 0 2.28 5 0.58 0 2.11 6 0.48 0 2.00 7 0.42 0.08 1.92 8 0.37 0.14 1.86 9 0.34 0.18 1.82 10 0.31 0.22 1.78 11 0.29 0.26 1.74 12 0.27 0.28 1.72
- Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces ± 0.60 ounces. The target process capability ratio is 1.33. The current distribution of the filling process is centered on 8.054 ounces with a standard deviation of 0.192 ounces. Compute the process capability ratios Cp and Cpk to assess whether the filling process is capable and set properly.Based on the following regression ANOVA table, what is R2? Source df SS MS F Regression 4 1793.2356 448.3089 7.4854 Residual 45 2695.0996 59.8911 Total 49 4488.3352 0.6005 0.3995 0.8807 0.7230Given the attached data. Answer the following questions for a 7 period moving average. t At 1 5477 2 3338 3 4799 4 2852 5 6332 6 8385 7 4382 8 4184 9 14229 10 10385 11 11676 12 5271 13 8938 14 9279 15 13990 16 2890 17 13027 18 8970 19 6700 20 7065 21 12491 22 8748 23 8396 24 6102 25 7559 26 7997 27 11981 28 10680 29 5783 30 2501 31 7620 32 10194 33 7908 34 11985 35 5863 36 3024 37 8758 38 1640 39 6569 40 11728 41 6659 42 2830 43 10018 44 8448 45 12409 46 2083 47 6374 48 7438 49 4576 50 4795 MAD = Average(|A-F|) TS =SUM(A-F)/MAD MSE = Average(A-F)2 1. Compute your forecast for period 51. The potential answers are: A: 11221.6 units. B: 12461 units. C: 8020.83 units. D: 4414 units. E: 6589 units.
- The following table contains the measurements of the key length dimension from a fuel injector. These samples of size five were taken at one-hour intervals. Use three-sigma control limits. Use Exhibit 10.13. SAMPLE NUMBER OBSERVATIONS 1 2 3 4 5 1 0.486 0.499 0.493 0.511 0.481 2 0.499 0.506 0.516 0.494 0.529 3 0.496 0.500 0.515 0.488 0.521 4 0.495 0.506 0.483 0.487 0.489 5 0.472 0.502 0.526 0.469 0.481 6 0.473 0.495 0.507 0.493 0.506 7 0.495 0.512 0.490 0.471 0.504 8 0.525 0.501 0.498 0.474 0.485 9 0.497 0.501 0.517 0.506 0.516 10 0.495 0.505 0.516 0.511 0.497 11 0.495 0.482 0.468 0.492 0.492 12 0.483 0.459 0.526 0.506 0.522 13 0.521 0.512 0.493 0.525 0.510 14 0.487 0.521 0.507 0.501 0.500 15 0.493 0.516 0.499 0.511 0.513 16 0.473 0.506 0.479 0.480 0.523 17 0.477 0.485 0.513 0.484 0.496 18 0.515 0.493 0.493 0.485 0.475 19 0.511 0.536 0.486 0.497 0.491 20 0.509 0.490 0.470 0.504 0.512 Calculate the mean and range for the above samples.…c) A process manufactures ball bearing whose diameters are normally distributed with mean 2.505cm and standard deviation 0.008cm.Specifications call for the diameter to be in the interval 2.5 ‡ 0.01 cm. estimate the proportion of the ball bearings will meet the specification?Rajan wanted a radio from Jacky's. It appears that Rajan is particularly interested in purchasing a Radio from the inexpensive retailer's sales manager for consumer electronics. He tells Jacky, the salesperson at the inexpensive store where he thinks he'll get the best deal, that his old radio died and he wants to listen to his favourite tunes. He wants a replacement radio as quickly as possible. In three and a half weeks, Rajan's favourite model will be 10% off.He suspects Rajan won't wait and will find another job. Jacky will earn less on the lowered price. He thinks that telling Rajan of the sale makes little sense.When Jacky tells Rajan that the radio set he wants is no longer available and won't be for another week, Rajan is enraged. Fearing losing the business, Jacky begs his sales manager, Michelle, to speed up delivery. Michelle says it's impossible and suggests Jacky tell Rajan the store can get the set in 24 hours and sell him the demo model. Michelle says the sample is new…