Data for excel
Week | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |
# of defects per lot | 9 | 2 | 19 | 14 | 18 | 7 | 14 | 13 | 17 | 12 |
Parameter | Quantitative Value | Remark |
Upper specification Limit or USL in Fluid Ounces | 5.40 | Given |
Lower specification Limit or LSL in Fluid Ounces | 5.20 | Given |
Normally Distributed Mean in Fluid Ounces | 5.23 | Given |
Standard Deviation in Fluid Ounces | 0.03 | Given |
Process Capability or Cp | 1.11 | It can Estimate the capability of the process if the process Mean needs to be centered between the specification limits, as here, value of Cp is less than 1.33, thus we can conclude that Process has Many variation within, and Process is not capable of 3 sigma Performance based on Cp, |
Upper Bound for Process Capability Index or Cpu | 1.89 | Cpu value can estimate that how is the capability of the process to meet the Upper specification limit or It depicts that How process mean is close to the Upper specification Limit |
Lower Bound for Process Capability Index or Cpl | 0.33 | Cpl value can estimate that how is the capability of the process to meet the lower specification limit or it depicts that How process mean is close to the Lower specification Limit |
Process Capability Index or Cpk | 0.33 | Cpk is defined as Minimum Value of Cpu and Cpl, it can estimate the capability of the process and in case Cpk is equal to Cp, then the process is centered at the middle point of the specification limits, value of Cpk with ref. to Cp can determine that how the process is off centered and whether process need potential Improvement or not, here, in actual, Case, as Cpk is not equal to Cp or Cpk is less than Cp, thus we can conclude that the process is off centered from the middle point of the specification limits, Most of the company is looking for Cpk value at Minimum 1.33 and Cpk should be equal to Cp, thus we can conclude that process is Improved and centered, However, here, Cpk value is far less than 1.33, thus Process is not improved and the process is also off centered around the mean |
Formula
1 | B | C |
2 | Upper specification Limit or USL in Fluid Ounces | 5.4 |
3 | Lower specification Limit or LSL in Fluid Ounces | 5.2 |
4 | Normally Distributed Mean in Fluid Ounces | 5.23 |
5 | Standard Deviation in Fluid Ounces | 0.03 |
6 | Process Capability or Cp | =(C2-C3)/(6*C5) |
7 | Upper Bound for Process Capability Index or Cpu | =(C2-C4)/(3*C5) |
8 | Lower Bound for Process Capability Index or Cpl | =(C4-C3)/(3*C5) |
9 | Process Capability Index or Cpk | =MIN(C7:C8) |
based on the Excel data given in the Question, let me prepare C Chart or Defect Chart considering Each Lot contains Constant Number of Item, here, I have taken help from Minitab 2017, using Minitab 2017, Go to Stat > Control Chart > Attribute charts > C - chart > Selecting the Column data for Number of Defects per lot, then go to C Chart Options and selecting Perform all test for special Causes, then Click OK, we will get below control chart and Control Chart options are shown for Reference,
Interpretation: As per above chart, as all defect data points are following and satisfying the control chart options, and No special causes are observed, thus, the process is in Statistical Control,
Fest ALEKS Ca'p. Week 3 Homework < 17 18 19 20 21 22 23 24 Question 34 of 37 (1 point) Attempt 1 of 3 Evaluate the combination. 10 = Х 5
Consider the following data. 14 21 23 20 16 19 22 26 15 16 23 25 24 20 15 20 19 20 21 13 17 17 18 23 26 21 22 15 20 18 25 24 15 23 25 19 21 24 21 19 (a) Develop a frequency distribution using classes of 12–14, 15–17, 18–20, 21–23, and 24–26. Class Frequency 12–14 15–17 18–20 21–23 24–26 Total (b) Develop a relative frequency distribution and a percent frequency distribution using the classes...
14 24 18 23 21 18 16 14 23 17 15 13 19 23 24 14 16 26 21 14 15 22 16 12 20 23 19 26 20 25 21 19 21 25 23 25 25 19 20 15 (a) Develop a frequency distribution using classes of 12-14, 15-17, 18-20, 21-23, and 24-26. Class Frequency 12-14 15-17 18-20 21-23 24-26 Total (b) Develop a relative frequency distribution and a percent frequency distribution using the classes in part (a). If...
O A. Ов. О с. 18, 19, 20, 21, 22, 23, 24 18, 18, 19, 19, 19, 20, 20, 20, 20, 20, 21, 22, 22, 23, 24 118, 119, 120. 121. 122, 123, 124. 218, 219, 220, 222, 319, 320, 420, 520 O D. 18, 1.8, 1.9, 1.9, 1.9,20,20, 20,20, 2.0, 2.1, 22, 22, 23, 24 The macimum data entry is 17 18 19 20 21 22 23 24 25
Sales (1000s of gallons) 18 Week 19 23 17 16 20 17 23 19 10 12 a. Compute four-week and five-week moving averages for the time series. If required, round your answers to two decimal places. 4 Period Period Moving ge Average Moving Averag WeekSales 18 19 23 17 16 19.6x 20.25 V 1911 19.4| 20 19.2x 19.25x 23 18.81 × 1019 1918.8 19.2x 11 14 20 ]X 18.75 12 b. Compute the MSE for the four-week and five-week moving...
Simplying radicals 16. 12 17. 18 18. 32 19. -2 27 20. 9x 21. -8y 22. 3 25m 23. 45w 24. 9,/20p 25. 50x 26, ?48m2n :-30
Week Sales (100s of gallons) 1 18 2 22 3 19 4 24 5 17 6 16 7 21 8 19 9 23 10 21 11 15 12 21 a. using a weight of 1/2 for the most recent observation, 1/3 for the second most recent observation, and 1/6 for the third most recent observation, compute a three-week weighted moving average for the time series (to 2 decimals). Enter negative values as negative numbers. b. compute the MSE for the...
16 16 25 39 | 40 17 18 19 JL 20 21 22 wi Questi A Moving to the next question prevents changes to this answer. te Question 7 1 points Determine the value of AG for the reaction at 298 K. CH,(g) +20,(g) → CO, (g) + 2H, 0(1) Given Substance 5° (J/mol · K)) AH :(kJ/mol) -74.8 0.0 -393.5 -285.8 186.2 205.0 213.8 69.9 CH4(g) O2(g) CO2(g) H20(1) a.-817 kJ b. +817 kJ c. -573 kJ d. +573...
Actual Demand 14 15 16 15 18 17. Forecasted Demand 20 18 22 20 24 21 Based on the information given in the table above, we can conclude that, in general, the forecasting method is underestimating demand. the forecasting method is overestimating demand. we cannot determine whether the predictions are underestimating or overestimating demand.
17 18 19 Direction of disk rotation 14 15 19 20 21 22 23 24 Figure on the right shows the tracks and sectors on a disk. The size of File A, B, and C is 1028 bytes, 2521 bytes, and 8211 bytes respectivelv. Answer the following questions. 17 18 1 27 28/29 2212324252 If data block size is 1KB, how many data blocks does this disk contain? Answer: How many data blocks does File A take? Answe How many...