Minitab Exercise -Sample size for Variables Charts (Rev. 2_14) Objectives To increase understanding of and errors when using control charts To practice ARL and ATS calculations To determine sample size considering o - What risk do you want to take of incorrectly stating a process shift has occurred? o or power - What risk do you take of not detecting a difference? o Difference you want to detect from the mean. o - What is the variation in the characteristic being measured To gain insight as to why 3 limits are used in industry. Part A. Practice Problem (work by hand) A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. 1. What is the standard deviation of the sample averages? (Remember Central Limit Theorem. To 4 decimal places) = 2. When taking samples of size 6, what are the 3 limits? (Assume target is process average). = = When the process is in control, what is the probability of a point being outside the control limits causing a Type I error? For 3 limits: = The average run length is: ARL = The average time to signal is: ATS = 3. If 2 limits are used, what are the limits? = = When the process is in control, what is the probability of a point being outside the control limits causing a Type I error? For 2 limits: = The average run length is: ARL = The average time to signal is: ATS = 4. If the process shifts to 2.05, what is the probability that it will not be detected on the next sample using the 3 limits? (Assume the standard deviation stays the same.) 5. What is the power of the control chart to detect the shift to 2.05 using the 3 limits? 6. How many points on average (ARL), would have to be taken before a point falls out of the 3 limits before the shift to 2.05 is detected? What is the average time (ATS) between out of limit points? Part B. Minitab 1. Check answers for previous power calculation Stat>Power and Sample Size>1 sample Z Specify values for any two of the following Sample size: 6 Difference: .05 Power: leave blank. This is what we are looking for. Standard Deviation: .05 Sigma for the process, not the distribution of sample averages Options: Check Not equal We have upper and lower control limits, so it is a two sided test Significance level: .0027 This is the alpha risk for the three sigma limits. OK Graph Check Display power curve OK. OK In the session window, you will find the power number to compare with your calculation. Now look at the power curve. We want to use it to estimate the power to detect other levels of shifts, but the scale for the difference and power makes it difficult to use. To change the power scale, click on the current scale numbers. You should get the dialog box Edit Scale. Select the Number of ticks option and enter 11 in the box. Click OK. See how the axis changed. Change the scale on the x axis in the same manner to get increments of .025 rather than .05. You can either use number of ticks, or specify the major tick locations. Edit the title of the chart to indicate that it reflects the power curve for a control chart with 3 sigma limits. From the chart you should now be able to see that power of the control chart to detect a difference of 0.075 is approximately 0.75. What is the probability that the next point will fall out of the control limits if the process has shifted .025? __________ What is the probability of a Type II error when samples of size 6 are taken from a process with a standard deviation of .05 and a shift of .025 has occurred? __________ (You can see how close you came to the power for the .025 shift with Minitab. You might want to turn off the graph if you do that. But turn it back on for the next step.) 2. For comparison, run the chart for alpha = .0456 (associated with the 2 sigma limits). Give this chart an appropriate title. Edit the scales so that the increments are the same as on the 3 sigma graph. Compare the two power charts you have run and complete the following statements. The curve for both charts used n=____ and a process standard deviation of _____. Using the 2 sigma control chart, the probability of detecting a shift of .025 on the 2sigma on the next sample taken is approximately _________. The power to detect a shift of .025 on the 2 sigma chart is greater / less than the power to detect on the shift on the 3 sigma chart. Using the 2 sigma control chart, the probability of detecting a shift of .05 on the 2sigma on the next sample taken is approximately _________. Back to our original problem: A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. How many points on average (ARL), would have to be taken before a point falls out of the 2 limits before the shift to 2.05 is detected? ____ What is the average time (ATS) between out of limit points? ______ 3. We now want to compare power curves for different sample sizes. To generate multiple power curves for different sample sizes, enter multiple sample sizes separated by a space. Generate the power curve for a 3 sigma control chart, with sigma = .05 entering multiple sample sizes of 6, 12 and 18 parts. Determine the power to detect a shift of .05. (Be sure to change sigma back.) From the curve and session window output, you should see that the power to detect a shift of .05 using sample size 12 is 0.68. For n=18, what is the probability of the next sample being outside the control limits when a shift of .05 occurs? ____ In general, if the sample size increases, the probability that the shift will not be detected on the next shift will increase / decrease. Back to our original problem again: A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. When the sample size is 12, how many points on average (ARL), would have to be taken before a point falls out of the 3 limits before the shift to 2.05 is detected? ____ What is the average time (ATS) between out of limit points? ______ Part C. Comparing the plans Complete the following table with the data from the previous sections. For n=12 With 3 limits n=6 With 3 limits n=6 With 2 limits For a type I error ARL to signal ATS For a .05 process shift Power to detect shift ARL to signal ATS In the United States, three sigma limits are most often used as the action limits. A typical scenario would be as follows: If a point is outside the limits the operator would notify the supervisor. The supervisor would question the operator then call a process engineer who would then investigate to search for an assignable cause. If the process is a critical one, it would be shut down during the investigation, likely delaying down-line production activities. The investigation could be as easily identifiable as a broken tool bit or as complicated as a slightly altered material composition. Investigations can become very costly and in general, with a Type I error, when there is really nothing wrong, a lot of time and effort/money, can be spent to no avail. On the other hand, if a process really has shifted, we want to detect and correct it. How costly it is for a process to shift will often depend on how close the process is to meeting specifications and whether or not a part can be reworked. Setting up a plan for sampling involves setting the sample size, determining the control limits, and determining the frequency of sampling. To illustrate costs involved, we will compare the three plans in the table above. (We will be holding the frequency of sampling constant, but be aware that further comparisons could be done in this area.) Assume the following: The machine runs 1000 hours per year and produces 50 parts per hour. On average, it costs $200 to investigate an out of control signal. (Typically a \"hidden\" cost of quality) A shift from a mean of 2.00 to 2.05 will result in approximately 25% of the parts having to be scrapped. This happens about 4 times per year. Assume the shift happens halfway between sample times The material cost of the part is $5.00 and after this operation it has $5.00 of labor cost in it. So each scrapped part represents a $10.00 loss. Samples are taken at the end of every two hours. The cost to inspect one unit is $1.00 n=12 With 3 limits n=6 With 3 limits Cost to investigate false out of control signals Average time to signal ATS 740.7 Work hours per year B 1000 Signals per year C=B/ ATS 1.35 Cost per investigation D $200 Cost per year E=C*D $270 Cost of Scrapped parts when shift occurs Average time to signal shift ATS 6.87 Shift occurs halfway F 0.5 Hours production, after shift, before signal G= ATS*F Parts made/hour H Parts made, after shift, before signal I=G*H Percent of parts scrapped J 25% Parts scrapped per occurrence K=I*J 43.0 Occurrences per year L Parts scrapped per year N=K*L Cost per scrapped part P Scrap cost per year Q=N*P $1718 Production hours per year B 1000 Hours between inspections S 2 Samples per year T=B/S Sample size n Total parts inspected per year U=T*n Cost to inspect one part V Total inspection costs W=U*V 3000 Total all costs X=E+Q+W $4988 3.44 50 171.8 4 171.8 $10 Cost of Inspection 500 6 3000 $1 n=6 With 2 limits Part D. Questions 1) Why are the cost associated with false signals so much higher for 2 sigma than for 3 sigma limits? 2) Why are costs associated with scrap so much lower where 2 sigma limits would be used instead of 3 sigma? 3) In comparing the cost data for the 3 sigma limit plans with sample sizes n=6 and n=12, T / F The costs associated with false signals would be the same for both plans because they have the same risk of a false out of control signal. T / F Increasing the sample size reduced the cost of scrap because the shift was identified and could be corrected more quickly. T / F In this example, the biggest cost driver in choosing between the n=6 and n=12 plans would be the cost of sampling. 4) In the above scenario, how seriously do you think the quality engineer, production supervisor, or machine operator would take a quality control system that gave frequent alarms? Note: There are some products (such as medicines) where it is extremely important to detect the shift, and therefore 2 sigma limits are used and taken very seriously. 5) A finance manager has suggested that you use 2 sigma limits rather than 3 sigma limits to reduce the scrap rate in the show. Explain to him how the scrap rates are not the only cost to consider. Part E Quiz. You will need this completed worksheet plus the graphs for the quiz. Concept questions from the associated lecture will also be included on the quiz. Minitab Exercise -Sample size for Variables Charts (Rev. 2_14) Objectives To increase understanding of and errors when using control charts To practice ARL and ATS calculations To determine sample size considering o - What risk do you want to take of incorrectly stating a process shift has occurred? o or power - What risk do you take of not detecting a difference? o Difference you want to detect from the mean. - What is the variation in the characteristic being measured To gain insight as to why 3 limits are used in industry. Part A. Practice Problem (work by hand) A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. What is the standard deviation of the sample averages? (Remember Central Limit Theorem. To 4 decimal places) = 2. When taking samples of size 6, what are the 3 limits? (Assume target is process average). = = When the process is in control, what is the probability of a point being outside the control limits causing a Type I error? For 3 limits: = The average run length is: ARL = The average time to signal is: ATS = 3. If 2 limits are used, what are the limits? = = When the process is in control, what is the probability of a point being outside the control limits causing a Type I error? For 2 limits: = The average run length is: ARL = The average time to signal is: ATS = If the process shifts to 2.05, what is the probability that it will not be detected on the next sample using the 3 limits? (Assume the standard deviation stays the same.) What is the power of the control chart to detect the shift to 2.05 using the 3 limits? How many points on average (ARL), would have to be taken before a point falls out of the 3 limits before the shift to 2.05 is detected? What is the average time (ATS) between out of limit points? Part B. Minitab 1. Check answers for previous power calculation Stat>Power and Sample Size>1 sample Z Specify values for any two of the following Sample size: 6 Difference: .05 Power: leave blank. This is what we are looking for. Standard Deviation: .05 Sigma for the process, not the distribution of sample averages Options: Check Not equal We have upper and lower control limits, so it is a two sided test Significance level: .0027 This is the alpha risk for the three sigma limits. OK Graph Check Display power curve OK. OK In the session window, you will find the power number to compare with your calculation. Now look at the power curve. We want to use it to estimate the power to detect other levels of shifts, but the scale for the difference and power makes it difficult to use. To change the power scale, click on the current scale numbers. You should get the dialog box Edit Scale. Select the Number of ticks option and enter 11 in the box. Click OK. See how the axis changed. Change the scale on the x axis in the same manner to get increments of .025 rather than .05. You can either use number of ticks, or specify the major tick locations. Edit the title of the chart to indicate that it reflects the power curve for a control chart with 3 sigma limits. From the chart you should now be able to see that power of the control chart to detect a difference of 0.075 is approximately 0.75. What is the probability that the next point will fall out of the control limits if the process has shifted .025? __________ What is the probability of a Type II error when samples of size 6 are taken from a process with a standard deviation of . 05 and a shift of .025 has occurred? __________ (You can see how close you came to the power for the .025 shift with Minitab. You might want to turn off the graph if you do that. But turn it back on for the next step.) For comparison, run the chart for alpha = .0456 (associated with the 2 sigma limits). Give this chart an appropriate title. Edit the scales so that the increments are the same as on the 3 sigma graph. Compare the two power charts you have run and complete the following statements. The curve for both charts used n=____ and a process standard deviation of _____. Using the 2 sigma control chart, the probability of detecting a shift of .025 on the 2sigma on the next sample taken is approximately _________. The power to detect a shift of .025 on the 2 sigma chart is greater / less than the power to detect on the shift on the 3 sigma chart. Using the 2 sigma control chart, the probability of detecting a shift of .05 on the 2sigma on the next sample taken is approximately _________. Back to our original problem: A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. How many points on average (ARL), would have to be taken before a point falls out of the 2 limits before the shift to 2.05 is detected? ____ What is the average time (ATS) between out of limit points? ______ 3. We now want to compare power curves for different sample sizes. To generate multiple power curves for different sample sizes, enter multiple sample sizes separated by a space. Generate the power curve for a 3 sigma control chart, with sigma = .05 entering multiple sample sizes of 6, 12 and 18 parts. Determine the power to detect a shift of .05. (Be sure to change sigma back.) From the curve and session window output, you should see that the power to detect a shift of .05 using sample size 12 is 0.68. For n=18, what is the probability of the next sample being outside the control limits when a shift of .05 occurs? ____ In general, if the sample size increases, the probability that the shift will not be detected on the next shift will increase / decrease. Back to our original problem again: A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. When the sample size is 12, how many points on average (ARL), would have to be taken before a point falls out of the 3 limits before the shift to 2.05 is detected? ____ What is the average time (ATS) between out of limit points? ______ Part C. Comparing the plans Complete the following table with the data from the previous sections. For n=12 n=6 n=6 With 3 limits With 3 limits With 2 limits For a type I error ARL to signal ATS For a .05 process shift Power to detect shift ARL to signal ATS In the United States, three sigma limits are most often used as the action limits. A typical scenario would be as follows: If a point is outside the limits the operator would notify the supervisor. The supervisor would question the operator then call a process engineer who would then investigate to search for an assignable cause. If the process is a critical one, it would be shut down during the investigation, likely delaying down-line production activities. The investigation could be as easily identifiable as a broken tool bit or as complicated as a slightly altered material composition. Investigations can become very costly and in general, with a Type I error, when there is really nothing wrong, a lot of time and effort/money, can be spent to no avail. On the other hand, if a process really has shifted, we want to detect and correct it. How costly it is for a process to shift will often depend on how close the process is to meeting specifications and whether or not a part can be reworked. Setting up a plan for sampling involves setting the sample size, determining the control limits, and determining the frequency of sampling. To illustrate costs involved, we will compare the three plans in the table above. (We will be holding the frequency of sampling constant, but be aware that further comparisons could be done in this area.) Assume the following: The machine runs 1000 hours per year and produces 50 parts per hour. On average, it costs $200 to investigate an out of control signal. (Typically a \"hidden\" cost of quality) A shift from a mean of 2.00 to 2.05 will result in approximately 25% of the parts having to be scrapped. This happens about 4 times per year. Assume the shift happens halfway between sample times The material cost of the part is $5.00 and after this operation it has $5.00 of labor cost in it. So each scrapped part represents a $10.00 loss. Samples are taken at the end of every two hours. The cost to inspect one unit is $1.00 n=12 n=6 n=6 With 3 limits With 3 limits With 2 limits Cost to investigate false out of control signals Average time to signal ATS 740.7 Work hours per year B 1000 Signals per year C=B/ ATS 1.35 Cost per investigation D $200 Cost per year E=C*D $270 Cost of Scrapped parts when shift occurs Average time to signal shift ATS 6.87 Shift occurs halfway F 0.5 Hours production, after shift, G= ATS*F 3.44 before signal Parts made/hour H 50 Parts made, after shift, before I=G*H 171.8 signal Percent of parts scrapped J 25% Parts scrapped per occurrence K=I*J 43.0 Occurrences per year L 4 Parts scrapped per year N=K*L 171.8 Cost per scrapped part P $10 Scrap cost per year Q=N*P $1718 Cost of Inspection Production hours per year B 1000 Hours between inspections S 2 Samples per year T=B/S 500 Sample size n 6 Total parts inspected per year U=T*n 3000 Cost to inspect one part V $1 Total inspection costs W=U*V 3000 Total all costs X=E+Q+W $4988 Part D. Questions Why are the cost associated with false signals so much higher for 2 sigma than for 3 sigma limits? Why are costs associated with scrap so much lower where 2 sigma limits would be used instead of 3 sigma? In comparing the cost data for the 3 sigma limit plans with sample sizes n=6 and n=12, T / F The costs associated with false signals would be the same for both plans because they have the same risk of a false out of control signal. T / F Increasing the sample size reduced the cost of scrap because the shift was identified and could be corrected more quickly. T / F In this example, the biggest cost driver in choosing between the n=6 and n=12 plans would be the cost of sampling. In the above scenario, how seriously do you think the quality engineer, production supervisor, or machine operator would take a quality control system that gave frequent alarms? Note: There are some products (such as medicines) where it is extremely important to detect the shift, and therefore 2 sigma limits are used and taken very seriously. A finance manager has suggested that you use 2 sigma limits rather than 3 sigma limits to reduce the scrap rate in the show. Explain to him how the scrap rates are not the only cost to consider. Part E Quiz. You will need this completed worksheet plus the graphs for the quiz. Concept questions from the associated lecture will also be included on the quiz. Minitab Exercise -Sample size for Variables Charts (Rev. 2_14) Objectives To increase understanding of and errors when using control charts To practice ARL and ATS calculations To determine sample size considering o - What risk do you want to take of incorrectly stating a process shift has occurred? o or power - What risk do you take of not detecting a difference? o Difference you want to detect from the mean. - What is the variation in the characteristic being measured To gain insight as to why 3 limits are used in industry. Part A. Practice Problem (work by hand) A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. What is the standard deviation of the sample averages? (Remember Central Limit Theorem. To 4 decimal places) = 2. When taking samples of size 6, what are the 3 limits? (Assume target is process average). = = When the process is in control, what is the probability of a point being outside the control limits causing a Type I error? For 3 limits: = The average run length is: ARL = The average time to signal is: ATS = 3. If 2 limits are used, what are the limits? = = When the process is in control, what is the probability of a point being outside the control limits causing a Type I error? For 2 limits: = The average run length is: ARL = The average time to signal is: ATS = If the process shifts to 2.05, what is the probability that it will not be detected on the next sample using the 3 limits? (Assume the standard deviation stays the same.) What is the power of the control chart to detect the shift to 2.05 using the 3 limits? How many points on average (ARL), would have to be taken before a point falls out of the 3 limits before the shift to 2.05 is detected? What is the average time (ATS) between out of limit points? Part B. Minitab 1. Check answers for previous power calculation Stat>Power and Sample Size>1 sample Z Specify values for any two of the following Sample size: 6 Difference: .05 Power: leave blank. This is what we are looking for. Standard Deviation: .05 Sigma for the process, not the distribution of sample averages Options: Check Not equal We have upper and lower control limits, so it is a two sided test Significance level: .0027 This is the alpha risk for the three sigma limits. OK Graph Check Display power curve OK. OK In the session window, you will find the power number to compare with your calculation. Now look at the power curve. We want to use it to estimate the power to detect other levels of shifts, but the scale for the difference and power makes it difficult to use. To change the power scale, click on the current scale numbers. You should get the dialog box Edit Scale. Select the Number of ticks option and enter 11 in the box. Click OK. See how the axis changed. Change the scale on the x axis in the same manner to get increments of .025 rather than .05. You can either use number of ticks, or specify the major tick locations. Edit the title of the chart to indicate that it reflects the power curve for a control chart with 3 sigma limits. From the chart you should now be able to see that power of the control chart to detect a difference of 0.075 is approximately 0.75. What is the probability that the next point will fall out of the control limits if the process has shifted .025? __________ What is the probability of a Type II error when samples of size 6 are taken from a process with a standard deviation of . 05 and a shift of .025 has occurred? __________ (You can see how close you came to the power for the .025 shift with Minitab. You might want to turn off the graph if you do that. But turn it back on for the next step.) For comparison, run the chart for alpha = .0456 (associated with the 2 sigma limits). Give this chart an appropriate title. Edit the scales so that the increments are the same as on the 3 sigma graph. Compare the two power charts you have run and complete the following statements. The curve for both charts used n=____ and a process standard deviation of _____. Using the 2 sigma control chart, the probability of detecting a shift of .025 on the 2sigma on the next sample taken is approximately _________. The power to detect a shift of .025 on the 2 sigma chart is greater / less than the power to detect on the shift on the 3 sigma chart. Using the 2 sigma control chart, the probability of detecting a shift of .05 on the 2sigma on the next sample taken is approximately _________. Back to our original problem: A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. How many points on average (ARL), would have to be taken before a point falls out of the 2 limits before the shift to 2.05 is detected? ____ What is the average time (ATS) between out of limit points? ______ 3. We now want to compare power curves for different sample sizes. To generate multiple power curves for different sample sizes, enter multiple sample sizes separated by a space. Generate the power curve for a 3 sigma control chart, with sigma = .05 entering multiple sample sizes of 6, 12 and 18 parts. Determine the power to detect a shift of .05. (Be sure to change sigma back.) From the curve and session window output, you should see that the power to detect a shift of .05 using sample size 12 is 0.68. For n=18, what is the probability of the next sample being outside the control limits when a shift of .05 occurs? ____ In general, if the sample size increases, the probability that the shift will not be detected on the next shift will increase / decrease. Back to our original problem again: A machine is set to drill holes to a depth of 2 inches. The historical standard deviation is 0.05. Samples of size 6 are taken every 2 hours. When the sample size is 12, how many points on average (ARL), would have to be taken before a point falls out of the 3 limits before the shift to 2.05 is detected? ____ What is the average time (ATS) between out of limit points? ______ Part C. Comparing the plans Complete the following table with the data from the previous sections. For n=12 n=6 n=6 With 3 limits With 3 limits With 2 limits For a type I error ARL to signal ATS For a .05 process shift Power to detect shift ARL to signal ATS In the United States, three sigma limits are most often used as the action limits. A typical scenario would be as follows: If a point is outside the limits the operator would notify the supervisor. The supervisor would question the operator then call a process engineer who would then investigate to search for an assignable cause. If the process is a critical one, it would be shut down during the investigation, likely delaying down-line production activities. The investigation could be as easily identifiable as a broken tool bit or as complicated as a slightly altered material composition. Investigations can become very costly and in general, with a Type I error, when there is really nothing wrong, a lot of time and effort/money, can be spent to no avail. On the other hand, if a process really has shifted, we want to detect and correct it. How costly it is for a process to shift will often depend on how close the process is to meeting specifications and whether or not a part can be reworked. Setting up a plan for sampling involves setting the sample size, determining the control limits, and determining the frequency of sampling. To illustrate costs involved, we will compare the three plans in the table above. (We will be holding the frequency of sampling constant, but be aware that further comparisons could be done in this area.) Assume the following: The machine runs 1000 hours per year and produces 50 parts per hour. On average, it costs $200 to investigate an out of control signal. (Typically a \"hidden\" cost of quality) A shift from a mean of 2.00 to 2.05 will result in approximately 25% of the parts having to be scrapped. This happens about 4 times per year. Assume the shift happens halfway between sample times The material cost of the part is $5.00 and after this operation it has $5.00 of labor cost in it. So each scrapped part represents a $10.00 loss. Samples are taken at the end of every two hours. The cost to inspect one unit is $1.00 n=12 n=6 n=6 With 3 limits With 3 limits With 2 limits Cost to investigate false out of control signals Average time to signal ATS 740.7 Work hours per year B 1000 Signals per year C=B/ ATS 1.35 Cost per investigation D $200 Cost per year E=C*D $270 Cost of Scrapped parts when shift occurs Average time to signal shift ATS 6.87 Shift occurs halfway F 0.5 Hours production, after shift, G= ATS*F 3.44 before signal Parts made/hour H 50 Parts made, after shift, before I=G*H 171.8 signal Percent of parts scrapped J 25% Parts scrapped per occurrence K=I*J 43.0 Occurrences per year L 4 Parts scrapped per year N=K*L 171.8 Cost per scrapped part P $10 Scrap cost per year Q=N*P $1718 Cost of Inspection Production hours per year B 1000 Hours between inspections S 2 Samples per year T=B/S 500 Sample size n 6 Total parts inspected per year U=T*n 3000 Cost to inspect one part V $1 Total inspection costs W=U*V 3000 Total all costs X=E+Q+W $4988 Part D. Questions Why are the cost associated with false signals so much higher for 2 sigma than for 3 sigma limits? Why are costs associated with scrap so much lower where 2 sigma limits would be used instead of 3 sigma? In comparing the cost data for the 3 sigma limit plans with sample sizes n=6 and n=12, T / F The costs associated with false signals would be the same for both plans because they have the same risk of a false out of control signal. T / F Increasing the sample size reduced the cost of scrap because the shift was identified and could be corrected more quickly. T / F In this example, the biggest cost driver in choosing between the n=6 and n=12 plans would be the cost of sampling. In the above scenario, how seriously do you think the quality engineer, production supervisor, or machine operator would take a quality control system that gave frequent alarms? Note: There are some products (such as medicines) where it is extremely important to detect the shift, and therefore 2 sigma limits are used and taken very seriously. A finance manager has suggested that you use 2 sigma limits rather than 3 sigma limits to reduce the scrap rate in the show. Explain to him how the scrap rates are not the only cost to consider. Part E Quiz. You will need this completed worksheet plus the graphs for the quiz. Concept questions from the associated lecture will also be included on the quiz