Exam 15: Process Improvement Using Control Charts

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The number of estimated process standard deviations between The number of estimated process standard deviations between   and the closest specification limit is the _____________ of the process. and the closest specification limit is the _____________ of the process.

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When employing measurement data to study a process,the __________ monitors the process variation.

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If If   = 2.0144,   = .0972,and there are 25 subgroups of size 5,find the UCL and the LCL for the   chart. = 2.0144, If   = 2.0144,   = .0972,and there are 25 subgroups of size 5,find the UCL and the LCL for the   chart. = .0972,and there are 25 subgroups of size 5,find the UCL and the LCL for the If   = 2.0144,   = .0972,and there are 25 subgroups of size 5,find the UCL and the LCL for the   chart. chart.

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A motorcycle manufacturer produces the parts for its vehicles in different locations and transports them to its plant for assembly.In order to keep the assembly operations running efficiently,it is vital that all parts be within specification limits.One important part used in the assembly is the engine camshaft,and one important quality characteristic is the case hardness depth.Specifications state that the hardness depth must be between 3.0 mm and 6.0 mm.To investigate the process,the quality control engineer selected 25 daily subgroups of n = 5 and measured the hardness depth.The process yielded a mean of the means A motorcycle manufacturer produces the parts for its vehicles in different locations and transports them to its plant for assembly.In order to keep the assembly operations running efficiently,it is vital that all parts be within specification limits.One important part used in the assembly is the engine camshaft,and one important quality characteristic is the case hardness depth.Specifications state that the hardness depth must be between 3.0 mm and 6.0 mm.To investigate the process,the quality control engineer selected 25 daily subgroups of n = 5 and measured the hardness depth.The process yielded a mean of the means   = 4.50 and an average range = 1.01.Find the sigma level capability of the process. = 4.50 and an average range = 1.01.Find the sigma level capability of the process.

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If If   = 5.2,   = .3,and n = 4,calculate the natural tolerance limits. = 5.2, If   = 5.2,   = .3,and n = 4,calculate the natural tolerance limits. = .3,and n = 4,calculate the natural tolerance limits.

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If the ________ limits are within the ________ limits,then it can be concluded that the process is __________.

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A foreman wants to use an A foreman wants to use an   chart to control the average length of the bolts manufactured.He has recently collected the six samples given below.   Determine the LCL and the UCL for the R chart. chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. A foreman wants to use an   chart to control the average length of the bolts manufactured.He has recently collected the six samples given below.   Determine the LCL and the UCL for the R chart. Determine the LCL and the UCL for the R chart.

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Sources of process variations that are inherent to the process design are called ___________ causes of variation.

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The quality of an electronic component used in manufacturing cell phones is monitored with a p chart.In the last 20 days,daily samples of 75 units resulted in the following number of defective units per sample: 8,4,3,7,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,and 5.Find the LCL and UCL for the p chart.

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A fastener company produces a certain type of bolt for the automobile industry with a nominal (target)length of 2.00 inches.The specifications for the length of the bolt are 2.00 ± .006 inches.An automobile manufacturer will only purchase from this company if the sigma level of capability of the process is at least 4.If the process mean is equal to 2.001,determine the maximum process standard deviation necessary for the fastener manufacturing company in order to qualify as a supplier for the automobile manufacturing company.

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A sequence of steadily decreasing points on a control chart is called run down.

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If the process variability steadily increases,we would observe:

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A cause-and-effect diagram enumerates the potential causes of an undesirable effect on the process to discover sources of process variation and to identify opportunities for process improvement.

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A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 ± .005 inches.Assume that the process is in statistical control with A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 ± .005 inches.Assume that the process is in statistical control with   = 1.0002 inches,   = .003 inches,and subgroup size of 5.Calculate the control limits for the   chart. = 1.0002 inches, A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 ± .005 inches.Assume that the process is in statistical control with   = 1.0002 inches,   = .003 inches,and subgroup size of 5.Calculate the control limits for the   chart. = .003 inches,and subgroup size of 5.Calculate the control limits for the A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 ± .005 inches.Assume that the process is in statistical control with   = 1.0002 inches,   = .003 inches,and subgroup size of 5.Calculate the control limits for the   chart. chart.

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Suppose that a tire manufacturer uses Suppose that a tire manufacturer uses   and R charts based on subgroups of size 4 to monitor tire diameter.The   andR charts are found to be in statistical control,with   inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Find the sigma level capability of the process. and R charts based on subgroups of size 4 to monitor tire diameter.The Suppose that a tire manufacturer uses   and R charts based on subgroups of size 4 to monitor tire diameter.The   andR charts are found to be in statistical control,with   inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Find the sigma level capability of the process. andR charts are found to be in statistical control,with Suppose that a tire manufacturer uses   and R charts based on subgroups of size 4 to monitor tire diameter.The   andR charts are found to be in statistical control,with   inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Find the sigma level capability of the process. inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Find the sigma level capability of the process.

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Unusual sources of variation that can be attributed to specific causes are called the common causes of process variation.

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How well the design of the product meets and exceeds the needs and expectations of the customers is called the quality of performance.

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Sigma level capability is the number of estimated process standard deviations between the estimated process mean and the specification limit closest to the estimated process mean.

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Common causes of process variation:

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A manufacturer of windows produces one type that has a plastic coating.The specification limits for the plastic coating are 30 and 70.From time to time the plastic coating can become uneven.Therefore,in order to keep the coating as even as possible,thickness measurements are periodically taken at four different locations on the window.15 subgroups were observed,each consisting of four thickness measurements,with the following results: mean of the means = A manufacturer of windows produces one type that has a plastic coating.The specification limits for the plastic coating are 30 and 70.From time to time the plastic coating can become uneven.Therefore,in order to keep the coating as even as possible,thickness measurements are periodically taken at four different locations on the window.15 subgroups were observed,each consisting of four thickness measurements,with the following results: mean of the means =   = 50.05,and average range of 8.85.Calculate the control limits for the R chart. = 50.05,and average range of 8.85.Calculate the control limits for the R chart.

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