Exam 17: Process Improvement Using Control Charts

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A p chart is a control chart on which the proportions of nonconforming units are plotted versus time.

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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 X\overline { \overline { X } } = 1)0002 inches, R\overline { \mathrm { R } } = )003 inches,and subgroup size of 5.Calculate the estimated proportion of out-of-specification sleeve inside diameters.

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If X\overline { \overline { X } } = 2)0144, Rˉ\bar { R } = )0972,and there are 25 subgroups of size 5,find the UCL and the LCL for the xˉ\bar { x } Chart.

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If the process variability steadily increases,we would observe:

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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,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 = x\overline { \overline{ x } } = 50.05,and average range of 8.85.Is the process capable of meeting the specifications?

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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 x\overline { \overline{ x } } = 4)50 and an average range = 1.01.Calculate the center line for the R chart.

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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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A control chart is a graph whose purpose is to detect assignable causes of variation in a process.

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For a given control chart,zone boundaries consist of the UCL and LCL.

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In a manufacturing process,if the limits for a control chart are set too _____________,the risk of not investigating a special cause of variation and possibly not taking a corrective action ____________.

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Process leeway is the distance between natural tolerance limits and control limits.

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Suppose that X\overline { \mathrm { X } } And R charts based on subgroups of size 4 are being used to monitor the tire diameter of a new radial tire.The X\overline { \mathrm { X } } And R charts are found to be in statistical control,with X=50.5,Rˉ=1.25\overline { \overline { X } } = 50.5 , \bar { R } = 1.25 Inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Compute the natural tolerance limits for this process.

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If 20 samples of size 7 are drawn,with X\overline { \overline { X } } = 33.33 and Rˉ\bar { R } = 5)65,what are the LCL and the UCL for the Xˉ\bar { X } Chart?

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Rational subgrouping allows us to detect changes ____________ subgroups.

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A foreman wants to use an X\overline { \mathrm { X } } Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. Sample 1 1.99 2.01 2.02 2.02 2 2.00 2.00 2.01 2.01 3 1.98 1.99 2.01 1.98 4 2.01 2.02 2.02 1.99 5 1.99 1.99 2.01 1.99 6 2.03 1.98 2.03 2.04 Calculate the average range.

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If X\overline { \overline { X } } = 5)2, Rˉ\bar { R } = )3,and n = 4,calculate the natural tolerance limits.

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Common causes of variation represent the inherent variability of a given process.

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A fastener company produces bolts with a nominal (target)length of 2.00 inches.The specifications are 2.00 ± .006 inches.If the process mean is 2.001 and the process standard deviation is .0016,determine the estimated standard deviations of the leeway for this process.

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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 X\overline { \overline { X } } = 1)0002 inches, R\overline { \mathrm { R } } = )003 inches,and subgroup size of 5.What is the sigma level capability for this process?

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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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