Natural tolerance limit problem 2 25 points a process

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Natural Tolerance Limit
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Problem 2 (25 points): A process is in statistical control with . The control chart uses a sample size of . Specifications are at . The quality characteristic is normally distributed. (a) Estimate the potential capability of the process. (b) Estimate the actual process capability. (c) Calculate and compare the PCRs and . (d) How much improvement could be made in process performance if the mean could be centered at the nominal value? (Hint: compare the probabilities of the unit being out of the specification limits) USL = 40 + 5 = 45; LSL = 40 – 5 = 35 (a) Potential: (b) Actual: (c) The closeness of estimates for C p , C pk , C pm , and C pkm indicate that the process mean is very close to the specification target. (d) The current fraction nonconforming is: If the process mean could be centered at the specification target, the fraction nonconforming would be: Problem 3 (25 points): A normally distributed process has specifications of LSL = 75 and USL = 85 on the output. A random sample of 25 parts indicates that the process is centered at the middle of the specification band and the standard deviation is s = 1.4 (a) Find a point estimate of . (b) Find a 95% confidence interval on . Comment on the width of this interval. LSL = 75; USL = 85; n = 25; S = 1.4
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(a) (b) This confidence interval is wide enough that the process may either be capable or far from it. Problem 4 (25 points) Ten parts are measured three times by the same operator in a gauge capability study. The data are shown in the following Table. (a) Describe the measurement error that results from the use of this gauge. (b) Estimate the total variability and the product variability. (c) What percentage of the total variability is due to the gauge? (d) If specifications on the part are at 100 ± 15, find the P/T ratio for this gauge. Comment on the adequacy of the gauge.
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The chart has a couple out-of-control points, and the R chart is in control. This indicates that the operator is not having difficulty making consistent measurements. (b) (c) (d) USL = 100 + 15 = 115; LSL = 100 – 15 = 85
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