Assignment 4

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West Chester University of Pennsylvania *

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341

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Statistics

Date

Jan 9, 2024

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Assignment 4 (Please show your work, formulas, etc.) Problem 1 A metal fabricator produces connecting rods with an outer diameter that has a 1 ± 0.04 inch specification. A machine operator takes several sample measurements over time and determines the sample mean outer diameter to be 1.002 inches with a standard deviation of 0.009 inch. a. Calculate the process capability index for this example. (Round your answer to 3 decimal places.) LSL 1 – 0.04 = 0.96 USL 1 + 0.04 = 1.04 1.002- 0.96 / 3 (0.009), 1.04 – 1.002/ 3(0.009) 0.038/ 0.027= 1.407 , 0.042 /0.027 = 1.556 1.407 > 1 is capable b. Is the process capable of producing the desired quality? Explain Yes, it is capable of producing the desired quality given that it is greater than one. Problem 2 Ten samples of 15 parts each were taken from an ongoing process to establish a p -chart for control. The samples and the number of defectives in each are shown in the following table: 3/15= 0.2 3/15= 0.2 1/15=0.066 2/15=0.133 0 3/15= 0.2 3/15 = 0.2 2/15= 0.133 0 3/15= 0.2 20 a. Determine the p, S p , UCL and LCL for a p -chart of 95 percent confidence (1.96 standard deviations). (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 3 decimal places. Round any negative LCL up to zero.) P-bar = 20/10 x 15= 20/150= 0.133 sp = √0.133(1-0.133) / 15 = 0.088 UCL= 0.133+1.96 x 0.088= 0.305 LCL=0.133-1.96 x 0.088= - 0.039 >> 0 SAMPLE n Number of Defective Items in the Sample 1 15 3 2 15 3 3 15 1 4 15 2 5 15 0 6 15 3 7 15 3 8 15 2 9 15 0 10 15 3
[LCL, UCL] = [0, 0.305] b. What comments can you make about the process? Why? The Process is in statistical control given that no sample proportions exceed the UCL or LCL
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