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Unformatted text preview: ts to use an chart to control the average length of the bolts manufactured. He has recently collected the six samples given below. Calculate the average range. A. .036 B. .18 C. .030 D. .05 0.03 AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 4 Topic: X-bar Control Charts 1-1730 Chapter 01 - An Introduction to Business Statistics 90. 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 A. (1.975 2.035) B. (1.915 2.095) C. (1.983 2.027) D. (1.991 2.019) chart. 1.983 to 2.027. All six sample means fall within the control limits. 2.005 ± (.483)(.03) = 1.983 to 2.027 All six sample means fall within the control limits. AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 4 Topic: X-bar Control Charts 1-1731 Chapter 01 - An Introduction to Business Statistics 91. 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. A. (0 .0685) B. (0 .076) C. (0 .03) D. (0 .0601) 0 and .0685. All six sample ranges fall within the control limits. UCLR= (2.004) (.03) = .0685 LCLR = none All six sample ranges fall within the control limits. AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 4 Topic: Range Control Charts 1-1732 Chapter 01 - An Introduction to Business Statistics 92. A fastener company produces bolts with a nominal (target) length of 2.00 inches. The specifications are 2.00 ± .006 inches. Determine the upper specification limit and the lower specification limit for this process. A. (1.982 2.018) B. (1.94 2.06) C. (1.994 2.006) D. (1.99 2.01) 1.994 and 2.006 USL = 2 + .006 = 2.006 LSL = 2 - .006 = 1.994 AACSB: Analytic Bloom's: Application Difficulty: Easy Learning Objective: 6 Topic: Process capability 1-1733 Chapter 01 - An Introduction to Business Statistics 93. 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 equal to the target value and the process standard deviation is .0016 determine the value of Cpk. A. 1.250 B. 3.750 C. 2.00 D. 1.998 1.25 USL = 2 + .006 = 2.006 LSL = 2 - .006 = 1.994 AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 6 Topic: Process capability 1-1734 Chapter 01 - An Introduction to Business Statistics 94. 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 value of Cpk. A. 1.042 B. 1.25 C. 2.001 D. 1.458 1.042 USL = 2 + .006 = 2.006 LSL = 2 - .006 = 1.994 AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 6 Topic: Process capability 1-1735 Chapter 01 - An Introduction to Business Statistics 95. 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 leeway for this process. A. .0016 B. .0048 C. 1.375 D. .125 0.125 USL = 2 + .006 = 2.006 LSL = 2 - .006 = 1.994 Estimated number of standard deviations of process leeway = 3.125 - 3 = 0.125 AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 6 Topic: Process capability 1-1736 Chapter 01 - An Introduction to Business Statistics 96. 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 manufacturing company 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. A. .0125 B. .005 C. .00125 D. .00025 .00125 AACSB: Analytic Bloom's: Application Difficulty: Hard Learning Objective: 6 Topic: Process capability 1-1737 Chapter 01 - An Introduction to Business Statistics 97. Suppose that and R charts are based on subgroups of size four are being used to monitor the tire diameter size of a new radial tire manufactured by a tire company. The and R charts are found to be in statistical control with inches. A histogram of the tire diameter measurements indicates that distribution of these measurements is approximately normally distributed. If the tire diameter specifications are 50 inches ± 1 inch, is the process capable of meeting the specifications? A. Yes B. No No, Natural tolerance limits are not within the specification limits. LSL = 50 - 1 = 49 USL = 50 + 1 = 51 No, 51.42 > 51. Natural tolerance limits are not within the specification limits. AACSB: Analytic Bloom's: Application Difficulty: Medium Learning Objective: 6 Topic: Process capabil...
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This document was uploaded on 01/20/2014.

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