Unformatted text preview: that the process is in statistical control with = 1.0002 inches, = .003 inches and subgroup size of 5.
Calculate the estimated proportion of out of specification sleeve inside diameters.
A. .0471
B. .0217
C. .4861
D. .4783
E. .0139 52. If a control chart is used correctly and the necessary corrective actions are taken, then as
the control limits get close to each other, the potential quality of the product _____________.
A. decreases
B. increases
C. stays the same
D. fluctuates 11673 Chapter 01  An Introduction to Business Statistics 53. A _____ is a set of process observations that are examined for the purpose of constructing
control charts.
A. Subgroup
B. Sample
C. Zone
D. Common causes 54. _____ is the sigma level capability divided by three.
A. Process leeway
B. Cpk index
C. R/d2
D. p 55. A control chart on which subgroup means are plotted versus time is a(n) _____ chart.
A.
B. R
C. p
D. C 56. A control chart on which subgroup ranges are plotted versus time is a(n) _____ chart.
A.
B. R
C. p
D. C 57. A control chart on which the proportions of nonconforming units in subgroups of size n
are plotted versus time is a _____ chart.
A.
B. R
C. p
D. C 11674 Chapter 01  An Introduction to Business Statistics 58. Sources of process variation that are inherent to the process design are called _____
causes of variation.
A. common
B. assignable
C. unusual
D. expected 59. Auxiliary lines drawn on a control chart for accomplishing pattern analysis are called
_____ boundaries.
A. sigma
B. range
C. zone
D. mean 60. _____ causes of variation may be remedied by local supervision.
A. Common
B. Assignable
C. Usual
D. Expected 61. A sequence of steadily increasing plot points on a control chart is referred to as a _____.
A. funneling
B. run up
C. cycle
D. out of control 62. A systematic method for analyzing process data in which we monitor process variation is
called _____.
A. control charting
B. sigma level capability
C. statistical process control
D. rational subgrouping 11675 Chapter 01  An Introduction to Business Statistics 63. The distance between natural tolerance limits and customer specifications is called _____.
A. process leeway
B. sigma level
C. Cpk Index
D. zone 64. A(n) _____ process has the ability to produce products or services that meet customer
requirements.
A. in control
B. assignable
C. capable
D. statistical 65. Unusual sources of process variation that can be attributed to specific reasons are called
_____ causes of variation.
A. common
B. assignable
C. usual
D. expected 66. The number of estimated process standard deviations between
specification limit is the ______ of the process.
A. sigma level capability
B. leeway
C. Cpk Index
D. capability and the closest 67. How well a product or a service performs in the market place is called the quality of
______________.
A. performance
B. conformance
C. design
D. control 11676 Chapter 01  An Introduction to Business Statistics 68. How well a process is able to meet the requirements set forth by the process design is
called the quality of _____________.
A. performance
B. conformance
C. design
D. control 69. If
and R charts are used to control a manufacturing process, ______ chart is analyzed
first and brought into a state of statistical control before preparing the _______ chart.
A. R,
B. ,R 70. A unit that fails to meet specifications is called a _____ unit.
A. conforming
B. capable
C. defective
D. common cause 71. _________ is the international standards on quality management and quality assurance
systems. It establishes processes for assuring that goods and services offered by the company
meet a consistent level of quality acceptable to customers.
A. Cpk Index
B. SQC
C. ISO 9000
D. ASQC 72. If
= 5.2,
A. (5.054 5.346)
B. (4.3 6.1)
C. (4.806 5.594)
D. (4.763 5.637) = .3, and n = 4, calculate natural tolerance limits. 11677 Chapter 01  An Introduction to Business Statistics 73. If
= 5.2,
capable?
A. Yes
B. No = .3, and n = 4, and if specifications are (4.6, 5.8), is the process 74. If
= 16.1,
A. (16.065 16.135)
B. (16.088 16.112)
C. (15.745 16.455)
D. (16.055 16.145) = .03, and n = 6, calculate natural tolerance limits. 75. If
= 16.1,
capable?
A. Yes
B. No = .03, and n = 6, and if specifications are (15.9, 16.3), is the process 76. If = 2.0144, = .0972, and 25 subgroups of size 5. Find the UCL and the LCL for the
A. (1.999 2.029)
B. (1.958 2.071)
C. (1.917 2.112)
D. (1.437 2.591) chart. 77. If
= 2.0144,
= .0972, and 25 subgroups of size 5.
Find the UCL and the LCL for the Rchart.
A. (.045 .1498)
B. (.0972 .2055)
C. (0 .2055)
D. (0.2, 114) 11678 Chapter 01  An Introduction to Business Statistics 78. If
= .0139 and 25 shipments of 20 items each were examined, what is the UCL and the
LCL for the pchart?
A. (.0118 .0159)
B. (0 .0401)
C. (.008 .0198)
D. (0 .0924) 79. If 20 samples of size 7 are drawn with
What are the UCL and the LCL for the
A. (32.313 34.347)
B. (30.693 35.697)
C. (27.68 38,98)
D. (32.911 33.749) = 33.33 and = 5.65. chart? 80. If 20 samples of si...
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 Frequency, Frequency distribution, Histogram, AACSB, Statistical charts and diagrams

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