Homework1 Solution

Homework1 Solution - 4.1. n = 15; x = 8.2535 cm; = 0.002 cm...

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Unformatted text preview: 4.1. n = 15; x = 8.2535 cm; = 0.002 cm (a) = 8.25, = 0.05 Test H : = 8.25 vs. H 1 : 8.25. Reject H if | Z | > Z /2 . 8.2535 8.25 6.78 0.002 15 x Z n -- = = = Z /2 = Z 0.05/2 = Z 0.025 = 1.96 Reject H : = 8.25, and conclude that the mean bearing ID is not equal to 8.25 cm. (b) P-value = 2[1 - ( Z )] = 2[1 - (6.78)] = 2[1 - 1.00000] = 0 (c) sqrt means square root. 8.2535-1.96(0.002/sqrt(15)) <= <= 8.2535+1.96(0.002/sqrt(15)) 8.2525 <= <= 8.2545 5.1 Chance or common causes of variability represent the inherent, natural variability of a process - its background noise. Variation resulting from assignable or special causes represents generally large, unsatisfactory disturbances to the usual process performance. Assignable cause variation can usually be traced, perhaps to a change in material, equipment, or operator method. A Shewhart control chart can be used to monitor a process and to identify occurrences of assignable causes. assignable causes....
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This note was uploaded on 10/25/2010 for the course ISE 426 taught by Professor Palmer during the Fall '06 term at USC.

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Homework1 Solution - 4.1. n = 15; x = 8.2535 cm; = 0.002 cm...

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