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Homework1 Solution

# Homework1 Solution - 4.1 n = 15 x = 8.2535 cm = 0.002 cm(a...

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4.1. n = 15; x = 8.2535 cm; σ = 0.002 cm (a) μ 0 = 8.25, α = 0.05 Test H 0 : μ = 8.25 vs. H 1 : μ 8.25. Reject H 0 if | Z 0 | > Z α /2 . 0 0 8.2535 8.25 6.78 0.002 15 x Z n μ σ - - = = = Z α /2 = Z 0.05/2 = Z 0.025 = 1.96 Reject H 0 : μ = 8.25, and conclude that the mean bearing ID is not equal to 8.25 cm. (b) P -value = 2[1 - Φ ( Z 0 )] = 2[1 - Φ (6.78)] = 2[1 - 1.00000] = 0 (c) sqrt means square root. 8.2535-1.96(0.002/sqrt(15)) <= μ <= 8.2535+1.96(0.002/sqrt(15)) 8.2525 <= μ <= 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. There is a high probability that an assignable cause has occurred when a plot point is outside the chart's control limits. By promptly identifying these occurrences and acting

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

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