The amount of lateral expansion (mils) was determined for a sample of n = 5 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% CI for ² and for o. (Round your answers to two decimal places.) CI for ² mils² CI for o mils

A First Course in Probability (10th Edition)
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Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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The amount of lateral expansion (mils) was determined for a sample of n = 5 pulsed-power gas metal arc welds used in LNG ship containment
tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% CI for ² and for 6. (Round your answers to
two decimal places.)
CI for ²
mils2
CI for o
mils
Transcribed Image Text:The amount of lateral expansion (mils) was determined for a sample of n = 5 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% CI for ² and for 6. (Round your answers to two decimal places.) CI for ² mils2 CI for o mils
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