The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power...

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.82 mils. Assuming normality, derive a 95% CI for σ² and for σ. (Round your answers to two decimal places.)

CI for σ²

(_______,_______)mils²

CI for σ

(_______, _____) mils

Expert Solution

orchestra answered 2 years ago

The amount of lateral expansion (mils) was determined for a sample of n = 6 pulsed-power...

The amount of lateral expansion (mils) was determined for a sample of n = 6 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.81 mils. Assuming normality, derive a 95% CI for σ2 and for σ. (Round your answers to two decimal places.) CI for σ2 ( __________ , __________ ) mils2 CI for σ ( __________ , __________ ) mils2

The amount of lateral expansion (mils) was determined for a sample of n = 9

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