Question

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

Answer 1

SOLUTION:

From given data,

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.

sample standard deviation was s = 2.81 mils

sample of n = 6

degree of freedom = n-1 = 6-1 = 5

95% written as a decimal is 0.95.

1 – 0.95 = 0.05 = α and α/2 = 0.025.

.....using chi square probability table.

We can use here,confidence interval formula.

Assuming normality, derive a 95% CI for σ² and for σ.

just take square root for both sides