The amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils. Assuming normality, derive a 95% CI for σ2 and for σ. (Round your answers to two decimal places.)

95% Confidence interval for σ2 and for σ is (3.33 , 38.85) and (1.82 , 6.23) respectively.

Step-by-step explanation:

We are given that the amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils.

Assuming data follows normal distribution.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

        P.Q. = ~

where, s = sample standard deviation = 2.83 mils

          = population variance

            = population standard deviation

           n = sample size = 7

So, 95% confidence interval for population variance, is;

P(1.237 < < 14.45) = 0.95 {As the table of at 6 degree of freedom

                                                     gives critical values of 1.237 & 14.45}

P(1.237 < < 14.45) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = (  ,  )

                                                  = ( , )

                                                  = (3.33 , 38.85)

95% C.I. for population standard deviation,  = ( , )

                                                                            = (1.82 , 6.23)

Therefore, 95% confidence interval for the population variance (σ2) and population standard deviation (σ) are (3.33 , 38.85) and (1.82 , 6.23) respectively.

the answer to your question a or b but i am pretty sure it is b. i hope this : )

step-by-step explanation:

the correct answer to your question is a. 4^4

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