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.

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above