Question

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.)

Answer 1

Answer:

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. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = 2.83 mils

          $\sigma^{2}$ = population variance

           $\sigma$ = population standard deviation

           n = sample size = 7

So, 95% confidence interval for population variance, $\sigma^{2}$ is;

P(1.237 < $\chi^{2} __n_-_1$ < 14.45) = 0.95 {As the table of at 6 degree of freedom

                                                     gives critical values of 1.237 & 14.45}

P(1.237 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.45) = 0.95

P( $\frac{ 1.237}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.45}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{ (n-1)s^{2}}{14.45 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{1.237 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.45 }$  , $\frac{ (n-1)s^{2}}{1.237 }$  )

                                                  = ( $\frac{ (7-1) \times 2.83^{2}}{14.45 }$ , $\frac{ (7-1) \times 2.83^{2}}{1.237 }$ )

                                                  = (3.33 , 38.85)

95% C.I. for population standard deviation, $\sigma$  = ( $\sqrt{3.33}$ , $\sqrt{38.85}$ )

                                                                            = (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.