Question

# pe Grades Course Contents » ... » Assignments » chisqr3.problem • Timer R Evaluate Feedback Print Info The amount of lateral expansion (mils) was determined for a sample of n=11 pulsed-power gas metal ore welds used in LNG ship containment tanks. The resulting sample standard deviation was s=2.98 mils. Assuming normality, derive (give answers to 3 places past decimal) : a) a 95% confidence interval for o2 upper limit = lower limit = * Tries 0/5 a) a 95% confidence interval for o upper limit = lower limit = * Tries 0/5

Answer 1

Chi square critical values at 0.05 level with 10 df = $\chi^{2}$ L = 3.247 , $\chi^{2}$ U = 20.483

95% confidence interval for $\sigma^{2}$ is

(n-1) * S² / $\chi^{2}$ U < $\sigma^{2}$ < (n-1) * S² / $\chi^{2}$ L

10 * 2.98² / 20.483 < $\sigma^{2}$ < 10 * 2.98² / 3.247

4.335 < $\sigma^{2}$ < 27.350

Lower limit = 4.335 , Upper limit = 27.350

b)

Chi square critical values at 0.05 level with 10 df = $\chi^{2}$ L = 3.247 , $\chi^{2}$ U = 20.483

95% confidence interval for $\sigma$ is

sqrt [ (n-1) * S² / $\chi^{2}$ U ] < $\sigma$ < sqrt [ (n-1) * S² / $\chi^{2}$ L]

sqrt [ 10 * 2.98² / 20.483] < $\sigma$ < sqrt [ 10 * 2.98² / 3.247]  

2.082 < $\sigma$ < 5.230

Lower limit = 2.082 , Upper limit = 5.230