Answer :
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. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, s = sample standard deviation = 2.83 mils
[tex]\sigma^{2}[/tex] = population variance
[tex]\sigma[/tex] = population standard deviation
n = sample size = 7
So, 95% confidence interval for population variance, [tex]\sigma^{2}[/tex] is;
P(1.237 < [tex]\chi^{2} __n_-_1[/tex] < 14.45) = 0.95 {As the table of at 6 degree of freedom
gives critical values of 1.237 & 14.45}
P(1.237 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 14.45) = 0.95
P( [tex]\frac{ 1.237}{(n-1)s^{2} }[/tex] < [tex]\frac{1 }{\sigma^{2} }[/tex] < [tex]\frac{ 14.45}{(n-1)s^{2} }[/tex] ) = 0.95
P( [tex]\frac{ (n-1)s^{2}}{14.45 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{ (n-1)s^{2}}{1.237 }[/tex] ) = 0.95
95% confidence interval for [tex]\sigma^{2}[/tex] = ( [tex]\frac{ (n-1)s^{2}}{14.45 }[/tex] , [tex]\frac{ (n-1)s^{2}}{1.237 }[/tex] )
= ( [tex]\frac{ (7-1) \times 2.83^{2}}{14.45 }[/tex] , [tex]\frac{ (7-1) \times 2.83^{2}}{1.237 }[/tex] )
= (3.33 , 38.85)
95% C.I. for population standard deviation, [tex]\sigma[/tex] = ( [tex]\sqrt{3.33}[/tex] , [tex]\sqrt{38.85}[/tex] )
= (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.