Please answer with an explanation on why whoever's method is wrong thank you
2 answers:
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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. = ~
where, s = sample standard deviation = 2.83 mils
= population variance
= population standard deviation
n = sample size = 7
<em>So, 95% confidence interval for population variance, </em> <em>is;</em>
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.