I need help trying yo figure out how to find the confidence interval for varianc

Question

I need help trying yo figure out how to find the confidence interval for variance and for SD. Can you do a step by step on how to do so.

The amount of lateral expansion (in mils) was determined for a sample of n = 16 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% confidence interval for , the population standard deviation. (Hint: You will first need to come up with a 95% confidence interval for the variance)

Explanation / Answer

CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.03
^2 left = 1 - ^2 right = 1 - 0.025 = 0.98
the two critical values ^2 left, ^2 right at 15 df are 27.488 , 6.5
s.d( s )=2.84
sample size(n)=16
confidence interval for ^2= [ 15 * 8.066/27.488 < ^2 < 15 * 8.066/6.5 ]
= [ 120.984/27.488 < ^2 < 120.984/6.503 ]
[ 4.401 < ^2 < 19.32 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (4.401) < < sqrt(19.32), ]
= [ 2.098 < < 4.395 ]

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