SOLUTION: Find the 95% confidence interval for the standard deviation of the lengths of pipes if a sample of 11 pipes has a standard deviation of 12 inches.
A.
70.3 < < 443.5
B.
13.0
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Question 1040410: Find the 95% confidence interval for the standard deviation of the lengths of pipes if a sample of 11 pipes has a standard deviation of 12 inches.
A.
70.3 < < 443.5
B.
13.0 << 11.0
C.
8.4 << 21.1
D.
133.0 << 155.0
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let's set up some notation.
XL = left chi-square critical value
XR = right chi-square critical value
The X will stand for the greek letter chi (pronounced "kai", rhymes with "sky"). R for "right" and L for "left.
XL and XR are the chi-square critical values such that P(XL < chi-square < XR) = 0.95
basically the area between XL and XR is 0.95
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Use a chi-square table of values. Highlight the df = 10 row (n = 11, so df = n-1 = 11-1 = 10).
This row is marked in red.
Keep in mind that if the confidence level is 95%, then alpha = 1-CL = 1-0.95 = 0.05. Cut this in half to get 0.05/2 = 0.025
Mark the column that has 0.025 at the top. I marked in blue. The blue and red boxes intersect at the value 20.483. Let's call this XR, so XR = 20.483 approximately
Then we compute 1 - 0.025 = 0.975
Mark the column that has 0.975 at the top. This is marked in green. The green and red boxes intersect at 3.247
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So we just found that
XL = 3.247
XR = 20.483
both are approximate. These are the chi-square critical values.
we will plug in
n = 11 (sample size)
s = 12 (standard deviation
Use the values to compute the confidence interval
*s^2}{XR}} < \sigma < \sqrt{\frac{(n-1)*s^2}{XL}})
*12^2}{20.483}} < \sigma < \sqrt{\frac{(11-1)*12^2}{3.247}})
where 
is the greek letter sigma and it is the population standard deviation
So the answer is choice C
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