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 14.6 inches.
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Question 1086484: 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 14.6 inches.
Found 2 solutions by rothauserc, jim_thompson5910:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
We use a chi-squared distribution which gives us a confidence interval that is not symmetric about the sample standard deviation
:
We assume that the population of pipes is normally distributed, then
:
chi-squared(1 - a/2) < or = (n-1)s^2/(std. dev.)^2 < or = chi-squared(a/2)
:
where a is the alpha, s is the sample standard deviation, and std. dev. is the population standard deviation
:
a = (1 - 95/100) = 0.05
:
for this problem a = 0.05, degrees of freedom = 11 - 1 = 10, then
:
chi-squared(0.975) = 20.5 and chi-squared(0.025) = 3.25
:
we evaluate (n-1)s^2 / chi-squared for 20.5 and 3.25
:
(11-1) * 14.6^2 / 20.5 = 103.98
(11-1) * 14.6^2 / 3.25 = 655.88
:
we take the square root of 103.98 and 655.88
:
square root(103.98) = 10.197 approx 10.2
square root(655.88) = 25.610 approx 25.6
:
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the 95% confidence interval is (10.2, 25.6)
*******************************************
:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Given Info:
Confidence Level = C = 95% = 0.95
sample size = n = 11
sample standard deviation = s = 14.6
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Based on the sample size, the degrees of freedom (df) is
df = n-1 = 11-1 = 10
Let
L = left chi-square critical value
R = right chi-square critical value
We need to find the L & R values that make  = 0.95)
true. The symbol 
is the Chi-Square symbol
Using a table or calculator those approximate values are
L = 3.247
R = 20.483
For this problem, I used a table. Specifically I used this table. Look at the df = 10 row. Highlight the columns that have 0.975 and 0.025 up top. These values are areas to the right of the values marked in the table. So for example,  = 0.975)
. The middle area has 95% area. The two tails combine to 5% so each tail has area of 0.05/2 = 0.025
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Now we can compute the confidence interval
*s^2}{R}} \le \sigma \le \sqrt{\frac{(n-1)*s^2}{L}})
*(14.6)^2}{20.483}} \le \sigma \le \sqrt{\frac{(11-1)*(14.6)^2}{3.247}})
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The 95% confidence interval for the standard deviation is approximately (10.2, 25.6)
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