Question 1086484
We use a chi-squared distribution which gives us a confidence interval that is not symmetric about the sample standard deviation
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We assume that the population of pipes is normally distributed, then
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chi-squared(1 - a/2) < or = (n-1)s^2/(std. dev.)^2 < or = chi-squared(a/2)
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where a is the alpha, s is the sample standard deviation, and std. dev. is the population standard deviation
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a = (1 - 95/100) = 0.05
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for this problem a = 0.05, degrees of freedom = 11 - 1 = 10, then
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chi-squared(0.975) = 20.5 and chi-squared(0.025) = 3.25
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we evaluate (n-1)s^2 / chi-squared for 20.5 and 3.25
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(11-1) * 14.6^2 / 20.5 = 103.98
(11-1) * 14.6^2 / 3.25 = 655.88
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we take the square root of 103.98 and 655.88
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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)
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