document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #700689 by rothauserc(4718)\"\" \"About 
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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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\n" ); document.write( "We assume that the population of pipes is normally distributed, then
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\n" ); document.write( "chi-squared(1 - a/2) < or = (n-1)s^2/(std. dev.)^2 < or = chi-squared(a/2)
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\n" ); document.write( "where a is the alpha, s is the sample standard deviation, and std. dev. is the population standard deviation
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\n" ); document.write( "a = (1 - 95/100) = 0.05
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\n" ); document.write( "for this problem a = 0.05, degrees of freedom = 11 - 1 = 10, then
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\n" ); document.write( "chi-squared(0.975) = 20.5 and chi-squared(0.025) = 3.25
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\n" ); document.write( "we evaluate (n-1)s^2 / chi-squared for 20.5 and 3.25
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\n" ); document.write( "(11-1) * 14.6^2 / 20.5 = 103.98
\n" ); document.write( "(11-1) * 14.6^2 / 3.25 = 655.88
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\n" ); document.write( "we take the square root of 103.98 and 655.88
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\n" ); document.write( "square root(103.98) = 10.197 approx 10.2
\n" ); document.write( "square root(655.88) = 25.610 approx 25.6
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\n" ); document.write( "the 95% confidence interval is (10.2, 25.6)
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