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