document.write( "Question 1196689: A hardware manufacturer produces bolts used to assemble various machines. Assume that the diameter of bolts produced by this manufacturer has an unknown population mean, μ, and the standard deviation is 0.1 mm. Suppose the average diameter of a simple random sample of 50 bolts is 5.11 mm.\r
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document.write( "What is the width of a 95% confidence interval for μ? \n" );
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Algebra.Com's Answer #829659 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Given info: \n" ); document.write( "n = 50 = sample size \n" ); document.write( "xbar = 5.11 = sample mean \n" ); document.write( "sigma = 0.1 = population standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Side note: In many real world situations, we often won't know what sigma is because it's a population parameter. The goal of statistics is to estimate such parameters. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "At 95% confidence, the z critical value is roughly z = 1.96 \n" ); document.write( "Use a table like this \n" ); document.write( "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf \n" ); document.write( "to get that value. Look at the bottom row labeled \"Z\" and above the 95% confidence level.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Use those values to find the following. \n" ); document.write( "E = margin of error \n" ); document.write( "E = z*sigma/sqrt(n) \n" ); document.write( "E = 1.96*0.1/sqrt(50) \n" ); document.write( "E = 0.02771858582251 \n" ); document.write( "E = 0.027719 \n" ); document.write( "The result is approximate\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The value of E is the distance from the center to either endpoint. \n" ); document.write( "Think of it as the radius of interval. \n" ); document.write( "It's half the confidence interval width, so the full width is 2*E = 2*0.027719 = 0.055438\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "You could also follow these steps \n" ); document.write( "L = lower bound \n" ); document.write( "L = xbar - E \n" ); document.write( "L = 5.11 - 0.027719 \n" ); document.write( "L = 5.082281 \n" ); document.write( "U = upper bound \n" ); document.write( "U = xbar + E \n" ); document.write( "U = 5.11 + 0.027719 \n" ); document.write( "U = 5.137719 \n" ); document.write( "The width of the confidence interval is U - L = 5.137719 - 5.082281 = 0.055438 which is the distance from the lower bound L to the upper bound U. \n" ); document.write( "These steps are handy if you are given the confidence interval in the format (L,U) or the format L < mu < U. \n" ); document.write( "Note how U - L = (xbar+E)-(xbar-E) = 2E, so the xbars cancel out and the E's double up.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: 0.055438 approximately \n" ); document.write( "Round that value however needed. \n" ); document.write( " \n" ); document.write( " |