document.write( "Question 1168750: You measure 30 textbooks' weights, and find they have a mean weight of 78 ounces. Assume the population standard deviation is 2.3 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight.\r
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document.write( "Give your answers as decimals, to two places\r
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document.write( "< μ < \n" );
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Algebra.Com's Answer #793361 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Given information: \n" ); document.write( "mean = xbar = 78 \n" ); document.write( "standard deviation = sigma = 2.3 \n" ); document.write( "sample size = n = 30 \n" ); document.write( "confidence level = 99%\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A 99% confidence level means the critical value is approximately z = 2.58 which you find using a calculator or table.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The lower value of the confidence interval is \n" ); document.write( "L = xbar - z*sigma/sqrt(n) \n" ); document.write( "L = 78 - 2.58*2.3/sqrt(30) \n" ); document.write( "L = 78 - 1.083395 \n" ); document.write( "L = 76.916605 \n" ); document.write( "L = 76.92\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The upper value of the confidence interval is \n" ); document.write( "U = xbar + z*sigma/sqrt(n) \n" ); document.write( "U = 78 + 2.58*2.3/sqrt(30) \n" ); document.write( "U = 78 + 1.083395 \n" ); document.write( "U = 79.083395 \n" ); document.write( "U = 79.08\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The 99% confidence interval is therefore \n" ); document.write( "76.92 < μ < 79.08\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |