document.write( "Question 1196386: You measure 21 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 8.3 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight.\r
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document.write( "Give your answer as a decimal, to two places\r
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document.write( " ounces \n" );
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Algebra.Com's Answer #829221 by Theo(13342)![]() ![]() You can put this solution on YOUR website! sample size = 21 \n" ); document.write( "sample mean = 45 \n" ); document.write( "population standard deviation = 8.3 \n" ); document.write( "sample standard error = 8.3 / sqrt(21) = 1.811208489. \n" ); document.write( "confidence level = .99 \n" ); document.write( "critical alpha = .01 / 2 = .005. \n" ); document.write( "critical z-score = plus or minus 2.575829303. \n" ); document.write( "z-score formula = z = (x - m) / s \n" ); document.write( "z is the critical z-score. \n" ); document.write( "x is the critical raw score \n" ); document.write( "m is the sample mean \n" ); document.write( "s is the standard error. \n" ); document.write( "solving for x, you get: \n" ); document.write( "x1 = z1 * se + 45 = -2.57582903 * 1.811208489 + 45 = 40.3346361. \n" ); document.write( "x2 = z2 * se + 45 = 2.57582903 * 1.811208489 + 45 = 49.6653639. \n" ); document.write( "maximum margin of error = (x2 - x1) / 2 = 4.6653639. \n" ); document.write( "your solution is 4.67 rounded to two decimal places. \n" ); document.write( "confirmation that the confidence level = 99% is shown below. \n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( " |