document.write( "Question 1119834: The level of nitrogen oxides (NOX) in the exhaust of a particular car model varies with mean 0.9 grams per mile and standard deviation 0.2 grams per mile .\r
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document.write( "(a) What sample size is needed so that the standard deviation of the sampling distribution is 0.01 grams per mile ? \n" );
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Algebra.Com's Answer #735495 by Theo(13342)![]() ![]() You can put this solution on YOUR website! mean = .9\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "standard deviation = .2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "standard error = standard deviation of sampling distribution.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "formula for that is s = standard deviation / square root of sample size.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to make s = .01, the formula becomes .01 = .2 / sqrt(n)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n is the sample size.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "multiply both sides of this equation by sqrt(n) and divide both sides of this equation by .01 to get:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "sqrt(n) = .2 / .01.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for sqrt(n) to get sqrt(n) = 20\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for n to get n = 20^1 = 400.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when n = 400, the standard error will be .2 / sqrt(400) = .2 / 20 = .01.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |