document.write( "Question 1026086: There are estimates of the margin of error for confidence levels other than 95%. For a 90% confidence level and a sample size of n, the margin of error is approximately
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document.write( "82/ sq root of n %
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document.write( "Suppose we wish to have a margin of error of 2% with a 90% confidence level. Approximately how many people should we interview?
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Algebra.Com's Answer #641338 by stanbon(75887) ![]() You can put this solution on YOUR website! Suppose we wish to have a margin of error of 2% with a 90% confidence level. Approximately how many people should we interview? \n" ); document.write( "--- \n" ); document.write( "n = [1.645/0.02]^2*(1/2)^2 = 1691 when rounded up \n" ); document.write( "----------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "---------- \n" ); document.write( " |