document.write( "Question 1001735: Find the sample size.
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document.write( "Margin of error = $125, confidence level = 95%, standard deviation = $500 \n" );
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Algebra.Com's Answer #618847 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Assuming I can use a normal distribution, z*SE=half width of confidence interval \n" ); document.write( "125=1.96*(s/sqrt(n)) \n" ); document.write( "125sqrt(n)=1.96*(500)=980 \n" ); document.write( "sqrt(n)=980/125 \n" ); document.write( "n=(980/125)^2 \n" ); document.write( "61.47 rounded upward to 62. \n" ); document.write( "I could go back to the t-table, where t df=60(0.95)=2 \n" ); document.write( "That would give a sample size of 64 (1000/125)^2 \n" ); document.write( " |