document.write( "Question 1001734: Find the margin of error and confidence interval.
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document.write( "95% confidence, n=100, xbar=9500, standard deviation = 12,345 \n" );
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Algebra.Com's Answer #618855 by Boreal(15235) You can put this solution on YOUR website! The SE is s/sqrt(n)=1234.5 \n" ); document.write( "n=100 is large enough to use the z-distribution, and z(0.95)-1.96 \n" ); document.write( "1.96*SE=2419.62, That is the half-width of the CI and the margin of error when said +/-. \n" ); document.write( "(9500-2419.6, 9500+2419.6) \n" ); document.write( "(7080.4, 11919.6) \n" ); document.write( " |