document.write( "Question 1151506:  A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.04 with 95% confidence if
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document.write( "(a) she uses a previous estimate of 0.52?
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document.write( "(b) she does not use any prior estimates?
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| Algebra.Com's Answer #773280 by rothauserc(4718)     You can put this solution on YOUR website! Margin of error(ME) = 0.04 \n" ); document.write( ": \n" ); document.write( "critical value(CV) for a 95% confidence is 1.96 \n" ); document.write( ": \n" ); document.write( "sample size(n) = ( CV^2 * p * (1-p) ) / ME^2 \n" ); document.write( ": \n" ); document.write( "(a) n = ( 1.96^2 * 0.52 * (1-0.52) ) / 0.04^2 = 599.2896 is approximately 600 \n" ); document.write( ": \n" ); document.write( "(b) when proportion is not known use p = 0.50 \n" ); document.write( ": \n" ); document.write( "n = ( 1.96^2 * 0.50 * (1-0.50) ) / 0.04^2 = 600.25 is approximately 601 \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |