document.write( "Question 1137977: A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 2 points with 99 % confidence assuming s equals 19.2 based on earlier studies? Suppose the doctor would be content with 90 % confidence. How does the decrease in confidence affect the sample size required?\r
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document.write( "A 99% confidence level requires....BLANK....subjects? \n" );
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Algebra.Com's Answer #755857 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 99%CI is tdf=?, 0.995*s/sqrt(n); 2=t*19.2/sqrt(n) \n" ); document.write( "square both sides and move the sqrt(n) over \n" ); document.write( "4n=t^2*19.2^2 \n" ); document.write( "n=t^2(368.64/4)=t^2*92.16 \n" ); document.write( "Since the t-value will be at least 2 and probably larger, the sample size will be a minimum of 370. At that size, z can be used\r \n" ); document.write( "\n" ); document.write( "half-interval of 2=1.96*19.2/sqrt(n) \n" ); document.write( "4n=1.96^2*19.2^2 \n" ); document.write( "n=354.04 or 355.\r \n" ); document.write( "\n" ); document.write( "The decrease in the confidence interval t 90% changes the t/z value to 1.645, and that is about 5/8 as much as the 99% confidence \n" ); document.write( "Because we are dealing with squares here, that will be about 25/64 or just under 40% of the prior sample size. Lower confidence allows smaller samples, all else remaining constant.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |