document.write( "Question 427710: A researcher collected sample data for 12 women ages 18 to 24. The sample had a mean serum cholesterol level (measured in mg/100 mL) of 189.9 , with a standard deviation of 5.5. Assuming that serum cholesterol levels for women ages 18 to 24 are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. \n" ); document.write( "
Algebra.Com's Answer #297433 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A researcher collected sample data for 12 women ages 18 to 24. \n" ); document.write( "The sample had a mean serum cholesterol level (measured in mg/100 mL) of 189.9 , with a standard deviation of 5.5. \n" ); document.write( "Assuming that serum cholesterol levels for women ages 18 to 24 are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. \n" ); document.write( "--- \n" ); document.write( "x-bar = 189.9 \n" ); document.write( "ME = 2.5758*5.5/sqrt(12) = 4.40897 \n" ); document.write( "--- \n" ); document.write( "99%CI: 189.9-4.40897 < u < 189.9+4.40897 \n" ); document.write( "99%CI: 185.81 < u < 194.30897 \n" ); document.write( "================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "========= \n" ); document.write( " |