document.write( "Question 1054503: 91 adults selected randomly from a town,61 have health insurance. Find the 90% confidence interval for the true proportion of all adults in the town who have onsurance. \n" ); document.write( "
Algebra.Com's Answer #669741 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! 91 adults selected randomly from a town,61 have health insurance. Find the 90% confidence interval for the true proportion of all adults in the town who have insurance. \n" ); document.write( "---- \n" ); document.write( "sample proportion = p-hat = 61/91 = 0.67 \n" ); document.write( "------ \n" ); document.write( "ME = z*sqrt(p*q/n) = 1.645*sqrt(0.67*0.33/91) = 1.645*0.05 = 0.08 \n" ); document.write( "------------------------ \n" ); document.write( "90% CI:: 0.67-0.08 < p < 0.67+0.08 \n" ); document.write( "90% CI:: 0.59 < p < 0.75 \n" ); document.write( "----------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "------------------- \n" ); document.write( " |