document.write( "Question 440970: . In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Construct a 95% confidence interval for our estimate of the percentage of T.V. viewers who watch network news programs \n" ); document.write( "
Algebra.Com's Answer #304560 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Construct a 95% confidence interval for our estimate of the percentage of T.V. viewers who watch network news programs \n" ); document.write( "------------------------------ \n" ); document.write( "p-hat = 0.4 \n" ); document.write( "ME = z*sqrt(pq/n) \n" ); document.write( "ME = 1.96*sqrt(0.4*0.6/5100) = 0.0134 \n" ); document.write( "----------------------- \n" ); document.write( "95% CI: 0.4-0.0134 < p < 0.4+0.0134 \n" ); document.write( "======================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |