document.write( "Question 1102141: a survey of 97 randomly selected homeowners found the the mean amount spent on lawn service was $720 per year. construct a 98% confidence interval for the mean amount of money spent on lawn service per household per year. assume the population standard deviation was $123 \n" ); document.write( "
| Algebra.Com's Answer #716788 by stanbon(75887)      You can put this solution on YOUR website! a survey of 97 randomly selected homeowners found the the mean amount spent on lawn service was $720 per year. construct a 98% confidence interval for the mean amount of money spent on lawn service per household per year. assume the population standard deviation was $123 \n" ); document.write( "---------------------------- \n" ); document.write( "sample mean = $720 \n" ); document.write( "ME = 2.3263*123/sqrt(97) = 29.05 \n" ); document.write( "----- \n" ); document.write( "98% CI:: 720-29.05 < u < 720+29.05 \n" ); document.write( "98% CI:: 690.95 < u < 749.05 \n" ); document.write( "---------------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "------------- \n" ); document.write( " |