document.write( "Question 925537: Suppose that the standard deviation of the tube life for a particular brand of TV picture tube is known to be σ = 500, but the mean operating life is not known. Overall, the operating life of the tubes is assumed to be approximately normally distributed. For a sample of n = 15, the mean operating life is x = 8900 hours. Construct (a) the 95 percent and (b) the 90 percent confidence intervals for estimating the population mean. \n" ); document.write( "
Algebra.Com's Answer #561587 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Suppose that the standard deviation of the tube life for a particular brand of TV picture tube is known to be σ = 500, but the mean operating life is not known. Overall, the operating life of the tubes is assumed to be approximately normally distributed. For a sample of n = 15, the mean operating life is x = 8900 hours. Construct \n" ); document.write( "---- \n" ); document.write( "sample mean = 8900 \n" ); document.write( "--------------------- \n" ); document.write( "(a) the 95 percent CI \n" ); document.write( "ME = 1.96*500/sqrt(15) = 253 \n" ); document.write( "---- \n" ); document.write( "95%CI:: 8900- 253 < u < 8900+253 \n" ); document.write( "----------------------------------- \n" ); document.write( "and (b) the 90 percent confidence intervals for estimating the population mean. \n" ); document.write( "ME = 1.645*500/sqrt(15) = 212 \n" ); document.write( "90% CI:: 8900-212 < u < 8900+212 \n" ); document.write( "--------------------------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "----------------- \n" ); document.write( " |