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)\"\" \"About 
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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
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\n" ); document.write( "sample mean = 8900
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\n" ); document.write( "(a) the 95 percent CI
\n" ); document.write( "ME = 1.96*500/sqrt(15) = 253
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\n" ); document.write( "95%CI:: 8900- 253 < u < 8900+253
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\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
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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