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.
Answer by stanbon(75887)   (Show Source):
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
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sample mean = 8900
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(a) the 95 percent CI
ME = 1.96*500/sqrt(15) = 253
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95%CI:: 8900- 253 < u < 8900+253
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and (b) the 90 percent confidence intervals for estimating the population mean.
ME = 1.645*500/sqrt(15) = 212
90% CI:: 8900-212 < u < 8900+212
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Cheers,
Stan H.
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