Answer to Question #305078 in Statistics and Probability for mohamed MD

Question #305078

Suppose that the standard deviation of the tube life of a particular brand of TV picture

tube is known to be 500, the population of tube life cannot be assumed to be normally

distributed. However, the sample mean of x = 8900 is based on a sample of n = 35.

Construct the 95% confidence interval for estimating the population mean.


1
Expert's answer
2022-03-03T12:48:30-0500

By the Central Limit Theorem if nn is sufficiently large, Xˉ\bar{X} has approximately a normal distribution with μXˉ=μ\mu_{\bar{X}}=\mu and σXˉ2=σ2/n.\sigma_{\bar{X}}^2=\sigma^2/n.

The larger the value of n,n, the better the approximation.

The Central Limit Theorem can generally be used if n>30.n>30.

The critical value for α=0.05\alpha = 0.05 is zc=z1α/2=1.96.z_c = z_{1-\alpha/2} = 1.96.

The corresponding confidence interval is computed as shown below:


CI=(Xˉzc×sn,Xˉ+zc×sn)CI=(\bar{X}-z_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{s}{\sqrt{n}})

=(89001.96×50035,8900+1.96×50035)=(8900-1.96\times\dfrac{500}{\sqrt{35}}, 8900+1.96\times\dfrac{500}{\sqrt{35}})

=(8734.35,9065.65)=(8734.35, 9065.65)

Therefore, based on the data provided, the 95% confidence interval for the population mean is 8734.35<μ<9065.65,8734.35<\mu<9065.65, which indicates that we are 95% confident that the true population mean μ\mu is contained by the interval  (8734.35,9065.65).(8734.35, 9065.65).



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