Solution to Suppose that the standard deviation of the tube life of a particular brand of TV … - Sikademy
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Mirian Woke

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.

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

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

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

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

The corresponding confidence interval is computed as shown below:

CI=(\bar{X}-z_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{s}{\sqrt{n}})

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

=(8734.35, 9065.65)

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

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