204 8.1%2c8.2 dist. sample mean CLT

204 8.1%2c8.2 dist. sample mean CLT - 8.1 Mean And Standard...

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Assume we have a population with the values 2,3,4,5,6. This population has μ = 4 and σ 2 = 2 = = We now take all possible non ordered samples of size two from the population and calculate the mean of each pair. pair 2,3 2,4 2,5 2,6 3,4 3,5 3,6 4,5 4,6 5,6 x 2.5 3 3.5 4 3.5 4 4.5 4.5 5 5.5 The statistic x also has a mean, μ, and variance σ 2 . For the above pairs μ= [2.5 + 3 + 2(3.5) + 2(4) + 2(4.5) + 5 + 5.5]/10= 4 σ 2 = .75 Note that μ = μ= 4. In contrast, the variance of is less than the variance of the population (2). The variances can be equated through the following formulas: for infinte populations σ 2 = n 2 σ σ= for finite populations σ 2 = 1 N n N n 2 - - σ = Since our population has N = 5 and n = 2 we obtain: σ 2 = o.75 = . For practical purposes we assume a population is infinite if n<.05N The following is a Histogram of the above distribution of sample means. 2
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204 8.1%2c8.2 dist. sample mean CLT - 8.1 Mean And Standard...

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