Question #303181

A population consist of the values 2,4,6 and 8

a) List all the possible sample size 2 when drawing with replacement and compute the mean of each sample.

b)Construct the sampling distribution of the mean.

c) Compute the mean, variance and standard deviation of the sample mean then compare these with the mean, Variance and standard deviation of the population.


1
Expert's answer
2022-03-01T13:13:57-0500

We have population values 2,4,6,82,4,6,8 population size N=4N=4 and sample size n=2.n=2.

Thus, the number of possible samples which can be drawn with replacement is 42=16.4^2=16.

a.


Sample valuesSample mean(Xˉ)2,222,432,642,854,234,444,654,866,246,456,666,878,258,468,678,88\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 2,2 & 2\\ \hdashline 2,4 & 3\\ \hdashline 2,6 & 4\\ \hdashline 2,8 & 5\\ \hdashline 4,2 & 3\\ \hdashline 4,4 & 4\\ \hdashline 4,6 & 5\\ \hdashline 4,8 & 6\\ \hdashline 6,2 & 4\\ \hdashline 6,4 & 5\\ \hdashline 6,6 & 6\\ \hdashline 6,8 & 7\\ \hdashline 8,2 & 5\\ \hdashline 8,4 & 6\\ \hdashline 8,6 & 7\\ \hdashline 8,8 & 8\\ \hdashline \end{array}




b.The sampling distribution of the sample mean Xˉ\bar{X} is


Xˉff(Xˉ)Xf(Xˉ)X2f(Xˉ)211/162/164/16322/166/1618/16433/1612/1648/16544/1620/16100/16633/1618/16108/16722/1614/1698/16811/168/1664/16Sum=161555/2\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 2 & 1 & 1/16 & 2/16 & 4/16\\ \hdashline & 3 & 2 & 2/16 & 6/16 & 18/16 \\ \hdashline & 4 & 3 & 3/16 & 12/16 & 48/16\\ \hdashline & 5 & 4 & 4/16 & 20/16 & 100/16 \\ \hdashline & 6 & 3 & 3/16 & 18/16 & 108/16 \\ \hdashline & 7 & 2 & 2/16 & 14/16 & 98/16 \\ \hdashline & 8 & 1 & 1/16 & 8/16 & 64/16\\ \hdashline Sum= & & 16 & 1 & 5 & 55/2\\ \hdashline \end{array}

c.

μ=2+4+6+84=5\mu=\dfrac{2+4+6+8}{4}=5

σ2=14((25)2+(45)2+(65)2+(85)2)\sigma^2=\dfrac{1}{4}((2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2)

=5=5

Check

The mean of the sample means is


μXˉ=E(Xˉ)=5=μ\mu_{\bar{X}}=E(\bar{X})=5=\mu




Var(Xˉ)=σXˉ2=E(Xˉ2)(E(Xˉ))2Var(\bar{X})=\sigma^2_{\bar{X}}=E(\bar{X}^2)-(E(\bar{X}))^2

=552(5)2=52=\dfrac{55}{2}-(5)^2=\dfrac{5}{2}

The standard deviation of the sample means is


σXˉ=σXˉ2=5/2\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{5/2}

σ2n=52=σXˉ2\dfrac{\sigma^2}{n}=\dfrac{5}{2}=\sigma^2_{\bar{X}}

μXˉ=μ\mu_{\bar{X}}=\mu

σXˉ=σn\sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}}


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