Question #108663
Consider a population of 6 units with values 1, 2, 3, 4, 5 and 6. Write down all possible samples of size 2 from this population. Obtain the sampling distribution of the sample mean. Compute mean and variance of the sampling distribution obtained.
1
Expert's answer
2020-04-13T09:18:00-0400

We randomly choose two unts, without replacement. The order doesn't matter.


Samplesample mean, xˉ1,21.51,321,42.51,531,63.52,32.52,432,53.52,643,43.53,543,64.54,54.54,655,65.5\begin{matrix} Sample & sample\ mean, \ \bar{x} \\ 1,2 & 1.5 \\ 1,3 & 2\\ 1,4 & 2.5 \\ 1,5 & 3 \\ 1,6 & 3.5 \\ 2,3 & 2.5 \\ 2,4 & 3 \\ 2,5 & 3.5 \\ 2,6 & 4\\ 3,4 & 3.5 \\ 3,5 & 4 \\ 3,6 & 4.5 \\ 4,5 & 4.5 \\ 4,6 & 5 \\ 5,6 & 5.5 \end{matrix}


xˉfprobability1.511/152.011/152.522/153.022/153.531/54.022/154.522/155.011/155.511/15\begin{matrix} \bar{x} & f & probability \\ 1.5 & 1 & 1/15 \\ 2.0 & 1 & 1/15\\ 2.5 & 2 & 2/15\\ 3.0 & 2 & 2/15 \\ 3.5 & 3 & 1/5\\ 4.0 & 2 & 2/15\\ 4.5 & 2 & 2/15\\ 5.0 & 1 & 1/15\\ 5.5 & 1 & 1/15\\ \end{matrix}


μxˉ=1.5(115)+2.0(115)+2.5(215)+3.0(215)+3.5(15)+\mu_{\bar{x}}=1.5({1\over15})+2.0({1\over15})+2.5({2\over15})+3.0({2\over15})+3.5({1\over5})+

+4.0(215)+4.5(215)+5.0(115)+5.5(15)=3.5+4.0({2\over15})+4.5({2\over15})+5.0({1\over15})+5.5({1\over5})=3.5


(1.5)2(115)+(2.0)2(115)+(2.5)2(215)+(3.0)2(215)+(3.5)2(15)+(1.5)^2({1\over15})+(2.0)^2({1\over15})+(2.5)^2({2\over15})+(3.0)^2({2\over15})+(3.5)^2({1\over5})+

+(4.0)2(215)+(4.5)2(215)+(5.0)2(115)+(5.5)2(15)=+(4.0)^2({2\over15})+(4.5)^2({2\over15})+(5.0)^2({1\over15})+(5.5)^2({1\over5})=

=16112={161\over 12}

Var(xˉ)=σxˉ2=16112(3.5)2=761.166667Var(\bar{x})=\sigma_{\bar{x}}^2={161\over 12}-(3.5)^2={7\over 6}\approx1.166667


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