Question #340289

Consider the population consisting of the values 1, 3 and 4. List all the possible samples of size 2 that can be drawn from the population with replacement. Then, compute the mean x— for each sample. Lastly, find the mean of the sampling distribution of means and the mean of the population.

1
Expert's answer
2022-05-13T02:30:56-0400

We have population values 1,3,4, population size N=3 and sample size n=2.

The number of possible samples which can be drawn with replacement is

Nn=32=9.N^n=3^2=9.

S={(1,1),(1,3),(1,4),(3,1),(3,3),(3,4),S=\{(1,1), (1,3), (1,4),(3,1), (3,3), (3,4),

(4,1),(4,3),(4,4)}(4,1), (4,3), (4,4)\}


noSampleSamplemean (xˉ)11,12/221,34/231,45/243,14/253,36/263,47/274,15/284,37/294,48/2\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,1 & 2/2 \\ \hdashline 2 & 1,3 & 4/2 \\ \hdashline 3 & 1,4 & 5/2 \\ \hdashline 4 & 3,1 & 4/2 \\ \hdashline 5 & 3,3 & 6/2 \\ \hdashline 6 & 3,4 & 7/2 \\ \hdashline 7 & 4,1 & 5/2 \\ \hdashline 8 & 4,3 & 7/2 \\ \hdashline 9 & 4,4 & 8/2 \\ \hdashline \end{array}



C.

Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)2/21/92/184/364/22/98/1816/365/22/910/1825/366/21/96/1836/367/22/914/1849/368/21/98/18648/36\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 2/2 & 1/9 & 2/18 & 4/36 \\ \hdashline 4/2 & 2/9 & 8/18 & 16/36 \\ \hdashline 5/2 & 2/9 & 10/18 & 25/36\\ \hdashline 6/2 & 1/9 & 6/18 & 36/36\\ \hdashline 7/2 & 2/9 & 14/18 & 49/36 \\ \hdashline 8/2 & 1/9 & 8/18 & 648/36 \\ \hdashline \end{array}


Mean of population 

μ=1+3+43=83\mu=\dfrac{1+3+4}{3}=\dfrac{8}{3}


Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=83=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{8}{3}=\mu




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