Question #339730

Consider the population of the values (1,2,3,4).




A. List all the possible samples of size 2 with replacement




B. Compute the mean of each sample.




C. Identify the probability of each sample.




D. Compute the mean of the sampling distribution of the means.

Expert's answer

A.We have population values 1,2,3,4, population size N=5 and sample size n=2.

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

Nn=42=16.N^n=4^2=16.

S={(1,1),(1,2),(1,3),(1,4),S=\{(1,1), (1,2), (1,3), (1,4),

(2,1),(2,2),(2,3),(2,4),(2,1), (2,2), (2,3), (2,4),

(3,1),(3,2),(3,3),(3,4),(3,1), (3,2), (3,3), (3,4),

(4,1),(4,2),(4,3),(4,4)}(4,1), (4,2), (4,3), (4,4)\}

B.


noSampleSamplemean (xˉ)11,12/221,23/231,34/241,45/252,13/262,24/272,35/282,46/293,14/2103,25/2113,36/2123,47/2134,15/2144,26/2154,37/2164,48/2\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 & 1,2 & 3/2 \\ \hdashline 3 & 1,3 & 4/2 \\ \hdashline 4 & 1,4 & 5/2 \\ \hdashline 5 & 2,1 & 3/2 \\ \hdashline 6 & 2,2 & 4/2 \\ \hdashline 7 & 2,3 & 5/2 \\ \hdashline 8 & 2,4 & 6/2 \\ \hdashline 9 & 3,1 & 4/2 \\ \hdashline 10 & 3,2 & 5/2 \\ \hdashline 11 & 3,3 & 6/2 \\ \hdashline 12 & 3,4 & 7/2 \\ \hdashline 13 & 4,1 & 5/2 \\ \hdashline 14 & 4,2 & 6/2 \\ \hdashline 15 & 4,3 & 7/2 \\ \hdashline 16 & 4,4 & 8/2 \\ \hdashline \end{array}



C.

Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)2/21/162/324/643/22/166/3218/644/23/1612/3248/645/24/1620/32100/646/23/1618/32108/647/22/1614/3298/648/21/168/3264/64\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 & 1/16 & 2/32 & 4/64 \\ \hdashline 3/2 & 2/16 & 6/32 & 18/64 \\ \hdashline 4/2 & 3/16 & 12/32 & 48/64 \\ \hdashline 5/2 & 4/16 & 20/32 & 100/64 \\ \hdashline 6/2 & 3/16 & 18/32 & 108/64 \\ \hdashline 7/2 & 2/16 & 14/32 & 98/64 \\ \hdashline 8/2 & 1/16 & 8/32 & 64/64 \\ \hdashline \end{array}



D.

Mean of population 

μ=1+2+3+44=2.5\mu=\dfrac{1+2+3+4}{4}=2.5


Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=2.5=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=2.5=\mu


Variance of population 


σ2=Σ(xixˉ)2n=14(2.25+0.25+0.25\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{4}(2.25+0.25+0.25


+2.25)=1.25+2.25)=1.25

The variance of sampling distribution 

Var(Xˉ)=σXˉ2=Xˉi2f(Xˉi)[Xˉif(Xˉi)]2Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2=44064(2.5)2=0.625=σ2n=\dfrac{440}{64}-(2.5)^2=0.625= \dfrac{\sigma^2}{n}


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