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.

1
Expert's answer
2022-05-12T02:30:07-0400

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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