Answer to Question #342948 in Statistics and Probability for Yoongi

Question #342948

Consider all possible samples of size 2 that can be drawn with replacement from the population 1, 5, and 8. Compute the following:


1. Population mean


2. Population variance


3. Population standard deviation


4. Illustrate the probability histogram of the sampling distribution of the means

1
Expert's answer
2022-05-23T09:26:14-0400

We have population values 1,5,8 population size N=3 and sample size n=2.

1.Population mean

μ=1+5+83=143\mu=\dfrac{1+5+8}{3}=\dfrac{14}{3}


2. Population variance 


σ2=Σ(xixˉ)2n=1219+19+10093=749\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{\dfrac{121}{9}+\dfrac{1}{9}+\dfrac{100}{9}}{3}=\dfrac{74}{9}



3. Population standard deviation


σ=σ2=749=7432.867442\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{74}{9}}=\dfrac{\sqrt{74}}{3}\approx2.867442

4. The number of possible samples which can be drawn with replacement is Nn=32=9.N^n=3^2=9.

noSampleSamplemean (xˉ)11,12/221,56/231,89/245,16/255,510/265,813/278,19/288,513/298,816/2\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 & 1,5 & 6/2 \\ \hdashline 3 & 1,8 & 9/2 \\ \hdashline 4 & 5,1 & 6/2 \\ \hdashline 5 & 5,5 & 10/2 \\ \hdashline 6 & 5,8 & 13/2 \\ \hdashline 7 & 8,1 & 9/2 \\ \hdashline 8 & 8,5 & 13/2 \\ \hdashline 9 & 8,8 & 16/2 \\ \hdashline \end{array}




Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)2/21/92/184/366/22/912/1872/369/22/918/18162/3610/21/910/18100/3613/22/926/18338/3616/21/916/18256/36\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 2/2 & 1/9 & 2/18 & 4/36\\ \hdashline 6/2 & 2/9 & 12/18 & 72/36\\ \hdashline 9/2 & 2/9 & 18/18 & 162/36\\ \hdashline 10/2 & 1/9 & 10/18 & 100/36\\ \hdashline 13/2 & 2/9 & 26/18 & 338/36\\ \hdashline 16/2 & 1/9 & 16/18 & 256/36\\ \hdashline \end{array}


Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=143=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{14}{3}=\mu


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=93236(143)2=379=σ2n=\dfrac{932}{36}-(\dfrac{14}{3})^2=\dfrac{37}{9}= \dfrac{\sigma^2}{n}

The standard error of the mean

σXˉ=3792.0276\sigma_{\bar{X}}=\sqrt{\dfrac{37}{9}}\approx2.0276

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