Question #335066

Activity 2. Construct Me!




A population consists of the five numbers 2, 3, 6, 8 and 11.




Consider samples of size 2 that can be drawn from this population.




a. List all the possible samples and the corresponding mean.




Sample




Mean




19




b. Construct the sampling distribution of the sample means.




Sample Mean




Probability




Frequency








c. Draw a histogram of the sampling distribution of the




meets.

1
Expert's answer
2022-05-06T06:00:06-0400

a. We have population values 2, 3, 6, 8 and 11 population size N=5 and sample size n=2.

Mean of population (μ)(\mu) = 

2+3+6+8+115=6\dfrac{2+3+6+8+11}{5}=6


Variance of population 


σ2=Σ(xixˉ)2n\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}


=16+9+0+4+255=10.8=\dfrac{16+9+0+4+25}{5}=10.8

σ=σ2=10.83.2863\sigma=\sqrt{\sigma^2}=\sqrt{10.8}\approx3.2863

The number of possible samples which can be drawn without replacement is NCn=5C2=10.^{N}C_n=^{5}C_2=10.

noSampleSamplemean (xˉ)12,35/222,68/232,810/242,1113/253,69/263,811/273,1114/286,814/296,1117/2108,1119/2\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,3 & 5/2 \\ \hdashline 2 & 2,6 & 8/2 \\ \hdashline 3 & 2,8 & 10/2 \\ \hdashline 4 & 2,11 & 13/2 \\ \hdashline 5 & 3,6 & 9/2 \\ \hdashline 6 & 3,8 & 11/2 \\ \hdashline 7 & 3,11 & 14/2 \\ \hdashline 8 & 6,8 & 14/2 \\ \hdashline 9 & 6,11 & 17/2 \\ \hdashline 10 & 8,11 & 19/2 \\ \hdashline \end{array}



b.


Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)5/21/105/2025/408/21/108/2064/409/21/109/2081/4010/21/1010/20100/4011/21/1011/20121/4013/21/1013/20169/4014/22/1028/20392/4017/21/1017/20289/4019/21/1019/20361/40\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 5/2 & 1/10 & 5/20 & 25/40\\ \hdashline 8/2 & 1/10 & 8/20 & 64/40\\ \hdashline 9/2 & 1/10 & 9/20 & 81/40\\ \hdashline 10/2 & 1/10 & 10/20 & 100/40\\ \hdashline 11/2 & 1/10 & 11/20 & 121/40\\ \hdashline 13/2 & 1/10 & 13/20 & 169/40\\ \hdashline 14/2 & 2/10 & 28/20 & 392/40\\ \hdashline 17/2 & 1/10 & 17/20 & 289/40\\ \hdashline 19/2 & 1/10 & 19/20 & 361/40\\ \hdashline \end{array}


Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=6=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=6=\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=160240(6)2=16240=8120=σ2n(NnN1)=\dfrac{1602}{40}-(6)^2=\dfrac{162}{40}=\dfrac{81}{20}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})

σXˉ=81202.01246\sigma_{\bar{X}}=\sqrt{\dfrac{81}{20}}\approx2.01246

c.

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