Answer to Question #333904 in Statistics and Probability for shiira

Question #333904

Consider a population consisting of 3, 6, 7, 9, 10, 4, and 8. Suppose samples of size 2 are drawn from this population. Describe the sampling distribution of sample means. What is the standard deviation of the sampling distribution? NOTE: EXPRESS YOUR ANSWER INTO 4 DECIMAL PLACE VALUES

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Expert's answer
2022-04-27T02:08:10-0400

Here population size : N=7N =7 and we have to draw a sample of size 2

So, there are NCn^{N}C_n possible samples that is, 7C2=21^{7}C_2=21

Mean of population (μ)(\mu) = 3+6+7+9+10+4+87=4776.7143\dfrac{3+6+7+9+10+4+8}{7}=\dfrac{47}{7}\approx6.7143


Variance of population 


(σ2)=Σ(xixˉ)2n=17(67649+2549+449(\sigma^2)=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{7}(\dfrac{676}{49}+\dfrac{25}{49}+\dfrac{4}{49}


+25649+52949+36149+8149)+\dfrac{256}{49}+\dfrac{529}{49}+\dfrac{361}{49}+\dfrac{81}{49})

=1932343=\dfrac{1932}{343}




noSampleSamplemean (xˉ)13,43.523,64.533,7543,85.553,9663,106.574,6584,75.594,86104,96.5114,107126,76.5136,87146,97.5156,108167,87.5177,98187,108.5198,98.5208,109219,109.5\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 3,4 & 3.5 \\ \hdashline 2 & 3,6 & 4.5 \\ \hdashline 3 & 3,7 & 5 \\ \hdashline 4 & 3,8 & 5.5 \\ \hdashline 5 & 3,9 & 6 \\ \hdashline 6 & 3,10 & 6.5 \\ \hdashline 7 & 4,6 & 5 \\ \hdashline 8 & 4,7 & 5.5 \\ \hdashline 9 & 4,8 & 6 \\ \hdashline 10 & 4,9 & 6.5 \\ \hdashline 11 & 4,10 & 7 \\ \hdashline 12 & 6,7 & 6.5 \\ \hdashline 13 & 6,8 & 7 \\ \hdashline 14 & 6,9 & 7.5 \\ \hdashline 15 & 6,10 & 8 \\ \hdashline 16 & 7,8 & 7.5 \\ \hdashline 17 & 7,9 & 8 \\ \hdashline 18 & 7,10 & 8.5 \\ \hdashline 19 & 8,9 & 8.5 \\ \hdashline 20 & 8,10 & 9 \\ \hdashline 21 & 9,10 & 9.5 \\ \hdashline \end{array}



Mean of sampling distribution 

μxˉ=E(xˉ)=xˉif(xˉi)=477=μ\mu_{\bar{x}}=E(\bar{x})=\sum\bar{x}_if(\bar{x}_i)=\dfrac{47}{7}=\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=3327220949=11549=σ2n(NnN1)=\dfrac{332}{7}-\dfrac{2209}{49}=\dfrac{115}{49}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})

σxˉ=115491.5320\sigma_{\bar{x}}=\sqrt{\dfrac{115}{49}}\approx1.5320


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