Answer to Question #345111 in Statistics and Probability for jn cepeda

Question #345111

Consider a population consist of 9, 5, 6, 12, and 15. Suppose that sample size of 2were drawn from this population (without replacement), describe the sampling distribution of the sample means

1
Expert's answer
2022-05-29T17:44:38-0400

We have population values 5,6,9,12,15, population size N=5 and sample size n=2.

Mean of population "(\\mu)" = "\\dfrac{5+6+9+12+15}{5}=9.4"

Variance of population 


"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}=\\dfrac{1}{5}(19.36+11.56"




"+0.16+6.76+31.36)=13.84"




"\\sigma=\\sqrt{13.84}\\approx3.720215"

Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

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

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 5,6 & 11\/2 \\\\\n \\hdashline\n 2 & 5,9 & 14\/2 \\\\\n \\hdashline\n 3 & 5,12 & 17\/2 \\\\\n \\hdashline\n 4 & 5,15 & 20\/2 \\\\\n \\hdashline\n 5 & 6,9 & 15\/2 \\\\\n \\hdashline\n 6 & 6,12 & 18\/2 \\\\\n \\hdashline\n 7 & 6,15 & 21\/2 \\\\\n \\hdashline\n 8 & 9,12 & 21\/2 \\\\\n \\hdashline\n 9 & 9,15 & 24\/2 \\\\\n \\hdashline\n 10 & 12,15 & 27\/2 \\\\\n \\hdashline\n\\end{array}"





"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f(\\bar{X}) &\\bar{X} f(\\bar{X}) & \\bar{X}^2f(\\bar{X})\n\\\\ \\hline\n 11\/2 & 1\/10 & 11\/20 & 121\/40 \\\\\n \\hdashline\n 14\/2 & 1\/10 & 14\/20 & 196\/40 \\\\\n \\hdashline\n 15\/2 & 1\/10 & 15\/20 & 225\/40 \\\\\n \\hdashline\n 17\/2 & 1\/10 & 17\/20 & 289\/40 \\\\\n \\hdashline\n 18\/2 & 1\/10 & 18\/20 & 324\/40 \\\\\n \\hdashline\n 20\/2 & 1\/10 & 20\/20 & 400\/40 \\\\\n \\hdashline\n 21\/2 & 2\/10 & 42\/20 & 882\/40 \\\\\n \\hdashline\n 24\/2 & 1\/10 & 24\/20 & 576\/40 \\\\\n \\hdashline\n 27\/2 & 1\/10 & 27\/20 & 729\/40 \\\\\n \\hdashline\n\\end{array}"



Mean of sampling distribution 



"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=\\dfrac{188}{20}=9.4=\\mu"



The variance of sampling distribution 



"Var(\\bar{X})=\\sigma^2_{\\bar{X}}=\\sum\\bar{X}_i^2f(\\bar{X}_i)-\\big[\\sum\\bar{X}_if(\\bar{X}_i)\\big]^2""=\\dfrac{3742}{40}-(5.4)^2=5.19= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})"

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