Answer to Question #347501 in Statistics and Probability for kia

Question #347501

Consider a population consisting of 1 2 3 4 and 5 suppose samples of size 2 are drawn from this population. Describes the sampling distribution of the sample mean INSTRUCTION 1.Determaine the number of possible samples of size n=2 2.List all possible sample and their corresponding means 3.Construct the sampling distribution of the sample means 4.Compute the mean of the distribution of the sample means.

1
Expert's answer
2022-06-03T03:09:09-0400

1. We have population values 1,2,3,4,5, population size N=5 and sample size n=2.

Mean of population "(\\mu)" = "\\dfrac{1+2+3+4+5}{5}=3"

Variance of population 



"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}=\\dfrac{4+1+0+1+4}{5}=2"



2. 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 & 1,2 & 3\/2 \\\\\n \\hdashline\n 2 & 1,3 & 4\/2 \\\\\n \\hdashline\n 3 & 1,4 & 5\/2 \\\\\n \\hdashline\n 4 & 1,5 & 6\/2 \\\\\n \\hdashline\n 5 & 2,3 & 5\/2 \\\\\n \\hdashline\n 6 & 2,4 & 6\/2 \\\\\n \\hdashline\n 7 & 2,5 & 7\/2 \\\\\n \\hdashline\n 8 & 3,4 & 7\/2 \\\\\n \\hdashline\n 9 & 3,5 & 8\/2 \\\\\n \\hdashline\n 10 & 4,5 & 9\/2 \\\\\n \\hdashline\n\\end{array}"



3.


"\\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 3\/2 & 1\/10 & 3\/20 & 9\/40 \\\\\n \\hdashline\n 4\/2 & 1\/10 & 4\/20 & 16\/40 \\\\\n \\hdashline\n 5\/2 & 2\/10 & 10\/20 & 50\/40 \\\\\n \\hdashline\n 6\/2 & 2\/10 & 12\/20 & 72\/40 \\\\\n \\hdashline\n 7\/2 & 2\/10 & 14\/20 & 98\/40 \\\\\n \\hdashline\n 8\/2 & 1\/10 & 8\/20 & 64\/40 \\\\\n \\hdashline\n 9\/2 & 1\/10 & 9\/20 & 81\/40 \\\\\n \\hdashline\n\\end{array}"



4. Mean of sampling distribution 


"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=\\dfrac{60}{20}=3=\\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{390}{40}-(3)^2=\\dfrac{3}{4}= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})"

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