Answer to Question #342345 in Statistics and Probability for max

Question #342345

Construct the sampling distribution of the sample mean for the population that consists of the scores 15, 18, 11, 14, 17, with a sample size of 3. Compute as well the mean of the sampling distribution using the expected value of the distribution and verify the result by computing for the population mean.


1
Expert's answer
2022-05-19T09:33:12-0400

We have population values 15, 18, 11, 14, 17, population size N=5 and sample size n=3.

Mean of population "(\\mu)" = "\\dfrac{15+18+11+14+17}{5}=15"

Variance of population 


"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}=\\dfrac{0+9+16+1+4}{5}=6"


"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{6}\\approx2.4495"

Select a random sample of size 3 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_3=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 & 11,14,15 & 40\/3 \\\\\n \\hdashline\n 2 & 11,14,17 & 42\/3 \\\\\n \\hdashline\n 3 & 11,14,18 & 43\/3 \\\\\n \\hdashline\n 4 & 11,15,17 & 43\/3 \\\\\n \\hdashline\n 5 & 11,15,18 & 44\/3 \\\\\n \\hdashline\n 6 & 11,17,18 & 46\/3 \\\\\n \\hdashline\n 7 & 14,15,17 & 46\/3 \\\\\n \\hdashline\n 8 & 14, 15,18 & 47\/3 \\\\\n \\hdashline\n 9 & 14,17,18 & 49\/3 \\\\\n \\hdashline\n 10 & 15, 17,18 & 50\/3 \\\\\n \\hdashline\n\\end{array}"



B.


"\\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 40\/3 & 1\/10 & 40\/30 & 1600\/90 \\\\\n \\hdashline\n 42\/3 & 1\/10 & 42\/30 & 1764\/90 \\\\\n \\hdashline\n 43\/3 & 2\/10 & 86\/30 & 3698\/90 \\\\\n \\hdashline\n 44\/3 & 1\/10 & 44\/30 & 1936\/90 \\\\\n \\hdashline\n 46\/3 & 2\/10 & 92\/30 & 4232\/90 \\\\\n \\hdashline\n 47\/3 & 1\/10 & 47\/30 & 2209\/90 \\\\\n \\hdashline\n 49\/3 & 1\/10 & 49\/30 & 2401\/90 \\\\\n \\hdashline\n 50\/3 & 1\/10 & 50\/30 & 2500\/90 \\\\\n \\hdashline\n\\end{array}"



Mean of sampling distribution 

"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=\\dfrac{450}{30}=15=\\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{20340}{90}-(15)^2=1= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})"

"\\sigma_{\\bar{X}}=\\sqrt{1}=1"


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