A group of students got the following scores in an achievement test:9,12,15,18,21 and 24 . consider the sample of size 3 that can be drawn from this population. Construct a sampling distribution of the resulting means and find the probability
We have population values 9,12,15,18,21, 24, population size N=6 and sample size n=3.
Mean of population "(\\mu)" = "\\dfrac{9+12+15+18+21+24}{6}=16.5"
Variance of population
"+2.25+20.25+56.25)=26.25"
"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{26.25}\\approx5.123475"
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=^{6}C_3=20."
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 9,12,15 & 12 \\\\\n \\hdashline\n 2 & 9,12,18 & 13 \\\\\n \\hdashline\n 3 & 9,12,21 & 14 \\\\\n \\hdashline\n 4 & 9,12,24 & 15 \\\\\n \\hdashline\n 5 & 9,15,18 & 14 \\\\\n \\hdashline\n 6 & 9,15,21 & 15 \\\\\n \\hdashline\n 7 & 9,15,24 & 16 \\\\\n \\hdashline\n 8 & 9,18,21 & 16 \\\\\n \\hdashline\n 9 & 9,18,24 & 17 \\\\\n \\hdashline\n 10 & 9,21,24 & 18 \\\\\n \\hdashline \n 11 & 12,15,18 & 15 \\\\\n \\hdashline \n 12 & 12,15,21 & 16 \\\\\n \\hdashline \n 13 & 12,15,24 & 17 \\\\\n \\hdashline \n 14 & 12,18,21 & 17 \\\\\n \\hdashline \n 15 & 12,18,24 & 18 \\\\\n \\hdashline \n 16 & 12,21,24 & 19 \\\\\n \\hdashline \n 17 & 15,18,21 & 18 \\\\\n \\hdashline \n 18 & 15,18,24 & 19 \\\\\n \\hdashline \n 19 & 15,21,24 & 20 \\\\\n \\hdashline \n 20 & 18,21,24 & 21 \\\\\n \\hdashline \n\\end{array}"d.
Mean of sampling distribution
"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=\\dfrac{330}{20}=16.5=\\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{5550}{20}-(16.5)^2=5.25= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})""\\sigma_{\\bar{X}}=\\sqrt{5.25}\\approx2.291288"
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