Consider all possible samples of size 2 that can be drawn with replacement from the population 1, 5, and 8. Compute the following:
1. Population mean
2. Population variance
3. Population standard deviation
4. Illustrate the probability histogram of the sampling distribution of the means
We have population values 1,5,8 population size N=3 and sample size n=2.
1.Population mean
"\\mu=\\dfrac{1+5+8}{3}=\\dfrac{14}{3}"2. Population variance
3. Population standard deviation
4. The number of possible samples which can be drawn with replacement is "N^n=3^2=9."
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 1,1 & 2\/2 \\\\\n \\hdashline\n 2 & 1,5 & 6\/2 \\\\\n \\hdashline\n 3 & 1,8 & 9\/2 \\\\\n \\hdashline\n 4 & 5,1 & 6\/2 \\\\\n \\hdashline\n 5 & 5,5 & 10\/2 \\\\\n \\hdashline\n 6 & 5,8 & 13\/2 \\\\\n \\hdashline\n 7 & 8,1 & 9\/2 \\\\\n \\hdashline\n 8 & 8,5 & 13\/2 \\\\\n \\hdashline\n 9 & 8,8 & 16\/2 \\\\\n \\hdashline \n\\end{array}"Mean of sampling distribution
"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=\\dfrac{14}{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{932}{36}-(\\dfrac{14}{3})^2=\\dfrac{37}{9}= \\dfrac{\\sigma^2}{n}"
The standard error of the mean
"\\sigma_{\\bar{X}}=\\sqrt{\\dfrac{37}{9}}\\approx2.0276"
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