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
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
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}"Mean of sampling distribution
The variance of sampling distribution
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