Answer to Question #180897 in Statistics and Probability for Amira

Question #180897

A random sample is size n = 2 are drawn from a finite population consisting of the numbers 3,4,5,6 and 7


1
Expert's answer
2021-04-15T06:30:42-0400

1. Mean


"\\mu=\\dfrac{3+4+5+6+7}{5}=5"

Variance


"\\sigma^2=\\dfrac{1}{5}\\big((3-5)^2+(4-5)^2+(5-5)^2"

"(6-5)^2+(7-5)^2\\big)=2"


Standard deviation


"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{2}\\approx1.4142"



2. We have population values "3,4,5,6,7," population sizeΒ "N=5" and sample sizeΒ "n=2." Thus, the number of possible samples which can be drawn without replacement is


"\\dbinom{N}{n}=\\dbinom{5}{2}=10""\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample & Sample & Sample \\ mean \\\\\n No. & values & (\\bar{X}) \\\\ \\hline\n 1 & 3,4 & 3.5 \\\\\n \\hdashline\n 2 & 3,5 & 4 \\\\\n \\hdashline\n 3 & 3,6 & 4.5 \\\\\n \\hdashline\n 4 & 3,7 & 5 \\\\\n \\hdashline\n 5 & 4,5 & 4.5 \\\\\n \\hdashline\n 6 & 4,6 & 5 \\\\\n \\hdashline\n 7 & 4,7 & 5.5 \\\\\n \\hdashline\n 8 & 5,6 & 5.5 \\\\\n \\hdashline\n 9 & 5,7 & 6 \\\\\n \\hdashline\n 10 & 6,7 & 6.5 \\\\ \\hline\n\\end{array}"

3. The sampling distribution of the sample means.


"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f & f(\\bar{X}) & \\bar{X}f(\\bar{X})& \\bar{X}^2f(\\bar{X}) \\\\ \\hline\n 7\/2 & 1& 1\/10 & 7\/20 & 49\/40 \\\\\n \\hdashline\n 4 & 1 & 1\/10 & 8\/20 & 64\/40 \\\\\n \\hdashline\n 9\/2 & 2 & 2\/10 & 18\/20 & 162\/40 \\\\\n \\hdashline\n 5 & 2 & 2\/10 & 20\/20 & 200\/40 \\\\\n \\hdashline\n 11\/2 & 2 & 2\/10 & 22\/20 & 242\/40 \\\\\n \\hdashline\n 6 & 1 & 1\/10 & 12\/20 & 144\/40 \\\\\n \\hdashline\n 13\/2 & 1& 1\/10 & 13\/20 & 169\/40 \\\\\n \\hdashline\n \n Total & 10 & 1 & 100\/20 & 1030\/40 \\\\ \\hline\n\\end{array}"

4.


"E(\\bar{X})=\\sum\\bar{X}f(\\bar{X})=\\dfrac{100}{20}=5"

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.


"E(\\bar{X})=5=\\mu"

5.

"Var(\\bar{X})=\\sum\\bar{X}^2f(\\bar{X})-(\\sum\\bar{X}f(\\bar{X}))^2"

"=\\dfrac{1030}{40}-(\\dfrac{100}{20})^2=\\dfrac{3}{4}=0.75"

"\\sqrt{Var(\\bar{X})}=\\sqrt{\\dfrac{3}{4}}=\\dfrac{\\sqrt{3}}{2}\\approx0.8660"

Verification:

"Var(\\bar{X})=\\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})=\\dfrac{2}{2}(\\dfrac{5-2}{5-1})"

"=\\dfrac{3}{4}=0.75, True"


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