1.
We have population values "5,6,7,8,9" population size "N=5" and sample size "n=2."Thus, the number of possible samples which can be drawn without replacement is
"\\dbinom{N}{n}=\\dbinom{5}{2}=10"
2.
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample & Sample & Sample \\ mean \\\\\n No. & values & (\\bar{X}) \\\\ \\hline\n 1 & 5,6 & 5.5 \\\\\n \\hdashline\n 2 & 5,7 & 6.0 \\\\\n \\hdashline\n 3 & 5,8 & 6.5 \\\\\n \\hdashline\n 4 & 5,9 & 7.0\\\\\n \\hdashline\n 5 & 6,7 & 6.5 \\\\\n\\hdashline\n 6 & 6,8 & 7.0 \\\\\n \\hline\n7 & 6,9 & 7.5 \\\\\n \\hline\n 8 & 7,8 & 7.5 \\\\\n \\hline\n9 & 7,9 & 8.0 \\\\\n \\hline\n10 & 8,9 & 8.5 \\\\\n \\hline\n\\end{array}"
"\\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 5.5 & 1 & 0.1 & 0.55 & 3.025 \\\\\n \\hdashline\n 6.0 & 1 & 0.1 & 0.60 & 3.600 \\\\\n \\hdashline\n 6.5 & 2 & 0.2 & 1.30 & 8.450 \\\\\n \\hdashline\n 7.0 & 2 & 0.2 & 1.40 & 9.800 \\\\\n \\hdashline\n 7.5 & 2 & 0.2 & 1.50 & 11.250 \\\\\n \\hdashline\n 8.0 & 1& 0.1 & 0.80 & 6.400\\\\\n \\hdashline\n 8.5 & 1& 0.1 & 0.85 & 7.225 \\\\\n \\hdashline\n Total & 10 & 1 & 7 & 49.75 \\\\ \\hline\n\\end{array}"
3.
Mean
"\\mu=\\dfrac{5+6+7+8+9}{5}=7"
"E(\\bar{X})=\\sum\\bar{X}f(\\bar{X})=7"The mean of the sampling distribution of the sample means is equal to the
the mean of the population.
"E(\\bar{X})=3.4=\\mu"
4.
Variance
"\\sigma^2=\\dfrac{1}{5}\\big((5-7)^2+(6-7)^2+(7-7)^2""(8-7)^2+(9-7)^2\\big)=2"
Standard deviation
"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{2}\\approx1.41421356"
"Var(\\bar{X})=\\sum\\bar{X}^2f(\\bar{X})-(\\sum\\bar{X}f(\\bar{X}))^2"
"=49.75-(7)^2=0.75"
"\\sqrt{Var(\\bar{X})}=\\sqrt{\\dfrac{118}{75}}\\approx1.254326"Verification:
"Var(\\bar{X})=\\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})=\\dfrac{2}{2}(\\dfrac{5-2}{5-1})"
"=0.75, True"
Comments
Leave a comment