Answer to Question #333904 in Statistics and Probability for shiira

Question #333904

Consider a population consisting of 3, 6, 7, 9, 10, 4, and 8. Suppose samples of size 2 are drawn from this population. Describe the sampling distribution of sample means. What is the standard deviation of the sampling distribution? NOTE: EXPRESS YOUR ANSWER INTO 4 DECIMAL PLACE VALUES

1
Expert's answer
2022-04-27T02:08:10-0400

Here population size : "N =7" and we have to draw a sample of size 2

So, there are "^{N}C_n" possible samples that is, "^{7}C_2=21"

Mean of population "(\\mu)" = "\\dfrac{3+6+7+9+10+4+8}{7}=\\dfrac{47}{7}\\approx6.7143"


Variance of population 


"(\\sigma^2)=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}=\\dfrac{1}{7}(\\dfrac{676}{49}+\\dfrac{25}{49}+\\dfrac{4}{49}"


"+\\dfrac{256}{49}+\\dfrac{529}{49}+\\dfrac{361}{49}+\\dfrac{81}{49})"

"=\\dfrac{1932}{343}"




"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 3,4 & 3.5 \\\\\n \\hdashline\n 2 & 3,6 & 4.5 \\\\\n \\hdashline\n 3 & 3,7 & 5 \\\\\n \\hdashline\n 4 & 3,8 & 5.5 \\\\\n \\hdashline\n 5 & 3,9 & 6 \\\\\n \\hdashline\n 6 & 3,10 & 6.5 \\\\\n \\hdashline\n 7 & 4,6 & 5 \\\\\n \\hdashline\n 8 & 4,7 & 5.5 \\\\\n \\hdashline\n 9 & 4,8 & 6 \\\\\n \\hdashline\n 10 & 4,9 & 6.5 \\\\\n \\hdashline\n 11 & 4,10 & 7 \\\\\n \\hdashline\n 12 & 6,7 & 6.5 \\\\\n \\hdashline\n 13 & 6,8 & 7 \\\\\n \\hdashline\n 14 & 6,9 & 7.5 \\\\\n \\hdashline\n 15 & 6,10 & 8 \\\\\n \\hdashline\n 16 & 7,8 & 7.5 \\\\\n \\hdashline\n 17 & 7,9 & 8 \\\\\n \\hdashline\n 18 & 7,10 & 8.5 \\\\\n \\hdashline\n 19 & 8,9 & 8.5 \\\\\n \\hdashline\n 20 & 8,10 & 9 \\\\\n \\hdashline\n 21 & 9,10 & 9.5 \\\\\n \\hdashline\n\\end{array}"



Mean of sampling distribution 

"\\mu_{\\bar{x}}=E(\\bar{x})=\\sum\\bar{x}_if(\\bar{x}_i)=\\dfrac{47}{7}=\\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{332}{7}-\\dfrac{2209}{49}=\\dfrac{115}{49}= \\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})"

"\\sigma_{\\bar{x}}=\\sqrt{\\dfrac{115}{49}}\\approx1.5320"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS