Answer to Question #197671 in Statistics and Probability for Jack lakos

Question #197671

1) The following data is obtained from a research


a) Find the best equation fitting this data by using regression analyses method

b) Find the correlation coefficient of the equation

c) Find the y value for the x=5.5

X Y

5 784

10 1043

15 1666

20 2478

25 4060

30 6069



1
Expert's answer
2021-05-24T18:57:31-0400
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n X & Y & XY & X^2 & Y^2 \\\\ \\hline\n 5 & 784 & 3920 & 25 & 614656 \\\\ \\hdashline\n 10 & 1043 & 10430 & 100 & 1087849 \\\\ \\hdashline\n15 & 1666 & 24990 & 225 & 2775556 \\\\ \\hdashline\n20 & 2478 & 49560 & 400 & 6140484 \\\\ \\hdashline\n25 & 4060 & 101500 & 625 & 16483600 \\\\ \\hdashline\n30 & 6069 & 182070 & 900 & 36832761 \\\\ \n \\hline\n\\sum=\\ \\ \\ 105 & 16100 & 372470 & 2275 & 63934906 \\\\\n\\end{array}"



"\\bar{X}=\\dfrac{1}{n}\\displaystyle\\sum_{i=1}^nX_i=\\dfrac{105}{6}=17.5"

"\\bar{Y}=\\dfrac{1}{n}\\displaystyle\\sum_{i=1}^nY_i=\\dfrac{16100}{6}=2683.333333"

"SS_{XX}=\\displaystyle\\sum_{i=1}^nX_i^2-\\dfrac{1}{n}(\\displaystyle\\sum_{i=1}^nX_i)^2"

"=2275-\\dfrac{105^2}{6}=437.5"

"SS_{YY}=\\displaystyle\\sum_{i=1}^nY_i^2-\\dfrac{1}{n}(\\displaystyle\\sum_{i=1}^nY_i)^2"

"=63934906-\\dfrac{16100^2}{6}=20733239.333333"

"SS_{XY}=\\displaystyle\\sum_{i=1}^nX_iY_i-\\dfrac{1}{n}(\\displaystyle\\sum_{i=1}^nX_i)(\\displaystyle\\sum_{i=1}^nY_i)"

"=372470-\\dfrac{105\\times16100}{6}=90720"


"m=\\dfrac{SS_{XY}}{SS_{XX}}=\\dfrac{90720}{437.5}=207.36"

"n=\\bar{Y}-m\\cdot \\bar{X}=2683.333333.2-207.36\\cdot17.5"


"=\u2212945.466667"

The regression equation is:


"Y=\u2212945.466667+207.36X"

b)


"r=\\dfrac{SS_{XY}}{\\sqrt{SS_{XX}}\\sqrt{SS_{YY}}}"

"=\\dfrac{90720}{\\sqrt{437.5}\\sqrt{20733239.333333}}\\approx0.952534"

"0.7<0.952534\\leq1"

Strong positive correlation,


c)

"X=5.5"

"Y=\u2212945.466667+207.36(5.5)=195"






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