Question #46539

For a given data set
x 5 15 25 35 45 50
y 10 18 20 25 32 45
i. Find the value of , and
ii. Write the normal equations and estimate the equation of line of regression of Y on x.
iii. Find the value of sample correlation coefficient for the above data.
1

Expert's answer

2014-09-17T13:53:45-0400

Answer on Question #46539 – Math – Statistics and Probability

Problem.

For a given data set

x 5 15 25 35 45 50

y 10 18 20 25 32 45

i. Find the value of, and

ii. Write the normal equations and estimate the equation of line of regression of Y on x.

iii. Find the value of sample correlation coefficient for the above data.

Remark: The part of question is missed. I suppose that the correct statement is

" For a given data set



i. Find the value of x2\sum x^{2} , y2\sum y^{2} and xy\sum xy

ii. Write the normal equations and estimate the equation of line of regression of Y on x.

iii. Find the value of sample correlation coefficient for the above data."

(see http://www.bits-pilani.ac.in/uploads/ReadPDFDOC/AAOC C111 515 C 2009 2.doc)

Solution:

i.



Then x2=6625,y2=4498\sum x^{2} = 6625, \sum y^{2} = 4498 and xy=5385\sum xy = 5385 .

ii. The normal equation for aa and bb are ax+nb=ya\sum x + nb = \sum y and ax2+bx=xya\sum x^2 + b\sum x = \sum xy . Hence

175a+6b=150175a + 6b = 150 and 6625a+175b=53856625a + 175b = 5385 . Then a=12121825a = \frac{1212}{1825} and b=41173b = \frac{411}{73} . Hence the equation of

the regression line is y=ax+b=12121825x+41173y = ax + b = \frac{1212}{1825} x + \frac{411}{73} .

Answer: y=12121825x+41173y = \frac{1212}{1825} x + \frac{411}{73} .

iii. The sample correlation coefficient


r=i=i6(xixˉ)(yiyˉ)i=i6(xixˉ)2i=i6(yiyˉ)20.947.r = \frac {\sum_ {i = i} ^ {6} (x _ {i} - \bar {x}) (y _ {i} - \bar {y})}{\sqrt {\sum_ {i = i} ^ {6} (x _ {i} - \bar {x}) ^ {2}} \sqrt {\sum_ {i = i} ^ {6} (y _ {i} - \bar {y}) ^ {2}}} \approx 0. 9 4 7.


Answer: r0.947r \approx 0.947 .

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