Question #65454

Compute the appropriate regression equation for the following data:
X Y
2 18
4 12
5 10
6 8
8 7
11 5

Where X is the independent variable and Y is the dependent variable.
Also find the correlation coefficient between X and Y and infer about the
relationship between X and Y.
1

Expert's answer

2017-02-22T09:58:05-0500

Question #65454, Math / Statistics and Probability

Compute the appropriate regression equation for the following data:



Where XX is the independent variable and YY is the dependent variable.

Also find the correlation coefficient between XX and YY and infer about the relationship between XX and YY.

**Answer.**


X=36,Y=60,X2=266,Y2=706,XY=293.\sum X = 36, \quad \sum Y = 60, \quad \sum X^2 = 266, \quad \sum Y^2 = 706, \quad \sum XY = 293.


Regression equation y=a+bxy = a + bx.

Where a=YX2XXYnX2(X)2=6026636293626636218.04,a = \frac{\sum Y\sum X^2 - \sum X\sum XY}{n\sum X^2 - (\sum X)^2} = \frac{60*266 - 36*293}{6*266 - 36^2} \approx 18.04,

b=nXYXYnX2(X)2=6293366062663621.34.b = \frac{n\sum XY - \sum X\sum Y}{n\sum X^2 - (\sum X)^2} = \frac{6*293 - 36*60}{6*266 - 36^2} \approx -1.34.


So y=18.041.34xy = 18.04 - 1.34x.

**Correlation coefficient**


r=nXYXYnX2(X)2nY2(Y)2=62933660626636267066020.9203.r = \frac{n\sum XY - \sum X\sum Y}{\sqrt{n\sum X^2 - (\sum X)^2} \sqrt{n\sum Y^2 - (\sum Y)^2}} = \frac{6*293 - 36*60}{\sqrt{6*266 - 36^2} \sqrt{6*706 - 60^2}} \approx -0.9203.


There is a strong negative linear correlation between XX and YY.

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