Question

Given the following data: X 1 4 6 7 Y 9 7 8 1 a) Find the coefficient of correlation. b) Find the equation of the regression line. c) What is the predicted value for Y, if X = 3?

Answer 1

Answer:

a) r=-0.719

b) y=10.642-0.976x

c)Predicted y=7.714

Step-by-step explanation:

a)

sumx=1+4+6+7=18

sumy=9+7+8+1=25

sumxy=1*9+4*7+6*8+7*1=92

sumx²=1²+4²+6²+7²=102

sumy²=9²+7²+8²+1²=195

n=number of observation=4

The correlation coefficient is computed by following formula

$r=\frac{nsumxy-(sumx)(sumy)}{\sqrt{[nsumx^{2} -(sumx)^2][nsumy^2-(sumy)^2]}}$

$r=\frac{4(92)-(18)(25)}{\sqrt{[4(102) -(18)^2][4(195)-(25)^2]}}$

$r=\frac{368-450}{\sqrt{[408 -324][780-625]}}$

$r=\frac{-82}{\sqrt{[84][155]}}$

$r=\frac{-82}{\sqrt{13020}}\\r=\frac{-82}{114.1052}\\r=-0.7186$

By rounding r to 3 decimal places we get r=-0.719.

b)

The regression equation can be written as y=a+bx

We have to find "a" and "b" for regression equation

$b=\frac{nsumxy-(sumx)(sumy)}{{nsumx^{2} -(sumx)^2}}$

$a=ybar-b(xbar)$

$b=\frac{4(92)-(18)(25)}{{4(102) -(18)^2}}$

$b=\frac{-82}{{84}}$

b=-0.976

xbar=sumx/n

xbar=18/4=4.5

ybar=sumy/n

ybar=25/4=6.25

a=ybar-b(xbar)

a=6.25-(-0.976)4.5

a=6.25+4.392

a=10.642

Thus, the regression equation is

y=a+bx

y=10.642-0.976x

c)

The predicted value of y for x=3 can be computed by putting the value of x in regression equation

y=10.642-0.976(3)

y=10.642-2.928

y=7.714

Hence, the predicted y-value for x=3 is 7.714.