Question

a-Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 100 mm Hg (using Regression line).

Right Arm	102       101       94        79           79
Left Arm	175       169      182      146        144

b-Find the Linear correlation coefficient between the blood pressure in right arm (x) and the blood pressure in left arm (y) using the data given in part a

Answer 1

Solution:

Regression equation can be calculated as

Slope = n*summation(XY)-summation(X)*summation(Y) / n*(summation(X^2) -(summation(X)^2))

X Y XY X2 Y2

102 175 17850 10404 30625

101 169 17069 10201 28561

94 182 17108 8836 33124

79 146 11534 6241 21316

79 144 11376 6241 20736

455 816 74937 41923 134362

Slope =( 5*74937)-(455*816)/(5*41923)-(455*455)

= 374685-371280/ 209615-207025

= 3405/2590

= 1.3146

Intercept = summation(Y) -b*summation(X)/n

= (816-1.3146*455)/5

= 43.57

So regression line Y = 43.57 + 1.3146*X

At X = 100

Than Y = 43.57+1.3146×100 = 43.57 + 131.46 = 175

Correlation coefficient can be calculated as

r = n*summation(XY) - summation(X)*summation(Y) / sqrt(n*summation(X^2) -(summation(X)^2)((n*summation(Y^2)- (summation(Y)^2))

= 5*74937 - (455*816)/sqrt(((5*41923)-(455*455))((5*134362)-(816*816)))

After calculation we found

= 0.8671