Question

Listed below are systolic blood pressure measurements​ (in mm​ Hg) obtained from the same woman. Find the regression​ equation, letting the right arm blood pressure be the predictor​ (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm Hg. Use a significance level of 0.05. Right Arm 101 100 94 76 77 Left Arm 175 169 143 141 143

Answer 1

Sum of X = 448
Sum of Y = 771
Mean X = 89.6
Mean Y = 154.2
Sum of squares (SS_X) = 601.2
Sum of products (SP) = 662.4

Regression Equation = ŷ = bX + a

b = SP/SS_X = 662.4/601.2 = 1.1018

a = M_Y - bM_X = 154.2 - (1.1*89.6) = 55.4790

ŷ = 1.1018X + 55.4790

X Values
∑ = 448
Mean = 89.6
∑(X - M_x)² = SS_x = 601.2

Y Values
∑ = 771
Mean = 154.2
∑(Y - M_y)² = SS_y = 1076.8

X and Y Combined
N = 5
∑(X - M_x)(Y - M_y) = 662.4

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 662.4 / √((601.2)(1076.8)) = 0.8233

The sample size is n=5, so then the number of degrees of freedom is df=n−2=5−2=3

The corresponding critical correlation value rc​ for a significance level of α=0.05, for a two-tailed test is:

rc​=0.878
Observe that in this case, the null hypothesis is rejected if ∣r∣>rc​=0.878.

As here r<rc, test is not significant

or x=85, ŷ = (1.1018*85) + 55.4790=149.132