Question

The data show systolic and diastolic blood pressure of certain people. Find the regression​ equation, letting the systolic reading be the independent​ (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 137. Is the predicted value close to 97.1​, which was the actual diastolic​ reading? Use a significance level of 0.05. Systolic 129 119 126 145 126 125 142 110 Diastolic 94 64 68 84 93 91 104 79

What is the regression​ equation?

Answer 1

Sum of X = 1022
Sum of Y = 677
Mean X = 127.75
Mean Y = 84.625
Sum of squares (SS_X) = 907.5
Sum of products (SP) = 554.25

Regression Equation = ŷ = bX + a

b = SP/SS_X = 554.25/907.5 = 0.6107

a = M_Y - bM_X = 84.63 - (0.61*127.75) = 6.6025

ŷ = 0.6107X + 6.6025

X Values
∑ = 1022
Mean = 127.75
∑(X - M_x)² = SS_x = 907.5

Y Values
∑ = 677
Mean = 84.625
∑(Y - M_y)² = SS_y = 1307.875

X and Y Combined
N = 8
∑(X - M_x)(Y - M_y) = 554.25

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

r = 554.25 / √((907.5)(1307.875)) = 0.5087

The sample size is n = 8n=8, so then the number of degrees of freedom is df=n−2=8−2=6

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

rc​=0.707

Observe that in this case, the null hypothesis is rejected if ∣r∣>rc​=0.707.

Here r>rc, so test is significant

For x=137,

ŷ = (0.6107*137) + 6.6025=90.2684

It is different from 91.7