In: Statistics and Probability
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
Sum of X = 448
Sum of Y = 771
Mean X = 89.6
Mean Y = 154.2
Sum of squares (SSX) = 601.2
Sum of products (SP) = 662.4
Regression Equation = ŷ = bX + a
b = SP/SSX = 662.4/601.2 =
1.1018
a = MY - bMX = 154.2 -
(1.1*89.6) = 55.4790
ŷ = 1.1018X + 55.4790
X Values
∑ = 448
Mean = 89.6
∑(X - Mx)2 = SSx = 601.2
Y Values
∑ = 771
Mean = 154.2
∑(Y - My)2 = SSy = 1076.8
X and Y Combined
N = 5
∑(X - Mx)(Y - My) = 662.4
R Calculation
r = ∑((X - My)(Y - Mx)) /
√((SSx)(SSy))
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