Question

Listed below are systolic blood pressure measurements​ (in mm​ Hg) obtained from the same woman. ​a) Find the regression​ equation, letting the right arm blood pressure be the predictor​ (x) variable. ​b) Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 90 mm Hg. Right Arm 101 100 94 79 80 Left Arm 175 169 144 145 143 ​a) The regression equation is ModifyingAbove y with caret equalsnothingplusnothing x. ​(Round to one decimal place as​ needed.) ​b) Given that the systolic blood pressure in the right arm is 90 mm​ Hg, the best predicted systolic blood pressure in the left arm is nothing mm Hg.

Answer 1

X	Y	XY	X²	Y²
101	175	17675	10201	30625
100	169	16900	10000	28561
94	144	13536	8836	20736
79	145	11455	6241	21025
80	143	11440	6400	20449

Ʃx =	454
Ʃy =	776
Ʃxy =	71006
Ʃx² =	41678
Ʃy² =	121396
Sample size, n =	5
x̅ = Ʃx/n = 454/5 =	90.8
y̅ = Ʃy/n = 776/5 =	155.2
SSxx = Ʃx² - (Ʃx)²/n = 41678 - (454)²/5 =	454.8
SSyy = Ʃy² - (Ʃy)²/n = 121396 - (776)²/5 =	960.8
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 71006 - (454)(776)/5 =	545.2

Slope, b = SSxy/SSxx = 545.2/454.8 = 1.1987687

y-intercept, a = y̅ -b* x̅ = 155.2 - (1.19877)*90.8 = 46.351803

a) Regression equation :

ŷ = 46.4 + (1.2) x

b) Predicted value of y at x = 90

ŷ = 46.3518 + (1.1988) * 90 = 154.2