Question

# Listed below are systolic blood pressure measurements​ (in mm​ Hg) obtained from the same woman. ​a)...

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.

 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

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