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 90 mm Hg. Use a significance level of 0.05 .

Right arm: 100, 99, 92, 80, 79

Left arm: 176, 170, 143, 145, 144

a. The regression equation is?

b. Given that the systolic blood pressure in the right arm is 90mm​ Hg, the best predicted systolic blood pressure in the left arm is _ mm hg

Answer 1

	x	y	(x-xbar)^2	(y-ybar)^2	(x-xbar)*(y-ybar)
	100	176	100	416.16	204
	99	170	81	207.36	129.6
	92	143	4	158.76	-25.2
	80	145	100	112.36	106
	79	144	121	134.56	127.6
Total	450	778	406	1029.2	542
Mean	90	155.6

Slope	beta1	1.334975
Intercept	beta0	35.45222

a) Regression equation is			y=beta0+beta1*x
			B.P. of Left Arm (y)=35.45222+1.334975*B.P. of right arm (x)

b)

B.P. of right arm is 90 so estimated B.P. of left arm is,

155.6

Formula Sheet:

	x	y	(x-xbar)^2	(y-ybar)^2	(x-xbar)*(y-ybar)
	100	176	=(B2-$B$8)^2	=(C2-$C$8)^2	=(B2-$B$8)*(C2-$C$8)
	99	170	=(B3-$B$8)^2	=(C3-$C$8)^2	=(B3-$B$8)*(C3-$C$8)
	92	143	=(B4-$B$8)^2	=(C4-$C$8)^2	=(B4-$B$8)*(C4-$C$8)
	80	145	=(B5-$B$8)^2	=(C5-$C$8)^2	=(B5-$B$8)*(C5-$C$8)
	79	144	=(B6-$B$8)^2	=(C6-$C$8)^2	=(B6-$B$8)*(C6-$C$8)
Total	=SUM(B2:B6)	=SUM(C2:C6)	=SUM(D2:D6)	=SUM(E2:E6)	=SUM(F2:F6)
Mean	=B7/5	=C7/5
Slope	beta1	=F7/D7
Intercept	beta0	=C8-C10*B8

a) Regression equation is			y=beta0+beta1*x
			B.P. of Left Arm (y)=35.45222+1.334975*B.P. of right arm (x)
b)	B.P. of right arm is 90 so estimated B.P. of left arm is,					=C11+C10*90