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 95 mm Hg. Use a significance level of 0.05. Right Arm 100 99 93 76 75 Left Arm 176 170 148 146 145 Click the icon to view the critical values of the Pearson correlation coefficient The regression equation isy - 66.5 -1.0 x (Round to one decimal place as needed) Given that the systolic blood pressure in the right arm is 95 mm Hg, the best predicted systolic blood pressure in the left arm is (Round to one decimal place as needed.) mm Hg.

Answer 1

	Right Arm X	Left Arm Y	X * Y	X2	Y2
	100	176	17600	10000	30976
	99	170	16830	9801	28900
	93	148	13764	8649	21904
	76	146	11096	5776	21316
	75	145	10875	5625	21025
Total	443	785	70165	39851	124121

Equation of regression line is Ŷ = a + bX

b = ( 5 * 70165 - 443 * 785 ) / ( 5 * 39851 - ( 443 )²)
b = 1.0

((X3) – 2X3u)/(43X3 – XX3U) = 9

a =( Σ Y - ( b * Σ X) ) / n
a =( 785 - ( 1.0213 * 443 ) ) / 5
a = 66.5
Equation of regression line becomes Ŷ = 66.5 + 1.0 X

When X = 95
Ŷ = 66.514 + 1.021 X
Ŷ = 66.5 + ( 1.0  * 95 )
Ŷ = 163.5