Question

The data show systolic and diastolic blood pressure of certain people. Find the regression​ equation, letting the systolic reading be the independent​ (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 150. Is the predicted value close to 66.2​, which was the actual diastolic​ reading? Use a significance level of 0.05.
Systolic
139
118
145
133
141
138
139
136
  
Diastolic
101
61
82
74
104
79
74
74
LOADING... Click the icon to view the critical values of the Pearson correlation coefficient r.

Answer 1

X	Y	XY	X²	Y²
139	101	14039	19321	10201
118	61	7198	13924	3721
145	82	11890	21025	6724
133	74	9842	17689	5476
141	104	14664	19881	10816
138	79	10902	19044	6241
139	74	10286	19321	5476
136	74	10064	18496	5476
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
1089	649	88885	148701	54131

Sample size, n =	8
x̅ = Ʃx/n = 1089/8 =	136.125
y̅ = Ʃy/n = 649/8 =	81.125
SSxx = Ʃx² - (Ʃx)²/n = 148701 - (1089)²/8 =	460.875
SSyy = Ʃy² - (Ʃy)²/n = 54131 - (649)²/8 =	1480.875
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 88885 - (1089)(649)/8 =	539.875

Slope, b = SSxy/SSxx = 539.875/460.875 = 1.1714131

y-intercept, a = y̅ -b* x̅ = 81.125 - (1.17141)*136.125 = -78.3336

Regression equation :

ŷ = -78.3336 + (1.1714) x

Predicted value of y at x = 150

ŷ = -78.3336 + (1.1714) * 150 = 97.3784

Residual = y - ŷ = 66.2 - 97.3784 = -31.1784

No, the predicted value is not close to the actual value 66.2.