Question

In: Statistics and Probability

The data show systolic and diastolic blood pressure of certain people. Find the regression​ equation, letting...

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.

Solutions

Expert Solution

X Y XY
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.


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