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 146 . Is the predicted value close to 91.0 ​, which was the actual diastolic​reading? Use a significance level of 0.05.Systolic 117 134 150 113 138 113 144 145Diastolic 85 74 86 55 75 80 106 841. What is the regression​ equation? ​ (Round to two decimal places as​ needed.)2. What is the best predicted diastolic pressure for a person with a systolic reading of 146 ​?3. Is the predicted value close to 91.0 ​, which was the actual diastolic​ reading?A. The predicted value is very close to the actual diastolic reading.B. The predicted value is not close to the actual diastolic reading.C. The predicted value is close to the actual diastolic reading.D. The predicted value is exactly the same as the actual diastolic reading.

Answer 1

ŷ = 6.69X - 706.18 ; ŷ = 270.56 ; B. The predicted value is not close to the actual diastolic reading.

Explanation:

Systolic reading(x) :

117

134

150

113

138

113

144

145

Diastolic (y) :

85

74

86

55

75

80

106

841

Using the online regression calculator, the regression equation obtained using the data values provided above is:

ŷ = 6.69X - 706.18

Comparing with the regression equation formula: ŷ = mX + c

Where :

ŷ = predicted variable (Diastolic)

m = 6.69 = gradient or slope of the regression line

c = - 706.18 = intercept ( where the regression line intersects the y-axis).

2.) What is the best predicted diastolic pressure for a person with a systolic reading of 146

X = 146

ŷ = 6.69(146) - 706.18

ŷ = 976.74 - 706.18

ŷ = 270.56

The predicted value, 270.56 is not close to the actual Diastolic reading of 91.0