4. The data show systolic and diastolic blood pressure of certain people. Find t
ID: 3064022 • Letter: 4
Question
4. The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 137. Use a significance level of 0.05 | 129 119 125 145 125 125 142 110 94 64 68 84 9391 04 79 Diastolic Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? x (Round to two decimal places as needed.) What is the best predicted value? (Round to one decimal place as needed.) 1: Critical Values of the Pearson Correlation CoefficientExplanation / Answer
x
y
x^2
y^2
xy
129
94
16641
8836
12126
119
64
14161
4096
7616
126
68
15876
4624
8568
145
84
21025
7056
12180
126
93
15876
8649
11718
125
91
15625
8281
11375
142
104
20164
10816
14768
110
79
12100
6241
8690
Ex
Ey
Ex^2
Ey^2
Exy
1022
677
131468
58599
87041
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.51
R Square
0.26
Adjusted R Square
0.14
Standard Error
12.71
Observations
8
ANOVA
df
SS
MS
F
Significance F
Regression
1
338.50
338.50
2.10
0.20
Residual
6
969.37
161.56
Total
7
1307.88
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
6.60
54.09
0.12
0.91
-125.75
138.95
x
0.61
0.42
1.45
0.20
-0.42
1.64
Correlation r: -
n(Exy)-(Ex)(Ey)/ÖnEx^2-(Ex)^2* nEy^2-(Ey)^2
8*87041-(1022)(677)/ Ö10*131468 -1022^2*10*58599^2-677^2
0.5087
b1= nE(xy)-ExEy/nE(x2)-(Ex2)
= 8*87041-1022*677/8*131468-(1022)^2
= 0.6107
b0=Ey-b1Ex/n
=677-(0.6107*1022)/8
=6.6
y=6.6+0.6107x
Predicted Value: -
Y=6.6+0.6107*137
Y=90.27
x
y
x^2
y^2
xy
129
94
16641
8836
12126
119
64
14161
4096
7616
126
68
15876
4624
8568
145
84
21025
7056
12180
126
93
15876
8649
11718
125
91
15625
8281
11375
142
104
20164
10816
14768
110
79
12100
6241
8690
Ex
Ey
Ex^2
Ey^2
Exy
1022
677
131468
58599
87041
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