residual(e)=y-y^=y-(a+bx)=y-(10.26+4.92*x)
for x=3, the estimated value of y is
y^=10.26+4.92*3=25.02
residual for this value is y-y^=22.3-25.02=-2.72,
negative residual means it is overestimated
Residual = Observed – Predicted
positive values for the residual mean the prediction
was too low, and negative values mean the prediction was
too high.when the model overestimates the
observation: the residual is negative.
here Damage(y) is dependent variable and Distance(x) is
independent variable.
the regression equation y^=a+bx=10.26+4.92*x
a=(sum(y)*sum(x2)-sum(x)*sum(xy))/(n*sum(x2)-(sum(x))2)=(396.16*196.16-49.2*1470.65)/(15*196.16-49.2*49.2)=10.26
b=(n*sum(xy)-sum(x)*sum(y))/
(n*sum(x2)-(sum(x))2)=(15*1470.65-49.2*396.2)/(15*196.16-49.2*49.2)=4.92
|
S.N. |
Distance(x) |
Damage(y) |
x2 |
y2 |
xy |
|
1 |
3.4 |
26.2 |
11.56 |
686.44 |
89.08 |
|
2 |
1.8 |
17.8 |
3.24 |
316.84 |
32.04 |
|
3 |
4.6 |
31.3 |
21.16 |
979.69 |
143.98 |
|
4 |
2.3 |
23.1 |
5.29 |
533.61 |
53.13 |
|
5 |
3.1 |
27.5 |
9.61 |
756.25 |
85.25 |
|
6 |
5.5 |
36 |
30.25 |
1296 |
198 |
|
7 |
0.7 |
14.1 |
0.49 |
198.81 |
9.87 |
|
8 |
3.0 |
22.3 |
9 |
497.29 |
66.9 |
|
9 |
2.6 |
19.6 |
6.76 |
384.16 |
50.96 |
|
10 |
4.3 |
31.3 |
18.49 |
979.69 |
134.59 |
|
11 |
2.1 |
24 |
4.41 |
576 |
50.4 |
|
12 |
1.1 |
17.3 |
1.21 |
299.29 |
19.03 |
|
13 |
6.1 |
43.2 |
37.21 |
1866.24 |
263.52 |
|
14 |
4.8 |
36.4 |
23.04 |
1324.96 |
174.72 |
|
15 |
3.8 |
26.1 |
14.44 |
681.21 |
99.18 |
|
sum= |
49.2 |
396.2 |
196.16 |
11376.5 |
1470.65 |
|
n= |
15 |
15 |
15 |
15 |
15 |
a=(sum(y)*sum(x2)-sum(x)*sum(xy))/(n*sum(x2)-(sum(x))2)=(396.16*196.16-49.2*1470.65)/(15*196.16-49.2*49.2)=
b=(n*sum(xy)-sum(x)*sum(y))/
(n*sum(x2)-(sum(x))2)=(15*1470.65-49.2*396.2)/(15*196.16-49.2*49.2)=
