Answer:
Y^= 1.767 + 0.6294X when rounded give s Y^= 1.77 +0.63X
b= 0.6294 rounded to 0.63
a= 1.77
The predicted lines are for each X and Y
3.340,3.96, 4.914, 5.228, and 5.543
Step-by-step explanation:
The data given is
Length (m) Speed (m/s) Predicted Line
2.5 3 3.340
3.5 4.5 3.96
5 4.8 4.914
5.5 5.2 5.228
6 5.5 5.543
The calculations are
Xsquare XY Y X
6.25 7.5 3 2.5
12.25 15.75 4.5 3.5
25 24 4.8 5
30.25 28.6 5.2 5.5
36 33 5.5 6
Total 109.75 108.85 23 22.5
The estimated regression line of Y on X is
Y^ = a +bX
and two normal equations are
∑Y = na + b∑X
∑XY= a∑X + b∑X²
Now X`= ∑X/ n= 22.5/5=4.5
Y`= ∑Y/ n= 23/5= 4.6
b= n∑XY- (∑X)(∑Y) / n∑X²- (∑X²)
Putting the values
b= 5(108.85) - (23)(22.5)/ 5(109.75)- (22.5)²
b= 544.25-517.5/ 548.75-506.25
b= 26.75 /42.5
b= 0.6294
and
a= Y`- bX~= 4.6- 0.6294(4.5)= 4.6-2.823= 1.767
Hence the
desired estimated regression line of Y on X is
Y^= 1.767 + 0.6294X
Y^= 1.77 +0.63X
The estimated regression co efficient b= 0.6294 indicates that the values of Y increase by 0.6294 units for a unit increase in X.
