Question

For questions 8 - 15, use the following set of data to find the equation of the line of

Answer 1

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.