Respuesta :
Solution :
a).
Given :
R = 0.636, [tex]$S_x = 99$[/tex], [tex]$S_y=113, M_x=108, M_y=129$[/tex]
Here R = correlation between the two variables
[tex]$S_x , S_y$[/tex] = sample standard deviations of the distance and travel time between the two train stops, respectively.
[tex]$M_x, M_y$[/tex] = means of the distance and travel between two train stops respectively.
The slope of the regression line is given by :
Regression line, [tex]b_1[/tex] [tex]$=R \times \left(\frac{S_y}{S_x}\right)$[/tex]
[tex]$=0.636 \times \left(\frac{113}{99}\right)$[/tex]
= 0.726
Therefore, the slope of the regression line [tex]b_1[/tex] is 0.726
The equation of the regression line is given by :
[tex]$\overline {y} = b_0+b_1 \overline x$[/tex]
The regression line also has to pass through the two means. That is, it has to pass through points (108, 129). Substituting these values in the equation of the regression line, we can get the value of the line y-intercept.
The y-intercept of the regression line [tex]$b_0$[/tex] is given by :
[tex]$b_0=M_y-(b_1 \times M_x)$[/tex]
= 129 - (0.726 x 108)
= 50.592
Therefore, the equation of the line is :
Travel time = 20.592 + 0.726 x distance
b).[tex]\text{ The slope of the line predicts that it will require 0.726 minutes}[/tex] for each additional mile travelled.
The intercept of the line, [tex]$b_0$[/tex] = 0.529 can be seen as the time when the distance travelled is zero. It does not make much sense in this context because it seems we have travelled zero distance in 50.529 minutes, but we could interpret it as that the wait time after which we start travelling and calculating the distance travelled and the additional time required per mile. Or we could view the intercept value as the time it takes to walk to the train station before we board the train. So this is a fixed quantity that will be added to travel time. It all depends on the interpretation.
c). [tex]$R^2=0.404$[/tex]
This means that the model accounts for around 40.4% variation in the travel time.
