A. Joshua Bentley
November 28, 2018
The Coast Starlight Amtrak train runs from Seattle to Los Angeles.
The mean travel time from one stop to the next on the Coast Starlight is 129 mins, with a standard deviation of 113 minutes.
The mean distance traveled from one stop to the next is 108 miles with a standard deviation of 99 miles.
The correlation between travel time and distance is 0.636.
1 of 3
Target | Mean | StDev | . | Other.Metrix | X_ |
---|---|---|---|---|---|
time (as y) | 129 | 113 | Correlation | 0.636 | |
distance (as x) | 108 | 99 | |||
Equation for slope using Standard Deviation
b-sub_1 = \( \frac{S_y}{S_x} \cdot \) R
b = \( \frac{113}{99} \cdot \) 0.636
Slope(b) = 0.726
2 of 3
Target | Mean | StDev | . | Other.Metrix | X_ |
---|---|---|---|---|---|
time (as y) | 129 | 113 | Correlation | 0.636 | |
distance (as x) | 108 | 99 | Slope | 0.726 | |
Slope-point equation for regression line
y - y-sub_0 = slope \( \cdot \) ( x - x-sub_0 )
129 - y = 0.726 \( \cdot \) (108 - x)
129 - y = 78.41 - 0.726x
3 of 3
Target | Mean | StDev | . | Other.Metrix | X_ |
---|---|---|---|---|---|
time (as y) | 129 | 113 | Correlation | 0.636 | |
distance (as x) | 108 | 99 | Slope | 0.726 | |
Slope-interept equation for regression line
129 - y = 78.41 - 0.726x
-y = -50.59 - 0.726x \( \Rightarrow \) y = 0.726x + 50.59
Target | Mean | StDev | . | Other.Metrix | X_ |
---|---|---|---|---|---|
time (as y) | 129 | 113 | Correlation | 0.636 | |
distance (as x) | 108 | 99 | Slope | 0.726 | |
y intercept | 50.590 |
The line has a slope of 0.726 and will cross the y axis when y = 50.59.
This means that the train travels at approximately 0.726 minutes per mile.
Correlation \( R \) is 0.6362 so \( R \)2 is .405, meaning that about 41% of travel time is explained by distance.
y = 0.726x + 50.59
y = (0.726)(103) + 50.59
y = 125.4 minutes
The residual for this point is 42.6 meaning that the predicted value was 42.6 minutes less than the actual observation.
In general we can see that the question is asking us to predict a value that lies outside the population from which samples are available and thus, ultimate, the answer is no, it would need to be extrapolated.