The line of best fit for the data of week and the corresponding miles run is evaluated to be: y = 4.699 + 1.671x
Firstly, there is a data set. Then, we try to fit a line which will tell about the linear trend. This line is made using the least squares method.
The following data for the Week and miles run variables are provided to construct linear regression model:
Week miles run
1 5
2 8
4 13
6 15
8 19
10 20
The independent variable is Week(represented by x), and the dependent variable is miles run(represented by y). In order to compute the regression coefficients, the following table needs to be used:
Let week = x, and miles run = y, then we get this table:
x y xy [tex]x^2[/tex] [tex]y^2[/tex]
----------------------------------------------------
1 5 5 1 25
2 8 16 4 64
4 13 52 16 169
6 15 90 36 225
8 19 152 64 361
10 20 200 100 400
-----------------------------------------------------
Sum = 31 80 515 221 1244
Based on the above table, the following is calculated:
[tex]\bar X = \frac{1}{n} \sum_{i=1}^{n} X_i = \frac{ 31}{ 6} \approx 5.167[/tex]
[tex]\bar Y = \frac{1}{n} \sum_{i=1}^{n} Y_i = \frac{ 80}{ 6} \approx 13.33[/tex]
[tex]\large SS_{XX} = \sum_{i=1}^{n} X_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right)^2 = 221 - 31^2/6 \approx 60.833[/tex]
[tex]\large SS_{YY} = \sum_{i=1}^{n} Y_i^2 - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} Y_i\right)^2 = 1244 - 80^2/6 \approx 177.33[/tex]
[tex]\large SS_{XY} = \sum_{i=1}^{n} X_i Y_i - \displaystyle\frac{1}{n}\left(\sum_{i=1}^{n} X_i\right) \left(\sum_{i=1}^{n} Y_i\right) = 515 - 31 \times 80/6 \approx 101.67[/tex]
Therefore, based on the above calculations, the regression coefficients (the slope m, and the y-intercept n) are obtained as follows:
[tex]m = \frac{SS_{XY}}{SS_{XX}} \approx \frac{ 101.67}{ 60.833} \approx 1.6712[/tex]
[tex]n = \bar Y - \bar X \cdot m = 13.33- 5.167 \times 1.6712 = 4.6986[/tex]
Therefore, we find that the regression equation is:
miles run = 4.6986 + 1.6712 Week
y = 4.6986 + 1.6712x (approximately)
To the thousandths place, it becomes:
y = 4.699 + 1.671x
Therefore, based on the information provided above, the following scatter plot and regression plot are obtained as the image attached below.
Thus, the line of best fit for the data of week and the corresponding miles run is evaluated to be: y = 4.699 + 1.671x
Learn more about linear regression here:
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