This document discusses linear regression analysis. It defines the equation of a linear regression line as ŷ = mx + b, where m is the slope and b is the y-intercept. The document provides the steps to compute the regression line from a dataset, including calculating the correlation coefficient, slope, y-intercept, and using the line to predict values and calculate errors. It works through an example dataset, computing the regression line and making predictions.

1. “ Equation of Linear Regression Line”
Presented by : Eng. Waleed Alzaghal
YouTube Channel: Waleed Alzaghal

2. Equation of a Straight Line
y = mx + b
y : the y-value for a given x-value
m : the slope of the line
b : the y-intercept

3. Equation of a Straight Line

4. Equation of a Straight Line

5. Equation of a Straight Line

6. The Equation of a Regression Line
ŷ = mx + b
ŷ : the predicted y-value for a given x-value
m : the slope of the line
b : the y-intercept
Obs. x i y j
1 x 1 y 1
2 x 2 y 2
3 x 3 y 3
… … …
… … …
… … …
… … …
… … …
n x n y n
Sum

7. ŷ = mx + b
m =
𝑛 𝑥𝑦 −( 𝑥)( 𝑦)
𝑛 𝑥2
−( 𝑥)2
𝑦 = m𝑥 + b
b = 𝑦 - m𝑥 where:
𝑥 : the mean of the x values.
𝑦 : the mean of the y values.
b =
𝑦
𝑛
- 𝑚
𝑥
𝑛
b =
𝑦 −𝑚 𝑥
𝑛
Obs. x i y j
1 x 1 y 1
2 x 2 y 2
3 x 3 y 3
… … …
… … …
… … …
… … …
… … …
n x n y n
Sum

8. For the following data:
1- Display the Scatter Plot.
2- Compute the Pearson’s Correlation Coefficient.
3- Find the Equation of the Regression Line.
4- Draw the Regression Line.
5- Predict the value of ŷ for x=14 and x=8
6- Compute the Error of Prediction for x=10
Obs. x y
1 1 1
2 2 3
3 3 2
4 3 3
5 5 3
6 5 5
7 7 4
8 7 7
9 9 7
10 10 5
11 11 7
12 11 8
13 13 8
14 13 9

9. Obs. x y
1 1 1
2 2 3
3 3 2
4 3 3
5 5 3
6 5 5
7 7 4
8 7 7
9 9 7
10 10 5
11 11 7
12 11 8
13 13 8
14 13 9
1- Display the scatter plot.

10. Obs. x y x 2
y 2
xy
1 1 1 1 1 1
2 2 3 4 9 6
3 3 2 9 4 6
4 3 3 9 9 9
5 5 3 25 9 15
6 5 5 25 25 25
7 7 4 49 16 28
8 7 7 49 49 49
9 9 7 81 49 63
10 10 5 100 25 50
11 11 7 121 49 77
12 11 8 121 64 88
13 13 8 169 64 104
14 13 9 169 81 117
Sum 100 72 932 454 638
𝑟 =
𝑛 𝑥𝑦 − ( 𝑥)( 𝑦)
(𝑛 𝑥2 − ( 𝑥)2). (𝑛 𝑦2 − ( 𝑦)2
𝑟 =
14 638 − (100)(72)
( 14 932 − 100 2)( 14 454 − 72 2)
𝑟 =
1732
(3048)(1172)
𝑟 =
1732
1890.04
r ≈ 0.92
(Interpretation: Strong Positive Correlation)
2- Compute the Pearson’s correlation coefficient.

11. ŷ = mx + b
m =
𝑛 𝑥𝑦 −( 𝑥)( 𝑦)
𝑛 𝑥2
−( 𝑥)2
m =
14 638 −(100)(72)
14 932 − 100 2
m =
1732
3048
m ≈ 0.57
3- Find the equation of the regression line.
Obs. x y x 2
y 2
xy
1 1 1 1 1 1
2 2 3 4 9 6
3 3 2 9 4 6
4 3 3 9 9 9
5 5 3 25 9 15
6 5 5 25 25 25
7 7 4 49 16 28
8 7 7 49 49 49
9 9 7 81 49 63
10 10 5 100 25 50
11 11 7 121 49 77
12 11 8 121 64 88
13 13 8 169 64 104
14 13 9 169 81 117
Sum 100 72 932 454 638

12. ŷ = mx + b
b =
𝑦 −𝑚 𝑥
𝑛
b =
(72) −(0.57)(100)
14
b =
15
14
b = 1.07
ŷ = 0.57 x + 1.07
Obs. x y x 2
y 2
xy
1 1 1 1 1 1
2 2 3 4 9 6
3 3 2 9 4 6
4 3 3 9 9 9
5 5 3 25 9 15
6 5 5 25 25 25
7 7 4 49 16 28
8 7 7 49 49 49
9 9 7 81 49 63
10 10 5 100 25 50
11 11 7 121 49 77
12 11 8 121 64 88
13 13 8 169 64 104
14 13 9 169 81 117
Sum 100 72 932 454 638
3- Find the equation of the regression line.

13. 4- Draw the regression line.

14. 5- Predict the value of ŷ for x=14 and x=8
For x=14 (Extrapolation)
ŷ = 0.57 x + 1.07
ŷ = (0.57)(14) + 1.07
ŷ = 9.05
For x=8 (Interpolation)
ŷ = 0.57 x + 1.07
ŷ = (0.57)(8) + 1.07
ŷ = 5.63

15. Predicted Value ŷ
ŷ = 0.57 x + 1.07
For x=10
ŷ = (0.57)(10) + 1.07 = 6.77
Actual Value = 5
Error of Prediction = Predicted Value – Actual Value
Error of Prediction = 6.77 – 5 = 1.77
6- Compute the Error of Prediction for x=10