Sure! Let’s consider another example of linear regression with a different set of sample data. Suppose we have data on the number of years of work experience (X) and the corresponding salary (Y) for a group of employees. We want to use linear regression to model the relationship between work experience and salary and make predictions for future employees.

Here is the sample data:

Years of Work Experience (X) | Salary (Y) |
---|---|

2 | 50000 |

3 | 60000 |

5 | 75000 |

7 | 90000 |

8 | 95000 |

Step 1: Calculate the Mean

Mean of X (X̄) = (2 + 3 + 5 + 7 + 8) / 5 = 5

Mean of Y (Ȳ) = (50000 + 60000 + 75000 + 90000 + 95000) / 5 = 74000

Step 2: Calculate the Deviations

Deviation of X (X – X̄):

(2 – 5) = -3

(3 – 5) = -2

(5 – 5) = 0

(7 – 5) = 2

(8 – 5) = 3

Deviation of Y (Y – Ȳ):

(50000 – 74000) = -24000

(60000 – 74000) = -14000

(75000 – 74000) = 1000

(90000 – 74000) = 16000

(95000 – 74000) = 21000

Step 3: Calculate the Covariance

Cov(X, Y) = (∑((X – X̄) * (Y – Ȳ))) / (n – 1)

Cov(X, Y) = ((-3 * -24000) + (-2 * -14000) + (0 * 1000) + (2 * 16000) + (3 * 21000)) / (5 – 1)

Cov(X, Y) = (72000 + 28000 + 0 + 32000 + 63000) / 4

Cov(X, Y) = 195000 / 4

Cov(X, Y) = 48750

Step 4: Calculate the Variance of X

Var(X) = (∑((X – X̄)^2)) / (n – 1)

Var(X) = ((-3)^2 + (-2)^2 + (0)^2 + (2)^2 + (3)^2) / (5 – 1)

Var(X) = (9 + 4 + 0 + 4 + 9) / 4

Var(X) = 26 / 4

Var(X) = 6.5

Step 5: Calculate the Regression Coefficients

β1 = Cov(X, Y) / Var(X)

β1 = 48750 / 6.5

β1 = 7500 (approximately)

β0 = Ȳ – (β1 * X̄)

β0 = 74000 – (7500 * 5)

β0 = 74000 – 37500

β0 = 36500 (approximately)

So, the regression equation is: Y = 36500 + 7500X

Step 6: Make Predictions

Using the regression equation, we can make predictions for salaries based on the number of years of work experience. For example, if an employee has 6 years of work experience, the predicted salary would be:

Y = 36500 + 7500 * 6

Y = 36500 + 45000

Y = 81500 (approximately)

So, based on the linear regression model, an employee with 6 years of work experience is predicted to have a salary of approximately 81500.

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