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Salary Regression Example

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Submitted By zoomboom
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John Caldock was studying for his Master’s in 1990 and at the same time he worked a the University as a research assistant. In 1995, he got his first job as a developer in a software company and he continue advancing in his career path and in 2003 he became a senior IT manager in a different IT company. In the following table you are provided with his yearly salary for these 13 years.

|1990 |7,000 |
|1991 |8,000 |
|1992 |9,200 |
|1993 |10,100 |
|1994 |11,000 |
|1995 |48,000 |
|1996 |50,000 |
|1997 |52,000 |
|1998 |57,000 |
|1999 |63,000 |
|2000 |67,000 |
|2001 |72,000 |
|2002 |103,000 |
|2003 |108,000 |

Perform a linear regression on his salary, and provide the linear regression eequation, the correlation coefficient and the corresponding graph. Use the equation to predict his salary in the years 2008 and 2010, other factors staying equal. Is the correlation coefficient positive or negative and what this represents?

Perform the linear regression in both Excel and by hand, remember that technology many times truncates numbers and do not allow us to use the regression equation as given. Verify the equation and your results by hand.

To solve this problem we have to set up the problem in an MS Excel spreadsheet and use the chart Wizard to create the regression line. The solution provided by the Excel is the following.

As you can realize the solution provided by the Excel is y= 7935.2 x – 2E +07 while the R2 = 0.9256.

Many software, including Excel, truncate the final numbers when these are really long integer or decimal numbers. In the given problem the y-intercept that is provided is -2E+07=20000000 that it is truncated. The problem with this exercise is that if we use this number and plug in the years 2008 and 2010 to find the projected salaries for the given years, the values that we receive are not accurate – actually they are negative numbers.

If you solve the problem by hand the value for the y-intercept is 15791048.
Thus the actual regression line will look as:

y= 7935.2x – 15791048.

For x= 2008 year : y= 7935.2*2008 – 15791048 = $149,891.6

For x= 2010 year : y= 7935.2*2010 – 15791048 = $165,762

Since R2 = 0.9256, the R = 0.977548 the positive value for the correlation coefficient is extracted from the fact that the line is increasing.

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