## Saturday, November 6, 2010

### CH 1 Project - Modeling with Functions and Regression Equations

For this problem, I will have to graph and predict the hourly earnings of Canadian production workers.

First, I will have to enter this data into the statistics tables on my calculator (STAT > Edit). I will enter the years into L1 and the wages into L2.

Then, I will find the linear regression equation. I can do this by going to STAT > CALC > LinReg(a+bx). I entered LinReg(a+bx) L1,L2,Y1 so the equation in the list of equations when I hit the "y=" button. You can show Y1 on your calculator by going to VARS > Y-VARS > Function > Y1.

r is the correlation coefficient, or how closely this equation matches the data. Rounded to the nearest one thousandth, r is approximately 0.962. This value is appropriate for the model, although I would prefer a value closer to 1.

Next, I will find the quadratic regression, which is a model of the data using a quadratic equation. I can do this by using QuadReg, instead of LinReg(a+bx).

The value of R2 (rounded to the nearest thousandth is .948) also suggests that this model is appropriate, but also that the linear model is better.

Lastly, I will graph these models. I turned on "Plot1" in the "y=" screen so the scatter plot will also be graphed.

The bold line is the quadratic model, and the squares are the original scatter plot. The viewing window I used here is: xmin=1971.3, xmax=2011.3, ymin=2.4254, ymax=33.7546.

The original problem here was to predict the wages of Canadian workers in 2010. Now that my models are set up, all I have to do is view the table (2ND > GRAPH).

The estimates here are slightly different from each other, with the linear model predicting \$28.55 and the quadratic model predicting \$30.55/hr.

The regression equations here fit the data relatively well as shown in the graph above. The linear equation was much better though, since it's correlation efficient, r≈.962, is closer to 1 than the quadratic equation's correlation efficient, R2≈.948.