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Nonlinear Regression in SPSS In this example, we are going to look at a hypothetical example of “medical cost offsets” associated with psychotherapy. A “medical cost offset” is a reduction in medical costs that results from someone getting psychological treatment. In this example, the “Month” variable tells you how many months before the start of treatment or after the start of treatment an observation was made. (For example, “-1” on this variable means that the observation was made one month before the person started psychotherapy, and “0” means that it was made in the same month the person started therapy). “Medical cost” tells you how much that person’s medical treatment cost during that particular month. Let’s try a standard linear regression model to see if we can predict “medical cost” from “month.” Here’s how we’ll do it: Use the “Regression” : “Curve Estimation” command. (There are other ways to do standard regression in SPSS, but this is the method we’ll use this week). The following dialog box will appear. Put the variables in the appropriate places: “Medical Cost” is the criterion variable … … “Month” is the predictor … … and we want to test a standard linear regression model (y = β1x + β0) Make sure that the “Plot Models” box is also checked, so that you get a graph on your printout. Then hit “OK” to see the results: Curve Fit MODEL: MOD_1._ Independent: MONTH Dependent Mth Rsq d.f. F MED_COST LIN .012 20 .25 Sigf b0 b1 .622 169.864 -1.8045 This printout shows you (a) the beta coefficients (beta zero and beta one) for the regression equation, and also (b) the F-test value to show whether the model is a good fit or not. The “signif” column gives you a p-value. This one (.622) is greater than .05, so the model is a poor fit for the data. Another way to see if the model is a good fit for the data is to look at the graph in the printout. In this case, the straight line (in red) is the hypothetical model that you’re trying to fit to the data. The green jagged line is the actual pattern of the data. As you can see, they don’t look much alike. $ Spent on Medical Expenses This Month 300 200 100 Observed 0 Linear -6 -4 -2 0 2 4 Month Before (-) or After (+) Start of Psychotherapy 6 Now go back to the original dialog box. Un-check the “linear” model, and select the “quadratic” and “cubic” models instead. Hit “OK” to see the new output: Curve Fit MODEL: MOD_2._ Independent: MONTH Dependent Mth Rsq d.f. F MED_COST QUA MED_COST CUB .664 .666 19 18 18.74 11.94 Sigf b0 b1 b2 b3 .000 216.781 -1.8045 -4.6917 .000 216.781 -.1155 -4.6917 -.0949 This printout shows you two different models (based on the two boxes that you checked). You can see that the p-values for both the quadratic and cubic models are significant. In general, I recommend that you choose the simpler model (in this case, the quadratic one). Plug in the values of the beta coefficients to get this equation to describe the model: y = 216.781 – 1.8045x – 4.6917x2 Here’s the graph: $ Spent on Medical Expenses This Month 300 200 100 Observed Quadratic Cubic 0 -6 -4 -2 0 2 4 6 Month Before (-) or After (+) Start of Psychotherapy Looking at this graph gives you another reason for choosing the quadratic model (as opposed to the cubic one). Both the quadratic and the cubic models look like a parabola—so the equation that actually describes a parabola (the quadratic equation) is probably the best explanation for the patterns observed in the data.