Prediction/Regression Chapter 12 Prediction/Regression Part 2: Apr. 17, 2008

Drawing the Regression Line Draw and label the axes for a scatter diagram Figure predicted value on criterion for a low value on predictor variable You can randomly choose what value to plug in.. Maybe x=1, so Ŷ = .688 + .92(1) = 1.61 Repeat step 2. with a high value on predictor, maybe x=6, so Ŷ = .688 + .92(6) = 6.21 Draw a line passing through the two marks (1, 1.61) and (6, 6.21) Hint: you can also use (Mx, My) to save time as one of your 2 points. Reg line always passes through the means of x and y.

Regression Error Now that you have a regression line or equation built from one sample, you can find predicted y scores using a new sample of x scores (Sample 2)… Then, assume that you later collect data on Sample 2’s actual y scores You can compare the accuracy of predicted ŷ to the actual y scores for Sample 2 Sometimes you’ll overestimate, sometimes underestimate…this is ERROR. Can we get a measure of error? How much is OK?

Error in regression Proportionate Reduction in Error (PRE) Actual score minus the predicted score Proportionate Reduction in Error (PRE) Squared error using prediction (reg) model = SSError =  (y - ŷ)2 Compare this to amount of error w/o this prediction (reg) model. If no other model, best guess would be the mean. Total squared error when predicting

Error and Proportionate Reduction in Error Formula for proportionate reduction in error: compares reg model to mean baseline (predicting everyone’s y score will be at the mean) Want reg model to be much better than mean(baseline) – that would indicate fewer prediction errors So you want PRE to be large…

Find prediction using reg model: Reg model was ŷ = .688 + .92(x) Use mean model to find error (y-My)2 for each person & sum up that column  SStot Find prediction using reg model: plug in x values into reg model to get ŷ Find (y-ŷ)2 for each person, sum up that column  SSerror Find PRE

Proportionate reduction in error = r2 If our reg model no better than mean, SSerror = SStotal, so (0/ SStot) = 0. Using this regression model, we reduce error over the mean model by 0%….not good prediction. If reg model has 0 error (perfect), SStot-0/SStot = 1, or 100% reduction of error. Proportionate reduction in error = r2 aka “Proportion of variance in y accounted for by x”, ranges between 0-100%.

Computing Error Sum of the squared residuals = SSerror X Y 6 6 1 2 5 6 6 6 .688 + .92(6) = 6.2 1 2 .688 + .92(1) = 1.6 5 6 .688 + .92(5) = 5.3 3 4 .688 + .92(3) = 3.45 3 2 .688 + .92(3) = 3.45 mean 3.6 4.0

Computing SSerror Sum of the squared residuals = SSerror X Y 6 6 6.2 6 6 6.2 -0.20 0.04 1 2 1.6 0.40 0.16 5 6 5.3 0.70 0.49 3 4 3.45 0.55 0.30 3 2 3.45 -1.45 2.10 mean 3.6 4.0 0.00 3.09 SSERROR

Computing SStotal = (y-My)2 X Y My y-My (y-My)2 6 6 4 2 4 1 2 4 -2 4 5 6 4 2 4 3 4 4 0 0 3 2 4 -2 4 mean 3.6 4.0 Σ=0 Σ=16 (SStotal)

PRE PRE = 16 - 3.09 16 = .807 We have 80.7% proportionate reduction in error from using our regression model as opposed to the mean baseline model So we’re doing much better using our regression as opposed to just predicting the mean for everyone…

SPSS Reg Example Analyze Regression  Linear Note that terms used in SPSS are “Independent Variable”…this is x (predictor) “Dependent Variable”…this is y (criterion) Class handout of output – what to look for: “Model Summary” section - shows R2 ANOVA section – 1st line gives ‘sig value’, if < .05  signif This tests the significance of the R2 (is the whole regression equation significant or not? If yes  it does predict y) Coefficients section – 1st line gives ‘constant’ = a Other lines give ‘standardized coefficients’ = b or beta for each predictor For each predictor, there is also a significance test (if ‘sig’ if < .05, that predictor is significantly different from 0 and does predict y) If it is significant, you’d want to interpret the beta (like a correlation)