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How To Make Predictions Using Regression Equations?

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- 1. How-To Use Regression Equations to Make Predictions Page 1 How-To Make Predictions Using Regression Equations Objective: Learn how to use regression equations to make predictions. Keywords and Concepts 1. Predictions 4. Error in prediction 2. Regression equations 5. Best-fit straight line 3. Linear correlation 6. Correlation coefficent 7. r Regression equations (See; How-to compute the best fit straight line to a set of data, and How to compute and interpret Pearson’s product moment correlation) can be used to predict the value of one variable using the value of another variable. If the regression line fits the data well (relatively high correlation and low standard error of estimate), then it makes sense to use regression for predictions within the range of the data set. If a significant linear correlation Start exists between an X and Y variable, the best-predicted Y value (Y’) occurs by substituting its associated X-value into Calculate the the regression equation (see Figure 1). value of r and test it’s The regression line fits the data well as significance r approaches –1.00 or +1.00. As r Use the approaches 0 (no relationship between regression equation to X and Y), the regression line fits poorly make Is there a high, Yes predictions. significant linear Substitute the and should not be used for prediction correlation ? given value in the regression because it provides just about the same equation No predictive information of the Y variable Given any value of one as the mean of Y. variable, the best predicted value of the other variable is the mean
- 2. How-To Use Regression Equations to Make Predictions Page 2 Example Predict the corresponding Fahrenheit temperature when temperature Celsius equals 20ºC. Figure 2 shows the regression of Fahrenheit temperatures plotted as a Figure 1. Fig 2. Celsius versus Fahrenheit 140 120 Degree Fahrenheit 100 80 60 y = 1.8001x + 32.004 40 R2 = 1 20 0 0 10 20 30 40 50 Degree Celsius function of Celsius temperature. The correlation between Celsius and Fahrenheit temperatures equals r = 0.999 (rounded to 1.0) and the regression equation computes as: Y’ = mX + C (eq. #1) Y’ = 1.80X + 32.0 The high linear correlation (r = 1.0) between the two variables justifies use of the regression equation to predict degrees Fahrenheit temperature from degrees Celsius temperature. For example, when Celsius temperature equals 20º, predict the corresponding Fahrenheit temperature?
- 3. How-To Use Regression Equations to Make Predictions Page 3 Solution Substituting in equation #1: Y’ = 1.80X + 32.0 Y’ = 1.8 (20) + 32.0 Y’ = 68.0 Thus, predicted Fahrenheit temperature from 20ºC equals 68ºF. Guidelines for Using Regression Equations 1. Without a linear correlation between X and Y variables (e.g., curvilinear or U- shaped relationships), do not use the linear regression equation model for prediction. 2. Only use a regression equation to predict scores within the range of scores used to derive the equation. For example, if the equation to predict percentage body fat from skinfolds was derived on women within a normal body fat range, it would be inappropriate to use this specific equation to predict body fat for very under fat (anorexic) or massively overfat (obese) women. 3. Do not make predictions about a sample that differs from the group on which the regression equation was derived. For example, if a prediction equation for percentage body fat is derived from data on males do not use the equation to predict the percentage body fat of females. Error in Prediction: The Standard Error of Estimate (SEE) Every prediction has an associated error that indicates how close the predicted value comes to the observed value. This standard error of estimate (SEE) equals: SEE = SDy 1 − r2 (eq. #2)
- 4. How-To Use Regression Equations to Make Predictions Page 4 where, SDY equals the standard deviation of the Y-variable and r equals the Pearson- product moment correlation between the X and Y variables. SEE Example Calculation The correlation between Celsius temperature (Y variable) and Fahrenheit temperature (X variable) equals r=0.999 and SDcelsius equals 13.92º. SEE = SDY 1− r2 SEE = 13.92 1 − 0.992 SEE = 0.277º SEE Interpretations The interpretation of SEE is similar to the standard deviation’s “68-95-99 rule”. That is, 68% of the time the predicted Fahrenheit temperature will fall plus-or-minus (±) 1 x SEE of the actual score based on the regression equation; 95% of the time the predicted score will fall ±2 x SEE; 99% of the time the predicted score will be within ±3 x SEE. In the previous example, the Fahrenheit temperature of 68º predicted from a temperature of 20ºC has a corresponding SEE of ±0.277ºF. Thus, 68% of the time the predicted temperature will range between 67.7 and 68.3ºF; 95% of the time the predicted temperature will lie between 67.4 and 68.6ºF; and 98% of the time the predicted temperature will lie between 67.2 and 68.8ºF. A smaller SEE means the range of error when predicting will be smaller, and hence, the more confidence one can have in the prediction.

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