Chapter 5. Regression

Transcription

1 Topics covered in this chapter: Chapter 5. Regression Adding a Regression Line to a Scatterplot Regression Lines and Influential Observations Finding the Least Squares Regression Model Adding a Regression Line to a Scatterplot Example 5.: Does fidgeting keep you slim? The Problem: Why is it that some people find it easy to stay slim? Do people with larger increases in nonexercise activity (NEA) tend to gain less fat? NEA change in calories is the explanatory variable and Fat Gain in kg is the response variable. Make a scatterplot of fat gain and NEA change.. Open the data set eg05-0.por. 2. Construct a scatterplot of the data. Refer to the example in Chapter 4 for the instructions. 3. Add a least squares regression line to the scatterplot: a. Double-click on the graph and the Chart Editor opens up. b. Go to the Elements menu and click on Fit Line at Total. The following screen appears: 50

2 5 Chapter 5 c. Select Linear and click Apply. d. Click the Close button. e. Once everything has been done in the Chart Editor, close the editor by clicking the X at the upper right corner of the screen. The resulting scatterplot with the regression line will be in the output:

3 Regression 52 Regression Lines and Influential Observations Example (Apply Your Knowledge 5.0): Make a Scatterplot for the Data in Exercise 5.4 and Fit Regression Lines to Determine Which New Point is More Influential The Problem: Using the body mass data found in Exercise 5.4, make a scatterplot of the data and add two new points, adding one point at a time. Point A: mass 42 kg, metabolic rate 500 calories. Point B: mass 70 kg, metabolic rate 400 calories.. Prepare the data. a. Open the data set ex05-04.por. b. Click on Transform; scroll down and select Compute Variable. c. Under the Target Variable box, type MassPlusOne. d. Click on Mass and then click the arrow to the right of the box. This moves Mass into the Numeric Expression box. e. Click OK. f. Repeat the same process using target variable RatePlusOne which will be based on Rate. g. Repeat the process again, creating variables MassPlusTwo and RatePlusTwo. h. Now the Data View screen will have six columns of data. i. At the bottom of the RatePlusOne and MassPlusOne columns, add the values for point A. Put 42 in row 5 in the MassPlusOne column and 500 in row 5 of the RatePlusOne column.

4 53 Chapter 5 j. Add the values for point A to row 5 of the RatePlusTwo and MassPlusTwo columns as well. k. Type the values for point B in row 6 of the RatePlusTwo and MassPlusTwo columns. 2. Construct a scatterplot of the data which includes all three data sets: the original 2 women, the original women plus Point A, and the original women plus Point B: a. Click on Graphs, scroll down to Legacy Dialogs and then click on Scatter/Dot. b. Click to select Overlay Scatter. c. Click Define. d. The following screen appears and we have to place each Y-X pair into the Y- X Pairs box.

5 Regression 54 e. Click RatePlusTwo and click on the arrow to the left of the Y-X Pairs box. f. Click MassPlusTwo and click on the arrow to the left of the Y-X Pairs box. g. Click RatePlusOne and click on the arrow to the left of the Y-X Pairs box. h. Click MassPlusOne and click on the arrow to the left of the Y-X Pairs box. i. Click Rate and click on the arrow to the left of the Y-X Pairs box. j. Click Mass and click on the arrow to the left of the Y-X Pairs box. k. If the pairs are in X-Y order, click the swap pair (double arrow) button.

6 55 Chapter 5 l. Click OK. The scatterplot will appear in the output window. Notice that points A and B are a different color from the rest of the points. 3. Follow the steps in Example 5. above to add least squares regression lines to the scatterplot. Finding the Least Squares Regression Model Example (Apply Your Knowledge 5.0) (Continued): Analyze the data in Exercise 5.4 using Linear Regression. Open the data set ex05-04.por. 2. Construct a scatterplot of the data and fit regression line. Refer Chapter 4 and Example 5. to construct a scatterplot with a regression fit line. 3. Analyze the data using regression: a. Go to the Analyze menu, scroll down to Regression, and select Linear. The following screen should appear:

7 Regression 56 b. Put Rate in the Dependent Variable box and Mass in the Independent Variable box. Change any default options that you want, and click OK.

8 57 Chapter 5 The above four boxes will appear in the output window. The least squares regression equation along with r, the correlation coefficient, may be found in the fourth and final box labeled Coefficients. The least squares equation is in the form y ˆ = a + bx. The estimates of a, the y-intercept, and b, the slope, can both be found in the first column labeled B. The y-intercept is in the row labeled (Constant) and the slope is in the row labeled Mass. For this example, the least squares equation is yˆ = x, where ŷ is the predicted value of the metabolic rate and x is the body mass used for the prediction. The value of the correlation coefficient can be found under the column labeled Beta. The correlation coefficient is.876, which indicates a strong, positive linear association between metabolic rate and body mass. Chapter 5 Exercises 5.3 Ebola and gorillas. 5. Outsourcing by airlines Keeping water clean Our brains don t like losses Managing diabetes. 5.4 Managing diabetes, continued. 5.5 Beavers and beetles Saving energy with solar panels.

9 336 Chapter 5 SPSS Solutions 5.3 Open data file ex We can find the mean and standard deviation for both variables using Analyze, Descriptive Statistics, Descriptives. Click to enter both variables, then OK. Descriptive Statistics N Minimum Maximum Mean Std. Deviation Ranges Days Valid N (listwise) 6 To find the correlation, use Analyze, Correlate, Bivariate. Again, click to enter both variables and OK. Correlations Ranges Days Ranges Days Pearson Correlation ** Sig. (2-tailed).002 N Pearson Correlation.962 **.000 Sig. (2-tailed).002 N **. Correlation is significant at the 0.0 level (2-tailed). sy 6.33 To hand calculate the slope, we have b= r =.962 =.263. Similarly, we s x.378 find the intercept as a = y bx = *3.5 = To check this against the software calculation, compute the regression using Analyze, Regression, Linear. Click to enter Days as the Dependent variable and Ranges as the Independent variable. OK performs the calculations. We see our results agree to within rounding. Model Unstandardized Coefficients Coefficients B Std. Error Beta (Constant) t Sig. Ranges a. Dependent Variable: Days

10 The scatterplot was created in the solution to Exercise 4.5. See that solution for details. We ll compute the correlation and regression with all the data points, then delete Hawaiian and recomputed to see its influence. Click Analyze, Regression, Linear. (There is also an option of Analyze, Correlate, Bivariate that computes the correlation but using this would mean you still have to perform the regression.) Click to enter Delay as the Dependent and Outsource as the Independent. OK performs the regression. Model R R Square Adjusted R Square Std. Error of the Estimate.476 a a. Predictors: (Constant), Outsource (Constant) Outsource a. Dependent Variable: Delay The regression equation becomes DelayPct = *OutsourcePct. The correlation is r = Based on this line, an airline with 76% outsourcing should have about *76 = 34.% delays. Now, delete the Hawaiian Airlines data values. Use the same set of commands to find the new equation. Model R R Square Adjusted R Square Std. Error of the Estimate.484 a a. Predictors: (Constant), Outsource

11 338 (Constant) Outsource a. Dependent Variable: Delay Hawaiian was influential for the regression, but not for the correlation. The correlation increased from to The regression equation changed from DelayPct = *OutsourcePct to DelayPct = *OutsourcePct. The equation that included Hawaiian Airlines predicts 34.% delayed flights for the airline with 76% outsourcing. The prediction from the regression that did not include Hawaiian is 29.8% delayed flights a decrease of 4.3% To make a scatterplot of these data, click Graph, Legacy Dialogs, Scatter/Dot. The default is Simple. Click Define to continue on to the plot definition dialog box. Click to Enter Absorb as the Y variable and Nitrates as the X variable. Click Titles to give your graph a title, then Continue and OK to generate the graph. The plot is extremely linear. To find the correlation, use Analyze, Correlate, Bivariate. Click to enter both variable names and OK.

12 339 Correlations Nitrate Absorb Nitrate Absorb Pearson Correlation ** Sig. (2-tailed).000 N Pearson Correlation.000 **.000 Sig. (2-tailed).000 N **. Correlation is significant at the 0.0 level (2-tailed). The correlation is.000, which is perfect; this calibration will not need to be done again. To find the equation that estimates nitrate levels, use Analyze, Regression, Linear. Click to enter Nitrate as the Response and Absorb as the Predictor. Since we also want to predict nitrate level for an absorbence of 40, add a new observation in absorbence of 40 without a Nitrate value (this will be ignored when computing the regression). Now, click Save, and put a check in the Unstandardized Predicted Values box. (Constant) Absorb a. Dependent Variable: Nitrate The estimated nitrate level for an absorbence of 40 is *40 = Predictions should be very accurate due to the strength of the linear relationship. Returning to the data worksheet, we find that the predicted nitrate level with an absorbence of 40 is

13 Open data file ex The scatterplot was created in the solution to Exercise 4.29; see that solution if you need help recreating this. We ll calculate the regression without and with the outlier. Since the outlier seems to follow the pattern of the rest of the data (making the relationship appear stronger), we expect that point to have little impact on the regression equation, but to increase the correlation (and r 2 ). We first calculate with the outlier included using Analyze, Regression, Linear using Behave as the Dependent variable and Neural as the Independent to find the following results. Model R R Square Adjusted R Square Std. Error of the Estimate.849 a a. Predictors: (Constant), Neural (Constant) Neural a. Dependent Variable: Behave Now, delete the outlier (the last observation) and recompute the regression to find the following: Model R R Square Adjusted R Square Std. Error of the Estimate.70 a a. Predictors: (Constant), Neural (Constant) Neural a. Dependent Variable: Behave With the outlier included, the correlation is r = (the square root of 0.72). With the outlier deleted, the correlation has weakened to r = Notice that the outlier has little

14 34 impact on the regression equation: with the point included, we have BehaviorLoss = * NeuralLoss; with the point deleted, the equation is BehaviorLoss = * NeuralLoss Open data file ta Use Graphs, Legacy Dialogs, Scatter/Dot to create the scatterplot, click to enter Fpg as the Y variable and Hba as the X variable. Click Titles to give your graph an appropriate title; Continue and OK generates the graph. The data point furthest to the right (subject 8) and in the top center (subject 5) are outliers. We ll first calculate the regression with all the data values; then remove subject 8 and recompute. Finally, we ll add back subject 8 and remove subject 5. Use Analyze, Regression, Linear to fit the model. Click to enter Fpg as the Dependent variable and Hba as the Independent variable. OK performs the calculations. Model R R Square Adjusted R Square Std. Error of the Estimate.482 a a. Predictors: (Constant), Hba

15 342 (Constant) Hba a. Dependent Variable: Fpg With all the data, we have the equation Fpg = *Hba. The correlation is r = Delete subject 8 in the worksheet, and return to the regression dialog box to recompute. Model R R Square Adjusted R Square Std. Error of the Estimate.384 a a. Predictors: (Constant), Hba (Constant) Hba a. Dependent Variable: Fpg With subject 8 removed, we have Fpg = *Hba. The correlation has lowered to r = Finally, add subject 8 (9.3, 255) back in and delete subject 5 (move the cursor to each value and press the delete key). Model R R Square Adjusted R Square Std. Error of the Estimate.568 a a. Predictors: (Constant), Hba Model Unstandardized Coefficients Coefficients B Std. Error Beta (Constant) t Sig. Hba a. Dependent Variable: Fpg

16 343 With all but subject 5, we have Fpg = *Hba and the correlation has increased to r = Deleting subject 8 weakened the relationship. This is because this data point was far to the right in the graph (extreme in its x value); visually, one s eyes follow through to points like these they are influential in fitting a model. Subject 5 was relatively in the middle of the x range but was clearly unusually high in his/her Fpg level. This type of point really adds only scatter to a plot, so deleting this increased the correlation. 5.4 We can recreate the scatterplot of all the data and add the three regression lines into the plot using the Chart Editor. If you have just completed Exercise 5.39, add subject 5 s data back into the worksheet; otherwise, open data file ta Create the scatterplot as was done in Exercise Double-click in the graph to bring up the Chart Editor. Click Elements, Fit Line at Total to add the regression line for the entire data set. This also brings up a Properties Box. Close this box. To add the other two lines, click Options, Reference Line from Equation. Type one of the equations into the Custom Equation box, then Apply and Close. Repeat this to add the last equation. Close the Chart Editor.

17 344 The separation between the lines becomes apparent at the right side of the plot. This is due to the influence of Subject 8 at the far right. 5.5 Open data file ex05-5. First, create a scatterplot of the data using Graphs, Scatter/Dot. Enter Stumps as the X variable and Larvae as the Y variable. Give your graph an appropriate title using Titles. We see that these data indicate that there are more beetle larvae with more stumps. Use Analyze, Regression, Linear to fit the line using Stumps as the Independent and Larvae as the Dependent. Model R R Square Adjusted R Square Std. Error of the Estimate.98 a a. Predictors: (Constant), Stumps

18 345 (Constant) stumps a. Dependent Variable: larvae The regression equation is Larvae = * Stumps. The relationship is strong; the regression model explains 84.3% of the variability in larvae (the correlation is r = r =.843 =0.98). These data support the beavers benefit beetles idea Open data for ex Since the data in this file have all the observations in each column with the before and after indicators in the variable named Data, we ll modify the regression dialog box input to fit each line separately. Click Analyze, Regression, Linear. Enter DD as the Independent and Gas as the Dependent variables. Now, Click to enter Data in the Selection Variable box, then Rule. Type in before, then Continue and OK.

19 346 Model Data = before (Selected) R Square R Adjusted R Square Std. Error of the Estimate.995 a a. Predictors: (Constant), DD,b (Constant) DD a. Dependent Variable: Gas b. Selecting only cases for which Data = before The regression equation before the solar panels is Gas = *DD. The relationship is strong r = This regression predicts *45 = hundred cubic feet of gas for a 45 degree-day day. Now, repeat the regression, selecting the after data. R Data = after Adjusted R Std. Error of the Model (Selected) R Square Square Estimate.99 a a. Predictors: (Constant), DD,b (Constant) DD a. Dependent Variable: Gas b. Selecting only cases for which Data = after The after regression equation is Gas = *DD. This is also a very strong relationship; we have r = This regression predicts *45 = 7.98 hundred cubic feet of gas for a 45 degree-day day. The solar panels saved about 60 cubic feet of gas usage.