VIEWS: 0 PAGES: 52 POSTED ON: 2/11/2012
Answers to Exercises 1 10. Structural Equation Models with Latent Variables Exercise 10.2 I went back to the original article by Crosby, Evans and Cowles (1990) and obtained the standard deviations for the variables examined in this exercise. This makes it possible to analyze the covariance matrix rather than the correlation matrix. The data are shown below (in the format used by AMOS): row var type_ name_ Y1 Y2 Y3 Y4 X1 X2 X3 X4 X5 X6 n 151 151 151 151 151 151 151 151 151 151 corr Y1 1.00 corr Y2 0.63 1.00 corr Y3 0.28 0.22 1.00 corr Y4 0.23 0.24 0.51 1.00 corr X1 0.38 0.33 0.29 0.20 1.00 corr X2 0.42 0.28 0.36 0.39 0.57 1.00 corr X3 0.37 0.30 0.39 0.29 0.48 0.59 1.00 corr X4 0.30 0.36 0.21 0.18 0.15 0.29 0.30 1.00 corr X5 0.45 0.37 0.31 0.39 0.29 0.41 0.35 0.44 1.00 corr X6 0.56 0.56 0.24 0.29 0.18 0.33 0.30 0.46 0.63 1.00 stddev 0.78 1.32 0.83 1.01 1.36 1.09 1.28 0.73 1.29 1.17 The measurement equations and structural equations for the model of salesperson service outcomes are shown below. The command syntax is taken from AMOS. There are two independent constructs ("similarity," measured by X 1, X2, and X3) and ("interaction," measured by X4, X5, and X6) and one dependent construct ("attitude," measured by Y1 and Y2) in the model. The total number of observed variances and covariances available to estimate the model parameters is 36; the total number of parameters estimated is 19 (six factor loadings and six error variances for the two independent constructs, one covariance term between independent constructs, two structural equation coefficients, one parameter describing the variance of the error term in the structural equation, and one factor loading and two error variances for the dependent construct). Thus, there are 17 degrees of freedom associated with the model specified below. Measurement and Structural Equations for 10.2: "Y1 = (1) attitude + (1) eps1" "Y2 = attitude + (1) eps2" "X1 = (1) similarity + (1) eps3" "X2 = similarity + (1) eps4" "X3 = similarity + (1) eps5" "X4 = (1) interaction + (1) eps6" "X5 = interaction + (1) eps7" "X6 = interaction + (1) eps8" "attitude = similarity + interaction + (1) zeta" 2 Answers to Exercises The goodness-of-fit statistics for the calibrated model are shown below. The chi-square test is significant at the p=0.025 level: 2(17)=30.14. Strictly speaking, this suggests we should reject the model (i.e., if this were in fact the true model, we would expect a fit this discrepant less than three percent of the time). However, both GFI (0.95) and AGFI (0.90) indicate good levels of fit. Thus, we are inclined to accept the model fit. Summary of models Model NPAR CMIN DF P CMIN/DF ---------------- ---- --------- -- --------- --------- Default model 19 30.144 17 0.025 1.773 Saturated model 36 0.000 0 Independence model 8 468.748 28 0.000 16.741 Model RMR GFI AGFI PGFI ---------------- ---------- ---------- ---------- ---------- Default model 0.061 0.954 0.903 0.451 Saturated model 0.000 1.000 Independence model 0.475 0.453 0.297 0.352 The parameter estimates (non-standardized, estimated from the covariance matrix) are shown below. The column "C.R." reports the critical ratio, given by the estimate divided by its standard error (S.E.). This is effectively a z-score, since the maximum likelihood routine provides asymptotic standard errors. These results suggest (and the standardized solution confirms) that "interaction" has a greater impact on "attitude" (in terms of explaining variance in the construct) than "similarity." Maximum Likelihood Estimates and Standard Errors: Regression Weights: Estimate S.E. C.R. Label ------------------- -------- ------- ------- ------- attitude <-------- similarity 0.177 0.071 2.489 attitude <------- interaction 1.042 0.213 4.891 Y1 <---------------- attitude 1.000 Y2 <---------------- attitude 1.537 0.183 8.385 X1 <-------------- similarity 1.000 X2 <-------------- similarity 0.967 0.128 7.530 X3 <-------------- similarity 0.986 0.137 7.185 X4 <------------- interaction 1.000 X5 <------------- interaction 2.356 0.380 6.198 X6 <------------- interaction 2.487 0.384 6.472 Variances: Estimate S.E. C.R. Label ---------- -------- ------- ------- ------- similarity 0.858 0.200 4.281 interaction 0.162 0.048 3.379 zeta 0.146 0.044 3.350 eps1 0.185 0.044 4.192 eps2 0.741 0.125 5.920 eps3 0.979 0.144 6.799 eps4 0.377 0.088 4.265 Answers to Exercises 3 eps5 0.793 0.124 6.370 eps6 0.368 0.046 7.919 eps7 0.756 0.117 6.444 eps8 0.360 0.093 3.888 Covariances: Estimate S.E. C.R. Label ------------ -------- ------- ------- ------- similarity <----> interaction 0.191 0.051 3.728 Standardized Solution: Regression Weights: Estimate -------------------------------- -------- attitude <-------- similarity 0.254 attitude <------- interaction 0.647 Y1 <---------------- attitude 0.833 Y2 <---------------- attitude 0.756 X1 <-------------- similarity 0.683 X2 <-------------- similarity 0.825 X3 <-------------- similarity 0.716 X4 <------------- interaction 0.552 X5 <------------- interaction 0.737 X6 <------------- interaction 0.857 Correlations: Estimate ------------- -------- similarity <----> interaction 0.513 For those using SAS, I have included the relevant portions of the SAS output for the purposes of comparison with AMOS. Results from PROC CALIS Model in SAS: Goodness of Fit Index (GFI) 0.9542 GFI Adjusted for Degrees of Freedom (AGFI) 0.9030 Root Mean Square Residual (RMR) 0.0448 Parsimonious GFI (Mulaik, 1989) 0.5793 Chi-Square 30.1438 Chi-Square DF 17 Pr > Chi-Square 0.0253 Standardized Solution: Manifest Variable Equations: y1 = 0.8329 f_sales + 0.5535 e1 y2 = 0.7564*f_sales + 0.6541 e2 c2 x1 = 0.6835*f_sim + 0.7300 e3 c3 x2 = 0.8249*f_sim + 0.5653 e4 c4 x3 = 0.7162*f_sim + 0.6979 e5 c5 4 Answers to Exercises x4 = 0.5524*f_int + 0.8336 e6 c6 x5 = 0.7366*f_int + 0.6764 e7 c7 x6 = 0.8573*f_int + 0.5149 e8 c8 Latent Variable Equations: f_sales = 0.2536*f_sim + 0.6467*f_int + 0.5910 d g1 g2 Correlations Among Exogenous Variables: Var1 Var2 Parameter Estimate f_sim f_int phi 0.51311 Exercise 10.4 Using AMOS, it is possible to perform a multiple group analysis on these data (where one group is comprised of 154 men and the other group is comprised of 125 women). We begin by specifying a model of the impact of personableness (denoted "person") and quality of argument (denoted "quality") on the perceived outcome of the debate (denoted "success"). We allow all coefficients of the model to differ across groups; that is, we allow different factor loadings, error variances, path coefficients, etc., for men and women. The equations describing this model are shown below: Measurement and Structural Equations for Two Groups: Men: "S1 = (1) success + (1) eps1" "S2 = success + (1) eps2" "S3 = success + (1) eps3" "P1 = (1) person + (1) eps4" "P2 = person + (1) eps5" "Q1 = (1) quality + (1) eps6" "Q2 = quality + (1) eps7" "success = person + quality + (1) zeta" Women: "S1 = (1) success + (1) eps1" "S2 = success + (1) eps2" "S3 = success + (1) eps3" "P1 = (1) person + (1) eps4" "P2 = person + (1) eps5" "Q1 = (1) quality + (1) eps6" "Q2 = quality + (1) eps7" Answers to Exercises 5 The total number of observed variances and covariances we have to estimate the model parameters is equal to 56 (28 each for men and women). The number of parameters estimated is 34 (17 each for mean and women), leaving a total number of degrees of freedom of 22. The goodness of fit results are shown below. Note that the fit of the model is exceptionally good. Summary of Results: Model I Model NPAR CMIN DF P CMIN/DF ---------------- ---- --------- -- --------- --------- Default model 34 15.255 22 0.851 0.693 Model RMR GFI AGFI PGFI ---------------- ---------- ---------- ---------- ---------- Default model 0.207 0.985 0.961 0.387 Note that most of the parameter estimates are quite closely comparable across groups. For example, the factor loading for men and women seem remarkably similar. One noticeable difference is in the past coefficient that describes the impact of "person" on "success." The estimate appears to be much greater among women (1.879) than among men (0.955). This is something we can subject to a more rigorous statistical test. Results for group: men Regression Weights: Estimate S.E. C.R. Label ------------------- -------- ------- ------- ------- success <----- person 0.955 0.137 6.962 success <---- quality 1.116 0.154 7.227 S1 <--------- success 1.000 S2 <--------- success 0.996 0.045 22.347 S3 <--------- success 1.035 0.039 26.606 P1 <---------- person 1.000 P2 <---------- person 0.931 0.110 8.470 Q1 <--------- quality 1.000 Q2 <--------- quality 1.044 0.120 8.711 Covariances: Estimate S.E. C.R. Label ------------ -------- ------- ------- ------- person <----> quality 0.066 0.479 0.137 Results for group: women Regression Weights: Estimate S.E. C.R. Label ------------------- -------- ------- ------- ------- success <----- person 1.879 0.199 9.463 success <---- quality 1.231 0.229 5.374 S1 <--------- success 1.000 S2 <--------- success 0.994 0.031 31.635 S3 <--------- success 1.014 0.031 32.645 P1 <---------- person 1.000 P2 <---------- person 0.884 0.090 9.801 Q1 <--------- quality 1.000 Q2 <--------- quality 1.196 0.204 5.853 6 Answers to Exercises Covariances: Estimate S.E. C.R. Label ------------ -------- ------- ------- ------- person <----> quality 0.159 0.500 0.318 We are now in a position to test for significant differences in model parameters between groups. We do this by constraining parameter values to be the same across groups. By convention, we proceed by first testing for equality in the measurement models, and then test for equality in the structural equation models. Model II, shown below, constrains the factor loadings and measurement error variances for all three constructs to be the same across groups. (Note that we might proceed by testing one at a time; we've accelerated the process somewhat to save space). This is done in AMOS by labeling the parameters across groups with the same label (e.g., the correlation between "person" and "quality" is given the same label, phi, in the models for men and women). In all, we constrain 12 parameters to be the same across groups: four factor loadings (those that are not already set to 1.0 to identify the model), seven measurement error variances, and one covariance between the independent factors "person" and "quality." Thus, Model II has 34 degrees of freedom. As shown below, this constrained model still fits extremely well. The chi-square test is even less significant than in Model I. Thus, we are unable to reject the hypotheses that the factor structures are not different across groups. Summary of Results: Model II Model NPAR CMIN DF P CMIN/DF ---------------- ---- --------- -- --------- --------- Default model 22 22.829 34 0.927 0.671 Model RMR GFI AGFI PGFI ---------------- ---------- ---------- ---------- ---------- Default model 0.228 0.976 0.960 0.592 Results for group: men Regression Weights: Estimate S.E. C.R. Label ------------------- -------- ------- ------- ------- success <----- person 0.934 0.126 7.399 success <---- quality 1.164 0.154 7.569 S1 <--------- success 1.000 S2 <--------- success 0.995 0.026 37.737 f1 S3 <--------- success 1.022 0.024 42.491 f2 P1 <---------- person 1.000 P2 <---------- person 0.885 0.066 13.338 f3 Q1 <--------- quality 1.000 Q2 <--------- quality 1.097 0.105 10.399 f4 Covariances: Estimate S.E. C.R. Label ------------ -------- ------- ------- ------- person <----> quality 0.146 0.356 0.411 phi Results for group: women Regression Weights: Estimate S.E. C.R. Label ------------------- -------- ------- ------- ------- success <----- person 1.856 0.183 10.126 Answers to Exercises 7 success <---- quality 1.154 0.201 5.748 S1 <--------- success 1.000 S2 <--------- success 0.995 0.026 37.737 f1 S3 <--------- success 1.022 0.024 42.491 f2 P1 <---------- person 1.000 P2 <---------- person 0.885 0.066 13.338 f3 Q1 <--------- quality 1.000 Q2 <--------- quality 1.097 0.105 10.399 f4 Covariances: Estimate S.E. C.R. Label ------------ -------- ------- ------- ------- person <----> quality 0.146 0.356 0.411 phi We now move on to test for equality of the structural equation model coefficients across groups. (As before, we can test these parameters separately, but for our purposes it serves to test all parameters at once). In addition to constraining the parameters of the measurement model, we now also constrain the structural equation coefficients. The fit results (shown below) suggest that this Model III is only marginally significant. However, the change in fit from Model II to Model III is 26.3 on a change of only three degrees of freedom. This suggests a significant deterioration in fit, which leads us to conclude that there are in fact differences across groups in the values of the path coefficients. Summary of Results: Model III Model NPAR CMIN DF P CMIN/DF ---------------- ---- --------- -- --------- --------- Default model 19 49.057 37 0.089 1.326 Model RMR GFI AGFI PGFI ---------------- ---------- ---------- ---------- ---------- Default model 3.804 0.952 0.927 0.629 It is not now possible to perform a multiple group analysis using PROC CALIS in SAS. One approach we might use to test for a difference between groups is to calibrate the model on one set of data (say, men), and then use the estimated parameter values to predict to the other sample (i.e., women). If the only differences between the two groups are non-systematic sampling error, then the predicted values should not be significantly difference from the observed. If there is a difference, then the parameters estimated using the data from the men should not fit the women's data very well. The model fits the men's data very well: 2(11)=4.4, which is not significant. Goodness of Fit Index (GFI) 0.9917 GFI Adjusted for Degrees of Freedom (AGFI) 0.9790 Root Mean Square Residual (RMR) 0.1114 Parsimonious GFI (Mulaik, 1989) 0.5195 Chi-Square 4.4300 Chi-Square DF 11 Pr > Chi-Square 0.9556 Manifest Variable Equations with Estimates s1 = 1.0000 f_succ + 1.0000 e1 s2 = 0.9960*f_succ + 1.0000 e2 8 Answers to Exercises Std Err 0.0446 c2 t Value 22.3546 s3 = 1.0347*f_succ + 1.0000 e3 Std Err 0.0389 c3 t Value 26.6158 p1 = 2.3637*f_pers + 1.0000 e4 Std Err 0.2173 c4 t Value 10.8782 p2 = 2.1999*f_pers + 1.0000 e5 Std Err 0.1966 c5 t Value 11.1889 q1 = 2.1214*f_qual + 1.0000 e6 Std Err 0.1917 c6 t Value 11.0683 q2 = 2.2155*f_qual + 1.0000 e7 Std Err 0.1974 c7 t Value 11.2260 Latent Variable Equations with Estimates f_succ = 2.2570*f_pers + 2.3670*f_qual + 1.0000 d Std Err 0.3083 g1 0.3106 g2 t Value 7.3210 7.6209 Covariances Among Exogenous Variables Standard Var1 Var2 Parameter Estimate Error t Value f_pers f_qual phi 0.01315 0.09613 0.14 When we use the parameters estimated from the men's data to predict the data observed for the women, the correspondence is not very good. Note that there is no estimation going on at this stage: no model parameters are being fitted. Because the chi-square statistic is highly significant, we reject the model and conclude that the two groups are indeed different. Predicting WOMEN Using Model Calibrated on MEN: Goodness of Fit Index (GFI) 0.8216 GFI Adjusted for Degrees of Freedom (AGFI) 0.8216 Root Mean Square Residual (RMR) 8.9304 Parsimonious GFI (Mulaik, 1989) 1.0955 Chi-Square 86.9466 Chi-Square DF 28 Pr > Chi-Square <.0001 The problem is that we do not know exactly the basis of the difference between the two groups. One thing we can do is increase the generality of the model by relaxing the constraints on the structural equation model parameters. Allowing these three parameters (the two path coefficients and the error variance) to be different across groups, we get the following result. Answers to Exercises 9 Relaxing Constraints on Structural Equation Parameters: Goodness of Fit Index (GFI) 0.9406 GFI Adjusted for Degrees of Freedom (AGFI) 0.9335 Root Mean Square Residual (RMR) 0.6635 Parsimonious GFI (Mulaik, 1989) 1.1197 Chi-Square 26.0779 Chi-Square DF 25 Pr > Chi-Square 0.4035 Latent Variable Equations with Estimates f_succ = 4.3922*f_pers + 2.2710*f_qual + 1.0000 d Std Err 0.3848 g1 0.3854 g2 t Value 11.4127 5.8931 Standardized: f_succ = 0.7352*f_pers + 0.3801*f_qual + 0.5547 d g1 g2 10 Answers to Exercises 11. Analysis of Variance Exercise 11.4 As currently stated, this problem asks for an ANOVA that involves a within-subjects treatment (i.e., each subject is asked to taste a product and a modified form of that product, which means we observe two different treatment levels within the same subject). Since we have not discussed these designs, I changed the format of the problem and asked students to look at the differences in perceptions of the two products (across six attributes) and test to see if there is a difference due to order of taste. The modified data format is shown below (the two products are identified as 27 and 45; the first product tasted is listed in column 2; columns 3-8 contain the attribute rating for product 27 - product 45): 1 27 -2 -1 0 0 0 0 2 27 -1 1 0 -2 -3 -1 3 27 0 0 0 -1 0 0 4 27 2 1 -1 -1 0 -1 5 27 1 1 0 0 0 0 6 27 -1 -1 1 -2 2 -1 7 27 -2 0 1 -1 0 0 8 27 -4 0 1 -1 1 -1 9 27 1 -2 -1 0 0 0 10 27 2 -1 0 0 0 0 < 89 rows omitted > 100 27 1 0 1 1 0 2 101 45 -1 -1 0 1 -2 0 102 45 1 -1 1 0 0 0 103 45 0 0 0 1 0 0 104 45 -1 -1 -1 -1 -1 0 105 45 1 0 0 -1 -1 2 106 45 1 2 0 0 0 1 107 45 0 0 0 0 0 0 108 45 -1 0 0 1 0 0 109 45 -2 2 0 -1 0 0 110 45 -2 0 1 -1 0 -1 < 90 rows omitted > The appropriate test is a MANOVA across all six attributes (one can look at the correlation matrix and verify that there is a high level of collinearity across attributes; Bartlett's test of sphericity could also be used here). As shown below, the null hypothesis is rejected, which means there are significant differences. This is interesting, since one would not necessarily expect the perceptions of the products to depend on which is tasted first. However, there are studies in the marketing literature that substantiate the existence of this type of first-mover reference effect. Answers to Exercises 11 MANOVA Test Criteria and Exact F Statistics for the Hypothesis of No Overall first Effect H = Type III SSCP Matrix for first E = Error SSCP Matrix S=1 M=2 N=95.5 Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.82829345 6.67 6 193 <.0001 Pillai's Trace 0.17170655 6.67 6 193 <.0001 Hotelling-Lawley Trace 0.20730159 6.67 6 193 <.0001 Roy's Greatest Root 0.20730159 6.67 6 193 <.0001 Treatment group means Level of ------------a1------------ ------------a2------------ first N Mean Std Dev Mean Std Dev 27 100 -0.55000000 1.69595478 -0.20000000 1.29490064 45 100 0.37000000 1.66153756 0.21000000 1.17460503 Level of ------------a3------------ ------------a4------------ first N Mean Std Dev Mean Std Dev 27 100 0.21000000 0.70057696 0.08000000 1.07007033 45 100 -0.01000000 0.67412495 -0.32000000 0.91981553 Level of ------------a5------------ ------------a6------------ first N Mean Std Dev Mean Std Dev 27 100 0.09000000 1.15553127 -0.02000000 1.01483939 45 100 -0.11000000 1.01399301 -0.18000000 1.08599905 A canonical correlation analysis helps to facilitate the interpretation. Note that the test (based on Wilks's is exactly the same. What we also can see is the canonical structure (i.e., the correlations between the attributes and their canonical variable. It shows that the product tasted first benefits on the first two attributes, but does more poorly on the last four. Canonical Correlation Analysis Canonical Structure Correlations Between the VAR Variables and Their Canonical Variables V1 first 1.0000 Correlations Between the WITH Variables and Their Canonical Variables W1 a1 0.6407 a2 0.3967 a3 -0.3832 a4 -0.4766 a5 -0.2222 a6 -0.1841 12 Answers to Exercises Exercise 11.6 These data are meant to mimic the results of an experiment exploring the impact of competitive expectations. The results are not particularly exciting. Single ANOVAs reveal that both share and profit are influenced by the experimental manipulation. Dependent Variable: share Sum of Source DF Squares Mean Square F Value Pr > F Model 2 10290.70000 5145.35000 12.76 <.0001 Error 57 22982.90000 403.20877 Corrected Total 59 33273.60000 Dependent Variable: profit Sum of Source DF Squares Mean Square F Value Pr > F Model 2 10157.70000 5078.85000 7.05 0.0018 Error 57 41043.90000 720.06842 Corrected Total 59 51201.60000 This results from the single ANOVAs is supported by the MANOVA. The pattern of means shows that the "competitive" treatment group performed relatively better in terms of profitability, while the "cooperative" treatment group performed relatively better on share. Both treatment groups seemed to perform better than the control. MANOVA Test Criteria and F Approximations for the Hypothesis of No Overall treat Effect H = Type III SSCP Matrix for treat E = Error SSCP Matrix Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.45856613 13.35 4 112 <.0001 Pillai's Trace 0.62548024 12.97 4 114 <.0001 Hotelling-Lawley Trace 0.99742972 13.88 4 66.174 <.0001 Roy's Greatest Root 0.75451889 21.50 2 57 <.0001 NOTE: F Statistic for Roy's Greatest Root is an upper bound. NOTE: F Statistic for Wilks' Lambda is exact. Level of -----------share---------- ----------profit---------- treat N Mean Std Dev Mean Std Dev Answers to Exercises 13 1 20 92.750000 18.8815560 91.600000 21.7894228 2 20 99.550000 16.5449149 122.650000 30.9078001 3 20 123.300000 24.0702918 113.350000 27.0209957 Almost nothing changes if we include past work experience as a covariate: Dependent Variable: share Sum of Source DF Squares Mean Square F Value Pr > F Model 3 13707.42043 4569.14014 13.08 <.0001 Error 56 19566.17957 349.39606 Corrected Total 59 33273.60000 Source DF Type III SS Mean Square F Value Pr > F treat 2 9929.133346 4964.566673 14.21 <.0001 past_exp 1 3416.720431 3416.720431 9.78 0.0028 Dependent Variable: profit Sum of Source DF Squares Mean Square F Value Pr > F Model 3 13252.53660 4417.51220 6.52 0.0007 Error 56 37949.06340 677.66185 Corrected Total 59 51201.60000 Source DF Type III SS Mean Square F Value Pr > F treat 2 9949.292723 4974.646362 7.34 0.0015 past_exp 1 3094.836603 3094.836603 4.57 0.0370 MANOVA Test Criteria and F Approximations for the Hypothesis of No Overall treat Effect H = Type III SSCP Matrix for treat E = Error SSCP Matrix Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.44174276 13.88 4 110 <.0001 Pillai's Trace 0.64961118 13.47 4 112 <.0001 Hotelling-Lawley Trace 1.05695747 14.45 4 64.974 <.0001 Roy's Greatest Root 0.79771155 22.34 2 56 <.0001 NOTE: F Statistic for Roy's Greatest Root is an upper bound. NOTE: F Statistic for Wilks' Lambda is exact. Least Squares Means profit treat share LSMEAN LSMEAN 1 93.015415 91.852604 2 99.583177 122.681575 14 Answers to Exercises 3 123.001408 113.065821 Canonical correlation also helps the interpretation of the MANOVA. In this case, we have three treatment groups and two dependent variables, so there are two pairs of canonical correlations. As shown by the sequential test based on Wilks's , both pairs are significant. Canonical Correlation Analysis Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.655777 0.637219 0.074202 0.430043 2 0.442083 . 0.104745 0.195437 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.45856613 13.35 4 112 <.0001 2 0.80456294 13.85 1 57 0.0005 Looking at the canonical structure, we see that the first pair of variables differentiates between the "competitive" and "cooperative" treatments (note that comp loads positively on V1, while coop loads negatively). Looking at W1, we see that it reflects the difference in performance on the two variables (loading positively on profit and negatively on share). The second pair of canonical variables reflects the difference in performance between the two treatment groups and the control group (which does worse in terms of profit and share). Canonical Structure Correlations Between the VAR Variables and Their Canonical Variables V1 V2 comp 0.7937 0.6083 coop -0.9236 0.3833 Correlations Between the WITH Variables and Their Canonical Variables W1 W2 share -0.6966 0.7175 profit 0.1121 0.9937 Answers to Exercises 15 Exercise 11.7 It is possible to approach this problem using MANOVA to examine the impact of the experimental factors on all four dependent variables (Y1, Y2, Y3 and Y4). However, it seems clear that these four dependent measures are designed to measure two distinct constructs: liking (Y1 and Y2) and purchase intention (Y3 and Y4). We begin by using exploratory factor analysis to examine the factor structure of the four dependent variables. As shown below, we find that a two-factor solution accounts for most of the variation in the data: the sum of the communalities is 3.23, which suggests that two common factors account for 3.23/4.00 or over 80 percent of the variation in the data. Exploratory Factor Analysis of Four Dependent Variables in Exercise 11.7 Prior Communality Estimates: SMC like1 like2 buy1 buy2 0.80643919 0.80162491 0.71857006 0.69248843 Eigenvalues of the Reduced Correlation Matrix: Total = 3.0191226 Average = 0.75478065 Eigenvalue Difference Proportion Cumulative 1 2.74173543 2.24935193 0.9081 0.9081 2 0.49238350 0.58158550 0.1631 1.0712 3 -.08920200 0.03659233 -0.0295 1.0417 4 -.12579433 -0.0417 1.0000 2 factors will be retained by the NFACTOR criterion. Factor Pattern Factor1 Factor2 like1 0.86169 -0.33046 like2 0.85272 -0.34438 buy1 0.81587 0.34354 buy2 0.77875 0.38284 Variance Explained by Each Factor Factor1 Factor2 2.7417354 0.4923835 Final Communality Estimates: Total = 3.234119 like1 like2 buy1 buy2 0.85170968 0.84573766 0.78365811 0.75301347 An oblique rotation of the factor solution results in two clear factors: “product liking” and “purchase intention.” The factor loadings exhibit simple structure (Y1 and Y2 load clearly on the first factor and Y3 and Y4 load clearly on the second). The two factors exhibit a correlation of 0.62 (i.e., product liking is strongly positively associated with purchase intention). 16 Answers to Exercises Results of Oblique Rotation for Exercise 11.7 Rotation Method: Promax (power = 3) Inter-Factor Correlations Factor1 Factor2 Factor1 1.00000 0.61950 Factor2 0.61950 1.00000 Rotated Factor Pattern (Standardized Regression Coefficients) Factor1 Factor2 like1 0.88132 0.06483 like2 0.89175 0.04397 buy1 0.09325 0.82444 buy2 0.02698 0.85079 Using the factor score coefficients from the factor analysis, we calculated factor scores and used these as the dependent variables in a MANOVA. (Note that the results would be similar if we simply took the sum of Y1 and Y2 and the sum of Y3 and Y4 and subjected them to the same analysis). The results of the simple ANOVAs of Factor1 (“Liking”) and Factor 2 (“Purchase Intent”) are shown below. Interestingly, we see that the inclusion of a uniqueness claim has a significant impact on liking (p < 0.05) and the inclusion of a competitive claim has a significant impact on purchase intent (p < 0.05). There is also a significant interaction between claims on liking (p < 0.05) and purchase intent (p < 0.05). Simple ANOVA Results for Exercise 11.7 Dependent Variable: Factor1 Sum of Source DF Squares Mean Square F Value Pr > F Model 3 9.12005460 3.04001820 3.66 0.0150 Error 96 79.64772132 0.82966376 Corrected Total 99 88.76777592 R-Square Coeff Var Root MSE Factor1 Mean 0.102741 2.56384E17 0.910859 3.5527E-16 Source DF Type III SS Mean Square F Value Pr > F compet 1 0.00543779 0.00543779 0.01 0.9356 unique 1 4.53521680 4.53521680 5.47 0.0215 compet*unique 1 4.57940001 4.57940001 5.52 0.0209 Dependent Variable: Factor2 Sum of Source DF Squares Mean Square F Value Pr > F Model 3 9.34817426 3.11605809 4.03 0.0096 Error 96 74.24853373 0.77342223 Answers to Exercises 17 Corrected Total 99 83.59670799 R-Square Coeff Var Root MSE Factor2 Mean 0.111825 8.25139E17 0.879444 1.0658E-16 Source DF Type III SS Mean Square F Value Pr > F compet 1 3.94340484 3.94340484 5.10 0.0262 unique 1 0.59738878 0.59738878 0.77 0.3817 compet*unique 1 4.80738064 4.80738064 6.22 0.0144 The statistical tests from the MANOVA of Factor1 and Factor2 yield the same conclusions: all effects significant at p < 0.05. MANOVA Results for Exercise 11.7 Multivariate Analysis of Variance COMPETITIVE CLAIM: Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.90767677 4.83 2 95 0.0100 UNIQUENESS CLAIM: Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.93481584 3.31 2 95 0.0407 INTERACTION: Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.93300518 3.41 2 95 0.0371 The effects themselves are revealed by the mean values of Factor1 and Factor2, shown below: Mean Values on Factor1 and Factor 2 Level of ----------Factor1--------- ----------Factor2--------- compet N Mean Std Dev Mean Std Dev 0 50 0.00737414 0.94288753 0.19858008 0.89527570 1 50 -0.00737414 0.96043709 -0.19858008 0.90777697 Level of ----------Factor1--------- ----------Factor2--------- unique N Mean Std Dev Mean Std Dev 0 50 0.21296048 1.00187542 0.07729093 0.96192245 1 50 -0.21296048 0.84574078 -0.07729093 0.87668059 Level of Level of ---------Factor1--------- ---------Factor2--------- compet unique N Mean Std Dev Mean Std Dev 0 0 25 0.00633929 1.07020031 0.05661361 1.00625374 18 Answers to Exercises 0 1 25 0.00840898 0.81840458 0.34054655 0.76282283 1 0 25 0.41958167 0.90280717 0.09796825 0.93579176 1 1 25 -0.43432995 0.82974666 -0.49512841 0.78964385 Note that the main effect of the inclusion of a competitive claim is largest (and negative) with respect to purchase intent. It suggests that a competitive claim has little effect on product liking and a pronounced negative effect on purchase intent. Similarly, the effect of the inclusion of a uniqueness claim is largest (and negative) with respect to product liking. It suggests that a uniqueness claim has little effect on purchase intent and a pronounced negative effect on product liking. These seemingly counterintuitive results make more sense when we examine the interaction effect revealed by the cell means. These suggest that the inclusion of a competitive claim alone has a positive effect on liking (but not purchase intent) relative to no competitive claim, and that the inclusion of a uniqueness claim alone has a positive effect on purchase intent (but not liking) relative to no uniqueness claim. However, when both claims are included, the effect on both liking and purchase intent is substantially negative relative to no claims at all. We can compare the results above to those from a MANOVA of all four dependent variables separately. As shown below, the while the main effects of competitive claim and uniqueness claim are both significant (at p < 0.05), the interaction effect is not significant. This seems unusual, in light of the fact that the interaction effect is significant in each of the simple ANOVAs. MANOVA for all four dependent variables in Exercise 11.7 Multivariate Analysis of Variance COMPETITIVE CLAIM: Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.81605120 5.24 4 93 0.0007 UNIQUENESS CLAIM: Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.87563562 3.30 4 93 0.0141 INTERACTION: Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.93077476 1.73 4 93 0.1501 The reason may be due to the fact that the pattern of the mean values differs across the four dependent variables. As shown below, the combination of competitive claim and uniqueness claim results in the lowest average score for each of the four dependent measures. But there appears to be more noise in the measures across four variables, and the pattern of interaction does not hold up in this analysis. Answers to Exercises 19 Mean Values on Four Dependent Variables for Exercise 11.7 Level of Level of ----------like1---------- ----------like2---------- compet unique N Mean Std Dev Mean Std Dev 0 0 25 49.6400000 9.34469546 49.4000000 10.0332780 0 1 25 49.2400000 8.00666389 49.5200000 7.4056285 1 0 25 53.3200000 8.24984848 53.3600000 8.3260635 1 1 25 45.4000000 8.50490055 46.0400000 7.1032856 Level of Level of -----------buy1---------- -----------buy2---------- compet unique N Mean Std Dev Mean Std Dev 0 0 25 51.0000000 8.96753403 50.3600000 8.37595766 0 1 25 52.8800000 6.55311631 53.9200000 7.69696910 1 0 25 52.6000000 8.91160292 48.4800000 7.94837510 1 1 25 46.5200000 7.93263302 45.4000000 6.74536878 20 Answers to Exercises SAS PROGRAM FOR EXERCISE 11.7 options ls=72; data claims; infile 'AD_CLAIM_TEST.txt'; input subject compet $ unique $ like1 like2 buy1 buy2; run; proc factor data=claims method=principal priors=smc n=2 rotate=promax score outstat=stat; var like1 like2 buy1 buy2; run; proc score data=claims score=stat out=scores; var like1 like2 buy1 buy2; run; proc glm data=scores; class compet unique; model factor1 factor2 = compet unique compet*unique; manova h=compet unique compet*unique; means compet unique compet*unique; run; proc glm data=claims; class compet unique; model like1 like2 buy1 buy2 = compet unique compet*unique; manova h=compet unique compet*unique; means compet unique compet*unique; run; Answers to Exercises 21 Exercise 11.8 Ever wondered how different methods of cooking fish impact the aroma, flavor, texture and moisture of the final product? Here is your chance to find out! All four of these measures are relatively highly correlated, so MANOVA is appropriate here. Pearson Correlation Coefficients, N = 36 Prob > |r| under H0: Rho=0 aroma flavor texture moisture aroma 1.00000 0.73024 0.49060 0.39795 <.0001 0.0024 0.0162 flavor 0.73024 1.00000 0.36140 0.41800 <.0001 0.0303 0.0112 texture 0.49060 0.36140 1.00000 0.59300 0.0024 0.0303 0.0001 moisture 0.39795 0.41800 0.59300 1.00000 0.0162 0.0112 0.0001 The results of the MANOVA show a significant difference across cooking methods. Looking at the patterns of means, we see the larges differences across groups in terms of flavor. MANOVA Test Criteria and F Approximations for the Hypothesis of No Overall method Effect H = Type III SSCP Matrix for method E = Error SSCP Matrix Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.24182119 7.75 8 60 <.0001 Pillai's Trace 0.85684365 5.81 8 62 <.0001 Hotelling-Lawley Trace 2.72727944 10.04 8 40.602 <.0001 Roy's Greatest Root 2.56842429 19.91 4 31 <.0001 NOTE: F Statistic for Roy's Greatest Root is an upper bound. NOTE: F Statistic for Wilks' Lambda is exact. Level of -----------aroma---------- ----------flavor---------- method N Mean Std Dev Mean Std Dev 1 12 5.38333333 0.58439298 5.70833333 0.44201673 2 12 5.25833333 0.76331613 5.23333333 0.57419245 3 12 4.97500000 0.54292976 4.83333333 0.45990776 Level of ----------texture--------- ---------moisture--------- method N Mean Std Dev Mean Std Dev 1 12 5.52500000 0.65244296 5.98333333 0.69652819 22 Answers to Exercises 2 12 5.30833333 0.59460962 5.87500000 0.51720402 3 12 5.90833333 0.51071845 6.23333333 0.45593726 Canonical correlation provides some additional interpretation. Again, with three methods and four variables, we have two pairs of canonical correlations. In this case, however, only the first is significant, so we only bother to attempt to interpret this. Looking at the canonical structure, we see that methods 1 and 2 load positively on the first canonical variable (so we may interpret this variable as being associated with these two cooking methods and not with method 3). Looking at the loadings on W1, we see that flavor (especially) and to some extent aroma load positively, while texture and moisture load negatively. Putting these together, we can conclude that methods 1 and 2 tend to lead to higher evaluations on flavor and aroma, but lower on texture and moisture than method 3. This is borne out by the patterns of treatment group means. Canonical Correlation Analysis Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.848389 0.832288 0.047368 0.719764 2 0.370242 0.312605 0.145860 0.137079 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.24182119 7.75 8 60 <.0001 2 0.86292062 1.64 3 31 0.1999 Canonical Structure Correlations Between the VAR Variables and Their Canonical Variables V1 V2 m1 0.7259 0.6878 m2 0.2327 -0.9726 Correlations Between the WITH Variables and Their Canonical Variables W1 W2 aroma 0.3177 0.0106 flavor 0.6811 0.4560 texture -0.3770 0.6613 moisture -0.2618 0.3999 Answers to Exercises 23 Exercise 11.11 This problem calls for a test of an orientation program on outcome measures of anxiety, depression, and anger. Presumably, the goal of the program is to increase the effectiveness of psychotherapy and lead to lower measures on these variables. Pearson Correlation Coefficients, N = 46 Prob > |r| under H0: Rho=0 anxiety depress anger anxiety 1.00000 0.85554 0.65612 <.0001 <.0001 depress 0.85554 1.00000 0.63084 <.0001 <.0001 anger 0.65612 0.63084 1.00000 <.0001 <.0001 A MANOVA shows that the results of the program are not significant at the 0.05 level. Thus, even though the patterns of means are consistent with the goals of the program, we cannot reject the null hypothesis that these differences might have arisen due to chance. MANOVA Test Criteria and Exact F Statistics for the Hypothesis of No Overall treat Effect H = Type III SSCP Matrix for treat E = Error SSCP Matrix Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.85025028 2.47 3 42 0.0754 Pillai's Trace 0.14974972 2.47 3 42 0.0754 Hotelling-Lawley Trace 0.17612428 2.47 3 42 0.0754 Roy's Greatest Root 0.17612428 2.47 3 42 0.0754 Level of ----------anxiety--------- ----------depress--------- treat N Mean Std Dev Mean Std Dev 0 26 158.769231 80.173466 182.884615 124.205419 1 20 104.350000 105.812782 154.250000 137.158715 Level of ------------anger------------ treat N Mean Std Dev 0 26 66.0000000 51.1890613 1 20 51.0500000 51.1967053 Exercise 11.11 This problem is a relatively straightforward application of MANOVA. This is a one factor design with three factor levels: control group, behavioral rehearsal treatment group, and behavioral rehearsal with cognitive restructuring. There are four dependent variables in the study: anxiety, social skills, 24 Answers to Exercises appropriateness, and assertiveness. All four variables are highly intercorrelated (see below). It might be argued that all four are indicators of one underlying factor. Correlation matrix for four dependent variables in Exercise 11.12 Pearson Correlation Coefficients, N = 33 Prob > |r| under H0: Rho=0 anxiety social approp assert anxiety 1.00000 -0.82209 -0.85925 -0.89866 <.0001 <.0001 <.0001 social -0.82209 1.00000 0.87183 0.83709 <.0001 <.0001 <.0001 approp -0.85925 0.87183 1.00000 0.93552 <.0001 <.0001 <.0001 assert -0.89866 0.83709 0.93552 1.00000 <.0001 <.0001 <.0001 We first analyze the four dependent measures using MANOVA. The results (shown below) show a statistically significant effect (p < 0.001). Thus, we reject the null hypothesis that the means of the dependent variables are the same across treatment groups. MANOVA test of all four dependent variables for Exercise 11.12 Multivariate Analysis of Variance Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.38164971 4.18 8 54 0.0006 Examining the means of the variables across groups, we see that the most dramatic differences are between the control group and the two treatment groups (behavioral rehearsal and behavioral rehearsal plus cognitive restructuring). Treatment group means for four dependent variables Level of ----------anxiety--------- ----------social---------- group N Mean Std Dev Mean Std Dev 1 11 4.27272727 0.64666979 4.27272727 0.64666979 2 11 4.09090909 0.30151134 4.27272727 0.90453403 3 11 5.45454545 0.82019953 2.54545455 1.03572548 Answers to Exercises 25 Level of ----------approp---------- ----------assert---------- group N Mean Std Dev Mean Std Dev 1 11 4.18181818 0.60302269 3.81818182 0.75075719 2 11 4.27272727 0.78624539 4.09090909 0.70064905 3 11 2.54545455 0.93419873 2.54545455 0.93419873 This raises an interesting question: is there any incremental effect of cognitive restructuring over and above behavioral rehearsal? To test this, we test the contrast between the two treatment groups (1 versus 2). The result (shown below) is not significant. MANOVA: Contrast of Group 1 versus Group 2 Multivariate Analysis of Variance Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.94238950 0.41 4 27 0.7980 We also analyzed the data by factor analyzing the four dependent variables and extracting one factor (which accounted for almost 85 percent of the variation in the data) and conducting a simple ANOVA. The results are similar: the treatment effects are significant (relative to control), and the impact of cognitive restructuring and behavioral rehearsal (relative to behavioral rehearsal alone) is not significant. ANOVA of single factor extracted from four dependent variables in Exercise 11.12 Dependent Variable: Factor1 Sum of Source DF Squares Mean Square F Value Pr > F Model 2 16.22951626 8.11475813 16.60 <.0001 Error 30 14.66709800 0.48890327 Corrected Total 32 30.89661426 R-Square Coeff Var Root MSE Factor1 Mean 0.525285 6.49479E17 0.699216 1.0766E-16 Contrast DF Contrast SS Mean Square F Value Pr > F 1 versus 2 1 0.14401863 0.14401863 0.29 0.5913 GROUP MEANS: Level of -----------Factor1----------- group N Mean Std Dev 1 11 0.41277050 0.59258357 2 11 0.57458893 0.60827970 26 Answers to Exercises 3 11 -0.98735943 0.86345255 SAS PROGRAM FOR EXERCISE 11.12 options ls=72; data skills; infile 'SOC_SKILLS.txt'; input group $ anxiety social approp assert; run; proc corr data=skills; var anxiety social approp assert; proc glm data=skills; class group; model anxiety social approp assert = group; contrast '1 versus 2' group -1 1 0; manova h=group; means group; proc factor data=skills method=principal priors=smc n=1 score outstat=stat; var anxiety social approp assert; run; Title Title Answers to Exercises 27 12. Discriminant Analysis Exercise 12.7 The data are collected from two groups of patients: 15 classified as ill and 30 classified as well. There is no information accompanying the problem to say whether this constitutes a representative sample from any population, so it is hard to say how to generalize this analysis. In the absence of any other specific information, we will assume that the sample proportions (1/3, 2/3) represent our priors. We can test for a difference between group centroids (i.e., the mean value on each of the five measures of everyday functionality) using discriminant analysis. Because discriminant analysis is a special case of canonical correlation, the results from PROC CANCORR in SAS are shown below. The correlation is 0.70 and the associated F-test of significance (based on Wilks's ) is significant at the 0.0001 level. Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.696192 0.666624 0.077687 0.484683 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.51531669 7.34 5 39 <.0001 We can look at the canonical loadings (i.e., the correlations between original variables and canonical variates) to interpret the results. They show that four of the five measures load highly (i.e., all but "feeling capable of making decisions"). This suggests that this one variable is perhaps a less reliable indicator of illness than the other four. Note that the standardized canonical coefficients (i.e., the weights used to form the linear combinations that are the canonical variates) suggest the same pattern (i.e., closest to zero for the decision-making variable). Total Canonical Structure Variable Can1 useful 0.875057 content 0.663438 decide 0.381177 nostart 0.778799 dread 0.789652 Total-Sample Standardized Canonical Coefficients Variable Can1 28 Answers to Exercises useful 0.6069552778 content 0.2857822573 decide -.0828879014 nostart 0.3842507350 dread 0.4922908730 Class Means on Canonical Variables ill Can1 1 -1.340710160 2 0.670355080 To test the predictive performance of the discriminant function, we use a one-at-a-time holdout cross- validation (in this case, using PROC DISCRIM in SAS). The results, shown below, suggest a hit rate of 38/45 = 84.4 percent. The proportional chance criterion is only (1/3) 2 + (2/3)2 = 55.6 percent. How likely are we to achieve a hit rate of 38 out of 45 if the true performance of our linear discriminant function were no better than 0.556? The standard deviation of the number of expected hits under this null hypothesis is equal to sqrt( 45 x 0.556 x 0.444 ) = 3.33. The t-ratio is t = (38 - 25) / 3.33 = 3.9, which is significant. Linear Discriminant Function _ -1 _ -1 _ Constant = -.5 X' COV X + ln PRIOR Coefficient = COV X j j j Vector j Variable 1 2 Constant -11.12483 -18.48019 useful 0.18715 1.48700 content 3.02631 3.72234 decide 5.22426 4.94589 nostart -0.79384 -0.10000 dread 3.17374 4.54374 Cross-Validation: Number of Observations and Percent Classified into ill From ill 1 2 Total 1 12 3 15 80.00 20.00 100.00 2 4 26 30 13.33 86.67 100.00 Total 16 29 45 35.56 64.44 100.00 We also test the assumption that the within-group covariance matrices are the same across ill and well patients. As shown below, this assumption clearly does not hold; among ill patients, two or more variables are perfectly collinear within sample. However, if we use a quadratic discriminant function instead of linear, we find the prediction performance goes down (when evaluated using one-at-a-time Answers to Exercises 29 holdout cross-validation) compared to the linear function. This suggests that the results are too sensitive to the differences in the estimated within-group sample covariance matrices across groups. Within Covariance Matrix Information Natural Log of the Covariance Determinant of the ill Matrix Rank Covariance Matrix 1 4 -27.23757 2 5 -2.80739 Pooled 5 -4.06452 Chi-Square DF Pr > ChiSq 245.651366 15 <.0001 Cross-Validation: Number of Observations and Percent Classified into ill From ill 1 2 Total 1 15 0 15 100.00 0.00 100.00 2 15 15 30 50.00 50.00 100.00 Total 30 15 45 66.67 33.33 100.00 30 Answers to Exercises Exercise 12.8 These data are from an exercise that appeared in the original text by Green and Carroll, developed back in the day when students were asked to do some of these calculations by hand. There is a typo in the statement of the problem in the textbook: the data are in fact coded Y=1 in favor and Y=0 against the control bill. The data are hypothetical. Since the data are not drawn from an actual legislative body, we cannot say that they are representative of any true population of voters. We will assume that the sample proportions reflect our priors about voting behavior. A plot of the data (shown below) suggests that the dividing line between those for and those against the gun control bill might be positively sloped: those voting in favor of the bill (Y=1) fall below this positively sloped line. Plot of age*guns$vote. Symbol points to label. age | | 70 + | | | 60 + > 0 | > 0 | > 0 | > 0 50 + | > 0 > 0 | 0 2 1 | > 0 40 + > 1 | > 0 > 1 | | 30 + 2 1 | > 1 | | 20 + | ---+---------+---------+---------+---------+---------+---------+-- 0 1 2 3 4 5 6 guns The results from applying Fisher's approach to discriminant analysis (given by PROC CANDISC in SAS) are shown below. The outcome is just significant at the 0.01 level; Wilks's = 0.468, with an F-test significant at 0.0105. Answers to Exercises 31 Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.729673 0.718972 0.124965 0.532423 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.46757711 6.83 2 12 0.0105 Fisher's discriminant function coefficients are proportional to the standardized canonical coefficients shown below. These are the weights we use to form the discriminant function score t = Xk. Note that the coefficient for age is positive and the coefficient for guns is negative. This is consistent with the interpretation from the scatter plot above: if the line dividing the two groups of observations (i.e., the Mahalanobis locus of points) is positively sloped, then the Fisher discriminant function axis must be negatively sloped. By looking at the group means, we can see that those voting against the legislation (i.e., vote = 0) have a positive mean discriminant score, while those voting in favor (vote = 1) have a negative score). This suggests that holding the number of guns constant, the older the legislator, the more likely he/she is to vote against the bill. Holding the age of the legislator constant, the greater the number of guns owned, the more likely he/she is to vote in favor of the gun control bill. For an interpretation of the discriminant function, we look at the loadings. Here, we see (due to the positive correlation between age and number of guns owned) that both are positively correlated with the discriminant function. Total Canonical Structure Variable Can1 age 0.986597 guns 0.818619 Total-Sample Standardized Canonical Coefficients Variable Can1 age 1.868967847 guns -0.531004155 Class Means on Canonical Variables vote Can1 0 0.811114481 1 -1.216671722 A test of the homogeneity of within-group covariance matrices across groups cannot be rejected, so we find it is appropriate to pool the estimates and use a linear discriminant function. 32 Answers to Exercises Within Covariance Matrix Information Natural Log of the Covariance Determinant of the vote Matrix Rank Covariance Matrix 0 2 3.81676 1 2 3.27740 Pooled 2 3.72112 Chi-Square DF Pr > ChiSq 1.193126 3 0.7547 PROC DISCRIM in SAS gives the coefficients of the Mahalanobis discriminant functions (i.e., the distance calculation for group 1 and group 2). Note that age has a bigger impact on the score for those voting against the gun control bill (coefficient equal to 2.51 for vote = 0) than for those voting in favor (coefficient = 2.14 for vote = 1). Thus, holding all else constant, the greater the age, the greater the age of the legislator, the greater the discriminant function score for vote = 0 (i.e., those voting against the bill). Linear Discriminant Function _ -1 _ -1 _ Constant = -.5 X' COV X + ln PRIOR Coefficient = COV X j j j Vector j Linear Discriminant Function for vote Variable 0 1 Constant -50.04961 -35.72468 age 2.51012 2.13764 guns -8.34471 -7.80113 These Mahalanobis distances can be used to calculate posterior probabilities of group membership. This is done below using one-at-a-time holdout validation (which means each observation is classified using the discriminant function coefficients calculated using only the data from the remaining n-1 observations). Posterior Probability of Membership in vote From Classified Obs vote into vote 0 1 1 1 1 0.0343 0.9657 2 1 1 0.2757 0.7243 3 1 0 * 0.8221 0.1779 4 1 1 0.0220 0.9780 5 1 1 0.0721 0.9279 Answers to Exercises 33 6 1 0 * 0.9309 0.0691 7 0 0 0.8801 0.1199 8 0 0 0.9608 0.0392 9 0 0 0.9981 0.0019 10 0 0 0.9800 0.0200 11 0 1 * 0.2168 0.7832 12 0 0 0.9783 0.0217 13 0 1 * 0.4903 0.5097 14 0 0 0.5108 0.4892 15 0 0 0.8576 0.1424 * Misclassified observation The "hits and misses" table below summarizes the results from the classification shown above. The hit rate is 11/15 = 73.3 percent. While this seems an improvement over the proportional chance hit rate of 55.6 percent, we difference is not statistically significant. Number of Observations and Percent Classified into vote From vote 0 1 Total 0 7 2 9 77.78 22.22 100.00 1 2 4 6 33.33 66.67 100.00 Total 9 6 15 60.00 40.00 100.00 Priors 0.6 0.4 Exercise 12.12 This data set is used in the SYSTAT manual (Wilkinson) in the chapter on “Discriminant Analysis” by Englemann. I do not know about the sampling scheme; it seems unlikely that these countries are a representative sample of the countries of the world. For the purposes of this exercise, proportional priors are assumed. a) Are the differences across groups of countries significant? The differences are clearly significant. Wilks’s is 0.051 (p < 0.0001); in fact, each of the independent variables in the data set is by itself a significant discriminator. (Note that the results below are based on including 11 of the independent variables; the 12th, which is a ratio of birth to death rate, is left out because it is a combination of information already present in the model). Because there are three groups of countries, there are two discriminant functions. We can test the second discriminant function alone and we find that it is also significant: Wilks’s = 0.402 (p < 0.0001) 34 Answers to Exercises Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.933842 0.920284 0.017410 0.872061 2 0.773599 0.735910 0.054643 0.598456 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.05137337 13.03 22 84 <.0001 2 0.40154436 6.41 10 43 <.0001 b) Interpret each discriminant function. For the purpose of interpretation, it is probably best to look at the discriminant function loadings (called the “canonical structure” in PROC CANDISC in SAS). The discriminant function coefficients are interpretable as partial regression coefficients; i.e., they describe the impact of each dependent variable holding all other variables in the model constant. As shown below, the first discriminant function is negatively correlated with birth rate and infant mortality, and positively correlated with life expectancy, literacy, per capita GDP, and spending on health and education. As shown by the group means, this discriminant function separates countries in Europe (which score most highly) from New World and Islamic countries. The second discriminant function is strongly positively correlated with death rate, and negatively correlated with literacy and life expectancy. Unlike the first discriminant function, the second is negatively correlated with percent of population living in cities. As shown by the group means, the second discriminant function serves to separate the Islamic countries (which score the highest) from the New World countries. Total Canonical Structure Variable Can1 Can2 citypop 0.644525 -0.427281 birth -0.912000 0.351380 death -0.140933 0.752922 infdeath -0.794393 0.453549 gdppc 0.860334 0.132150 educ 0.678699 0.124903 health 0.757760 0.127102 military 0.498877 0.368253 lifex_m 0.719396 -0.477567 lifex_f 0.779220 -0.465591 literacy 0.773787 -0.560221 Class Means on Canonical Variables group Can1 Can2 Europe 3.381723396 0.411490370 Islamic -2.719775968 1.462927569 NewWorld -1.116957381 -1.417249074 Answers to Exercises 35 Plot of Can2*Can1$group. Symbol points to label. Can2 | | 3 + Islamic Islamic | ^ ^ > Islamic | Islamic > Islamic 2 + ^ ^ ^ > Islamic Europe | Islamic < > NewWorld ^ Europe | > Islamic Europe <v ^ Europe 1 + > Islamic EuropevEurope^ | > Islamic Europe < Europ^ v | Europe <^ ^ Europe 0 + NewWorld > Islamic NewWorld 2 Europe v> Europe | Islamic ^^ > Islamic Europe < Europe | ^ > NewWorld 2 Europe -1 + > NewWorld < > NewWorld > Europe | NewWorld <> NewWorld NewWorld | NewWorld < > NewWorld ^ > NewWorld -2 + > NewWorld | NewWorld < NewWorld > NewWorld | NewWorld < ^ > NewWorld -3 + > NewWorld | --+------+------+------+------+------+------+------+------+------+- -4 -3 -2 -1 0 1 2 3 4 5 Can1 c) How well does linear discriminant analysis perform in correctly classifying countries by group? To answer this question, we assume that it is appropriate to pool across groups in estimating the within-group covariance matrix. (We address the question of whether this assumption is appropriate in part d below). We use the cross-validation option in SAS to assess predictive validity and calculate the hit rate. The result is that 45 out of 55 countries (two have missing data) are correctly categories, a hit rate of 82 percent. This compares favorably to the proportional chance hit rate of (19/55) 2 + (15/55)2 + (21/55)2 = 34 percent. Cross-Validation: Number of Observations and Percent Classified into group From New group Europe Islamic World Total Europe 18 0 1 19 94.74 0.00 5.26 100.00 Islamic 0 12 3 15 0.00 80.00 20.00 100.00 NewWorld 2 4 15 21 9.52 19.05 71.43 100.00 Total 20 16 19 55 36.36 29.09 34.55 100.00 36 Answers to Exercises d) Test the assumption that linear discriminant analysis is appropriate. This we do by using Box’s test of the equality of covariance matrices across groups. Shown below are the log-determinants of each group covariance matrix. Note that the Islamic group is smaller than the others; this difference turns out to be highly significant. This calls into question our decision in part c to pool across groups in estimating within-group covariance The results from the quadratic discriminant analysis are also shown below. Clearly, the quadratic analysis does not outperform the linear analysis in classification accuracy when using a cross- validation approach. Part of the problem is that the procedure assigns too few countries to the holdout group and too many to the New World group. This is because the log-determinant of the within-group covariance matrix is smallest for the Islamic group (and largest for the New World group). Within Covariance Matrix Information Natural Log of the Covariance Determinant of the group Matrix Rank Covariance Matrix Europe 11 52.25294 Islamic 11 49.66450 NewWorld 11 55.19725 Pooled 11 64.02932 Chi-Square DF Pr > ChiSq 412.706444 132 <.0001 Cross-Validation (Quadratic Discriminant Function) Number of Observations and Percent Classified into group From New group Europe Islamic World Total Europe 17 0 2 19 89.47 0.00 10.53 100.00 Islamic 0 7 8 15 0.00 46.67 53.33 100.00 NewWorld 2 1 18 21 9.52 4.76 85.71 100.00 Total 19 8 28 55 34.55 14.55 50.91 100.00 Exercise 12.13 Marketers are often concerned with the accuracy of new product testing methods. Here is an example data set in which each of 24 new products is testing using two methods: a concept test and a panel test. Although the question is not posed explicitly in the exercise, it might be interesting to investigate Answers to Exercises 37 which of these two tests is more valuable in discriminating successful new products from failures. One could also ask whether it is worth purchasing the second test in addition to the first (i.e., is the improvement in discrimination worthwhile?). Of course, to answer these questions, one would need to know something about the costs of the tests, the cost of product development, the profits from a successful new product, and the cost of launching a failure. The graph below shows that the combined information from the panel and concept tests clearly help discriminate between successes and failures. Plot of p_score*c_score$success. Symbol points to label. p_score | | 20 + | | > 1 | > 1 | > 1 15 + > 0 | > 1 > 1 > 1 | > 1 | > 0 > 1 > 0 > 1 | > 0 > 0 > 1 10 + > 1 | > 0 > 0 > 0 | > 0 | > 1 > 0 | > 0 5 + > 0 | ---+------------+------------+------------+------------+-- 20 40 60 80 100 c_score A single canonical discriminant function based on both concept and panel scores proves to be a significant discriminator between successes and failures. Wilks’s = 0.562, significant at p < 0.01. The canonical structure suggests that concept score and panel score load almost equally on the discriminant function. Multivariate Statistics and Exact F Statistics Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.56156799 8.20 2 21 0.0023 Pillai's Trace 0.43843201 8.20 2 21 0.0023 Hotelling-Lawley Trace 0.78072828 8.20 2 21 0.0023 Roy's Greatest Root 0.78072828 8.20 2 21 0.0023 Total Canonical Structure 38 Answers to Exercises Variable Can1 c_score 0.849651 p_score 0.820445 Class Means on Canonical Variables success Can1 0 -.8459713895 1 0.8459713895 The performance of the discriminant function based on the two tests in correctly classifying successes and failures is reasonably good. Due to the small sample size, there is some capitalization on chance. Using resubstitution, the hit rate is 20/24 = 83 percent; using cross-validation, the estimate of the hit rate drops to 16/24 = 67 percent. There is, of course, some question about the appropriate priors for such an analysis. The literature suggests that despite best efforts to the contrary, “most new products fail.” There are also asymmetric costs of misclassification that should also be taken into account when trying to decide whether a new product should be classified as potential success or failure. Linear Discriminant Function Variable 0 1 Constant -9.18461 -17.57076 c_score 0.16171 0.23381 p_score 0.97063 1.33842 Resubstitution Summary using Linear Discriminant Function Number of Observations and Percent Classified into success From success 0 1 Total 0 10 2 12 83.33 16.67 100.00 1 2 10 12 16.67 83.33 100.00 Total 12 12 24 50.00 50.00 100.00 Cross-validation Summary using Linear Discriminant Function Number of Observations and Percent Classified into success From success 0 1 Total 0 7 5 12 58.33 41.67 100.00 1 3 9 12 25.00 75.00 100.00 Total 10 14 24 41.67 58.33 100.00 Answers to Exercises 39 Exercise 12.14 These data on vehicle ownership require multiple discriminant analysis because there are three categories of ownership: car only, van only, and both car and van. We seek to discriminate among these three categories of ownership using data on income, family size, and age of head of household. The results of canonical discriminant analysis shows that there are indeed significant differences among these groups. Wilks’s = 0.556, which is significant at the 0.01 level. In fact, we can form two discriminant functions, both of which are significant. As shown in the table below, after removing the first discriminant function, Wilks’s for the second is equal to 0.797, which is significant at the 0.05 level. The discriminant function loadings suggest that the first function is primarily associated with higher income families with older heads of household. The second function is primarily large families that tend to be younger and have lower income. Multivariate Statistics and F Approximations Statistic Value F Value Num DF Den DF Pr > F Wilks' Lambda 0.55620611 3.52 6 62 0.0046 Pillai's Trace 0.50511605 3.60 6 64 0.0039 Hotelling-Lawley Trace 0.68764390 3.50 6 39.604 0.0071 Roy's Greatest Root 0.43305620 4.62 3 32 0.0085 Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.549719 0.449059 0.117951 0.302191 2 0.450472 . 0.134730 0.202925 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.55620611 3.52 6 62 0.0046 2 0.79707461 4.07 2 32 0.0265 Total Canonical Structure Variable Can1 Can2 income 0.771759 -0.631940 fam_size 0.362713 0.839026 age_hh 0.607869 -0.430959 A plot of the observations in the discriminant function space shows that the first discriminant function is separating families who own both car and van from families who own only a car or only a van. Thus, it appears that older, wealthier families are more likely to own both types of vehicle. The second discriminant function is separating families who own only a van (i.e., the larger, younger, lower income families). 40 Answers to Exercises Plot of Can2*Can1$own. Symbol points to label. Can2 | | 3 + | | > 0 2 + | > 1 > 2 | > 1 2 > 1 1 + > 0 > 1 ^> 1 | > 1 0 > 1 > 2 | 0 0 1 2> 2 > 1 > 2 0 + ^^> 1 > 2 | > 1 > 2 > 2 | > 0 -1 + > 0 > 0 > 2 2 > 2 | 0 < > 0 < ^ > 2 | -2 + > 0 > 0 | ---+-------+-------+-------+-------+-------+-------+-------+-- -4 -3 -2 -1 0 1 2 3 Can1 Class Means on Canonical Variables own Can1 Can2 0 -.5276820278 -.4991924912 1 -.3413484420 0.6796258485 2 0.8845582686 -.0821984957 The two discriminant functions do a reasonable job of accurately classifying households into ownership categories. Based on cross-validation, the estimated hit rate is 21/36 = 58.3 percent, which compares favorably to the proportional chance hit rate of 33.5 percent. Answers to Exercises 41 Linear Discriminant Function _ -1 _ -1 _ Constant = -.5 X' COV X + ln PRIOR Coefficient = COV X j j j Vector j Variable 0 1 2 Constant -25.98898 -27.44166 -36.75413 income 0.61400 0.53149 0.67633 fam_size 3.01115 3.74437 3.90843 age_hh 0.18752 0.23891 0.26387 Cross-validation Summary using Linear Discriminant Function Number of Observations and Percent Classified into own From own 0 1 2 Total 0 7 3 3 13 53.85 23.08 23.08 100.00 1 3 7 1 11 27.27 63.64 9.09 100.00 2 2 3 7 12 16.67 25.00 58.33 100.00 Total 12 13 11 36 33.33 36.11 30.56 100.00 If we test the homogeneity of within-group covariance matrices across groups, the result is marginally significant: Box’s chi-square test is 2(12) = 21.4, which is significant at the 0.05 level but not at the 0.01 level. Within Covariance Matrix Information Natural Log of the Covariance Determinant of the own Matrix Rank Covariance Matrix 0 3 9.40027 1 3 8.10301 2 3 8.79143 Pooled 3 9.54972 Test of Homogeneity of Within Covariance Matrices Chi-Square DF Pr > ChiSq 21.351007 12 0.0455 Since the Chi-Square value is significant at the 0.1 level, the within covariance matrices will be used in the discriminant function. 42 Answers to Exercises If we run a quadratic discriminant analysis, we fit that the fitted classification accuracy (based on resubstitution) improves over linear, but the predictive accuracy actually declines to a hit rate of 16/36 = 44.4 percent. Thus, even though the test suggests that there are differences across groups in their covariance structure, we are better off (from a classification accuracy standpoint) pooling our estimates across groups and using linear (rather than non-linear) discriminant analysis. Resubstitution Summary using Quadratic Discriminant Function Number of Observations and Percent Classified into own From own 0 1 2 Total 0 9 3 1 13 69.23 23.08 7.69 100.00 1 2 8 1 11 18.18 72.73 9.09 100.00 2 2 2 8 12 16.67 16.67 66.67 100.00 Total 13 13 10 36 36.11 36.11 27.78 100.00 Cross-validation Summary using Quadratic Discriminant Function Number of Observations and Percent Classified into own From own 0 1 2 Total 0 5 5 3 13 38.46 38.46 23.08 100.00 1 5 4 2 11 45.45 36.36 18.18 100.00 2 2 3 7 12 16.67 25.00 58.33 100.00 Total 12 12 12 36 33.33 33.33 33.33 100.00 Answers to Exercises 43 Exercise 12.15 Once again, we revisit the famous Iris data. The data set is comprised of 50 specimens of each of three different species of iris: Iris setosa (1), Iris versicolor (2), and Iris virginica (3). Four characteristics of each are measured: sepal length, sepal width, petal length, and petal width. The results from a multiple discriminant analysis indicate highly significant differences across species of iris. Wilks’s = 0.02, which is significant at p < 0.0001. In fact, two discriminant functions are significant. For the second function only, Wilks’s = 0.78, also significant at p < 0.0001. However, while both are significant, the first is by far the more important, accounting for much more of the variance in the data. The first function, which is highly correlated with all but sepal width, clearly separates Iris setosa (1) from Iris versicolor (2) and Iris virginica (3): the difference in means is quite large. By contrast, the second function (which is correlated with sepal width) provides a subtle distinction between (2) and (3): the difference in group means is about 1.2 (versus 4.0 on the first function). Adjusted Approximate Squared Canonical Canonical Standard Canonical Correlation Correlation Error Correlation 1 0.984821 0.984508 0.002468 0.969872 2 0.471197 0.461445 0.063734 0.222027 Test of H0: The canonical correlations in the current row and all that follow are zero Likelihood Approximate Ratio F Value Num DF Den DF Pr > F 1 0.02343863 199.15 8 288 <.0001 2 0.77797337 13.79 3 145 <.0001 Total Canonical Structure Variable Can1 Can2 sep_l 0.791888 0.217593 sep_w -0.530759 0.757989 pet_l 0.984951 0.046037 pet_w 0.972812 0.222902 Class Means on Canonical Variables species Can1 Can2 1 -7.607599927 0.215133017 2 1.825049490 -0.727899622 3 5.782550437 0.512766605 Plot of Can2*Can1$species. Symbol points to label. Can2 | | 3 + 44 Answers to Exercises | > 1 > 3 | 3 <> 3 | > 3 > 3 2 + 3 < > 3 | 1 < > 1 > 3 3 |1 < 1 > 1 3 > 3 2 3 | 1 < ^^ > 1 > 2 ^^> 3 1 + 1 <v > 1 > 2 3 | 1 22 1 > 2 ^2 2 3 | 1 <2^^> 1 > 2 2 3 v> 3 | 1 <2 > 1 2 < ^^^^3 > 3 0 + 1 <2^1 2 <^^^>22 ^^3> 3 | 1 <32 1 2 2>22 2 v2 3 > 3 | 1 2> 1 2 24 > 2 > 3 | 1 <5 1 > 2 >^2 ^ > 3 > 3 -1 + 1 2^> 1 2 < v2 v> 2 ^ > 3 | > 2 <> 2 3 | 2 < 2 3 > 2 > 3 | 2 <3 2 > 2 -2 + > 1 > 2 > 2 | 2 3 | | 2 2 -3 + | ---+---------+---------+---------+---------+-- -10 -5 0 5 10 Can1 According to Box’s test, we reject the homogeneity of covariance matrices across groups. However, unlike previous examples in this chapter, the classification accuracy of the quadratic discriminant functions (hit rate of 97 percent based on holdout validation) is closely comparable to the linear discriminant functions (hit rate of 98 percent based on holdout validation). Linear Discriminant Function for species Variable 1 2 3 Constant -86.30847 -72.85261 -104.36832 sep_l 23.54417 15.69821 12.44585 sep_w 23.58787 7.07251 3.68528 pet_l -16.43064 5.21145 12.76654 pet_w -17.39841 6.43423 21.07911 Cross-Validation: Number of Observations and Percent Classified into species From species 1 2 3 Total 1 50 0 0 50 100.00 0.00 0.00 100.00 2 0 48 2 50 0.00 96.00 4.00 100.00 3 0 1 49 50 0.00 2.00 98.00 100.00 Answers to Exercises 45 Total 50 49 51 150 33.33 32.67 34.00 100.00 Within Covariance Matrix Information Natural Log of the Covariance Determinant of the species Matrix Rank Covariance Matrix 1 4 -13.06736 2 4 -10.87433 3 4 -8.92706 Pooled 4 -9.95854 Test of Homogeneity of Within Covariance Matrices Chi-Square DF Pr > ChiSq 140.943050 20 <.0001 Cross-validation: Number of Observations and Percent Classified into species From species 1 2 3 Total 1 50 0 0 50 100.00 0.00 0.00 100.00 2 0 47 3 50 0.00 94.00 6.00 100.00 3 0 1 49 50 0.00 2.00 98.00 100.00 Total 50 48 52 150 33.33 32.00 34.67 100.00 46 Answers to Exercises 13. Logit Choice Models Exercise 13.1 a. We first calibrate a logit choice model conditional on making a purchase in the category on the trip. That is, we only account for trips with a purchase in the category. The model is 1 S1t 1 exp( * ( price1t price2t )) Model Fit Statistics Intercept Intercept and Criterion Only Covariates AIC 1718.496 1311.916 SC 1723.621 1322.167 -2 Log L 1716.496 1307.916 Testing Global Null Hypothesis: BETA=0 Test Chi-Square DF Pr > ChiSq Likelihood Ratio 408.5798 1 <.0001 Score 382.6007 1 <.0001 Wald 329.0601 1 <.0001 Analysis of Maximum Likelihood Estimates Standard Wald Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 -0.2596 0.0693 14.0435 0.0002 Coke_rp 1 -1.2234 0.0674 329.0601 <.0001 b. For the conditional logit choice model calibrated in part a, we have the formula for price elasticity for Coke as ˆ 1t * (1 S1t ) * price1t The estimated elasticity each week is listed in the table below (eta) Obs week Coke_p eta Pep_p cv eta2 1 1 4.33 -2.99052 4.33 -4.72561 -5.03855 2 2 4.33 -2.99052 4.33 -4.72561 -5.03855 3 3 4.33 -2.99052 4.33 -4.72561 -5.03855 4 4 3.24 -1.00941 4.33 -3.92952 -3.40798 5 5 3.24 -1.00941 4.33 -3.92952 -3.40798 6 6 4.33 -4.40235 3.24 -3.77878 -5.11123 7 7 4.33 -4.40235 3.24 -3.77878 -5.11123 8 8 3.24 -1.00941 4.33 -3.92952 -3.40798 Answers to Exercises 47 9 9 3.24 -1.00941 4.33 -3.92952 -3.40798 10 10 3.24 -1.00941 4.33 -3.92952 -3.40798 11 11 4.33 -4.38400 3.26 -3.79908 -5.10995 12 12 4.33 -4.38400 3.26 -3.79908 -5.10995 13 13 3.24 -1.00941 4.33 -3.92952 -3.40798 14 14 4.33 -4.40235 3.24 -3.77878 -5.11123 15 15 4.33 -4.40235 3.24 -3.77878 -5.11123 16 16 3.24 -1.00941 4.33 -3.92952 -3.40798 17 17 3.24 -1.00941 4.33 -3.92952 -3.40798 18 18 4.33 -4.39321 3.25 -3.78894 -5.11059 19 19 3.15 -0.90303 4.33 -3.84632 -3.26703 20 20 4.33 -4.30764 3.34 -3.87938 -5.10482 c. We build a model that takes into account of both purchase incidence probability and conditional brand choice. That is: P(coke) P(coke | c)* P(c) 1 1 * 1 exp( *( price1t price2t )) 1 exp(a cv) where cv ln(exp( * price1t ) exp( * price2t )) After some calculation, we can find the derivative of purchase probability with respect to price is: P(coke) / price1t P(coke | c) P(c) P (c ) P(Coke | c) price1t price1t * P(coke | c)*(1 P(Coke | c))* P(c) * * P(coke | c) 2 * P(c)*(1 P(c)) * P(c)* P(coke | c)[1 Pr ob(coke | c) *(1 P(c))* P(coke | c)] price1 Then the elasticity for this case is 2 * . Since now we take into account category s1 purchase probability elasticity, we get a higher price elasticity (eta is the conditional price elasticity, eta2 is the unconditional price elasticity). Exercise 13.2 a. We estimate a simple binary logit model with age, income and mobility as explanatory variables. Model Fit Statistics Intercept Intercept and Criterion Only Covariates AIC 139.628 114.107 48 Answers to Exercises SC 142.233 124.528 -2 Log L 137.628 106.107 Testing Global Null Hypothesis: BETA=0 Test Chi-Square DF Pr > ChiSq Likelihood Ratio 31.5207 3 <.0001 Score 27.9401 3 <.0001 Wald 21.5713 3 <.0001 Analysis of Maximum Likelihood Estimates Standard Wald Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 0.2810 0.4312 0.4246 0.5147 age 1 1.0691 0.5550 3.7105 0.0541 income 1 -2.2221 0.5430 16.7481 <.0001 mobility 1 -0.9101 0.5218 3.0420 0.0811 As we can see, the model is jointly significant. Age and mobility are significant at 10% level, but not at 5% level. Income is highly significant. b. We do a stepwise model selection for models with interaction terms. As we can see from the output, none of the terms remain significant when we add all the interaction terms. Analysis of Maximum Likelihood Estimates Standard Wald Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 0.2777 0.4705 0.3483 0.5551 age 1 0.7576 0.6791 1.2444 0.2646 income 1 -1.7300 1.2267 1.9891 0.1584 mobility 1 -0.7703 1.3378 0.3316 0.5647 age*income 1 0.3109 1.3886 0.0501 0.8228 age*mobility 1 0.4076 1.4206 0.0823 0.7742 income*mobility 1 -1.4765 1.1666 1.6019 0.2056 Our stepwise selection model chooses the model with age, income and mobility as explanatory variables. There does not seem to be any significant interaction effects. After adding all the interaction terms, even income effect is not significant. Exercise 13.3 a. As we can see from the estimation result, after controlling for gun ownership effect, there is no statistically significant relationship between age and voting behavior. It is negative, but not statistically significant. Analysis of Maximum Likelihood Estimates Standard Wald Answers to Exercises 49 Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 17.0683 11.6918 2.1312 0.1443 age 1 -0.4372 0.3081 2.0134 0.1559 gun_own 1 0.5301 0.9835 0.2905 0.5899 b. With our logit model, we set the following rule: if the predicted probability of a subject voting for gun control is greater than 0.5, then we classify the subject as voting =1. Given this rule, we perform a cross validation. For the original data, we correctly predicted 12 out of 15 subjects’ behavior. Comparing with discrimination analysis, using simple linear discrimination analysis and prediction, the hitting rate is the same. Exercise 13.5 a. We estimate logit models with brand dummy only, and with price and display in addition to brand dummies as covariate. The output is as following: Model Fit Statistics Without With Criterion Covariates Covariates -2 LOG L 87.889 87.542 AIC 87.889 91.542 SBC 87.889 94.920 Testing Global Null Hypothesis: BETA=0 Test Chi-Square DF Pr > ChiSq Likelihood Ratio 0.3466 2 0.8409 Score 0.3500 2 0.8395 Wald 0.3489 2 0.8399 Analysis of Maximum Likelihood Estimates Parameter Standard Hazard Variable DF Estimate Error Chi-Square Pr > ChiSq Ratio dummya 1 0.14310 0.37893 0.1426 0.7057 1.154 dummyb 1 -0.08005 0.40032 0.0400 0.8415 0.923 Model Fit Statistics Without With Criterion Covariates Covariates -2 LOG L 87.889 68.241 AIC 87.889 76.241 SBC 87.889 82.997 Testing Global Null Hypothesis: BETA=0 Test Chi-Square DF Pr > ChiSq Likelihood Ratio 19.6475 4 0.0006 Score 19.0897 4 0.0008 Wald 14.1197 4 0.0069 Analysis of Maximum Likelihood Estimates 50 Answers to Exercises Parameter Standard Hazard Variable DF Estimate Error Chi-Square Pr > ChiSq Ratio dummya 1 2.98015 0.95903 9.6564 0.0019 19.691 dummyb 1 0.91253 0.56518 2.6069 0.1064 2.491 price 1 -5.49366 1.70215 10.4166 0.0012 0.004 disp 1 0.72674 0.68873 1.1134 0.2913 2.068 The information added by price and display can be computed using the following statistics: LLr 68.241 2 1 1 0.224 LL0 87.542 That is, about 22% of the uncertainty in choice in the intercept only model is explained by the full model with price and display covariates. This is not a bad fit. To test whether price and display are significant in explaining the choice behavior of the household, we use the likelihood ration test; that is, X2(2) = -2(LLR -LLf) = 87.542 - 68.241 = 19.301, which is significant at 0.05 level. This is consistent with what we estimated from the statistics. When we 2 look at price and display separately, we see that the t-statistics for price is highly significant, while the t-statistics for display is not significant. But our 2 statistics says that they are jointly significant. b. The probability of each brand chosen given price=[1.5, 0.5, 0.80] and disp=0 is exp(a * price _ A) SA exp(a * price _ A) exp(b * price _ B) exp( * price _ C ) 0.089571 exp(b * price _ B) SB exp(a * price _ A) exp(b * price _ B) exp( * price _ C ) 0.69761 exp( * price _ C ) SC exp(a * price _ A) exp(b * price _ B) exp( * price _ C ) 0.21282 when price of A change from 1.50 to 1.20, the probability that each brand will be chosen is: (0.33832, 0.50701, 0.15467). Exercise 13.6 Answers to Exercises 51 a. Simple logistic regression show that order of the samples does not really matter for the rating: Model Fit Statistics Intercept Intercept and Criterion Only Covariates AIC 278.759 278.328 SC 282.057 284.924 -2 Log L 276.759 274.328 Testing Global Null Hypothesis: BETA=0 Test Chi-Square DF Pr > ChiSq Likelihood Ratio 2.4310 1 0.1190 Score 2.4261 1 0.1193 Wald 2.4161 1 0.1201 Analysis of Maximum Likelihood Estimates Standard Wald Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 0.1201 0.2004 0.3596 0.5487 ord 1 -0.4429 0.2849 2.4161 0.1201 The coefficients for order of sample factor is not significant. As a whole, the model is not significant either. The probability modeled is for the choice of 45. b. Adding attributes into model does not change the overall preference. Model Fit Statistics Intercept Intercept and Criterion Only Covariates AIC 278.759 275.712 SC 282.057 321.888 -2 Log L 276.759 247.712 Testing Global Null Hypothesis: BETA=0 Test Chi-Square DF Pr > ChiSq Likelihood Ratio 29.0471 13 0.0064 Score 26.6749 13 0.0138 Wald 23.0145 13 0.0415 Standard Wald Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 1.3350 1.7544 0.5791 0.4467 ord 1 -0.4758 0.3151 2.2808 0.1310 rate271 1 -0.3982 0.1656 5.7805 0.0162 rate272 1 0.4840 0.1806 7.1846 0.0074 rate273 1 -0.6150 0.3300 3.4732 0.0624 rate274 1 0.3954 0.2536 2.4311 0.1189 rate275 1 -0.2127 0.2019 1.1095 0.2922 rate276 1 0.0970 0.2465 0.1549 0.6939 rate451 1 -0.2833 0.1561 3.2938 0.0695 rate452 1 -0.1333 0.2012 0.4389 0.5077 rate453 1 0.1378 0.2857 0.2326 0.6296 52 Answers to Exercises rate454 1 -0.3989 0.2107 3.5860 0.0583 rate455 1 0.4548 0.1954 5.4188 0.0199 rate456 1 0.2289 0.2299 0.9911 0.3195 c. To see the contribution of attributes rating to the preference formation, we use the following statistic: LLr 247.712 2 1 1 0.097 . LL0 274.328 This statistics also says that there is not much contribution by adding attribute rating to the model.