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ANOVA Results, Exploratory Factor Analysis - Assignment Example

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From the paper "ANOVA Results, Exploratory Factor Analysis" it is clear that the analysis is aimed at determining the factors that influence the other responses, exploratory analysis involves determining the correlation coefficient matrix and analyzing the values. …
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Extract of sample "ANOVA Results, Exploratory Factor Analysis"

Question 1: 5 way ANOVA (GLM univariate) Five variables are considered in this test, the “intent” variable which represents the buying intentions, has value that range from 1 to 10, the value 1 represents extremely unlikely and 10 represents extremely likely. The variable in this case is therefore considered as the dependent variable. Fixed factors which are the categorical variables that have a different effect on the dependent variable include “numstop” representing number of stops, the variable “deptime” which represents the time of departure and the variable “fooddrnk” which represent in fright food and drinks. The variable “fareraw” is considered as a covariate variable; this means that “fareraw” is assumed to have a linear correlation with the variable “intent”. Assumptions are that; the error terms are independent, they are normal distributed with mean value of zero and their variance remains constant across observations. Descriptive statistics are summarised in appendix 1 and they indicate that there is a difference in mean values across the categories. Appendix 2 indicates the Levene’s test for error term variance equality, the significance value of this test is 0.594, this value is greater than 0.05 the null hypothesis that variances are equal is accepted, therefore the assumptions of this test are not violated. Appendix 3 indicates the ANOVA results, values indicated include the sum of squares and factors partial Eta squared, the able below summarises these results: PartiaL Eta Squared: Source Partial Eta Squared Intercept 0.918968189 numstops 0.297073884 deptime 0.178478465 fooddrnk 0.05703781 fareraw 0.747659374 The partial Eta value indicates the amount of variation that is accounted for by the variable, it indicates the importance and significance of the variable and from the above: “fareraw” is of great importance because the value is relatively high (0.7477) while “fooddrink” is of less importance in the model given that the value is only (0.05). Omega squared: The partial Eta value may be biased when a sample is used and an alternative is the Omega squared value, This value is determined as follows: W squared ={ sum of squares T – [ (k-1)*(MSerror)]}/{sum of squares total + MSerror} the table below summarises the values: Source Type III Sum of Squares df Mean Square Omega squared numstops 223.725042 1 223.725042 0.011456765 deptime 115.0076353 3 38.33587844 0.005789845 fooddrnk 32.02051904 3 10.67350635 0.001525285 fareraw 1568.4707 1 1568.4707 0.080560836 The results indicate that the variable “foodrnk” variable is less important while “numstop” and “fareraw” are important factors in explaining variations in “intent”. Managerial implications: Implications of these results indicate that buying decisions will be greatly influenced by fares and number of stops. Another factor that will influence this is the departure time. Food and drinks however will have little effect on buying decisions as indicated by the partial Eta squared and omega squared statistics. Multiple regression: Appendix 4 summarises the multiple regression model estimated, according to these results the model is as follows: Parameter B Intercept 9.059 [numstops=0] 1.140 [numstops=1] 0(a) [deptime=1] -.448 [deptime=2] .637 [deptime=3] .377 [deptime=4] 0(a) [fooddrnk=1] -.460 [fooddrnk=2] .048 [fooddrnk=3] .060 [fooddrnk=4] 0(a) fareraw -.014 The model is as follows: Intent = 9.059 + 1.140 numstop0 –0.448 deptime1 + 0.637 deptime2 + 0.377 deptime3 – 0.460 fooddrnk1 + 0.48 fooddrnk2 + 0.060 fooddrnk3 – 0.014 fareraw Eta squared: Parameter Estimates Dependent Variable: Buying intention Parameter B Partial Eta Squared Noncent. Parameter Observed Power(a) Intercept 9.059 .857 63.884 1.000 [numstops=0] 1.140 .297 16.940 1.000 [numstops=1] 0(b) . . . [deptime=1] -.448 .032 4.705 .997 [deptime=2] .637 .0 62 6.691 1.000 [deptime=3] .377 .023 3.960 .977 [deptime=4] 0(b) . . . [fooddrnk=1] -.460 .033 4.826 .998 [fooddrnk=2] .048 .000 .507 .080 [fooddrnk=3] .060 .001 .627 .096 [fooddrnk=4] 0(b) . . . fareraw -.014 .748 44.853 1.000 Question 2: Exploratory Factor Analysis: This analysis is aimed at determining the factors that influence the other responses, exploratory analysis involves determining the correlation coefficient matrix and analysis the values. Factors whose correlation coefficient is high either negative or positive are more likely to be influenced by one factor, those that whose correlation coefficients are low are more likely to be influenced by different factors. The following tables summarises correlation coefficients, the high correlation coefficients are highlighted: C1 C2 C3 C4 C5 C1 1 0.73 0.57 0.62 0.45 C2 0.73 1 0.55 0.63 0.39 C3 0.57 0.55 1 0.56 0.48 C4 0.62 0.63 0.56 1 0.41 C5 0.45 0.39 0.48 0.41 1 C6 0.69 0.79 0.56 0.65 0.42 S1 S2 S3 S4 S5 C1 0.26 0.12 0.15 0.14 0.16 C2 0.16 0.23 0.14 0.19 0.3 C3 0.23 0.18 0.07 0.28 0.2 C4 0.12 0.06 0.11 0.11 0.09 C5 0.06 0.1 0.05 0.17 0.16 C6 0.12 0.21 0.09 0.11 0.33 S1 1 0.47 0.48 0.43 0.5 S2 0.47 1 0.49 0.37 0.7 S3 0.48 0.49 1 0.48 0.44 S4 0.43 0.37 0.48 1 0.44 S5 0.5 0.7 0.44 0.44 1 K1 K2 K3 K4 C1 0.08 0.14 0.07 0.07 C2 0.23 0.3 0.29 0.28 C3 0.13 0.16 0.1 0.13 C4 0.2 0.18 0.09 0.13 C5 0.09 0.13 0.15 0.11 C6 0.24 0.3 0.28 0.35 S1 -0.14 -0.16 -0.16 -0.14 S2 -0.02 -0.06 -0.01 -0.06 S3 0.02 0.02 0.02 -0.13 S4 0 0.03 0.05 -0.1 S5 -0.09 -0.03 -0.02 0.03 K1 1 0.79 0.76 0.68 K2 0.79 1 0.84 0.75 K3 0.76 0.84 1 0.78 K4 0.68 0.75 0.78 1 From the correlation coefficient matrix the value indicate that there is a strong correlation between variable c1, c2, c3, c4 and c5. A strong correlation is also evident between variable k1, K2, k3 and k4. Low correlation values are evident between C and S variables and also between K and C variable. As a result the conclusion is that there is a relatively high possibility that there are three factors. Confirmatory factor analysis This involves using the Principal Component Analysis the following are the results: Total Variance Explained Component Initial Eigenvalues Extraction Sums of Squared Loadings Rotation Sums of Squared Loadings Total % of Variance Cumulative % Total % of Variance Cumulative % Total % of Variance Cumulative % dimension0 1 4.740 31.597 31.597 4.740 31.597 31.597 3.850 25.664 25.664 2 3.368 22.450 54.047 3.368 22.450 54.047 3.387 22.581 48.245 3 2.095 13.966 68.014 2.095 13.966 68.014 2.965 19.768 68.014 4 .859 5.725 73.739 5 .759 5.062 78.801 6 .583 3.887 82.688 7 .532 3.550 86.238 8 .468 3.117 89.355 9 .373 2.484 91.839 10 .317 2.113 93.953 11 .229 1.528 95.481 12 .198 1.322 96.803 13 .188 1.254 98.056 14 .166 1.105 99.161 15 .126 .839 100.000 Extraction Method: Principal Component Analysis. According to the Kaiser criteria the number of factors should be the number of factors that are evident before the major drop in Eigen value or values greater than 1, the aboer results indicate that there are three Eigen values greater than 1 and therefore three factors are extracted. The rotated factor matrix is as follows: Rotated Component Matrixa Component 1 2 3 c1 .858 c6 .829 c2 .821 c4 .817 c3 .758 c5 .624 k3 .932 k2 .918 k1 .877 k4 .868 s2 .805 s5 .803 s3 .760 s1 .722 s4 .689 Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization. a. Rotation converged in 5 iterations. The 3 factors are as follows: Factor 1: C1, C2, C3, C4 and C5 These variables measure the level of expertise in microcomputer use, therefore the factor is proficiency in microcomputer use Factors 2: K1, K2, K3 and K4 These variables measure expertise in keyboarding; therefore the factor here is proficiency in keyboarding. Factor 3: S1, S2, S3 and S4 These variables measure expertise in statistics; therefore the factor here is proficiency in statistics. Therefore the factors are: Factor 1: Proficiency in microcomputer use Factor 2: Proficiency in keyboarding Factor 3: Proficiency in statistics Reliability: The table below summarises the goodness of fit values Goodness-of-fit Test Chi-Square df Sig. 362.414 90 .000 The significance value is 0.000, o.05 which is the significant level is greater thean b this value and the conclusion is that the model structure is a good fit. The above results indicate that only one factor influences responses in this study, the values are reliable as indicate by the chi square test of goodness of fit, the value is less than 0.05 indicating that the model is statistically significant and reliable. Question 3: Parallel model: Assumed path diagram: Two variables y1 and y2 are assume a parallel model if the true score values T1 = T2, where T1 = y1 +e1 and T2 = y2 + e2….. if the variance values for both are equal Var (y1) = Var (y2)= Var (y3) = Var (y4) (W1=w2=w3=w4) Variance errors are equal Var (e1) = Var (e2) =Var (e3) = Var (e4) (V1=v2=v3=v4) diagram And finally if the variance errors are uncorrelated: Corr (e1, e2) = 0 Tau equivalent: Two variables y1 and y2 are Tau equivalent if the true score values T1 = T2, where T1 = y1 +e1 and T2 = y2 + e2…. They are also Tau equivalent if the variance values for both are equal Var (y1) = Var (y2)= Var (y3) = Var (y4) (W1=w2=w3=w4) diagram Variance errors are unequal Var (e1) ≠ Var (e2) ≠ Var (e3) ≠ Var (e4) (V1 ≠ v2 ≠ v3 ≠ v4) diagram And finally if the variance errors are uncorrelated: Corr (e1, e2) = 0 Congeneric It is less constrained Two variable y1 and y2; are co-generic if the true score T1 ≠T2 They are also co-generic if the variance values for both are unequal Var (y1) ≠ Var (y2) ≠ Var (y3) ≠ Var (y4) (W1 ≠w2 ≠w3 ≠w4) diagram Variance errors are unequal Var (e1) ≠ Var (e2) ≠Var (e3) ≠ Var (e4) (V1≠v2≠v3≠v4) diagram And finally if the variance errors are uncorrelated: Corr (e1, e2) = 0 These assumptions are applied to the model and the following are the results: Results: Model Params AIC x2 p parallel 0 1006.402 67.093 0 congeneric 1 1007.669 71.834 0 tau equivalent 4 1011.197 91.2 0 Validity: Validity can be assessed using the AVE method, in this method when the AVE value is greater than 0.5 then the construct indicates convergent validity. The AVE value is determined as follows: AVE = S1/S1 + S2 Where S1 is the sum of the squared factor loadings and S2 is the sum of 1 minus squared for each factor loadings, the table below summarises the results: Standardized Factor Loading Factor loading Factor loading squared 1- factor loading squared 1 0.856 0.73274 0.26726 2 0.851 0.72420 0.27580 3 0.843 0.71065 0.28935 4 0.793 0.62885 0.37115 Sum 3.343 2.79644 1.203565 From the table S1 = 0.903 and S2 = 1.203565 AVE = S1/S1 + S2 AVE = 0.903 / 0.903 +1.203565 AVE = 0.699 This value is greater than 1 and this means that the construct has a good convergent validity Discriminant validity: When the AVE value is greater than any squared covariance value in the column then there is Discriminant validity, in this case the construct reliability value = s1 squared/ s1 squared plus s1 The value (CR) = 0.903 CR squared = 0.121 AVE > CR squared The AVE value is greater and this means that there is discriminate validity. The other conclusion is that if the reliability value is greater than 0.7 then the construct is reliable, in this case CR = 0.903 and this value is greater than 0.7 meaning that there is construct reliability Question 4: A Logit model is estimated to determine whether there were discriminatory issues related to the acceptance of the application process, variables included in this model include GPA, GMAT, Country and whether application was accepted or rejected, the following table summarises the results: SPSS output: Parameter Estimates Accept or reject the application(a) B Std. Error Wald df Sig. Reject Intercept 67.156 15.244 19.409 1 .000 gmat -.091 .021 18.678 1 .000 gpa -7.127 1.873 14.474 1 .000 country -1.034 .849 1.485 1 .223 a The reference category is: Accept. From the results model is stated as follows: Y = e [-(67.156 + -0.91 gmat – 7.127 gpa – 1.034 country)]/ (1 + e [-(67.156 + -0.91 gmat – 7.127 gpa – 1.034 country)]) The significance of the model is summarised below: Model Fitting Information Model -2 Log Likelihood Chi-Square df Sig. Intercept Only 182.224 Final 39.218 143.006 3 .000 From the table 0.05 is greater than the significance value 0.000 meaning that the fitted model is statistically significant. Pseudo R-Square The Pseudo r squared is an alternative to the R squared in regression models, the higher the value then the stronger the relationship between the explanatory and the dependent variable, the table below summarizes the results: Pseudo R-Square Cox and Snell .612 Nagelkerke .873 McFadden .785 The values are large meaning that more variations are explained by the model Contribution of variables to the model, Likelihood Ratio Tests Effect -2 Log Likelihood Chi-Square df Sig. Intercept 39.218(a) .000 0 . gmat 138.673 99.455 1 .000 gpa 72.192 32.974 1 .000 country 40.788 1.569 1 .210 When 0.05 is greater than the significance value then the variable contribution to the model is significant, therefore the country variable does not contribute to the acceptance or rejection of application, for this reason therefore it is evident that the country variable has not effect on the application acceptance, this means that discrimination is not practiced in the application approval process. The odd ratio, this is the probability that it is true divided by the probability that it is false, the value here is 618.7494 meaning that the probability that the results are true is greater than the probability that results are false. Determining whether application will be accepted or rejected: The estimated model takes the form: Y = e [g(x)]/ (1 + e [g(x)]) Where g(x) = - (b0 + b2x1 + b3x2 …bnxn) In this case given that the estimated values are as follow: Intercept 67.156 gmat -.091 gpa -7.127 country -1.034 Then g(x) = - (67.156 + -0.91 GMAT – 7.127 GPA – 1.034 country) Given that GMAT = 560 GPA = 3.33 Country = 0 Then the values will be estimated as follows: G(x) = - (67.156 + -0.91 (560) – 7.127 (3.33) – 1.034 (0)) G(x) = 7.53691 Y = e [g(x)]/ (1 + e [g(x)]) Y = e [7.53691]/ (1 + e [7.53691]) Y = 0.999467 The y value is approximately 1; therefore given these values the application will be accepted. References: Lemeshow, S. (2009). Applied Logistic Regression. New York: Wiley and Sons Mueller, W. (2008). Introduction to factor analysis, Newbury Park, CA: Sage. Read More
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