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MULTIPLE CHOICE Solutions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: D C A D B D B D C A B B D A D A B D A B B B C B D A C C D C A D D B C Note: This is a Chi-Square Goodness of Fit Test Note: This is a Chi-Square Goodness of Fit Test Note: This is a Chi-Square Test for Independence Recall: DF = (r-1) (c-1) Goodness of Fit Chi-Square test Interpret the slope (note: y is in 1000s) Test we must use in Multiple Regression (use the ANOVA) hint: no error deterministic model is a math model with no error Hint: no error once again. note: MSE = SSE/(n-2) Note: If you look at x =2 ,4, and 6 you can see that y goes down (weakly) Recall: t = b1 / Sb1 looking for b0 R2 is always SSR/SST for both the simple and multiple regressions interpret slope of X1 note: look at the coefficient values to get the equation note: SST = SSE = SSR same as #35 1 Short Answer Problems Solutions 1. a. Recall that we go to the values of coefficient and we find that the equation is: = 13.251 + 0.803x b. Since we want to run a hypothesis test on b1 we run the following test: (i) Hypotheses Ho : B1 = 0 Ha : B1 ≠ 0 (ii) Critical/Rejection Rule: Critical value is n -2 df which is 8-2 = 6 t -tα/2 =-2.447 μ tα/2 = 2.447 (iii) Test Stat: (0.803 - 0) / 0.385 ≈2.085 (iv) Conclusion: Fail to reject Ho and conclude that it is not significantly different from 0; So this tells us the value we have for b1 is not to be used. c. First we want to fill in the ANOVA Table Completely: Source of Degrees of Sum of Variation Freedom Squares Regression 1 (71.87541.674) = 30.201 Residual Error 6 41.674 Total 8 Mean Square MSR/df = 30.201 MSE/df ≈6.95 71.875 (1) Identify Hypotheses Ho: Bi = 0 for the all coefficients Ha: Bi ≠ 0 for at least one of the coefficients (2) Rejection Rule/Critical value Note: we have an F-stat with df = k-1 = 2-1=1 for the numerator and df n-(k+1) = 8 – 2 = 6 denominator degrees of freedom. So the critical value for F = 5.99 (3) Calculate the test statistic F = MSR/MSE = 4.35 2 F-Stat MSR/MSE ≈4.35 (4) Conclusion: Our test stat does not lie in the tail so we fail to reject Ho and conclude that our constant (b0) and slope (b1) are not any different from 0. So our estimated regression equation should not be used as a predictor of y. d. R2 = SSR/SST = 30.201/71.875 ≈ 0.42 ; So this tells us the regression equation can only explain 42% of the total variation in y. So it does not do a good job. 2. The first Step is to finish putting in all the values for the ANOVA table: Source of Degrees of Sum of Squares Variation Freedom Regression 2 822.088 7 736.012 Error Mean Square F SSR/df = 822.088/20 = 411.044 SSE/df = 736.012/7 ≈105.14 MSR/MSE =411.044/105.14 ≈3.91 2+7 = 9 822.088+736.012=1558.1 Total a. Now we want to run the following Hypothesis Test: (1) Identify Hypotheses Ho: Bi = 0 for the all coefficients Ha: Bi ≠ 0 for at least one of the coefficients (2) Rejection Rule/Critical value Note: we have an F-stat with df = 2 for the numerator and df = 7 denominator degrees of freedom. So the critical value for F = 4.74 (3) Calculate the test statistic F = MSR/MSE = 3.91 (4) Conclusion: Our test stat does not lie in the tail so we fail to reject Ho and conclude that our constant (b0) and slope (b1) are not any different from 0. So our estimated regression equation should not be used as a predictor of y. b.R2 = SSR/SST =822.088/1558.1≈0.528 So this tells us the regression equation can only explain around 53% of the total So it does not do a terribly good job. c. Adj-R2 = [R2 –k/(n-1)] [(n-1)/{n-(k+1)}] =[0.528 –2/(10-1)] [(10-1)/{10-(2+1)}] 0.393 d. The entire sample is n. Since the total degrees of freedom is n-1 = 9 n=10 3. a.When advertising was used, sales were higher by $18000; note: You only put in values of 0 or 1. b. Recall that = 10 - 4X1 + 7X2 + 18X3 ; So if we have X1 = 3 because of 3000 in sales previously, X2= 10 because of 10,000 miles, and X3= 1 since they used advertising we get that Sales ( ) = 10 – 4(3) + 7(10) + 18*1 = 86 or $86,000 in sales. 3 4. Has there been any significant change in the number of students in each major between the last school year and this school year? Use = 0.05. Recall 30% were Accounting majors, 24% Management majors, 26% Marketing majors, and 20% Economics majors with n=300 students Expected Major Actual 0.30(300)=90 0.24(300)72 0.26(300)=78 0.20(300)=60 300 Accounting Management Marketing Economics Total 83 68 85 64 300 49/90≈0.54 016/72≈0.22 49/78≈0.63 16/60≈0.27 1.66 So now given the table we created above we can run the goodness of fit test: (1) Hypotheses: Ho: That Paccountig = 0.30 PMGT = 0.24 PMKT = 0.26 PEconomics = 0.20 Ha: At least one of the probabilities is different from 0.33 (2) Critical Value: If we note that the df = k -1 = 4 -1 = 3 and go to table A.10 we find that the critical value is 7.81 = 7.81 (3) Test Stat: X2=1.66 (4) Conclusion: Since 1.66 is NOT in the tail we fail to reject Ho and conclude that the proportions are not significantly different from the expected values 5. Recall that 3 x 2 contingency table with observed values from a sample of 1,500. At 95% confidence, test for independence of the row and column factors. Table 1: Original Actual Frequencies Row Factor A B C Total Table2: Expected Frequency Row\Colum A B C Total X Column Factor Y Total 450 300 150 900 300 300 0 600 750 600 150 1,500 X (750)*(900)/1500=450 600(900)/1500 =360 150(900)/1500 = 90 900 Y 600(750)/1500=300 600(600)/1500=240 150(600)/1500 = 60 600 4 Total 750 600 150 1500 Chi-Square Test stat we take X2– Σ = (1) Hypotheses: Ho: Column Factor and Row Factor are independent from one another Ha: The variables are statistically dependent (2) Critical value – it is chi-square distributed with (r-1) (c-1) =2*1 = 2 with α=0.05 Critical value is -5.99147 =5.99147 (3) Test Statistic: X2– Σall cells = 125 (4) Conclusion: Reject Ho because 125>>5.99147 which implies the test-stat> critical value (i.e. it is in the tail). So we conclude with 95% confidence that Column and Row factor are statistically dependent (ie they are not independent) 6. Recall 2 x 3 contingency table with observed values from a sample of 500. At 95% confidence, test for independence of the row and column factors. Table 1: Original Observed Values Row Factor\Column Factor A B Total Table2: Expected Frequency Row\Colum A B Total X Y Z Total 40 60 100 50 100 150 110 140 250 200 300 500 X (200)*(100)/500=40 100(300)/500 =60 100 Chi-Square Test stat we take X2– Σ Y 150(200)/500=60 150(300)/500=90 150 = (1) Hypotheses: 5 Z 250(200)/500=100 250(300)/500=150 250 Total 200 300 500 Ho: Column Factor and Row Factor are independent from one another Ha: The variables are statistically dependent (2) Critical value – it is chi-square distributed with (r-1) (c-1) =1*2 = 2 with α=0.05 Critical value is -5.99147 =5.99147 (3) Test Statistic: X2– Σall cells = 4.45 (4) Conclusion: Reject Ho because 4.45<5.99147 which implies the test-stat< critical value (i.e. it is not in the tail). So we conclude with 95% confidence that Column and Row factor are statistically independent 6