12 14 98 252y9871 ECO252 QBA2 Name FINAL EXAM Hour of Class Registered Circle December 14 1998 MWF 10 11 TR 12 30 2 00 I 16 points Do all the following Hand in your fourth regression problem 2 points and answer the following questions If you did not do this problem you can have a copy of the problem output for a 2 point penalty 1 In the regression of price against oldpr chassets chincome and chsales which coefficients are significant at the 5 level Why 3 2 Which coefficients are significant at the 1 level Why 1 3 Comparing the regressions that you did noting t tests Rsq Rsq adj which seems best Why 2 4 Compare the regression against oldpr with the regression against oldpr chassets and chincome using an F test Do the added variables have additional explanatory power 4 5 Using the regression against oldpr and chassets observation give a confidence and prediction interval for price in the first observation 3 6 In the regression against oldpr chassets chincome and eps are the signs of the coefficients as expected Why 2 7 If a company has an old price of 20 a 25 change in assets in 1992 and a 30 change in income in 1992 what price does the regression against oldpr chassets and chincome predict 2 Solution 1 2 From the computer output VariableCoefficient t p Constant 36 240 4 14 003 S5 S1 Oldpr 2 4072 8 96 000 S5 S1 Chassets 0 1783 1 18 273 Chincome 0 15782 2 63 030 S5 Chinsales 0 3217 1 02 339 The easiest way to do this problem is to compare the p value to 05 and 01 Those significant at the 1 level are marked S1 and those significant at the 5 level are marked S5 You can also compare the t coefficient against t In particular t 8025 2 306 and t 8005 3 355 this yields the same results 2 3 The regression against Oldpr Chassets and Chincome has significant coefficients and a relatively large R 2 adjusted 87 0 and is preferred The addition of Chsales doesn t raise R 2 adjusted and creates two insignificant coefficients No other regression has as high an R 2 adjusted 4 The first two ANOVAs below come from the printout The last one is figured out from the difference between the Regression sums of squares in the first two Regression 1 Regression 2 Difference Source SS DF Source SS DF Source SS DF MS Regr 3904 7 1 Regr 4600 2 3 Regression 1 3904 7 1 Error 1195 7 11 Error 500 1 9 Regression 2 695 5 2 342 75 Total 5100 3 12 Total 5100 3 12 Error 500 1 9 55 57 Total 5100 3 12 342 75 6 258 If we take the ratio of the mean squares in the difference ANOVA we find that F 55 57 2 9 Since F 05 4 26 is less than the computed F we conclude that the effect of the new variables is significant 5 Reading the standard deviation and the s e from the printout the confidence interval is 2 2 12 Since no one 51 93 t 12 025 3 87 and the prediction interval is 51 93 t 025 3 87 9 182 got this I never finished getting the values 1 See bottom of page 3 for rest of answer 2 12 14 98 252y9871 II Do at least 4 of the following 6 Problems at least 15 each or do sections adding to at least 60 points Anything extra you do helps and grades wrap around Show your work State H 0 and H1 where applicable Use a significance level of 5 unless noted otherwise 1 The data below shows salary months employed and gender for 9 employees y is salary in thousands y x1 x2 x1 is months employed and x 2 is a dummy variable with 1 for female Salary Months Gender a Compute the simple regression equation of salary against gender 6 7 5 6 0 b Compute R 2 4 8 6 10 0 c Compute se 3 9 1 10 3 13 0 6 2 8 7 9 4 9 8 12 18 30 5 13 15 21 y 82 6 y x d Compute s b1 and do a significance test on b1 4 e Do a prediction interval for salary for an employee with 17 years service 4 0 0 0 1 1 1 1 2 786 64 x 1 130 0 x 2 1 2364 0 x 2 4 0 x 2 2 4 0 x y 34 1 x x 54 0 and n 9 You do not need all of these Solution No x1 y x y If you think this is true you didn t learn anything from me 1y 2 1 2 in ECO251 x1 y 82 6 9 17778 9 130 0 X1 14 44444 9 Our spare parts are Y 45 0 86 0 109 2 185 4 390 0 31 0 113 1 141 0 205 8 1306 5 x1 y 1306 5 So X X Y X Y nX Y X nX 1 1 2 1 1Y nX 1Y 1306 5 9 14 44444 9 17778 113 38889 2 1 2 nY 2 786 64 6 9 17778 2 28 55556 Using capital letter instead of small ones b1 nX 12 2364 0 9 14 44444 2 486 22222 a 2 1 113 38889 0 233204 486 22222 b0 Y b1 X 9 17778 0 233204 14 44444 5 80928 so that our equation is Y 5 8093 0 233204 X 1 XY nXY 113 38889 X nX Y nY 482 22222 28 55556 92601 2 b R 2 2 2 2 2 2 3 12 14 98 252y9871 c s e2 d Y 2 nY s b21 s e2 2 b XY nXY 1 n 2 X so that t 1 2 nX 2 28 55556 0 233204 113 38889 7 0 301831 s e 0 54939 1 0 301831 482 2222 0 00062077 s b1 0 0249152 b1 10 0 233204 n 2 7 9 35991 Since t 025 t 025 2 365 0 is significant s b1 0249151 H 0 is 1 0 We reject it e If X 0 17 Y 0 b0 b1 X 0 5 80928 0 233204 17 9 77375 so that 1 n 2 Y Y 0 t 025 s e2 0 n X 0 X 2 9 774 2 365 1 X 2 nX 2 9 774 2 365 0 58263 9 774 1 378 2 0 3018 1 17 14 4444 486 2222 9 Remainder of answer to I 6 Logic says that all the independent variables should be related to higher prices so that the coefficients should all be positive However the coefficient of Eps is negative 7 Using coefficients from the printout Price 33 4 2 32 20 0 282 25 147 30 24 46 Solution Continues in 252z9871 4 1
View Full Document
Unlocking...