Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(20 Points)
1. Consider the following EVIEWS output from the second homework problem. Use it to answer parts (a) through (d) below.
============================================================ LS // Dependent Variable is TENSILE Date: 03/28/98 Time: 17:04 Sample: 1 19 Included observations: 19 ============================================================ Variable CoefficienStd. Errort-Statistic Prob. ============================================================ C 21.32126 5.430178 3.926439 0.0011 HARDWOOD 1.770986 0.647814 2.733788 0.0141 ============================================================ R-squared 0.305374 Mean dependent var 34.18421 Adjusted R-squared 0.264513 S.D. dependent var 13.77777 S.E. of regression 11.81589 Akaike info criter 5.038191 Sum squared resid 2373.458 Schwarz criterion 5.137605 Log likelihood -72.82264 F-statistic 7.473597 Durbin-Watson stat 0.246890 Prob(F-statistic) 0.014140 ============================================================
a = .00001.
Do not reject the null hypothesis: P-Value > .00001.
a = .05.
t.025,17 = 2.11
Test Statistic = (1.770986 – 2)/.647814 = -.3535
Since –2.11 < -.3535 < 2.11, We do not reject the null hypothesis.
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
t.005,17 = 2.898
1.770986 ± 2.898*.647814, or (-.106379 , 3.648351)
A one unit increase in HARDWOOD produces a 1.770986 unit increase in TENSILE.
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
2. What numbers belong in the spaces marked (a) and (b) in the following EVIEWS output.
a) (5 Points)
t.1,17 = 1.333
b) (5 Points)
Std. Error = Coefficient/T-Stat = -0.668089/-2.661159 = .2510519
============================================================ LS // Dependent Variable is Y Date: 03/28/98 Time: 17:41 Sample: 1 20 Included observations: 20 ============================================================ Variable Coefficient Std. ErrorT-Statistic Prob. ============================================================ C 2.280695 0.475395 4.797476 0.0002 X1 0.349817 (a) 0.2000 X2 -0.668089 (b) -2.661159 0.0165 ============================================================ R-squared 0.298305 Mean dependent var 2.119493 Adjusted R-squared 0.215753 S.D. dependent var 2.377157 S.E. of regression 2.105156 Akaike info criter 1.626260 Sum squared resid 75.33860 Schwartz criterion 1.775620 Log likelihood -41.64137 F-statistic 3.613528 Durbin-Watson stat 1.332075 Prob(F-statistic) 0.049233 ============================================================
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
3. With reference to the EVIEWS output used in question 2:
a) (5 Points) State the null hypothesis that is tested by the F-statistic. Is this hypothesis rejected at the 5% significance level?
H0: b 1 = b 2 = 0
Yes, the null hypothesis is rejected, P-Value < .05.
b) (5 Points) What is SSY for this regression?
SSY = (2.377157)2*19 = 107.36663
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
a) (5 Points) Perform the Chow Breakpoint test at the 9th observation and report the p-value. Do you reject or not reject the null hypothesis and what does that mean substantively?
P-Value = .00062. The Null hypothesis here is that observations 1 – 8 have the same linear structure as observations 9 – 19. We reject the null hypothesis. Substantively, it appears as though different models fit these two sets of observations.
b) (10 Points) Perform the Ramsey Reset Test. Try 3, then 2, then 1 fitted terms and report the three p-values. What is your interpretation of these three p-values and what do they tell you about the relationship between TENSILE and HARDWOOD?
The P-Values are all 0.000000. This indicates that the simple linear specification is not correct. A good strategy to follow would be to try adding powers of HARDWOOD to the specification. In particular, the Rest Test at 2 shows that Y-Hat is significant squared and cubed. Reset Test at 3 shows all Y-Hat powers to be individually insignificant. Hence, trying:
LS TENSILE C HARDWOOD HARDWOOD^2 would be a good start.
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
SSEUR = 47.78610 + 64.04799 = 111.83409
SSER = 202.6925
F5,17,.05 = 2.81
Test Statistic = [(202.6925 – 111.83409)/5]/[111.83409/17] = 2.76
Since 2.76 < 2.81, Do not reject the null hypothesis.
============================================================ LS // Dependent Variable is Y Date: 03/28/98 Time: 19:24 Sample: 1 27 Included observations: 27 ============================================================ Variable Coefficient Std. ErrorT-Statistic Prob. ============================================================ C 2.911143 1.644876 1.769825 0.0906 X1 -0.114837 0.651213 -0.176344 0.8616 X2 0.908389 0.352389 2.577802 0.0172 X3 -1.919070 1.822874 -1.052772 0.3039 X4 -2.551981 0.750349 -3.401058 0.0026 ============================================================ R-squared 0.550582 Mean dependent var 1.309189 Adjusted R-squared 0.468869 S.D. dependent var 4.164922 S.E. of regression 3.035341 Akaike info criter 2.386224 Sum squared resid 202.6925 Schwartz criterion 2.626193 Log likelihood -65.52536 F-statistic 6.738045 Durbin-Watson stat 2.162684 Prob(F-statistic) 0.001066 ============================================================ ============================================================ LS // Dependent Variable is Y Date: 03/28/98 Time: 19:25 Sample: 1 13 Included observations: 13 ============================================================ Variable Coefficient Std. ErrorT-Statistic Prob. ============================================================ C -0.028963 2.405406 -0.012041 0.9907 X1 4.337600 2.063752 2.101803 0.0687 X2 1.915258 0.606572 3.157513 0.0134 X3 -4.357744 3.039306 -1.433796 0.1895 X4 3.035299 2.598807 1.167958 0.2765 ============================================================ R-squared 0.636701 Mean dependent var 1.430997 Adjusted R-squared 0.455051 S.D. dependent var 3.310762 S.E. of regression 2.444026 Akaike info criter 2.071016 Sum squared resid 47.78610 Schwartz criterion 2.288304 Log likelihood -26.90781 F-statistic 3.505105 Durbin-Watson stat 2.262544 Prob(F-statistic) 0.061787 ============================================================ ============================================================ LS // Dependent Variable is Y Date: 03/28/98 Time: 19:25 Sample: 14 27 Included observations: 14 ============================================================ Variable Coefficient Std. ErrorT-Statistic Prob. ============================================================ C 3.659305 2.427713 1.507305 0.1660 X1 -1.951862 0.859362 -2.271291 0.0493 X2 1.364739 0.519494 2.627053 0.0275 X3 -3.970569 2.592990 -1.531271 0.1601 X4 -3.521417 0.808209 -4.357062 0.0018 ============================================================ R-squared 0.799289 Mean dependent var 1.196081 Adjusted R-squared 0.710084 S.D. dependent var 4.954448 S.E. of regression 2.667666 Akaike info criter 2.234861 Sum squared resid 64.04799 Schwartz criterion 2.463096 Log likelihood -30.50917 F-statistic 8.960135 Durbin-Watson stat 1.758126 Prob(F-statistic) 0.003342 ============================================================
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
P-Value = .582785, thus we do not reject the null hypothesis that the two pairs of coefficients are equal. In other words, there is not a significant difference between a one point increase in the essay score and a one point increase in the letter rating score; and there is not a significant difference between a one point increase in verbal GMAT and a one point increase in quantitative GMAT.
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
7. Consider the model:
where:
E(e i )=0 for all i
E(e i2 ) = s 2 for all i
E(e i e j ) = 0 for all i and j
x fixed for all i
(a) (5 Points) Suppose a regression of y against x and a constant is run for the n observations from this model (i.e., LS Y C X). The estimated residuals (RESID) are then plotted against x (i.e., SCAT RESID X). Illustrate below the pattern of dots that you would expect to see in this scatter plot.
For Example:
(b) (5 Points) Write below the EVIEWS commands that you would use to estimate this as a linear model.
LS LOG(Y) C X1
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
SSY = [(1.765115)2]*24 = 74.775143
SSE = [(0.832516)2]*20 = 13.861658
R2 = (74.775143 – 13.861658)/74.775143 = .8146
============================================================ LS // Dependent Variable is Y Date: 03/28/98 Time: 16:16 Sample: 1 25 Included observations: 25 ============================================================ Variable Coefficient Std. ErrorT-Statistic Prob. ============================================================ C 0.585593 0.602372 0.972145 0.3426 X1 1.419253 0.199449 7.115882 0.0000 X2 2.658296 0.606784 4.380958 0.0003 X3 0.712843 0.222854 3.198703 0.0045 X4 0.563379 0.561315 1.003676 0.3275 ============================================================ R-squared ??? Mean dependent var 2.853512 Adjusted R-squared 0.777547 S.D. dependent var 1.765115 S.E. of regression 0.832516 Akaike info criter-0.189749 Sum squared resid ??? Schwartz criterion 0.054026 Log likelihood -28.10160 F-statistic 21.97194 Durbin-Watson stat 2.094706 Prob(F-statistic) 0.000000 ============================================================
Probability and Statistics Name_________________________________
Spring 1998 45-734
Midterm Exam
Keith Poole
(10 Points)
Yes, the signs are correct. Economic theory tells us that ice cream consumption should increase with income, decrease in price, and, common sense tells us that it should increase with temperature. However, the coefficient on price is not statistically significant.
Although the signs are correct, the fact that the price variable is not statistically significant merits some further investigation. A sensible first step would be to check the residuals by looking at scatterplots: e.g., SCAT RESID INCOME, SCAT RESID PRICE, SCAT RESID TEMP to see if there is anything suspicious.
Another useful step would be to drop PRICE and run the regression without it. That is:
LS CONSUMP C INCOME TEMP
Another route would be to run the Ramsey Reset Test to check for non-linear effects.
============================================================ LS // Dependent Variable is CONSUMP Date: 03/30/98 Time: 16:23 Sample: 1 30 Included observations: 30 ============================================================ Variable CoefficienStd. Errort-Statistic Prob. ============================================================ C 0.197315 0.270216 0.730212 0.4718 INCOME 0.003308 0.001171 2.823722 0.0090 PRICE -1.044414 0.834357 -1.251759 0.2218 TEMP 0.003458 0.000446 7.762213 0.0000 ============================================================ R-squared 0.718994 Mean dependent var 0.359433 Adjusted R-squared 0.686570 S.D. dependent var 0.065791 S.E. of regression 0.036833 Akaike info criter-6.479173 Sum squared resid 0.035273 Schwarz criterion -6.292346 Log likelihood 58.61944 F-statistic 22.17489 Durbin-Watson stat 1.021170 Prob(F-statistic) 0.000000 ============================================================