2. 7.1 Lecture #7 Studenmund(2006) Chapter 7 Objective: Applications of Dummy Independent Variables

3. 7.2 Qualitative information Gender: male and female Regional: HK Island, Kowloon & NT Zone: East, South, West, North, Center Time/period: peace and war, before & after crisis Age: young, middle, elder Education: Post-graduate, College, High, Element Others:

4. 7.3 obs Male Dummy Female Dummy Salary(K) Years of teaching 110231 20119.51 310242 401212 510253 601223 71026.54 80123.14 901255 1010285 111029.56 1201266 130127.57 141031.57 1501296 1610225 1701192 1810 2 190121.75 200118.52 2110 4 221020.54 2301171 240117.51 251021.25 Example: Gender issue of whether discrimination is existing for salary

5. 7.4 Separate sample of male Male Sample: (Gujarati- 1995, Table 15.1 & 15.5) 12 Total # obs: 12

6. 7.5 Female sample: (Gujarati-1995, Table 15.1 & 15.5) Separate sample of female 13 Total # obs: 13

7. 7.6 10 15 20 25 30 35 012345678 Male Linear (Male) Salary Y X teaching years Y =  0 +  1 X (male) ^ ^ ^ Linear (Female) Female Y =  * 0 +  2 X (female) ^ ^ ^ Two separate models: Y i =  0 +  1 X i +  i Y j =  * 0 +  2 X j +  j (male) (female)

8. 7.7 Assuming  1 =  2, same slope but different constant between Y and X. 1st model: Y i =  0 +  ’ 0 D i +  1 X i +  i Y i =  0 +  ’ 0 D i +  1 X i +  ’ 1 D i X i +  i Y i = annual salary X i = years of teaching experience D i = 1 if male = 0 otherwise (female) control variable Assuming  1   2, different slope and different constant between Y and X. 2nd model:

9. 7.8 Y =  * 0 +  * 1 X (whole) ^ ^ ^ Two separate models: Y i =  0 +  1 X i +  i Y j =  * 0 +  2 X j +  j (male) (female)

10. 7.9 D 1 + D 2 = 1 D 1 = 1 - D 2 Each dummy identify two different categories, but when sum up two dummies it cannot identify which is male or female

11. 7.10 (Dummy variable trap) If we introduce two dummy variables in one model to identify two categories of one qualitative variable such as Y i =  0 +  ’ 0 D1 i +  ’’ 0 D2 i +  1 X i +  i where D1 i = 1 if male = 0 otherwise where D2 i = 1 if female = 0 otherwise This model cannot be estimated because of perfect collinearity between D1 and D2 D1 = 1 - D2 or D2 = 1 - D1 or D1 + D2 = 1 ( Perfect collinearity )

12. 7.11 Use two dummy variables to identify two different qualitative categories in one model will be fall into the trap of perfect multicollinearity. General rule : To avoid the perfect multicollinearity If a qualitative variable has “m” categories, introduce only “m-1” dummy variables. 1 D 1 D 2 D 3 D 4 D 5 … D m-1 age 110203040 m Categories dummy => Qualitative variable

13. 7.12 Measure the estimated result for two groups: Male: ==> Y i = (  0 +  ’ 0 D i )+  1 X i D i = 1 ^ ^ ^ ^ Female: ==> Y i =  0 +  1 X i D i = 0 ^ ^ ^ Now consider different intercepts of two groups: Model: Y i =  0 +  ’ 0 D i +  1 X i +  i D i = 1 if male = 0 otherwise, (i.e. female) When a category is assigned the value of zero, this category is called a control category (or omitted group). 2

14. 7.13 In order to test whether there is any difference in the relationships between two categories Compare: Y i =  0 +  1 X i ^ ^ ^ Y i = (  0 +  ’ 0 D)+  1 X i ^ ^^ ^ If t-statistics is significant in  ’ 0, there is different in constant term. =>same  1 means two categories of X have the same relationship with Y ^ ^ Check the t-value

15. 7.14 H 0 :  ’ 0 = 0 H 1 :  ’ 0 > 0 or H 1 :  ’ 0  0 Appropriate test is the t-test on  ’ 0 ^ Comparet c and t *, N-K 2  If t * > t c ==> reject H 0 :  ’ 0 = 0 Y =  0 +  ’ 0 D+  1 X i +  ’ 1 DX ^^ ^ ^ ^ Check t-statistics = This part is testing the difference of intercept This part is testing The difference of slope in two categories Check t-statistics =

16. 7.15 Separate Examples for female and male: Female Male The two regression results performed differently in slope and intercept. But are they really statistically different? We cannot answer from these two separate regression results unless you test with the F*.

17. 7.16 Y i = (  0 +  ’ 0 D)+  X i ^ ^^ ^ = (17.937-1.2810) + 1.561X D1:Female =1 others = 0 D2:Male =1 others = 0 = (16.656+1.2810) + 1.561X Y i = (  0 +  ’ 0 D)+  X i ^ ^^ ^ =17.937=16.656  If the dummy were significant  Set two different dummies for the Example

18. 7.17 Y i =  0 +  1 X i ^ ^ ^ = 17.095+1.608X Whole Sample

19. 7.18 D1: Female =1 Male: Y =  0 +  1 X i ^ ^ = 18.689 + 1.373 X Female: Y = (  0 +  ’ 0 D)+ (  1 +  ’ 1 D)X ^ ^ ^ ^ = 16.255 +1.677 X =0 = 18.689 + 1.373 X

20. 7.19 D2: Male =1 Female: Y =  0 +  1 X i ^ ^ =16.255 + 1.677 X Male: Y = (  0 +  ’ 0 D)+(  1 +  ’ 1 D )X ^ ^ ^ ^ =18.689 + 1.373 X =0 =16.255 + 1.677 X

21. 7.20 One qualitative variable with more than two categories (Health care) =  0 +  ’ 0 D 2 +  ’’ 0 D 3 +  Income +  (Y) (X) D 2 = 1 if high school education = 0 otherwise D 3 = 1 if college education = 0 otherwise 2

22. 7.21 Health care income Less than high school education Y =  0 +  X ^ ^^ 00 ^ High school education Y = (  0 +  ’ 0 D 2 )+  X ^ ^ ^ ^ D 2 = 1 ’0’0 ^ D 3 = 1 College education Y = (  0 +  0 D’’ 3 )+  X ^ ^^ ^  ’’ 0 ^

23. 7.22 D 2 = 1 High school = 0 otherwise D 3 = 1 College = 0 otherwise ========================================= obsYXD2D3 ========================================= 16.00000040.000000.0000001.000000 23.90000031.000001.0000000.000000 31.80000018.000000.0000000.000000 41.90000019.000000.0000000.000000 57.20000047.000000.0000001.000000 63.30000027.000001.0000000.000000 73.10000026.000001.0000000.000000 81.70000017.000000.0000000.000000 96.40000043.000000.0000001.000000 107.90000049.000000.0000001.000000 111.50000015.000000.0000000.000000 123.10000025.000001.0000000.000000 133.60000029.000001.0000000.000000 142.00000020.000000.0000000.000000 156.20000041.000000.0000001.000000 =========================================

24. 7.23

25. 7.24 Less than high school:Y i = -1.2859 + 0.1722 X i ^ Y i = (-1.2859 - 0.068 ) + 0.1722 X i ^ = -1.3539 + 0.1722 X High school: When t-value of D 2 is statistically significant Y i = (-1.2859 + 0.447 ) + 0.1722 X i ^ = -0.8389 + 0.1722 X i College: When t-value of D 3 is statistically significant = -1.2859 + 0.1722 X When t-value is not statistically significant

26. 7.25 One Qualitative variable with many categories : Example :An estimate model on three different age’s medical care expenditure Y i =  0 +  ’ 0 A 1 +  ’’ 0 A 2 +  X i +  i (t-value) whereA 1 = 1 if 55 > age > 25 = 0 otherwise A 2 = 1 if age > 55 = 0 otherwise A 1 + A 2  1 A 2 =1A 1 =10 2555

27. 7.26 Qualitative variable with many categories :(Cont.) then the estimated models are : age below 25 Y =  0 +  X ^ ^ ^ Y = (  0 +  ’ 0 A 1 )+  X ^ ^^ ^ 25 < age < 55 age > 55 Y = (  0 +  ’’ 0 A 2 )+  X ^ ^^ ^ H 0 :  ’ 0 = 0,  ’’ 0 = 0t 1 * H 1 :  ’ 0  0,  ’’ 0  0t 2 * Compare to t c p, n-k

28. 7.27 ’0’0 ^ 25 < age < 55 Y = (  0 +  ’ 0 )+  X ^ ^^ ^ ^ age > 55 Y = (  0 +  ’’ 0 )+  X ^ ^^ ^  ’’ 0 In scatter diagram : 00 ^ Y X age < 25 Y = (  0 ) +  X ^ ^ ^

29. 7.28 One Qualitative variable with many categories : Example : An estimate model on four different age’s medical care expenditure Y =  0 +  ’ 0 A 1 +  ’’ 0 A 2 +  ’’’ 0 A 3 +  1 X +  where A 1 = 1 if age > 55 = 0 otherwise A 2 = 1 if 35 < age  55 = 0 otherwise A 3 = 1 if 15 < age  35 = 0 otherwise

30. 7.29 Qualitative variable with many categories :(Cont.) The estimated models are : age  15 Y =  0 +  1 X ^ ^ ^ 15 < age  35 Y = (  0 +  ’ 0 A 3 ) +  1 X ^^^ ^ 35 < age  55 Y = (  0 +  ’’ 0 A 2 )+  1 X ^^^ ^ age > 55 Y = (  0 +  ’’’ 0 A 1 )+  1 X ^^^ ^

31. 7.30 Two qualitative variables (Y) Salary =  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  1 X +  or Y =  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  1 X +  ’ 1 D 1 *X +  ’’ 1 D 2 *X +  ’ D 1 = 1 if male = 0 otherwise sex D 2 = 1 if white = 0 otherwise race (1) Mean salary for “black” female teacher: Y =  0 +  1 X that are D 1 = 0, D 2 = 0 ^ ^ ^ (2) Mean salary for “black” male teacher: Y = (  0 +  ’ 0 D 1 ) + (  1 +  1 D 1 )X that are D 1 = 1, D 2 = 0 ^ ^ ^ ^^

32. 7.31 (3) Mean salary for “white” female teacher: Y = (  0 +  ’’ 0 D 2 ) +  1 X +  1 D 2 X that are D 1 = 0, D 2 = 1 ^ ^ ^ (4) Mean salary for “white” male teacher: ^ Y = (  0 +  ’ 0 D 0 +  ’’ 0 D 2 )+ (  1 +  ’ 1 D 1 +  ’’ 1 D 2 )X that are D 1 = 1, D 2 = 1 ^^ ^^ ^ ^^

33. 7.32 D = 1 if 1946-1954 = 0 otherwise (1955-1963) 1. Identical regression: Y =  0 +  1 X +  ’ 0 D +  ’ 1 D*X H 0 :  ’ 0 = 0 and  ’ 1 = 0 2. Parallel regression: Y =  0 +  1 X +  ’ 0 D +  ’ 1 D*X H 0 :  ’ 1 = 0 4. Dissimilar regression: Y =  0 +  1 X +  ’ 0 D +  ’ 1 D*X H 0 :  ’ 0  0 and  ’ 1  0 3. Concurrent regression: Y =  0 +  1 X +  ’ 0 D +  ’ 1 D*X H 0 :  ’ 0 = 0 Different types of dummy regression:

34. 7.33 Reconstruction (46-54): Y t = A 0 + A 1 X t +  1t Pastreconstruction (55-63): Y t = B 0 + B 1 X t +  2t Y X A 0 = B 0 1 A 1 = B 1 Identical regressions Y X A0A0 1 A1A1 Parallel regressions A 0  B 0, A 1 = B 1 B1B1 1 B0B0

35. 7.34 Y X A 0 = B 0 1 B1B1 Concurrent regressions A1A1 1 A 0 = B 0, A 1  B 1 Y X A0A0 1 A1A1 dissimilar regressions A 0  B 0, A 1  B 1 B0B0 1 B1B1

36. 7.35 Interactive effects between the two qualitative variables Spending(Y) =  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  1 income(X) +  D 1 = 1 if female = 0 otherwise sex D 2 = 1 if college graduate = 0 otherwise education Spending(Y) =  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  ’’’ 0 D 1 *D 2 +  1 income(X) +  Interaction effect:  ’ 0 = different effect of being a female  ’’ 0 = different effect of being a college graduate  ’’’ 0 = different effect of being a female with college graduate

37. 7.36 Concurrent model (or Covariance, or Slope shift model) Example : how can we test the hypothesis that the gasoline spending is different between a new car and a used car ? Let us assume that at the begin mile, there is no different between used car and new car. gas spending miles running Y X 00 ^ * * * * * * * * * * * New car Y =  0 +  1 X ^ ^^ o o o o o o o o o o used car Y =  0 +  1 X ^ Y =  0 + (  1 +  ’ 1 )X ^ ^^^ ^^

38. 7.37 The estimated relations are : used car : Y i =  0 + (  1 +  ’ 1 D) X i where D = 1 ^ ^^^ new car : Y i =  0 +  1 X i ^ ^^ == Y i =  0 +  1 X i == ^ ^ ^ or If  ’ 1  0, means the estimated slopes for cars is different. ^ Let  1 =  1 +  ’ 1 D where D = 1 if used car = 0 otherwise Now in one model : multiplicative dummy variable Y i =  0 + (  1 +  ’ 1 D) X i +  i =  0 +  1 X i +  ’ 1 D*X i +  i =  0 +  1 X i +  ’ 1 Z i +  i

39. 7.38 Test whether  ’ 1 = 0 or not ? ^ (i) Compare : (a) Y =  0 +  1 X ^ ^ ^ (b) Y =  0 +  1 X ^ ^^ Two separate models (ii) use t-test on  ’ 1 :Y =  0 +  1 X i +  ’ 1 Z ^ ^^^^ compare t c P, N-3 and t * H 0 :  ’ 1 = 0 ^ H 1 :  ’ 1 > 0 ^ If t* > t c P, N-3 => reject H 0 or (  ’ 1  0)

40. 7.39 Check the t-value …... Y =  0 +  1 X i +  ’ 1 Z i ^ ^ ^^

41. 7.40 Shifts in both intercept and slope Example: Estimating Seasonal effects : E =  0 +  1 T +  E : electricity consumption T : temperature To capture effect of seasonal factors E =  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  ’’’ 0 D 3 +  1 T +  where D 1 = 1 if winter 0 otherwise D 2 = 1 if spring 0 otherwise D 3 = 1 if summer 0 otherwise springsummerfallwriter Q1Q2Q3Q4 Control group

42. 7.41 The estimated models : Fall E =  0 +  1 T ^ ^ ^ Spring E = (  0 +  ’’ 0 )+(  1 +  ’’ 1 ) T ^ ^^ ^ ^ Winter E = (  0 +  ’ 0 )+ (  1 +  ’ 1 ) T ^ ^ ^ ^ ^ Summer E = (  0  ’’’ 0 )+(  1 +  ’’’ 1 ) T ^ ^ ^ ^^ 00 ^ T E E =  0 +  1 T (Fall) ^ ^ ^ E = (  0 +  ’ 0 )+(  1 +  ’ 1 )T(winter) ^ ^^ ^^ ’0’0 ^ E = (  0 +  ’’ 0 )+(  1 +  ’’ 1 )T (Spring) ^ ^ ^ ^  ’’ 0 ^  ’’’ 0 E=(  0 +  ’’’ 0 )+(  1 +  ’’’ 1 )T(Summer) ^ ^ ^ ^ ^ ^

43. 7.42 Estimating Seasonal effects :(Cont.) Also consider the slope in different seasons Let  =  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  ’’’ 0 D 3 Thus, the full general specification is E = [  0 +  ’ 0 D 1 +  ’’ 0 D 2 +  ’’’ 0 D 3 ]+  1 T +  ’ 1 D 1 T+  ’’ 1 D 2 T +  ’’’ 1 D 3 T +  Z1Z1 Z2Z2 Z3Z3

44. 7.43 Quarterly effect is same as seasonal effect D1 = 11st Quarter = 0otherwise D2 = 12nd Quarter = 0otherwise D3 = 13rd Quarter = 0otherwise Control quarter is the 4th quarter

45. 7.44 1. Set the seasonal dummy = 1 if there is the 1st quarter = 0 otherwise

46. 7.45 How does the quarterly dummy variable look like?

47. 7.46 (2) Structural Test based on Dummy variables Basic model Y T =  0 +  1 X T +  T 1974 1960 1989 Define a dummy variable :D = 1 for the period 1974 onward = 0 otherwise To test whether the structures of two periods are different, the specification must assume that  * =  0 +  ’ 0 D  * =  1 +  ’ 1 D Dummy regression: Y T =  0 +  ’ 0 D +  1 X T +  ’ 1 D X T +  T

48. 7.47 The Chow test on the Unemployment rate-capacity utilization rate Dependent Var.ConstantCAP t R 2 FRSSn _ Sample : 60 - 89 unempl t 30.0-0.2930.76193.617.1530 (12.1) (9.7) RSS R ^ Sample : 60 - 73 unempl t 19.64-0.1750.5919.74.6914 (5.9) (4.4) RSS 1 ^ Sample : 74 - 89 unempl t 30.63-0.2960.871102.13.2916 (13.1) (10.1) RSS 2 ^ Note : t-values are in parentheses

49. 7.48 No structural change H 0 : No structural change Yes H 1 : Yes For the unrestricted model : RSS u = RSS 1 + RSS 2 = 4.69 + 3.29 = 7.98 F * = (RSS R - RSS u ) / k+1 RSS u / (N - 2k-2) = (17.15 - 7.98) / 2 7.98 / (30 - 4) = 14.9 F * > F c ==> reject H 0 F c 0.01, k, T -2k = F c 0.01 = 5.53 0.050.05, 2, 26 = 3.37 Restriction F-test procedures:

50. 7.49 The unemployment rate - capacity utilization rate Sample : 1960 - 1989 D t = 11974 to 1980 = 0prior to 1974 unempl = 19.6 + 11.0 D t - 0.175 CAP t - 0.121 (D t *CAP t ) ^ (6.7)(2.7)(5.0) (2.5) R 2 = 0.88 SEE = 0.554 F = 72.2 n = 30 _ The estimated of 1974-1980: unempl = (19.6+11.0) - (0.175+0.121)CAP = 30.6 - 0.296 CAP ^ ^ The estimated of 1960-1973: unempl = 19.6 - 0.175 CAP Using the dummy variable to identify the structural change

51. 7.50 D = 1 if t  74 = 0 otherwise Observed data YearU t CAP t D t D t *CAP t 604.205.70 0 0 61 0 0 62 0 0 63 0 0 … …... 68 0 0 69 0 0 70 0 0 71 0 0 72 0 0 73 0 0 74 1 75 1 76 1 77 1... 1 89 1 … ………….…… … …...….…….... 10.5 11.2 10.5 11.2 U t =  0 +  1 CAP t +  ’ 0 D t +  2 D t *CAP t

52. 7.51 (2) Structural stability test based on dummy variables The estimated models are : 1974 : 1 and onward Y =  0 +  1 X ^ ^ ^ Now the basic model becomes Y T =  0 +  ’ 0 D +  1 X T +  ’ 1 D X T +  T Y T =  0 +  ’ 0 D +  1 X T +  ’ 1 X * T +  T ==> ===== t t-test on  ’ 1 = 0 ^ 1974 : 1 1950 1995 Prior to 1974 : 1 Y = (  0 +  ’ 0 D)+(  1 +  ’ 1 D) X ^ ^^ ^ ^ ** **

53. 7.52 GENR DUMMY = 1 (sample 1970 - 1980) GENR DUMMY = 0 (sample 1981 - 1991) ================================================= obsSAVINGSINCOMEDUMMY D*INCOME ================================================= 197057.50000831.00001.000000831.0000 197165.40000893.50001.000000893.5000 197259.70000980.50001.000000980.5000 197386.100001098.7001.0000001098.700 197493.400001205.7001.0000001205.700 1975100.30001307.3001.0000001307.300 197693.000001446.3001.0000001446.300 197787.900001601.3001.000000 1601.300 1978107.80001807.9001.0000001807.900 1979123.30002033.1001.0000002033.100 1980153.80002265.4001.0000002265.400 1981191.80002534.7000.0000000.000000 1982199.50002690.9000.0000000.000000 1983168.70002862.5000.0000000.000000 1984222.00003154.6000.0000000.000000 1985189.30003379.8000.0000000.000000 1986187.50003590.4000.0000000.000000 1987142.00003802.0000.0000000.000000 1988155.70004075.9000.0000000.000000 1989175.60004664.2000.0000000.000000 1990175.60004664.2000.0000000.000000 1991199.60004828.3000.0000000.000000 =================================================

54. 7.53 Savings =  0 +  1 Income +  ’ 0 D +  ’ 1 D*Income +  D = 1 1970--1980 = 0 1981--1991 Estimated for 1970 - 1980 : D = 1 Savings = (  0 +  ’ 0 ) +(  1 +  ’ 1 ) Income ^^ ^^ 1 Estimated for 1981 - 1991 : D = 0 Savings =  0 +  1 Income ^^ 2 Dummy Regression Results: 1970 - 1991 : Savings = 217.81 - 203.19 D - 0.010 Income + 0.066 D*Income (7.96)(-6.19)(-1.39)(4.63)

55. 7.54 70 - 80: Savings = (217.81 - 203.19) + (-0.010 + 0.066) Income = 14.62 + 0.056 Income 81 - 91: Savings = 217.81 - 0.010 Income 1970 - 1991 : Savings = 57.63 + 0.031 Income (3.86)(5.95) 1970 - 1980 : Savings = 14.61 + 0.056 Income (1.40)(7.93) 1981 - 1991 : Savings = 217.81 + 0.010 Income (6.16)(-1.08)

56. 7.55 Savings = -1.250 + 0.091 Dummy + 0.125 Income (-3.42)(0.506)(7.04) LS // Dependent Variable is SAVINGS Date: 03/02/99 Time: 22:23 Sample: 1946 1963 Number of observations: 18 ===================================================== Variable Coefficient Std. Error t-Statistic. Prob. ===================================================== C-1.2509570.364879-3.4284190.0037 DUMMY 0.0918570.181244 0.5068160.6197 INCOME -0.125655 0.017837-7.044517 0.0000 ===================================================== R-squared 0.919909 Mean dependent var 0.773333 Adjusted R-squared 0.909230 S.D. dependent var 0.642806 S.E. of likelihood 0.193665 Akaike info criterion -3.132238 Sum squared resid 0.562593 Schwarz criterion -2.983843 Log likelihood 5.649250 F-statistic 86.14326 Durbin-Watson stat 0.976197 Prob(F-statistic) 0.000000 ===================================================== Only consider the difference in intercept

57. 7.56 LS // Dependent Variable is SAVINGS Date: 03/02/99 Time: 22:23 Sample: 1946 1963 Number of observations: 18 ===================================================== Variable Coefficient Std. Error t-Statistic. Prob. ===================================================== C-1.7501720.331888-5.2733770.0001 DUMMY 1.483923 0.470362 3.1548520.0070 INCOME 0.150450 0.016286 9.238172 0.0000 DINCOME -0.103422 0.033260-3.1094710.0077 ===================================================== R-squared 0.952626 Mean dependent var 0.773333 Adjusted R-squared 0.942475 S.D. dependent var 0.642806 S.E. of likelihood 0.154173 Akaike info criterion -3.546228 Sum squared resid 0.332771 Schwarz criterion -3.348367 Log likelihood 10.37516 F-statistic 93.84109 Durbin-Watson stat 1.468099 Prob(F-statistic) 0.000000 ===================================================== Whether intercept and slope change?

58. 7.57 1946 - 1954 : = -0.2662 + 0.047 Income D1 = 1 1955 - 1963 : = -1.750 + 0.150 Income D1 = 0 Savings = -1.750 + 1.483 D + 0.150 Income - 0.103 (Income * D) (-5.273) (3.154) (9.238) (-3.109) ^

59. 7.58 The use of Dummy variables in the Pooled data 2. Dummy variable method : (i) Y =  0 +  1 X 1 +  2 X 2 +  3 D +  H 0 :  0 =  0 ’ H 1 :  0   0 ’ D = 1 for GM = 0 otherwise == if t  3 > t c ==> reject H 0 ^ Panel = time-series + cross-section 1. For each firm, run the separated regression : GM : Y =  0 +  1 X 1 +  2 X 2 +  Y =  0 ’ +  1 ’ X 1 +  2 ’ X 2 +  ’ Westinghouse : H 0 :  1 =  1 ’,  2 =  2 ’,  0 =  0 ’

60. 7.59 (ii) Y =  0 +  1 X 1 +  2 X 2 +  3 D +  4 D X 2 +  5 D X 3 +  H 0 :  1 =  1 ’ H 1 :  1   1 ’ == if t  4 > t c ==> reject H 0 ^ H 0 :  2 =  2 ’ H 1 :  2   2 ’ == if t  5 > t c ==> reject H 0 ^

61. 7.60 “Chow Test” - structural stability Test Using dummy variables approach H 0 : no structural change H 1 : yes Procedures: Generate dummy variable Generate dummy variable : D1 = 0 for 1946-1954 D1 = 1 for 1955-1963 or D1 D1 = 1 for 1946-1954 D2 D2 = 0 for 1955-1963

62. 7.61 GENR dummy = 0 for 1946-1954 dummy = 1 for 1955-1963

63. 7.62

64. 7.63 Using the dummy variable to identify the structural instability Check the t-statistics Generate a dummy series “DUMMY”

65. 7.64

66. 7.65 Slope change? Intercept change?

67. 7.66 Read the estimated results from the dummy regressions: 1955-1963 For the period of 1955-1963: savings = (-0.2662 - 1.4839) + (0.0470 + 0.1034) = -1.7501 + 0.1504 Income 1946-1954 For the period of 1946-1954: Savings = -0.2662 + 0.0470 Income