? ? Some preliminaries to set up the data ? Some preliminaries to set up the data ? These commands obtain the group (farm) means of the 4 inputs, ? then computes the logs of the group means. ? SETPANEL ; Group = farm ; Pds = t \$ NAMELIST ; (new) ; means=cowsbar,landbar,laborbar,feedbar \$ NAMELIST ; factors = cows,land,labor,feed \$ CREATE ; means = Group Mean (factors,pds=t)\$ CREATE ; milkbar=Group Mean (milk,pds=t) \$ CREATE ; yb=log(milkbar) ; x1b=log(cowsbar) ;x2b=log(landbar) ; x3b=log(laborbar) ;x4b=log(feedbar)\$ NAMELIST ; cobbdgls = one,x1,x2,x3,x4 \$ NAMELIST ; quadrtic =x11,x22,x33,x44,x12,x13,x14,x23,x24,x34 \$ NAMELIST ; translog = cobbdgls,quadrtic \$ NAMELIST ; time = year93,year94,year95,year96,year97 \$ Omit year98 ? ? Part 1. Examine the data ? DSTAT ; Rhs = * \$ ? Part 2. Conventional and Robust standard errors. ? The standard errors are almost identical. But, in spite of ? this, the LM statistic is 65.07 with only 4 degress of freedom, ? so the hypothesis of homoscedasticity is rejected anyway. ? REGRESS ; Lhs = yit ; Rhs = cobbdgls ; Table = OLS\$ REGRESS ; Lhs = yit ; Rhs = cobbdgls ; Het ; Table = White\$ REGRESS ; Lhs = yit ; Rhs = cobbdgls ; Cluster = 6 ; Table = Cluster \$ MAKETABLE ; OLS, White, Cluster ; StandardErrors\$ ? ? Part 3. The LAD and QUANTILE estimators ? For the coefficients that are highly significant (ONE,X1,X4), ? the LAD results are almost the same. For the others, they ? differ quite a bit. Though this makes intuitive sense, it ? is hard to justify. The standard errors of the LAD estimator ? seem generally smaller, again, contrary to expectations. ? REGRESS ; Lhs = yit ; Rhs = cobbdgls; Alg = LAD ; NBT = 25 \$ QREG ; LHS = YIT ; RHS = COBBDGLS ; Qnt = .25 \$ ? Now repeat with different values for the quantile to see if the ? regression coefficients are sensitive to the choice of quantile. QREG ; LHS = YIT ; RHS = COBBDGLS ; Qnt = .25 ; table = quant25\$ QREG ; LHS = YIT ; RHS = COBBDGLS ; Qnt = .50 ; table = quant50\$ QREG ; LHS = YIT ; RHS = COBBDGLS ; Qnt = .75 ; table = quant75\$ MAKETABLE ; quant25,quant50,quant75 \$ ? ? Part 4. Hypothesis testing. ? First run the regression, putting the time variables first in ? the list. This makes it convenient, but it is not necessary. REGRESS ; Lhs = yit ; Rhs = time,cobbdgls \$ CALC ; ru = rsqrd ; dfn = col(time) ; dfd = n-kreg \$ REGRESS ; Lhs = yit ; Rhs = cobbdgls \$ CALC ; rl = rsqrd ; list ; f = ((ru-rl)/dfn)/((1-ru)/dfd) ; fc = ftb(.95,dfn,dfd) \$ ? ? Now use matrix algebra to compute the Wald statistic. The ? chi squared value is 9.377. The statistic has 5 degrees of freedom. ? Is the hypothesis rejected? Execute the command in lower case below ? to get the critical value from the chi squared table. ? What do you find? What is the conclusion. ? REGRESS ; Lhs = yit ; Rhs = time,cobbdgls \$ MATRIX ; bt = b(1:5) ; vt = varb(1:5,1:5) ; list ; w = bt'bt \$ CALC ; list ; cc = ctb(.95,5) \$ ? The built in calculation reports both F and chi squared. ? REGRESS ; Lhs = yit ; Rhs = time,cobbdgls ; Test: time \$ ? ? Part 5. Test the null hypothesis of the translog model against the ? alternative of the Cobb-Douglas (with time variables included.) ? The resulting chi squared statistic is 41.954. The critical ? value is 18.307, so the hypothesis is rejected. ? REGRESS ; Lhs = yit ; Rhs = translog,time ; Test: quadrtic \$ ? Part 6. Testing for constant returns to scale. The test is of the hypothesis ? that the four output elasticities sum to one. ? The first regression does not impose the restriction. ? The second computes the conventional F statistic. The value is ? 55.67. There is one restriction, so the critical value is 3.84 ? (square of critical t value with 1477 degrees of freedom which is ? standard normal, or 1.96). ? REGRESS ; LHS = YIT ; RHS = One,X1,X2,X3,X4 \$ REGRESS ; LHS = YIT ; RHS = One,X1,X2,X3,X4 ; Test: x1 + x2 + x3 + x4 = 1 \$ ? ? A second way is to use the WALD command for computing standard ? errors and hypothesis tests for functions of parameters. The two WALD ? commands do exactly the same thing. They show two ways to form the ? command. The chi squared value is 55.67, which is larger than the critical ? value of 3.84, so this hypothesis is rejected. ? WALD ; Fn1 = b_X1+b_X2 + b_X3 + b_X4 - 1 \$ WALD ; Start = B ; Var = VARB ; Labels = b0,b1,b2,b3,b4 ; Fn1 = b1+b2+b3+b4 - 1 \$ ? Constant returns in the translog model involves several restrictions. ? The first order terms must sum to 1.0 and each row of the matrix of ? second order terms must sum to zero. Here is how to impose the ? restrictions and test the hypothesis at the same time. REGRESS ; Lhs = yit ; Rhs = Translog ; CLS: x1+ x2+ x3+ x4 = 1, x11+x12+x13+x14 = 0, x12+x22+x23+x24 = 0, x13+x23+x33+x34 = 0, x14+x24+x34+x44 = 0 \$ ? Part 7. Look for normally distributed disturbances by examining residuals. ? ? First compute the residuals. Then, look at a kernel density estimate. ? The distribution looks somewhat bell shaped, but there is a noticeable skew. ? Maybe they are not normally distributed. The chi squared statistic ? is 81.98, with two degrees of freedom. The critical value ? is 5.99 (Find this with CALC;List;CTB(.95,2)\$) So, we will reject the ? hypothesis of normality. ? REGRESS ; Lhs = yit ; Rhs =translog; Res = u \$ KERNEL ; Rhs = u; normal ; Grid ; Title=OLS Residuals and Normal with Same Mean and SD\$ DSTAT ; Rhs=u ; all ; normal test\$ CREATE ; u2 = u*u ; u3 = u2*u ; u4 = u2*u2 \$ CALC ; su = sqr ( xbr ( u2 )) ; m3 = xbr( u3 ) ; m4 = xbr ( u4) \$ CALC ; List ; c = n *( ( m3/su^3)^2 / 6 + (m4/su^4 - 3)^2 /24 ) \$