Ordinary least squares regression lhslogcpf mean

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---------------------------------------------------------------------- Ordinary least squares regression ............ LHS=LOGC_PF Mean = -.31956 Standard deviation = 1.54236 Number of observs. = 158 Model size Parameters = 7 Degrees of freedom = 151 Residuals Sum of squares = 2.85625 (restricted) Residuals Sum of squares = 2.56313 (unrestricted) --------+------------------------------------------------------------- Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- Constant| -7.24078*** 1.01411 -7.140 .0000 LOGQ| .45889*** .12402 3.700 .0003 8.26549 LOGQSQ| .06008*** .00441 13.612 .0000 35.7913 LOGPK_PF| .38943** .16933 2.300 .0228 .85979 LOGPL_PF| .19130 .19530 .979 .3289 5.58162 LQ_LPKPF| -.02954 .02125 -1.390 .1665 7.15696 LQ_LPLPF| -.00462 .02364 -.195 .8454 46.1956 --------+------------------------------------------------------------- F Statistic = ((2.85625 – 2.56313)/2) / (2.56313/(158-9)) = 8.52 (This is a problem. The underlying theory requires linear homogeneity in prices.) ™    48/50
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Part 8: Hypothesis Testing Wald Test of the Restrictions Chi squared = J*F --> wald ; fn1 = b_k + b_l + b_f - 1 ; fn2 = b_qk + b_ql + b_qf – 0 $ ----------------------------------------------------------- WALD procedure. Estimates and standard errors for nonlinear functions and joint test of nonlinear restrictions. Wald Statistic = 17.03978 Prob. from Chi-squared[ 2] = .00020 --------+-------------------------------------------------- Variable| Coefficient Standard Error b/St.Er. P[|Z|>z] --------+-------------------------------------------------- Fncn(1)| -1.87546*** .45621 -4.111 .0000 Fncn(2)| .21921*** .05353 4.095 .0000 --------+-------------------------------------------------- Note: ***, **, * = Significance at 1%, 5%, 10% level. ----------------------------------------------------------- Recall: F Statistic = ((2.85625 – 2.56313)/2) / (2.56313/(158-9)) = 8.52 The chi squared statistic is JF. Here, 2F = 17.04 = the chi squared. ™    49/50
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Part 8: Hypothesis Testing Test of Homotheticity Cross Product Terms = 0 Homothetic Production: β7 = β8 = β9 = 0. With linearity homogeneity in prices already imposed, this is β6 = β7 = 0. ----------------------------------------------------------- WALD procedure. Estimates and standard errors for nonlinear functions and joint test of nonlinear restrictions. Wald Statistic = 2.57180 Prob. from Chi-squared[ 2] = .27640 --------+-------------------------------------------------- Variable| Coefficient Standard Error b/St.Er. P[|Z|>z] --------+-------------------------------------------------- Fncn(1)| -.02954 .02125 -1.390 .1644 Fncn(2)| -.00462 .02364 -.195 .8451 --------+-------------------------------------------------- The F test would produce F = ((2.90490 – 2.85625)/1)/(2.85625/(158-7)) = 1.285     50/50
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