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Q: Chow test on Cobb-Douglas production function ( No Answer,   1 Comment )
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 Subject: Chow test on Cobb-Douglas production function Category: Science > Social Sciences Asked by: permanent_student-ga List Price: \$5.00 Posted: 28 May 2006 11:54 PDT Expires: 27 Jun 2006 11:54 PDT Question ID: 733094
 ```I am modeling a production function based on a time series data for 40 periods. I have Capital and Labor inputs so I used a cobb-douglas function. I cannot use CES because it is difficult if not impossible to find rent and wage data. I would like to perform a Chow test in SAS to examine the data to see if the results are meaningful. I know how to do an autoregression in SAS and use the Chow option. The problem is that I don't know how to interpret the results. How do I interpret the results of a Chow test?``` Clarification of Question by permanent_student-ga on 28 May 2006 12:45 PDT ```Below is the output from SAS using proc autoreg /chow. The SAS System 12:37 Sunday, May 28, 2006 1 The AUTOREG Procedure Dependent Variable logoutput Ordinary Least Squares Estimates SSE 0.03573462 DFE 36 MSE 0.0009926 Root MSE 0.03151 SBC -151.14479 AIC -156.13547 Regress R-Square 0.5388 Total R-Square 0.5388 Durbin-Watson 1.5906 Structural Change Test Break Test Point Num DF Den DF F Value Pr > F Chow 20 3 33 3.39 0.0295 Standard Approx Variable DF Estimate Error t Value Pr > |t| Intercept 1 -11.0368 5.2099 -2.12 0.0411 logcows 1 2.1147 0.3272 6.46 <.0001 loglabor 1 0.2484 0.0891 2.79 0.0084```
 ```Very first, Chow test is a application of F test. If you know how to interpret F test, you should be able to interpret Chow test. F test hypothesis looks like H_0 : X1=0,X2=1,X3=3 H_1 : H_0 is false So F statistics tests... how untruthful is the null hypothesis. It is one tailed test. (If I remember correctly, I never got confused of one-tail test or two-tail test to operate. The F-table only refers two degrees of freedom - SAS apparently gives Num DF and DEN DF.) It is really difficult to see because the output you pasted was wrapped. But SAS gave you this result: Break Test Point Num DF Den DF F Value Pr > F Chow 20 3 33 3.39 0.0295 It says: "Test was selected to be 'Chow Test'. (I don't know what Break Point refers) Numerator Degrees of Freedom is 3. Denominator Degrees of Freedom is 33. F Value of the test was 3.39. Thus the probability, Pr(F>3.39)~0.0295." So, depending on your significance level, you will find the test reject or fail to reject the null hypothesis. In the beginning, there is some descriptive statistics about errors, thus goodness of fit. The last thing I suppose is the testing of significance of each variable you tested in Chow Test...```