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ARE106SU09HW3KEY

# ARE106SU09HW3KEY - 1 University of California Davis...

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1 University of California, Davis Managerial Economic(ARE) 106, Summer 2009 Instructor: John H. Constantine KEY —Problem Set #3: Due Monday, August 24, 2009 Problem 1 : This problem requires the use of SHAZAM. Consider the problem of estimating a production function that expresses the relationship between the level of output of a product ( y ) and the level of an input, or factor of production ( x ). The SHAZAM command file is given below. When necessary, assume α = 5%. sample 1 15 read(are106/table3-2.dat) x y print x y plot ols y x stop (a) Assume that the data can be described the simple linear regression model: y = β 1 + β 2 x + e . Use SHAZAM estimate b 1 and b 2 . State the associated t statistics. b 1 = 0.4656; t b1 = 3.692 b 2 = 0.29243; t b2 = 22.63 (b) Give an economic interpretation of the estimated parameters. Strictly speaking, the intercept coefficient, b 1 = 0.4656, gives the level of output when the feed input is zero. However, because a zero feed input is unrealistic, this interpretation must be treated with caution; it would be expected that, when the feed input is zero, output would be zero. The slope coefficient, b 2 = 0.29246, indicates that, for a 1 unit change in the feed input, there will be a corresponding change of 0.29 in the poultry food output. (c) Draw your estimated OLS line through the data plot.

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2 (d) (i) Find the confidence interval for b 2 . (ii) Draw a picture and show the regions of “Reject H 0 ” and “Do Not Reject H 0 ”. (iii) Clearly explain what this confidence interval means. The SHAZAM results for b 1 and b 2 are given here : VARIABLE ESTIMATED STANDARD T-RATIO PARTIAL STANDARDIZED ELASTICITY NAME COEFFICIENT ERROR 13 DF P-VALUE CORR. COEFFICIENT AT MEANS X 0.29246 0.1292E-01 22.63 0.000 0.988 0.9875 0.8340 CONSTANT 0.46562 0.1175 3.962 0.002 0.740 0.0000 0.1660 (i) [ ] % 100 * ) 1 ( ) b ( se t b ) b ( se t b Prob 2 c 2 HO 2 2 c 2 α = + β Or, in shorthand : ) b ( se t b 2 c 2 ± t-critical = 16 . 2 t t 13 025 . 0 df 2 / = = α The interval estimate for β 2 : 0.29246 ± 2.16(0.01292) = (0.2645, 0.3204) (ii) DRAW GRAPH HERE. (iii) If we use the interval estimators to compute a large number of interval estimates like these, in repeated samples, 95% of these intervals will contain β 2 as stated under H 0 . (e) Using the Confidence Interval approach, you are to test the following hypotheses. Draw β 2 HO on your picture in part (d, ii) for each hypothesis. (i) H 0 : β 2 = 0 (ii) H 0 : β 2 = 0.30 (iii) H 0 : β 2 = 0.50 (i) H 0 : β 2 = 0 vs. H A : β 2 0 . Since β 2 = 0 lies in the rejection portion of the graph, we reject that hypothesis that β 2 = 0. (ii) H 0 : β 2 = 0.30 vs. H A : β 2 0.30 . Since β 2 = 0.30 lies in the do not reject portion of the graph, we cannot reject that hypothesis that β 2 = 0.30. (Note: This does not mean that β 2 = 0.30 exactly but that we cannot reject that it is possible.) (iii) H 0 : β 2 = 0.50 vs. H A : β 2 0.50 . Since β 2 = 0.50 lies in the rejection portion of the graph, we reject that hypothesis that β 2 = 0.50. (f) Now, conduct the three hypotheses of part (e), but this time using the t-test approach.
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ARE106SU09HW3KEY - 1 University of California Davis...

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