i) the probability of type I error; ii) the 95% con dence interval; iii) the p value; iv) the probability of type II error; v) the power of a test.

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1 Problem Set 5. Questions:. Exlain what is: i) the robability of tye I error; ii) the 95% con dence interval; iii) the value; iv) the robability of tye II error; v) the ower of a test.. Solve exercise 3. in Johnston and Dinardo: Q: The following regression equation is estimated as a roduction function for ln(q) :37 + :63 ln(k) + :45 ln(l); (:57) (:9) R :98; Cov(b k ; b l ) :55; where the standard errors are given in arentheses. Test the following hyotheses: i) The caital and labor elasticities of outut are identical. ii) There are constant returns to scale. 3. Solve exercise 3. in Johnston and Dinardo: One asect of the rational exectation hyothesis involves the claim that exectations are unbiased, that is, that the average rediction is equal to the observed realization of the variable under investigation. This claim can be tested by reference to announced redictions and to actual values of the rate of interest on three-month U.S. Treasury Bills ublished in the Goldsmith-Nagan Bond and Money Market Letter. The results of least-squares estimation (based on 3 quarterly observations) of the regression of the actual on the redicted interest rates were as follows: r t :4 (:86) + :94 (:4) r t + e t ; RSS 8:56; where r t is the observed interest rate, and rt is the average exectation of r t held at the end of the receding quarter. Figures in arentheses are estimated standard errors. The samle data on r give P t 3 ; (r t r t ) 5: Carry out the test assuming that all basic assumtions of the classical regression model are satis ed.

2 4. Solve exercise 3.9 in Johnston and Dinardo: An economist is studying the variation in fatalities from road tra c in different states. She hyothesizes that the fatality rate deends on the average seed and the standard deviation of seed in each state. No data are available on the standard deviation, but in a normal distribution the standard deviation can be aroximated by the di erence between the 85th ercentile and the 5th ercentile. Thus, the seci ed model is Y b + b X + b 3 (X 3 X ) + u; where Y fatality rate; X average seed; X 3 85th ercentile seed. Instead of regressing Y on X and (X 3 X ) as in the seci ed model, the research assistant ts with the following results: Y a + a X + a 3 X 3 + u; by constant :4X + :X 3 ; with R :6. The (absolute) t statistics for the two sloe coe cients are :8 and :3, resectively, and the covariance of the regression sloes is :3. Use these regression results to calculate a oint estimate of b and test the hyothesis that average seed has no e ect on fatalities.

3 Answers:. i) It is the robability of rejecting the null when the null is true. iv) It is the robability of failing to reject the null when the null is false. v) It is one minus the robability of tye II error.. i) The caital(labour) elasticity, that is the elasticity of outut Q with resect to caital(labour) K(L), is given by e Q K d ln(q) d ln(k) b k; e Q L d ln(q) d ln(l) b l: Hence, to test the hyothesis that e Q K eq L that b k b l : The t-test is given by is equivalent to test the hyothesis t ( k l ) SE( k l ) : The standard error in the dominator is given by: SE( k l ) Var( k l ) Var( k ) + Var( l ) Cov(b k ; b l ) :57 + :9 :55: Thus t :63 :45 :57 + :9 :55 :846: In order to get the critical values we need the samle size. For a large samle size the 5% critical value is.96. Thus we reject the null hyothesis. ii) Similarly, to test the hyothesis that there are constant returns to scale is equivalent to test the hyothesis that b k + b l. The t-test is given by t ( k + l ) SE( k + l ) ( k + l ) Var(k ) + Var( l ) + Cov(b k ; b l ) (:63 + :45) :57 + :9 + :55 :78: Thus, we fail to reject the null hyothesis. 3. Under the rational hyothesis, the regression r t b + b r t + e t ; 3

4 should yield b and b. We can write these two hyotheses in a matrix form as Rb r; where R I; b b Thus we can carry out this join test by the F b ; r test : (R r) [s R(X X) R ] (R r) F (; n k): Since R I, that is R is an identity matrix, we have Next, ( r) [s (X X) ] r is given by ( r) F (; n k) or ( r) s (X X)( r) s F (; n k): :4 :94 :4 :6 : Moreover, X X is given by P N r P P t r t (r t ) : We know that: It follows that t 3 ; t r t ) 5: t 3; N 3; t r t ) t ) N(r t ) ) t ) t r t ) + N(r t ) : In addition, Thus :4 :6 s RSS N k 8:56 8 : ( r) s (X X)( r) s 3 3 : :6 8:568 :998: 4

5 The 5% critical value of F (; 8) is So we fail to reject the null hyothesis. 4. We rewrite the initial regression as Y b + (b b 3 )X + b 3 X 3 + u: This is the regression the research assistant estimated. Comaring the coe - cients, we have :4: To carry out the test we need an estimate of the variance of : Var( ) Var( + 3 ) The variance of is given by Similarly, Thus the variance of is Var( ) ( Var( ) + Var( 3 ) + Cov( ; 3 ): t SE( ) ) SE( ) t ) Var( ) ( t ) ( :4 :8 ) : Var( 3 ) ( 3 t 3 ) ( : :3 ) : :4 :8 ) + ( : :3 ) + :3 :33: The test statistic for the signi cance of b is then t SE( ) :4 :33 :6: This is not signi cant at conventional signi cance levels and for any samle size. We can not reject the hyothesis that average seed has no e ect on fatalities. 5