Question

Question 1: (August 2014/2015) Based on the data for 1995-15Q to 2013-4thQ (n=76), an economist obtained the following regression model to explain the personal consumption expenditure (PCE) in country XYZ. R2 = 0.9996 T, =-10.96 +0.93X 2: -2.09X3 t= (-3.33) (249.06) (-3.09) where Y = the PCE (RM in billions) A2 = the disposable income (RM in billions) 43 = the prime rate charged by banks (%) Determine the marginal propensity to consume (MPC). (b) Is the Marginal Propensity to Consume statistically different from 1 at a = 5%? Show the appropriate testing procedure. Discuss the rationale for the inclusion of prime rate variable in the model. A priori, would you expect a negative sign for this variable? (d) Is X, significantly different from zero at a = 1%? Explain. Compute the standard error of each coefficient. Calculate the adjusted coefficient of determination (R). Using t-statistic, test individually whether the slope coefficients are significantly different from zero at a. = 1%.

Answer 1

het - Pot P, Xt BX = -10-96 +0.93 X-2.09 X3t (a) MPC = 39- 8- 0.93 since Y=PCE amd X2=disposable income . (b) = 249.06 249.06 fer Ho B-0 and H: B, #0. > SECB) 20.003734. %3D 0.93 -> 0.93-0 SECPI) Now for Ho: MPC = B,=1 and Hi P,71 => SE CP) = 349. 06 249.06 0.93-1 S.E CA) - D.07 %3D = 18.7465 = - 18.746S -> O. 003734 The Criteca value at d =5% and df=n-2 76-2=74, is 1.99 Since the calaulated value (18.7465) > critical value (1.99), we reject Ho and conclude that MPC = B, is statistically dibbercent trom 1. (c) Consumption and saving consumplion falk and vice vensa. On the other hand, saving rises when preime rate falls and nice veresa. In other words, when paime rate increases, consumption falls omd vice veesa. Hence, we expect a negati re sign for Xg. (4) P Ho: B,=0, Hi Pe 70. Here, lt = 3.69. are complementary, i.e.cwhen saving reises, Ccritical value at d=1% with m-2=674 degnee of freedem = 2.64 >We reject Hto and conclude that P, is statishicaly dibberzent fromo at a=1%. (e) Ho: B.=P, - B =0 and Ite= 3:33, te= 249.06 , Itp, = 3.09. Now Itl lfol-0 SE (AD) lo 46 10.96 = 3-33 > SECP.) = 3. 291291. SECA)= 3.33 0.93 249.06=0.003734 (calculated in (b)), Similary, SE (Â)= Pi/ta, and SE CB)= P te, %3D -2.09 - 0.676375 %3D Cf) Adjusted R (R?)= (I-R°) (n-1)where n= # obserevations and Ln-p-1 [1-0.9996)(76-1) 999 = 0.999589 -1 P= # predictros 2) R. 76-2 -1 (9) Crikcal value at d = 1% with dff 76-2 )74 is 2 64 Since t-stats for all the sepe coefficients (P and B.) azei.e.249.06 and 3.09 are greatee than lhi eritical value, the Hb: Bi=P2=D0i rejected and concluded that they are statistically ditfurtent trom zezo at d = 1%