Question

help wih these question please

3. Consider the following autoregressive process Yt = Bo + B1yt-1 + B2Yt-2 + Ut, where E (UtYt-1, Yt-2, ...) = 0. You obtained the following parameter estimates: Bo= -0.2, B1 = 0.4 and B2 = -0.1. Furthermore, you have the following observations: 419 = -0.2 and Y20 = 0.3. What is the estimate for E Y 22 y 20,419)? (a) -0.3333 (b) -0.06 (C) 0.3 (d) -0.2857 (e) -0.254 4. You have estimated the following regression model on the basis of 213 observations: gri n 3 = 27.34 + 2.97 sthi - 2.04 awbci (2.64) (0.43) (1.15) = 104, R2 = 0.29 = 3.6, DW = 2.06 where gr is the grade in ECON 20110, sth is the average number of weekly hours spend on studying and awbc is the average weekly beer consumption measured in pints. In parenthesis (you can find the usual standard errors. On average all students study 8 hours and drink 3 pints per week. Jimmy studies on average 11 hours per week and drinks an average of 4.3 pints of beer a week. What is Jimmy's predicted grade in ECON 20110? (a) 28.27 (b) 27.34 (c) 44.98 (d) 54.84 (e) 51.24

Answer 1

3) Since the estimate of the parameters are given the estimated regression model is:

Ey)= y = -0.2 + 0.4 % yt-1- 0.1 % yt-2

Now it is given that y19 = – 0.2 and y20 = 0.3

Therefore,

EY21) = -0.2 + 0.4 * Y20 - 0.1 * 719

         Or,

Yz1 = -0.2 +0.4*0.3 -0.1 *(-0.2) = -0.06

Thus,     

EY22) = -0.2 + 0.4 * Y21 -0.1 * 720

        Or,

EY22) = -0.2 +0.4*(-0.06) -0.1*(0.3) = -0.254

So, option (e) - 0.254 is the correct one.

4) The estimated regression model is:

            

gr, = 27.34 + 2.97 * sthi - 2.04 * awbc;

Since Jimmy studies on average 11 hours per week so for Jimmy sth = 11 and he drinks on average 4.3 pints of beer a week so awbc = 4.3

Therefore, Jimmy’s predicted grade in ECON20110 is:

  

gr = 27.34 + 2.97 * 11 - 2.04 * 4.3 = 51.24

So, option (e) 51.24 is the correct one.