\fHello! From your answer I also do not know how you calculate Cov (e,y^)? Can just tell me how to calculate this one, ? 1) The number of number of observations, 30 is lower than the parameters in the regression equation, 40. It the amount of data, observations made is insufficient to explain the relationship between parameters that are more. This would result to errors in conclusion rendering the result of the model unviable. b) In the model y=0 + 2 x2+3 x3 + we can only determine the percentage of the dependent variable explained by the independent variable by neglecting for a while the percentage of the dependent variable explained by the x3 . This act however does not suggest that x3 does not explain the independent variable. In fact, it is possible that x3 explains a greater percentage of the dependent variable than x2. The test is only meant to give light on the significance of the variable x2 in the regression equation. c) The partial effect of x2 on y can only be equal to the total effect of x2 on y if i) x3 has no effect on y, that is if y is independent of x3 ii) both x2 and x3 have the same effect on y that is the percentage of y explained by both x2 and x3 is equal. d) Models with yt on the left hand side indicate that yt is dependent on some variables. It is possible to carry out the analysis of variance test and obtain the R 2 measure and also carry out an F-test, however when ln yt is dependent it means that lnyt is the dependent variable, it is not possible to carry out the analysis of variance and thus not possible to apply the concepts of R 2 and F-test. e) Both b2 and b3 as deduced from the regression equation are involved in explaining the dependent variable. The percentage explained can be partial all total, unless partial percentage explained by x2 is equal to the total percentage, it is clear that the two random variables are correlated because there is a percentage in which the jointly and dependently take part in explaining the dependent variable. Can u detailed introduce b and d for me I cannot understand your meaning. Thank u Question 2 Using y- for y bar and y' for y hat Opening up the right hand side of the equation, we have that Yi2 - 2yiy'2+ yi'2+ yi'2- 2y- I y' 2+y-2+2y I y'-2yy- -2 y' 2+ 2y- I y' The variables colored the same cancel out and the result becomes Yi2-2yy-+ y-2 which is equal to (y-y-)2 b) (Y-Y-)2= TSS (y-y')2= ExpSS (Y'-Y-)2= RSS 2(n-1)cov(e,y')= 2(y-y')(y-y-) Can u tell me how to get this one ??? I need detail about the solution of 2(n-1)cov(e,y') Thank u And therefore the relationship total sum of squares (TSS) = ExpSS +RSS + 2(n-1)cov(e,y') C) When the residual is uncorrelated with each of the regressor, then 2(y-y')(y-y -)=0 And consequently 2(n-1)cov(e,y')=0 But since 2 is not equal to 0 and also n is the number of observations and cannot be 0, then Cov(e,y')=0 d) From the equation given( y-y-)2= (y-y')2+ (y-y-)2+ 2(y-y-)(y-y') But since the residual is uncorrelated with the regressor, then 2(y-y')(y-y -)=0 Leaving as with the equation (y-y-) 2= (y-y') 2+ (y-y-) 2 Which is equivalent to TSS=ExpSS + RSS Question 2 Using y- for y bar and y' for y hat Opening up the right hand side of the equation, we have that Yi2 - 2yiy'2+ yi'2+ yi'2- 2y- I y' 2+y-2+2y I y'-2yy- -2 y' 2+ 2y- I y' The variables colored the same cancel out and the result becomes Yi2-2yy-+ y-2 which is equal to (y-y-)2 b) (Y-Y-)2= TSS (y-y')2= ExpSS (Y'-Y-)2= RSS 2(n-1)cov(e,y')= 2(y-y')(y-y-) Can u tell me how to get this one ??? I need detail about the solution of 2(n-1)cov(e,y') Thank u And therefore the relationship total sum of squares (TSS) = ExpSS +RSS + 2(n-1)cov(e,y') C) When the residual is uncorrelated with each of the regressor, then 2(y-y')(y-y -)=0 Why you can from residual is uncorrelated get 2(y-y')(y-y -)=0 ? Can u tell me detail ? And consequently 2(n-1)cov(e,y')=0 But since 2 is not equal to 0 and also n is the number of observations and cannot be 0, then Cov(e,y')=0 d) From the equation given( y-y-)2= (y-y')2+ (y-y-)2+ 2(y-y-)(y-y') But since the residual is uncorrelated with the regressor, then 2(y-y')(y-y -)=0 Leaving as with the equation (y-y-) 2= (y-y') 2+ (y-y-) 2 Which is equivalent to TSS=ExpSS + RSS b) (Y-Y-)2= TSS (y-y')2= ExpSS (Y'-Y-)2= RSS Error e= y-yCov(e, y')= cov( y-y-, y') Multiplying cov( y-y-, y') with 2(n-1) makes it equal to cov (y-y')(y-y -) and therefore 2(n-1) cov( y-y-, y')= 2(y-y')(y-y-) Can u give me this one more detailed solution, I mean step by step how to get 2(n-1) cov( y-y -, y')= 2(y-y') (y-y-)? Thus, 2(n-1)Cov(e, y') = 2(y-y')(y-y-) And therefore the relationship total sum of squares (TSS) = ExpSS +RSS + 2(n-1)cov(e,y') C) When the residual is uncorrelated with the regressor, it means that (y-y') and (y'-y -) have no correlation that is correlation is equal to zero. Since 2(y-y')(y-y-) is the correlation, between the residual and the regressor, then 2(y-y')(y-y-)=0 And consequently 2(n-1)cov(e,y')=0 But since 2 is not equal to 0 and also n is the number of observations and cannot be 0, then Cov(e,y')=0 d) From the equation given( y-y-)2= (y-y')2+ (y-y-)2+ 2(y-y-)(y-y') But since the residual is uncorrelated with the regressor, then 2(y-y')(y-y -)=0 Leaving as with the equation (y-y-) 2= (y-y') 2+ (y-y-) 2 Which is equivalent to TSS=ExpSS + RSS Hello! From your answer I also do not know how you calculate Cov (e,y^)? Can just tell me how to calculate this one