Question

linear regression

solve number 1 only

1. (a) Consider the model Y =Bo+BX +BX2+BX3 + B4X4+€. If it is suggested to you that the two variables Z = X1+ X4 and Z2 X+ X might be adequate to represent the data, what hypothesis, in the form CB 0, would you need to test? (Give the form of C) (b) For the data ((Xi, X2, Y) : (-1, -1,5.2). (-1,0,6.1). (0,0.7.8), (1,0, 10.3), (1.1,10.9)). fit the model Y Bo + 3,X+2X2+. Test the hypothesis Ho: 8 282 versus Hi: not so. 2. (a) Consider the data of Table 2.1 from the text book. Suppose you are told that the 23rd obser- vation (that is, the last observation (Y. X)(5.9,6.7)) has variance 9a rather than a Refit the equation nsing weighted least squares. (b) Repeat part (a) but with 2 olerve for the variance of the last oleervation. What changes do sou (Hint: See solutions for exercises A. B. D and E in Chapter 9). 3. Given the data below, apply thestepwise regreionand ckwar limination algoriths to select the est regresson lel Al cpste the Malln for all psible models ad plot the Cvles agast p Compare he sl What is th orect molel hp delete pt sc 141 & 7 5 6 3 3/4 2 1/4 3 2 P U Y R L K J H F D pause BNM V G 4

Answer 1

1(a) We need to test, B1 = 84 and B2 = B3 so (01 00 -1 0 0 1 -1 0 ) 10 10 0 -1 CB=0=1 O 0 1 -1 0 cs=o=(!! : 2 (0-0--------- 0 B1 - B4 = 0, B2 - B3 = 0 /1 -1 -1 1 -1 0 15 o o) 0.2 0.0 (6) X= 1 0 0 X'X = 0 4 2 ; (X'X)-1 = 0.0 0.5 11 1 0 10 2 4) 0.0 -0.5 1 1 1 / 8.06 Y = (5.2,6.1,7.8, 10.3, 10.9)' then B = (X'X)-'X'Y = | 2.10 ) ► 0.0 -0.5 1.0 / 0.75) 3 = 2.10, 3, = 0.75.

SSE = sum of squares due to error = Y'Y-EX'Y = 0.107 with degrees of freedom = 5 – 3 = 2 MSE = SSE/2 = mean sum of squares due to error = 0.0535 and MSE is an unbiased estimator of o”. Now, Var(1 - 282) = Var(1') = l'(X'X)-102 = 6.502 where, 1 = (0,1,-2) Since od is unknown so estimate of standard error of B1 – 282 = SE (– 282) = V6.5M SE = 0.5897 B1 - 282 910_2*0.75)/0.5897 = 1.0175 Test statistic =t=- SE (B1 – 232) p - value = 2P(t > 1.0175 t - tz) = 0.4160 > 0.05 So we fail to reject Ho at 5% level of significance.