chapter 1

Chapter 13: Nonlinear and Multiple Regression10.a. e y − ( β ˆ − βˆx ) = y − y − βˆ( x − x)i=i 0 1 i i1Σei, so( y − y) − βˆΣ( x − x) = 0 + βˆ⋅00i= Σi1 i1= .b. Since e = 0 always, the residuals cannot be independent. There is clearly a linearΣ irelationship between the residuals. If one e I is large positive, then al least one other e Iwould have to be negative to preserve e = 0 . This suggests a negative correlationΣ ibetween residuals (for fixed values of any n – 2, the other two obey a negative linearrelationship).c. Σxe = Σxy − Σxy − β Σx( x − x )iiiii( Σx)( Σy) ⎤ ⎡ ( Σx)ˆ ⎡i i⎥ −21 i i = ⎢Σxiyi−βˆ1⎢Σxi−⎣ n ⎦ ⎢⎣, but the first term in brackets is the numerator of ˆβ1, while the second term is thedenominator of ˆβ1, so the difference becomes (numerator of ˆβ1) – (numerator of ˆβ1) =0.*d. The five e i' sfrom Exercise 7 above are –1.55, .68, 1.25, -.05, and –1.06, which sum to-.73. This sum differs too much from 0 to be explained by rounding. In general it is nottrue that e * = 0 .Σ iin2⎤⎥⎥⎦400