chapter 1

Chapter 12: Simple Linear Regression and Correlation= 2 ˆ ˆˆ Σy − βˆ1Σx0 1β0= , SSE becomesn2Σy( Σy− βˆΣx) ( Σy) ˆ ΣxΣySSE = Σy2 1−− ˆ2 β1β1Σxy= Σy− + − β ˆ1Σxynn n2⎡2 ( Σy) ⎤x yy ˆ ⎡ Σ Σ ⎤= β1xy = S ˆ⎢Σ− ⎥ −yy − β1S xyn ⎢Σ−, as desired.n ⎥⎣ ⎦ ⎣ ⎦77. SSE Σy− β Σy− β Σxy. Substituting78. The value of the sample correlation coefficient using the squared y values would notnecessarily be approximately 1. If the y values are greater than 1, then the squared y valueswould differ from each other by more than the y values differ from one another. Hence, therelationship between x and y 2 would be less like a straight line, and the resulting value of thecorrelation coefficient would decrease.79.a. With =xx∑s ( − x)2x i, =s ( − y )yy∑y i2, note thatssyxssyy= ( since thefactor n-1 appears in both the numerator and denominator, so cancels). Thusy = β ˆ0+ βˆ1x= y + βˆ1sy= y + ⋅ r ⋅ ( x − x), as desired.sxsxyyy xy( x − x ) = y + ( x − x ) = y + ⋅ ( x − x )sxxssxxxxssxxsyyb. By .573 s.d.’s above, (above, since r < 0) or (since s y = 4.3143) an amount 2.4721 above.388