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Use the definitions bar{x} =(1)/(n)sum x_(i) and s_(x)^(2)=(1)/(n)sum (x_(i)-(bar{x} ))^(2) to show that if Y_(i)= a+bx_(i) , then (a) /bar (Y)=a+bbar{x} and s_(Y)^(2)=(1)/(n)sum (Y_(i)-(/bar
Use the definitions
\\\\bar{x} =(1)/(n)\\\\sum x_(i)
and
s_(x)^(2)=(1)/(n)\\\\sum (x_(i)-(\\\\bar{x} ))^(2)
to show that if
Y_(i)=
\
a+bx_(i)
, then\ (a)
/bar (Y)=a+b\\\\bar{x}
and
s_(Y)^(2)=(1)/(n)\\\\sum (Y_(i)-(/bar (Y)))^(2)=b^(2)s_(x)^(2)
; (3 pts)\ (b) If
Z_(i)=(x_(i)-(\\\\bar{x} ))/(s_(x))
, then
/bar (Z)=0
and
s_(z)^(2)=1.(3pts)
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