Introduction to Probability and Statistics for Engineers and Scientists

372 Chapter 9: Regression

variables and is thus itself normally distributed. Because we already know its mean, we
need only compute its variance, which is accomplished as follows:

Var(A+Bx 0 )=

∑n

i= 1

[

1

n

−c(xi−x)(x−x 0 )

] 2

Var(Yi)

=σ^2

∑n

i= 1

[

1

n^2

−c^2 (x−x 0 )^2 (xi−x)^2 − 2 c(xi−x)

(x−x 0 )

n

]

=σ^2

[

1

n

+c^2 (x−x 0 )^2

∑n

i= 1

(xi−x)^2 − 2 c(x−x 0 )

∑n

i= 1

(xi−x)

n

]

=σ^2

[

1

n

+

(x−x 0 )^2

Sxx

]

where the last equality followed from

∑n

i= 1

(xi−x)^2 =

∑n

i= 1

xi^2 −nx^2 =1/c=Sxx,

∑n

i= 1

(xi−x)= 0

Hence, we have shown that

A+Bx 0 ∼N

(

α+βx 0 ,σ^2

[

1

n

+

(x 0 −x)^2

Sxx

])

(9.4.4)

In addition, becauseA+Bx 0 is independent of

SSR/σ^2 ∼χn^2 − 2

it follows that
A+Bx 0 −(α+βx 0 )
√
1
n

+

(x 0 −x)^2

Sxx

√

SSR

n− 2

∼tn− 2 (9.4.5)

Equation 9.4.5 can now be used to obtain the following confidence interval estimator of
α+βx 0.

Confidence Interval Estimator ofα+βx 0

With 100(1−a) percent confidence,α+βx 0 will lie within

A+Bx 0 ±

√

1

n

+

(x 0 −x)^2

Sxx

√

SSR

n− 2

ta/2,n− 2