Introduction to Probability and Statistics for Engineers and Scientists

374 Chapter 9: Regression

the future response whose input level isx 0 and consider the probability distribution of the
response minus its predicted value — that is, the distribution ofY−A−Bx 0. Now,

Y∼N(α+βx 0 ,σ^2 )

and, as was shown in Section 9.4.3,

A+Bx 0 ∼N

(

α+βx 0 ,σ^2

[

1

n

+

(x 0 −x)^2

Sxx

])

Hence, becauseYis independent of the earlier data valuesY 1 ,Y 2 ,...,Ynthat were used
to determineAandB, it follows thatYis independent ofA+Bx 0 and so

Y−A−Bx 0 ∼N

(

0,σ^2

[

1 +

1

n

+

(x 0 −x)^2

Sxx

])

or, equivalently,

Y−A−Bx 0

σ

√

n+ 1

n

+

(x 0 −x)^2

Sxx

∼N(0, 1) (9.4.6)

Now, using once again the result thatSSRis independent ofAandB(and also ofY) and

SSR

σ^2

∼χn^2 − 2

we obtain, by the usual argument, upon replacingσ^2 in Equation 9.4.6 by its estimator
SSR/(n−2) that

Y−A−Bx 0

√

n+ 1

n

+

(x 0 −x)^2

Sxx

√

SSR

n− 2

∼tn− 2

and so, for any valuea,0<a<1,

p











−ta/2,n− 2 <

Y−A−Bx 0

√

n+ 1

n

+

(x 0 −x)^2

Sxx

√

SSR

n− 2

<ta/2,n− 2











= 1 −a

That is, we have just established the following.