*9.10Multiple Linear Regression 405
Since the value of the preceding statistic is (
√
10/19.262)(.075)/.616=.088, thep-value
of the test of the hypothesis thatβ 2 =0is
p-value=P{|T 10 |>.088}
= 2 P{T 10 >.088}
=.9316 by Program 5.8.2.A
Hence, the hypothesis is accepted (and, in fact, would be accepted at any significance level
less than .9316). ■
REMARK
The quantity
R^2 = 1 −
SSR
∑
i
(Yi−Y)^2
which measures the amount of reduction in the sum of squares of the residuals when using
the model
Y=β 0 +β 1 x 1 +···+βnxn+e
as opposed to the model
Y=β 0 +e
is called thecoefficient of multiple determination.
9.10.1 Predicting Future Responses
Let us now suppose that a series of experiments is to be performed using the input levels
x 1 ,...,xk. Based on our data, consisting of the prior responsesY 1 ,...,Yn, suppose we
would like to estimate the mean response. Since the mean response is
E[Y|x]=β 0 +β 1 x 1 +···+βkxk
a point estimate of it is simply
∑k
i= 0 Bixiwherex^0 ≡1.
To determine a confidence interval estimator, we need the distribution of
∑k
i= 0 Bixi.
Because it can be expressed as a linear combination of the independent normal random
variablesYi,i=1,...,n, it follows that it is also normally distributed. Its mean and
variance are obtained as follows:
E
∑k
i= 0
xiBi
=
∑k
i= 0
xiE[Bi] (9.10.8)
=
∑k
i= 0
xiβi sinceE[Bi]=βi