Applied Statistics Using SPSS, STATISTICA, MATLAB and R

282 7 Data Regression
Example 7.5
Q: Consider the ART(PRT) linear regression in Example 7.1. Is it valid to reject
the null hypothesis of a linear fit through the origin at a 5% level of significance?
A: The results of the respective t test are shown in the last two columns of Figure
7.3. Taking into account the value of p (p ≈ 0 for t * = −9.1), the null hypothesis is
rejected. This is a somewhat strange result, since one expects a null area
corresponding to a null perimeter. As a matter of fact an ART(PRT) linear
regression without intercept is also a valid data model (see Exercise 7.3).
7.1.3.3 Inferences About Predicted Values
Let us assume that one wants to derive interval estimators of E[ Y ˆ
k ] , i.e., one wants
to determine which value would be obtained, on average, for a predictor variable
level xk, and if repeated samples (or trials) were used.
The point estimate of E[ Y ˆ
k ] , corresponding to a certain value xk, is the
computed predicted value:
ˆ = + x .
yk b0
b1
k
The yˆ k value is a possible value of the random variableYk ˆ which represents all
possible predicted values. The sampling distribution for the normal regression
model is also normal (since it is a linear combination of observations), with:
– Mean: E[ Y ˆ
k ] = E[ b0 + b1x
k ] = E[
b0
] + xk
E[
b1
] = β 0 + β1x
k ;
⎛
2 ⎞
– Variance: ˆ 2 ⎜ 1 ( xk
− x)
V[ Y = +
⎟
k ] σ .
⎜
2 ⎟
⎝
n ∑ ( xi
− x)
⎠
Note that the variance is affected by how far xk is from the sample mean x . This
is a consequence of the fact that all regression estimates must pass through ( x , y)
.
Therefore, values xk far away from the mean lead to higher variability in the
estimates.
Since σ is usually unknown we use the estimated variance:
⎛
2 ⎞
ˆ ⎜ 1 ( xk
− x)
s[
Y = +
⎟
k ] MSE
. 7.17
⎜
2 ⎟
⎝
n ∑ ( xi
− x)
⎠
Thus, in order to make inferences about k Yˆ , we use the studentised statistic:
t
*
yˆ
ˆ
k − Ε[
Yk
]
= ~ tn−2. 7.18
s[
Yˆ
]
k