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# 23.

## TESTING WHETHER THE MODEL

IS USEFUL FOR PREDICTING Y
It is always possible that our explanatory variables are
completely useless for predicting y. We can formulate this as
the null hypothesis that all regression parameters (except the
intercept) are zero, that is,
H0 : 1 = 2 = = k = 0 .
Under H0, the true regression function does not depend at all
on our explanatory variables, although our estimates 1 ,L, k
will almost certainly be different from zero, due to natural
variability, i.e., noise, as opposed to signal.

## You might think that you could test H0 versus HA by checking

to see if any of the estimated coefficients is significantly
different from zero. The problem with this approach is that
you would be performing k different hypothesis tests at the
same time. Even if H0 is true, the probability of finding at
least one significant t-statistic at (let's say) the 5% level of
significance is actually much larger than 0.05.
The larger the value of k is, the worse the problem becomes.
The level of significance of the t-test gives the probability of a
Type I error for that test by itself. But in the scenario above,
we have many chances to find a significant t-statistic. This
affects the overall probability of a Type I error.

## To determine if the model is useful, we would like to perform

a hypothesis test of H0.
The alternative hypothesis HA is that H0 is false.
So HA says that at least one of 1,..., k is nonzero.
In other words, under HA, the model is not completely useless.
(But notice that HA does not say that all coefficients are nonzero).

## Here's an analogy: If we draw a card at random from a

shuffled deck, we only have a 1/52 chance of obtaining the
Ace of Spades. But if we repeat this experiment 100 times, the
probability that we will draw the Ace of Spades at least once
is much larger. In fact, it's almost 1.
A better way to test H0 above is to use the "F-test".
This allows you to test H0 versus HA, with an actual Type I
error rate of .
The test statistic is

F=

SSR k
.
SSE (n k 1)

## To understand the F-statistic better, we need to return to the

Analysis of Variance part of the Minitab output.

## We can write the F-statistic as

Analysis of Variance
Source
Regression
Error
Total

DF
3
11
14

SS
570744
52280
623024

MS
190248
4753

F-Value
40.03

P-Value
0.000

R-sq
91.61%

89.32%

R-sq(pred)
87.66%

Coefficients
Term
Constant
Size
Age
Lot Size

Coef
-161
41.46
-2.36
48.31

MSR
SSR
SSE
, where MSR =
and MSE =
.
MSE
k
n k 1

Model Summary
S
68.9399

F=

SE Coef
191
7.51
8.81
9.01

T-Value
-0.84
5.52
-0.27
5.36

P-Value
0.418
0.000
0.794
0.000

## Minitab gives three entries for degrees of freedom in the DF

column: DF(Regression), DF(Residual), and DF(Total).
We have already mentioned DF(Residual) = nk 1. This is
the DF we have been using to calculate confidence intervals
and tests on the individual regression parameters.
In fact, DF(Residual) is the number of degrees of freedom
available for estimating 2. The estimator, s2, is based on the
residuals, which have lost k + 1 degrees of freedom from the
original n because of the need to estimate k + 1 parameters in
order to calculate the residuals.

So MSE is the "Mean Square for Residuals", that is, the ratio of
SSE to the (Residual) degrees of freedom, nk 1. Note that MSE
is the same as s2.
In the housing example, from the Minitab output, we find that
MSE = 4753, which is the ratio SSE/DF(Residual) = 52280/11.

## DF(Regression) is k, the number of regression parameters, not

counting the intercept.
DF(Total) is n1, the same as the df we have used in
estimating the variance of the y's.
For the housing example, DF(Regression) = 3,
DF(Residual) = 11, and DF(Total) = 14.
Note: The value of n does not appear anywhere in the output,
but we can get it by adding one to DF(Total).

## The Mean Square for Regression is defined as

MSR = SSR/k = SSR/DF(Regression)
MSR represents the "average" squared fluctuation of the y
values about their mean, y .
In the housing example, MSR = 190248 = 570744 / 3.
The F statistic is the ratio F = MSR/MSE.
In the housing example, F = 40.03 = 190248 / 4753.
It is clear from its definition that the F statistic is always
positive.
If F = MSR/MSE is sufficiently large, we can reject H0 in favor
of HA .

## If H0 is true, then the F-statistic has an F distribution with k, and

nk 1 degrees of freedom.
Note that the F distribution has two different df: one for the
numerator (1), one for the denominator (2).
We have 1 = k, 2 = nk1.
The F distribution is skewed to the right, since the F-statistic can
never be negative. If the F-statistic exceeds F (Table 9), then we
reject H0 at level . This test is inherently right-tailed, since we
are looking for a large value of the F statistic as evidence that the
regression is useful. Thus, we use the critical value F , and not
F/2 .

ANOVA
Source
Regression

DF

SS

SSR

Residual
nk1
(Error)
Total

n1

SSE
SST=
SSR+SSE

MS
MSR= SSR
k
MSE = SSE
n k 1

F
P
pMSR
F=
MSE value

## Fortunately, we can avoid using tables here by just looking at

the p-value reported by Minitab.
In the housing example, Minitab gives a p-value for the F-test
of p = 0.000. So we can reject H0 at level 0.05, and also at
level 0.01. The model does seem to have predictive power.
For a manual calculation, suppose we wanted to do a formal
test at level 0.05. From Table 9 with 1 = 3, 2 = 11, we find
F0.05 = 3.59. Since the observed F statistic exceeds 3.59, we
reject H0 at level 0.05. (We already knew this had to happen.)

## Practical Interpretation of the F-Test

Most books recommend that in analyzing multiple regression
computer output we look first at the F-test. If the F-statistic is
not significant, we do not continue with the analysis, since the
regression is not useful for prediction.
From a practical point of view, there are some problems with
the F-test. If the F-statistic exceeds the critical value, then we
have some indication that at least one of the i is nonzero.
However, the test gives us no clue as to which of the i is (are)
nonzero.
Unfortunately, this is precisely what we will typically want to
know in practice.

## Going back to the individual t-statistics for each parameter and

picking those which are significant does not solve the problem,
due partly to the multiple testing issues we discussed earlier.
It is tempting to conclude that if H0 is rejected, the model, with
all k variables, must be "good". But this is not necessarily true.
In the housing example, the coefficient for age is not significant
(p=0.794).
So even though we have a highly significant F-statistic,
(p=0.000), it is not clear that the age variable has predictive
power, given that the other two variables are in the model.
Perhaps the age variable should be deleted. The F-statistic tells
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