`
The last stage of a regression analysis is to assess the adequacy of the model.
In order to do this, we need to examine the components of the model which
are purely statistical, i.e. the residual terms ei. To see why this is necessary, consider
the four data sets in the EXCEL worksheet “Tufte.xls” which you will find in the
EXCEL folder MBA Part 1. All four of these data sets lead to exactly the same
values of the regression coefficients and se. In addition the y values and x values all
have the same means and standard deviations. However when you do the regression
and examine the automatic residual graph you get very different results. Consider
the residual plot below:
Residual Plot
1.5
1
Residuals
0.5
0
3
4
5
6
7
-0.5
-1
-1.5
-2
-2.5
X
8
9
10
11
Now consider the next residual plot:
Residual Plot
3.5
3
2.5
Residuals
2
1.5
1
0.5
0
3
4
5
6
7
-0.5
-1
-1.5
X
8
9
10
11
Now consider the third plot:
Residual Plot
2.5
2
1.5
Residuals
1
0.5
0
5
6
7
8
9
-0.5
-1
-1.5
-2
X
10
11
12
13
Finally consider the fourth residual plot:
Residual Plot
2.5
2
1.5
Residuals
1
0.5
0
-0.5
3
4
5
6
7
-1
-1.5
-2
-2.5
X
8
9
10
11
For our electrical data, the residual plot looks like the following:
Residual Plot
1500
Residuals
1000
500
0
-500
0
200
400
600
-1000
-1500
DD
800
1000
EXCEL also provides a plot of the actual values of y and the predicted values
for each value of x. This is illustrated below:
KWH
DD Line Fit Plot
8,000
7,000
6,000
5,000
4,000
3,000
2,000
1,000
-
KWH
Predicted KWH
0
500
DD
1000
Since our model looks like it fits the data, we can try to use it to predict
kilowatt hour usage using degree days as a predictor.
The basic equation would be:
Predicted KWH usage = 903.515 + 7.089 x (degree days).
Thus for a billing period with 100 degree days (possibly a period in the
spring or autumn in Dallas), the predicted KWH usage would be 1,612 KWH.
This “point” estimate ignores variability. However we can use the results
discussed earlier about the “mound” rule and Chebyshev to incorporate variability
into our forecasts to get what is called an “interval” forecast. In this case, the
formula is given by:
Forecast  b̂0  b̂1 x  2 se
In the case discussed above for a billing period of 100 degree days, the
interval forecast would be:
1612 +/- 2*(605.9)
or approximately 400 KWH to 2,824 KWH.
Although the above method is the most practical way to assess the
adequacy of a model for forecasting purposes, it is common to use a single
descriptive measure called the “correlation coefficient” to describe the adequacy of
the regression fit.
The correlation coefficient attempts to quantify how useful x is as a predictor
of y.
If one were not to use x in the forecasting of y, then one would guess the mean
value of y as the forecast of kilowatt hours for each period. If one defines the error
made as:
i th error  yi  y
and,
SST   ( y i  y ) 2
i
SST is a measure of the total errors made not using x as a predictor.
If we use x, then we can define the error made as:
êi  yi  b̂0  b̂1 xi
and,
SSE   ê i
2
i
SSE is a measure of the total errors made using x as a predictor.
One can show that,
0  SSE  SST
Therefore,
0
SSE
1
SST
If we define,
R 2  1  SSE SST
then,
0  R2  1
By rewriting the definition of R2 as:
R 2  ( SST  SSE ) / SST
One can interpret R2 as
the proportion of the variability in y explained (or eliminated) by using x
as a predictor
As we shall see, all of the above generalizes to the case where one has more
than one x as a predictor. In the case however of a single x predictor, one usually
encounters the correlation coefficient r defined as:
r  sign( b̂1 ) R 2
In our case R2=.8538, and r=.9240. Both of these values can be found on the
EXCEL as highlighted below:
A natural question is how big does R2 have to be in order for the regression
analysis to be useful? I would suggest that the important measure is the usefulness
of the interval forecast.