STAT 350: Lecture 10 Example

Confidence intervals for

Refer to the polynomial regression example (data on insurance costs). I fit polynomials of degree 1 through 5. Each model gives a vector of fitted parameters and to predict the mean value of Y at time t we use when the fitted polynomial has degree p. The SAS code below computes both this fitted value and standard errors for each of the 5 models. Notice how I run proc glm 5 times to get the 5 different values.

data insure;
  infile 'insure.dat' firstobs=2;
  input year cost;
  code = year - 1975.5 ;
proc glm  data=insure;
   model cost = code ;
   estimate 'fit1982.25' intercept 1 code  6.75  / E;
run ;
proc glm  data=insure;
   model cost = code code*code;
   estimate 'fit1982.25' intercept 1 code  6.75 code*code 45.5625 / E;
run ;
proc glm  data=insure;
   model cost = code code*code code*code*code;
   estimate 'fit1982.25' intercept 1 code  6.75 code*code 45.5625 code*code*code 307.5469/ E;
run ;
proc glm  data=insure;
   model cost = code code*code code*code*code code*code*code*code;
   estimate 'fit1982.25' intercept 1 code  6.75 code*code 45.5625 code*code*code 307.5469 code*code*code*code  2075.9414 / E;
run ;
proc glm  data=insure;
   model cost = code code*code code*code*code code*code*code*code code*code*code*code*code;
   estimate 'fit1982.25' intercept 1 code  6.75 code*code 45.5625 code*code*code 307.5469 code*code*code*code  2075.9414 code*code*code*code*code 14012.6045/ E;
run ;

The line estimate ... is probably unfamiliar to you. You have to give the values of each column of the design matrix at the place where you want to estimate . Notice that I had to work out each power of 6.75 by hand.

Now have a look at the edited output. I show here only the 5th degree polynomial results.

                        General Linear Models Procedure
Coefficients for estimate fit1982.25
INTERCEPT                                         1
CODE                                           6.75
CODE*CODE                                   45.5625
CODE*CODE*CODE                             307.5469
CODE*CODE*CODE*CODE                       2075.9414
COD*COD*COD*COD*CODE                     14012.6045

Dependent Variable: COST
                                     Sum of            Mean
Source                  DF          Squares          Square   F Value     Pr > F

Model                    5     3935.2507732     787.0501546   2147.50     0.0001
Error                    4        1.4659868       0.3664967
Corrected Total          9     3936.7167600

                  R-Square             C.V.        Root MSE            COST Mean
                  0.999628         0.851438       0.6053897            71.102000

Source                  DF        Type I SS     Mean Square   F Value     Pr > F
CODE                     1     3328.3209709    3328.3209709   9081.45     0.0001
CODE*CODE                1      298.6522917     298.6522917    814.88     0.0001
CODE*CODE*CODE           1      278.9323940     278.9323940    761.08     0.0001
CODE*CODE*CODE*CODE      1        0.0006756       0.0006756      0.00     0.9678
COD*COD*COD*COD*CODE     1       29.3444412      29.3444412     80.07     0.0009

Source                  DF      Type III SS     Mean Square   F Value     Pr > F
CODE                     1       0.88117350      0.88117350      2.40     0.1959
CODE*CODE                1      20.86853994     20.86853994     56.94     0.0017
CODE*CODE*CODE           1      72.35876312     72.35876312    197.43     0.0001
CODE*CODE*CODE*CODE      1       0.00067556      0.00067556      0.00     0.9678
COD*COD*COD*COD*CODE     1      29.34444115     29.34444115     80.07     0.0009

                                        T for H0:    Pr > |T|   Std Error of
Parameter                  Estimate    Parameter=0                Estimate
fit1982.25               70.2630583           4.33     0.0123     16.2154539

                                        T for H0:    Pr > |T|   Std Error of
Parameter                  Estimate    Parameter=0                Estimate

INTERCEPT               64.88753906         176.14     0.0001     0.36839358
CODE                    -0.50238411          -1.55     0.1959     0.32399642
CODE*CODE                0.75623470           7.55     0.0017     0.10021797
CODE*CODE*CODE           0.80157430          14.05     0.0001     0.05704706
CODE*CODE*CODE*CODE     -0.00020251          -0.04     0.9678     0.00471673
COD*COD*COD*COD*CODE    -0.01939615          -8.95     0.0009     0.00216764

While we have this output notice the value of which is quite close to 1 and the t-tests of hypotheses that various parameters are 0.

Here is a table of the results of all the forecasts with associated standard errors:

Degree Estimate SE

1 113.98 7.04

2 142.04 12.06

3 204.74 9.45

4 204.50 25.24

5 70.26 16.22

Notice that the standard errors are so small that there is no way that the forecasts from various different degree fits can be reconciled. The problem is that it is very likely that a crucial assumption is not right for this problem, namely, the assumption that the mean of Y is exactly a polynomial of degree 5 (or 3 or whatever). Notice also that adding a term to the model without improving the fit, as in going from degree 3 to degree 4, increases the SE of the prediction greatly.

One final point. The calculations give a confidence interval for based on the distribution of . For the insurance the quantity of interest is . In this formula, is a future value associated with the covariate value x. The prediction can be split up, if the model is correct, as

which is a sum of two independent random variables. The first has variance while the second has variance . An estimate of the square root of the second quantity was printed out by SAS. The Mean Squared Error is an estimate of the first. The estimated standard deviation of is the square root of the sum of the squares of these two quantities which comes, for the 5th degree polynomial to which is only slightly larger, $16.23. Notice that this accuracy is spurious; the major source of error is in the model for the mean of Y which is surely not a 5th degree polynomial.

You can see the principal by deleting the observation for 1980 and then fitting the different polynomials:

Degree Estimate SE

1 91.50 4.26

2 98.22 7.34

3 121.39 4.74

4 132.40 6.73

5 472.34 66.15

The true value is actually $115.19 so most of the forecasts would have been dreadful. Again all the SE's are for errors in predicting the mean not an individual value but for the higher degree polynomials this makes no difference.