topic Re: Multiple regression in SAS Programming

Multiple regression

alot4315 — Fri, 12 Apr 2019 22:13:52 GMT

(How can I solve this problem):

I have two responses y1, y2 and 8 independent variables.

Im trying to run multiple linear regression on the data using proc reg either on y1 or y2:

And I got the following notes:

PROC IMPORT DATAFILE="/folders/myfolders/sasuser.v94/ENB2012_data.xlsx"

OUT=energy

DBMS=XLSX

REPLACE;

RUN;

PROC PRINT DATA=energy; RUN;

proc reg data=energy ;

model y1= x1 x2 x3 x4 x5 x6 x7 x8 / vif collin;

run;

The REG Procedure

Model: MODEL1

Dependent Variable: Y1 Y1

Number of Observations Read	1296
Number of Observations Used	768
Number of Observations with Missing Values	528

Analysis of Variance
Source	DF	Sum of Squares	Mean Square	F Value	Pr > F
Model	7	71546	10221	1187.06	<.0001
Error	760	6543.77041	8.61022
Corrected Total	767	78090

Root MSE	2.93432	R-Square	0.9162
Dependent Mean	22.30720	Adj R-Sq	0.9154
Coeff Var	13.15413

Note:Model is not full rank. Least-squares solutions for the parameters are not unique. Some statistics will be misleading. A reported DF of 0 or B means that the estimate is biased.

Note:The following parameters have been set to 0, since the variables are a linear combination of other variables as shown.

X4 =

2.97E-8 * Intercept - 1.63E-8 * X1 + 0.5 * X2 - 0.5 * X3 + 342E-12 * X5

Parameter Estimates
Variable	Label	DF	Parameter Estimate	Standard Error	t Value	Pr > \|t\|	Variance Inflation
Intercept	Intercept	B	84.01342	19.03361	4.41	<.0001	0
X1	X1	B	-64.77343	10.28945	-6.30	<.0001	105.52405
X2	X2	B	-0.08729	0.01708	-5.11	<.0001	201.53113
X3	X3	B	0.06081	0.00665	9.15	<.0001	7.49298
X4	X4	0	0	.	.	.	.
X5	X5	B	4.16995	0.33799	12.34	<.0001	31.20547
X6	X6	1	-0.02333	0.09470	-0.25	0.8055	1.00000
X7	X7	1	19.93274	0.81399	24.49	<.0001	1.04751
X8	X8	1	0.20378	0.06992	2.91	0.0037	1.0475

Also I tried to make dummy variables, since I have 2 categorical variables as follow and still the same:

DATA energy1;
SET energy;
IF (x6 = 2) THEN d61 = 1; ELSE d61 = 0;
IF (x6 = 3) THEN d62 = 1; ELSE d62 = 0;
IF (x6 = 4) THEN d63 = 1; ELSE d63 = 0;
IF (x6 = 5) THEN d64 = 1; ELSE d64 = 0;

IF (x8 = 0) THEN d81 = 1; ELSE d81 = 0;
IF (x8 = 1) THEN d82 = 1; ELSE d82 = 0;
IF (x8 = 2) THEN d83 = 1; ELSE d83 = 0;
IF (x8 = 3) THEN d84 = 1; ELSE d84 = 0;
IF (x8 = 4) THEN d85 = 1; ELSE d85 = 0;
IF (x8 = 5) THEN d86 = 1; ELSE d86 = 0;
run;

see the attached data or ( https://archive.ics.uci.edu/ml/machine-learning-databases/00242/ENB2012_data.xlsx )

Re: Multiple regression

PaigeMiller — Fri, 12 Apr 2019 22:26:46 GMT

@alot4315 wrote:

(How can I solve this problem):

proc reg data=energy ;

model y1= x1 x2 x3 x4 x5 x6 x7 x8 / vif collin;

run;

Analysis of Variance

Source

DF

Sum of
Squares

Mean
Square

F Value

Pr > F

Model

7

71546

10221

1187.06

<.0001

Error

760

6543.77041

8.61022

Corrected Total

767

78090

Root MSE

2.93432

R-Square

0.9162

Dependent Mean

22.30720

Adj R-Sq

0.9154

Coeff Var

13.15413

Note:Model is not full rank. Least-squares solutions for the parameters are not unique. Some statistics will be misleading. A reported DF of 0 or B means that the estimate is biased.

Note:The following parameters have been set to 0, since the variables are a linear combination of other variables as shown.

X4 =

2.97E-8 * Intercept - 1.63E-8 * X1 + 0.5 * X2 - 0.5 * X3 + 342E-12 * X5

You can't have x4 and the combination of x1 x2 x3 x5 in the model at the same time. Why? Because x4 does not provide additional information, if you know x1 x2 x3 x5, then you know x4 exactly. So the solution is to remove x4 from the model.

Re: Multiple regression

PGStats — Sat, 13 Apr 2019 05:05:51 GMT

This feature of the data results from its artificial nature.

http://people.maths.ox.ac.uk/tsanas/Preprints/ENB2012.pdf

Multiple regression is not a promissing analysis tool for such data. You should try something simpler (conceptually) first, such as regression trees (proc hpsplit).