# Doubt in Interpreting Regression Output in SPSS

#### bsaikrishna

##### New Member
I am trying to understand Regression through a small example. I want to predict the sales for the year 2011, and I’ve considered only two independent variables Production and Ad Expenditure for simplicity. Following is the data:

Table 1:
YEAR SALES(in lakhs) PRODUCTION(in lakhs) ADEXPENDITURE(GRPs)
2008 7 11 900
2009 8.5 12 900
2010 9.2 12.4 1000

Following are the Z-scores for the above:

Table 2:
-1.09729 -1.30854 -0.86603
0.23725 -0.11896 -0.86603
0.86004 0.35687 0.86603

I run Linear Regression on the data of Table1, with Sales as the dependent variable and Production and AdExpenditure as the independent variables.
Following is the output in SPSS:

Model Summary
Model R R Square Adjusted R Square Std. Error of the Estimate
1 1.000a 1.000 . .
a. Predictors: (Constant), ADSPEND, PRODUCTION

ANOVA
Model Sum of Squares df Mean Square F Sig.
1 Regression 2.527 2 1.263 . .a
Residual .000 0 .
Total 2.527 2
a. Predictors: (Constant), ADSPEND, PRODUCTION
b. Dependent Variable: SALES

Coefficient
Model Unstandardized Coefficients Standardized Coefficients
B Std. Error Beta
1 (Constant) -10.400 .000
PRODUCTION 1.500 .000 .962
ADSPEND .001 .000 .051

As per the coefficients, the regression equation is:
SALES = -10.4 + 1.5*PRODUCTION + .001*ADSPEND

Verification:
To verify the above linear equation, let us try to predict the sales of 2010,and 2011.
Sales (2010) = -10.4 +1.5*12.4 + 1
Sales (2010) = 9.8 (seems reasonable, atleast for demonstration)

Verification (Z Score Coefficients): (Constant is always zero in this case)
Sales (2010) = 0.962*0.35687 + 0.86603*0.051
Sales (2010) = 0.3433 + 0.04 = 0.3833 ( no where close to the actual value 0.86 from Table 2)

Please let me know why am I not able to get the Z Score coefficient for Sales 2010 correctly. This forum has been excellent in resolving my doubts.

I am sorry for the untidy math here:shakehead

#### Dason

##### Ambassador to the humans
You only have 3 observations. You have two linearly independent predictors (ie one is not a linear transformation of the other). You are perfectly predicting the observations. This will happen anytime you only have 3 observations and try to use two predictors. If you have 4 observations and try to use 3 predictors (as long as the predictors aren't linear combinations of each other) you will be able to perfectly predict the responses.

Take for example if you have two observations and a single predictor. It should be pretty obviously that you can perfectly fit a line through the two points. If you don't see this draw a few scatterplots using two observations.

What it boils down to is if you have n observations and try to use n-1 predictors that aren't linear combinations of each other then you will always be able to perfectly predict the response.

#### bsaikrishna

##### New Member
HTML:
[B]Verification (Z Score Coefficients[/B]): (Constant is always zero in this case)
Sales (2010) = 0.962*0.35687 + 0.86603*0.051
Sales (2010) = 0.3433 + 0.04 = 0.3833 ( no where close to the actual value 0.86 from Table 2)
Am I correct to verify the standardised coefficients like the above?
If so, why am I not getting the Z score of Sales 2010, which is 0.86