Significance of Regression Intercept (R lm model) [duplicate]

Question

Question:

Having performanced a linear regression in R with the lm function, I'm not sure how to interpret the results for the Intercept (as shown below).

It seems the probability of the intercept's relevance is low (i.e. Pr(>|t|) is 0.845, and higher that 0.05). Does this mean I should drop the intercept from the model by forcing it through zero? Alternatively, does it mean that I should still keep the intercept but recognise that it's not significant?

Output:

Call:
lm(formula = DI ~ II)

Residuals:
 Min       1Q   Median       3Q      Max 
-0.23960 -0.03306 -0.01116  008724  0.20568 

Coefficients:
            Estimate    Std. Error   t value   Pr(>|t|)
(Intercept) -0.07952    0.39953      -0.199   0.845
II           0.86381    0.04593      18.809   8.23e-11 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 0.1346 on 13 degrees of freedom
Multiple R-squared: 0.9646, Adjusted R-squared: 0.9618 
F-statistic: 353.8 on 1 and 13 DF,  p-value: 8.23e-11 

Additional Background Information My overall aim is to find a relationship between two datasets of mass that I have. So, if I have a value for DI, I am able to find out the corresponding value of II.

Answer 1

Let me say that the meaning of the intercept depends on the location of predictor variables, here II. The intercept is defined as the expected outcome when II is zero, but if an II value of zero is not meaningful, then the intercept is meaningless too. In such cases, the correlation between the intercept and slope will tend toward $-1.0$, i.e., the intercept is essentially determined by the slope - and will be not significant.

Centering the predictor variables is often the right solution. Look:

> x <- c(1,2,3)
> y <- c(2.9,4.8,6.9)
> summary(lm(y ~ x))

Call:
lm(formula = y ~ x)

Residuals:
       1        2        3 
 0.03333 -0.06667  0.03333 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  0.86667    0.12472   6.949   0.0910 .
x            2.00000    0.05774  34.641   0.0184 *

The intercept is not significant and could be as meaningless as x == 0. But if you center:

> cx <- x - mean(x)
> summary(lm(y ~ cx))

Call:
lm(formula = y ~ cx)

Residuals:
       1        2        3 
 0.03333 -0.06667  0.03333 

 Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
 (Intercept)  4.86667    0.04714  103.24  0.00617 **
 cx           2.00000    0.05774   34.64  0.01837 * 

Now the intercept is significant and you can say that $4.87$ is the expected value of y when x == mean(x).

Answer 2

Don't know about forcing it to zero. The result you got is allowing you to say that the intercept term is very probably not different from zero (the zero value is included into your confidence interval, if you look at it).

I don't think it is a good idea to force your model (DI ~ 0 + II, if I understand correctly), as models built that way will always give you a higher $R^2$ value.