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

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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.

marked as duplicate by gung♦, Andy, whuber♦May 19 '14 at 14:55

• Have you tried centering II? – Sergio May 19 '14 at 14:14
• In the particular context of the model, do you care if the intercept is significant or not? What kind of conclusions are you trying to make? – Aaron May 19 '14 at 14:15
• @Aaron - 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. – hoof_hearted May 19 '14 at 14:19
• @Sergio - No, I have not tried centring it. My overall aim is to find a relationship between two datasets (i.e. if I have a value for DI, I am able to find out the corresponding value of II). Is centring appropriate to use if I am trying to do this? – hoof_hearted May 19 '14 at 14:22
• @gung - Yes I did see this but I'm not sure it addresses my problem. I want to know if I should drop the Intercept or not, based on the output I have. The lm results suggest that the Intercept isn't significant. On the other hand, by forcing the model through zero (and dropping the Intercept), I am changing the interoperation of the results. – hoof_hearted May 19 '14 at 14:26

Ledt 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 can result 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).

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.