How to explain the result of ANCOVA and linear regression

Question

Currently I am learning ANCOVA, but I'm confused with the result.

I created a linear regression model using mtcars like this:

summary(lm(qsec ~ wt+factor(am), data = mtcars))

The output is:

Call:
lm(formula = qsec ~ wt + factor(am), data = mtcars)

Residuals:
    Min      1Q  Median      3Q     Max 
-2.6898 -1.3063  0.0167  1.1398  3.9917 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  22.5990     1.5596  14.490 8.17e-15 ***
wt           -1.1716     0.4025  -2.911  0.00685 ** 
factor(am)1  -2.4141     0.7892  -3.059  0.00474 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.582 on 29 degrees of freedom
Multiple R-squared:  0.267, Adjusted R-squared:  0.2165 
F-statistic: 5.283 on 2 and 29 DF,  p-value: 0.01106

As you see, the p value of wt showed 0.00685, which meaned a strong linear correlation between wt and qsec, as well as am.

But when I ran aov code:

summary(aov(qsec ~ wt+factor(am), data = mtcars))

With the output:

            Df Sum Sq Mean Sq F value  Pr(>F)   
wt           1   3.02   3.022   1.208 0.28081   
factor(am)   1  23.41  23.413   9.358 0.00474 **
Residuals   29  72.55   2.502                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

It seems like there was no effect from wt on qsec.

Does it mean that a strong linear correlation between wt and qsec could be confirmed but there is no great effect from wt on qsec? Is my explanation appropriate?

Answer 1

First drop the factor since am only has two values so making it a factor will not have any effect on the results.

Now regarding the tests to get the p values they are different. For lm the wt p value is based on the comparison of these two models

qsec ~ am
qsec ~ wt + am

so we have

anova(lm(qsec ~ am, mtcars), lm(qsec ~ mt + am, mtcars))
## Model 1: qsec ~ am
## Model 2: qsec ~ wt + am
##   Res.Df    RSS Df Sum of Sq      F   Pr(>F)   
## 1     30 93.758                                
## 2     29 72.554  1    21.204 8.4753 0.006854 ** <-- 0.00685

summary(lm(qsec ~ wt + am, mtcars))
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  22.5990     1.5596  14.490 8.17e-15 ***
## wt           -1.1716     0.4025  -2.911  0.00685 ** <-- 0.00685
## am           -2.4141     0.7892  -3.059  0.00474 ** 

whereas aov is really meant for balanced designs where the terms are orthogonal and if not, as here, then it conceptually orthogonalizes them sequentially so in that case the comparison is effectively between these two models

qsec ~ r.am
qsec ~ r.wt + r.am

where r.wt is the portion of wt orthogonal to the intercept and r.am is the portion of am orthogonal to wt and the intercept so we have

r.wt <- resid(lm(wt ~ 1, mtcars))
r.am <- resid(lm(am ~ wt, mtcars))

anova(lm(qsec ~ r.am, mtcars), lm(qsec ~ r.wt + r.am, mtcars))
## Model 1: qsec ~ r.am
## Model 2: qsec ~ r.wt + r.am
##   Res.Df    RSS Df Sum of Sq      F Pr(>F)
## 1     30 75.576                           
## 2     29 72.554  1    3.0217 1.2078 0.2808 <------- 0.2808

summary(aov(qsec ~ wt + am, mtcars))
##             Df Sum Sq Mean Sq F value  Pr(>F)   
## wt           1   3.02   3.022   1.208 0.28081 <------- 0.28081
## am           1  23.41  23.413   9.358 0.00474 **
## Residuals   29  72.55   2.502                   

It would also be possible to demonstrate this by performing Gram Schmidt on the cbind(1, wt, am) model matrix to make the columns orthogonal. The pracma package has a Gram Schmidt routine.