Why don't the following contrasts and linear models behave as I expected?

Question

I am trying to understand how contrasts work, so I ran a small simulation using the following R code:

hsb2 <- read_csv("http://www.ats.ucla.edu/stat/data/hsb2.csv")
hsb2$race.f <- factor(hsb2$race, labels=c("Hispanic", "Asian", "African-Am", "Caucasian"))

# method 1
(contrasts(hsb2$race.f) <- matrix(c(0,1,0,0,0,0,1,0,0,0,0,1),ncol = 3, byrow=FALSE))
     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    1    0    0
[3,]    0    1    0
[4,]    0    0    1
m1 <- lm(write ~ race.f, hsb2)


# method 2
(contrasts(hsb2$race.f) <- matrix(c(-1,1,0,0,-1,0,1,0,-1,0,0,1),ncol = 3, byrow=FALSE))
     [,1] [,2] [,3]
[1,]   -1   -1   -1
[2,]    1    0    0
[3,]    0    1    0
[4,]    0    0    1
m2 <- lm(write ~ 0 + race.f, hsb2)

I was expecting that method 1 and method 2 return the same set of coefficients because:

• In method 1, the three groups are compared to the reference group, and the contrast reflects that
• In method 2, I planned to take out the intercept in the linear model, so that all factors are in the model. To emulate the scenario in the method1, I encoded that information in the contrast.

But I didn't get the same set of coefficients:

summary(m1) returns:

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   46.458      1.842  25.218  < 2e-16 ***
race.f1       11.542      3.286   3.512 0.000552 ***
race.f2        1.742      2.732   0.637 0.524613    
race.f3        7.597      1.989   3.820 0.000179 ***

summary(m2) returns:

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
race.fHispanic    46.4583     1.8422   25.22   <2e-16 ***
race.fAsian       58.0000     2.7212   21.31   <2e-16 ***
race.fAfrican-Am  48.2000     2.0181   23.88   <2e-16 ***
race.fCaucasian   54.0552     0.7495   72.12   <2e-16 ***

Clearly, I am not understanding contrasts correctly. Where do my reasoning fail and what methods would have made the contrast 1 and contrast 2 produce the same results(or is it just a complete nonsense?)

Answer 1

Using the 0+ syntax in the model specification will bypass the defined contrasts, and give you separately estimated coefficients for each level. Consider the examples below - you get the same estimates when you remove the intercept (0+) no matter what contrasts you use:

#Dummy coding
> contr.treatment(4)
  2 3 4
1 0 0 0
2 1 0 0
3 0 1 0
4 0 0 1

> contrasts(hsb2$race.f) = contr.treatment(4)
#model with intercept
> summary(lm(write ~ race.f, data=hsb2))

Call:
lm(formula = write ~ race.f, data = hsb2)

Residuals:
     Min       1Q   Median       3Q      Max 
-23.0552  -5.4583   0.9724   7.0000  18.8000 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   46.458      1.842  25.218  < 2e-16 ***
race.f2       11.542      3.286   3.512 0.000552 ***
race.f3        1.742      2.732   0.637 0.524613    
race.f4        7.597      1.989   3.820 0.000179 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 9.025 on 196 degrees of freedom
Multiple R-squared:  0.1071,    Adjusted R-squared:  0.0934 
F-statistic: 7.833 on 3 and 196 DF,  p-value: 5.785e-05

# removing intercept
> summary(lm(write ~ 0 + race.f, data=hsb2))

Call:
lm(formula = write ~ 0 + race.f, data = hsb2)

Residuals:
     Min       1Q   Median       3Q      Max 
-23.0552  -5.4583   0.9724   7.0000  18.8000 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
race.fHispanic    46.4583     1.8422   25.22   <2e-16 ***
race.fAsian       58.0000     2.7212   21.31   <2e-16 ***
race.fAfrican-Am  48.2000     2.0181   23.88   <2e-16 ***
race.fCaucasian   54.0552     0.7495   72.12   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 9.025 on 196 degrees of freedom
Multiple R-squared:  0.9722,    Adjusted R-squared:  0.9717 
F-statistic:  1716 on 4 and 196 DF,  p-value: < 2.2e-16

# Deviation coding
> contr.sum(4)
  [,1] [,2] [,3]
1    1    0    0
2    0    1    0
3    0    0    1
4   -1   -1   -1

> contrasts(hsb2$race.f) = contr.sum(4)
# model with intercept
> summary(lm(write ~ race.f, data=hsb2))

Call:
lm(formula = write ~ race.f, data = hsb2)

Residuals:
     Min       1Q   Median       3Q      Max 
-23.0552  -5.4583   0.9724   7.0000  18.8000 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  51.6784     0.9821  52.619  < 2e-16 ***
race.f1      -5.2200     1.6314  -3.200  0.00160 ** 
race.f2       6.3216     2.1603   2.926  0.00384 ** 
race.f3      -3.4784     1.7323  -2.008  0.04602 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 9.025 on 196 degrees of freedom
Multiple R-squared:  0.1071,    Adjusted R-squared:  0.0934 
F-statistic: 7.833 on 3 and 196 DF,  p-value: 5.785e-05

#model without intercept
> summary(lm(write ~ 0 + race.f, data=hsb2))

Call:
lm(formula = write ~ 0 + race.f, data = hsb2)

Residuals:
     Min       1Q   Median       3Q      Max 
-23.0552  -5.4583   0.9724   7.0000  18.8000 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
race.fHispanic    46.4583     1.8422   25.22   <2e-16 ***
race.fAsian       58.0000     2.7212   21.31   <2e-16 ***
race.fAfrican-Am  48.2000     2.0181   23.88   <2e-16 ***
race.fCaucasian   54.0552     0.7495   72.12   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 9.025 on 196 degrees of freedom
Multiple R-squared:  0.9722,    Adjusted R-squared:  0.9717 
F-statistic:  1716 on 4 and 196 DF,  p-value: < 2.2e-16