Linear Regression and ANOVA shaken and stirred (Part 1)

Linear models

These concepts are understood as separate concepts most of the times. The truth is they are extremely related to each other.

Author

Mauricio “Pachá” Vargas S.

Published

March 20, 2017

Updated 2022-05-28: I am moving the blog to Quarto. The update was to load the packages directly instead of using “pacman::p_load”.

Motivation

Linear Regression and ANOVA concepts are understood as separate concepts most of the times. The truth is that they are extremely related to each other, being ANOVA a particular case of Linear Regression.

Even worse, it’s quite common that students do memorize equations and tests instead of trying to understand Linear Algebra and Statistics concepts that can keep you away from misleading results, but that is material for another entry.

Most textbooks present econometric concepts and algebraic steps and do emphasize about the relationship between Ordinary Least Squares, Maximum Likelihood and other methods to obtain estimates in Linear Regression.

Here I present a combination of algebra and R commands to try to clarify some concepts.

Linear Regression

Let \(\renewcommand{\vec}[1]{\boldsymbol{#1}} \newcommand{\R}{\mathbb{R}} \vec{y} \in \R^n\) be the outcome and \(X \in \mathbb{R}^{n\times p}\) be the design matrix in the context of a linear model with intercept: \[\vec{y} = X\vec{\beta} + \vec{e}\]

Being: \[ \underset{n\times 1}{\vec{y}} = \begin{pmatrix}y_0 \cr y_1 \cr \vdots \cr y_n\end{pmatrix} \text{ and } \underset{n\times p}{X} = \begin{pmatrix}1 & x_{11} & & x_{1p} \cr 1 & x_{21} & & x_{2p} \cr & \ddots & \cr 1 & x_{n1} & & x_{np}\end{pmatrix} = (\vec{1} \: \vec{x}_1 \: \ldots \: \vec{x}_p) \]

In linear models the aim is to minimize the error term by choosing \(\hat{\vec{\beta}}\). One possibility is to minimize the squared error by solving this optimization problem: \[\begin{gather}\label{min} \displaystyle \min_{\vec{\beta}} S = \|\vec{y} - X\vec{\beta}\|^2 \tag{*} \end{gather}\]

Books such as Baltagi discuss how to solve \(\eqref{min}\) and other equivalent approaches that result in this optimal estimator: \[\begin{gather} \label{beta} \hat{\vec{\beta}} = (X^tX)^{-1} X^t\vec{y} \tag{**} \end{gather}\]

With one independent variable and intercept, this is \(y_i = \beta_0 + \beta_1 x_{i1} + e_i\), equation \(\eqref{beta}\) means: \[\begin{gather}\label{beta2} \hat{\beta}_1 = cor(\vec{y},\vec{x}) \cdot \frac{sd(\vec{y})}{sd(\vec{x})} \text{ and } \hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{\vec{x}} \tag{***} \end{gather}\]

Coding example with mtcars dataset

Consider the model: \[mpg_i = \beta_1 wt_i + \beta_2 cyl_i + e_i\]

This is how to write that model in R notation:

lm(mpg ~ wt + cyl, data = mtcars)


Call:
lm(formula = mpg ~ wt + cyl, data = mtcars)

Coefficients:
(Intercept)           wt          cyl  
     39.686       -3.191       -1.508

Or written in matrix form:

y <- mtcars$mpg
x0 <- rep(1, length(y))
x1 <- mtcars$wt
x2 <- mtcars$cyl
X <- cbind(x0,x1,x2)

It’s the same to use lm or to perform a matrix multiplication because of equation \(\eqref{beta}\):

fit <- lm(y ~ x1 + x2)
coefficients(fit)

(Intercept)          x1          x2 
  39.686261   -3.190972   -1.507795

beta <- solve(t(X)%*%X) %*% (t(X)%*%y)
beta

        [,1]
x0 39.686261
x1 -3.190972
x2 -1.507795

Coding example with Galton dataset

Equation \(\eqref{beta2}\) can be verified with R commands:

library(HistData)

# read the documentation
# ??Galton

y <- Galton$child
x <- Galton$parent
beta1 <- cor(y, x) *  sd(y) / sd(x)
beta0 <- mean(y) - beta1 * mean(x)

c(beta0, beta1)

[1] 23.9415302  0.6462906

#comparing with lm results
lm(y ~ x)


Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
    23.9415       0.6463

Coding example with mtcars dataset and mean centered regression

Another possibility in linear models is to rewrite the observations in the outcome and the design matrix with respect to the mean of each variable. That will only alter the intercept but not the slope coefficients.

For a model like \(y_i = \beta_0 + \beta_1 x_{i1} + \beta_2 x_{i2} + e_i\) I can write the equivalent model: \[y_i - \bar{y} = \beta_0 + \beta_1 (x_{i1} - \bar{x}_{i1}) + \beta_2 (x_{i2} - \bar{x}_{i2}) + e_i\]

Another possibility is to consider that \(\bar{y}_i = \beta_0 + \beta_1 \bar{x}_{i1} + \beta_2 \bar{x}_{i2} + 0\) under the classical assumption \(\bar{e}_i = 0\) and substracting I obtain: \[y_i - \bar{y} = \beta_1 (x_{i1} - \bar{x}_{i1}) + \beta_2 (x_{i2} - \bar{x}_{i2}) + e_i\]

I’ll analyze the first case, without dropping \(\beta_0\) unless there’s statistical evidence to show its not significant.

In R notation the model \(y_i - \bar{y} = \beta_0 + \beta_1 (x_{i1} - \bar{x}_{i1}) + \beta_2 (x_{i2} - \bar{x}_{i2}) + e_i\) can be fitted in this way:

# read the documentation
# ??mtcars

new_y <- mtcars$mpg - mean(mtcars$mpg)
new_x1 <- mtcars$wt - mean(mtcars$wt)
new_x2 <- mtcars$cyl - mean(mtcars$cyl)

fit2 <- lm(new_y ~ new_x1 + new_x2)
coefficients(fit2)

  (Intercept)        new_x1        new_x2 
 6.721193e-16 -3.190972e+00 -1.507795e+00

new_X <- cbind(x0,new_x1,new_x2)
new_beta <- solve(t(new_X)%*%new_X) %*% (t(new_X)%*%new_y)
new_beta

                [,1]
x0      6.879624e-16
new_x1 -3.190972e+00
new_x2 -1.507795e+00

Here the intercept is close to zero, so I can obtain more information to check the significance:

summary(fit2)


Call:
lm(formula = new_y ~ new_x1 + new_x2)

Residuals:
    Min      1Q  Median      3Q     Max 
-4.2893 -1.5512 -0.4684  1.5743  6.1004 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  6.721e-16  4.539e-01   0.000 1.000000    
new_x1      -3.191e+00  7.569e-01  -4.216 0.000222 ***
new_x2      -1.508e+00  4.147e-01  -3.636 0.001064 ** 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.568 on 29 degrees of freedom
Multiple R-squared:  0.8302,    Adjusted R-squared:  0.8185 
F-statistic: 70.91 on 2 and 29 DF,  p-value: 6.809e-12

In this particular case I should drop the intercept because it’s not significant so I write:

fit3 <- lm(new_y ~ new_x1 + new_x2 - 1)
coefficients(fit3)

   new_x1    new_x2 
-3.190972 -1.507795

new_X <- cbind(new_x1,new_x2)
new_beta <- solve(t(new_X)%*%new_X) %*% (t(new_X)%*%new_y)
new_beta

            [,1]
new_x1 -3.190972
new_x2 -1.507795

Residuals

The total sum of squares is defined as the sum of explained and residual (or unexplained) sum of squares or, in other words, the sum of explained and unexplained variance in the model: \[TSS = ESS + RSS = \sum_i (\hat{y}_i - \bar{y})^2 + \sum_i (y_i - \hat{y}_i)^2 = \sum_i (y_i - \bar{y})^2 \] Being \(\hat{\vec{y}} = X\hat{\vec{\beta}}\).

Here \(TSS\) follows a \(F(p,n-1)\) distribution with \(n-1\) degrees of freedom. This is, \(ESS\) has \(p\) degrees of freedom and \(RSS\) has \(n-p-1\) degrees of freedom and the F-statistic is: \[ F = \frac{ESS/p}{RSS/(n-p-1)} \] This statistic tests the null hypothesis \(\vec{\beta} = \vec{0}\). This is, the F-statistic provides information about the joint effect of all the variables in the model together and therefore p-values are required to determine single coefficients’ significance.

ANOVA

The term analysis of variance refers to categorical predictors, therefore ANOVA is a particular case of the linear model that works around the statistical test just described and the difference in group means.

ANOVA is a particular case of the linear model where the predictors (or independent variables) are dummy variables that reflect if an observation belongs to a certain group. An example of this would be \(x_{i1} = 1\) if observation \(i\) belongs to a group of interest (e.g. the interviewed person is in the group of people who has a Twitter account) and \(x_{i1} = 0\) otherwise.

The null hypothesis in ANOVA is “group means are all equal” as I’ll explain with examples. This comes from the fact that regression coefficients in ANOVA measure the effect of belonging to a group, and as it was explained about the F test, you can examinate the associated p-value to a regression coefficient to check if the group effect is statistically different from zero. For example, if you have a group of people who uses social networks and a subgroup of people who use Twitter, then if the dummy variable that expresses that Twitter usage has a significative regression coefficient, then you have to evidence to state that group means are different.

An example with mtcars dataset

In the mtcars dataset, am can be useful to explain ANOVA as its observations are defined as: \[ am_i = \begin{cases}1 &\text{ if car } i \text{ is manual} \cr 0 &\text{ if car } i \text{ is automatic}\end{cases} \]

Case 1

Consider a model where the outcome is mpg and the design matrix is \(X = (\vec{x}_1 \: \vec{x}_2)\) so that the terms are defined in this way:

y <- mtcars$mpg
x1 <- mtcars$am 
x2 <- ifelse(x1 == 1, 0, 1)

This is: \[ x_1 = \begin{cases}1 &\text{ if car } i \text{ is manual} \cr 0 &\text{ if car } i \text{ is automatic}\end{cases} \quad \quad x_2 = \begin{cases}1 &\text{ if car } i \text{ is automatic} \cr 0 &\text{ if car } i \text{ is manual}\end{cases} \]

The estimates without intercept would be:

fit <- lm(y ~ x1 + x2 - 1)
fit$coefficients

      x1       x2 
24.39231 17.14737

Taking \(\eqref{beta}\) and replacing in this particular case would result in this estimate: \[ \hat{\vec{\beta}} = \begin{bmatrix}\bar{y}_1 \cr \bar{y}_2 \end{bmatrix} \] being \(\bar{y}_1\) and \(\bar{y}_2\) the group means.

This can be verified with R commands:

y1 <- y*x1; y1 <- ifelse(y1 == 0, NA, y1)
y2 <- y*x2; y2 <- ifelse(y2 == 0, NA, y2)

mean(y1, na.rm = TRUE)

[1] 24.39231

mean(y2, na.rm = TRUE)

[1] 17.14737

If you are not convinced of this result you can write down the algebra or use R commands. I’ll do the last with the notation \(U = (X^tX)^{-1}\) and \(V = X^t\vec{y}\):

X <- cbind(x1,x2)
U <- solve(t(X)%*%X)
V <- t(X)%*%y

U;V;U%*%V

           x1         x2
x1 0.07692308 0.00000000
x2 0.00000000 0.05263158

    [,1]
x1 317.1
x2 325.8

       [,1]
x1 24.39231
x2 17.14737

\(U\) entries are just one over the number of observations of each group and V entries are the sum of mpg observations of each group so that the entries of \(UV\) are the means of each group:

u11 <- 1/sum(x1)
u22 <- 1/sum(x2)

v11 <- sum(y1, na.rm = TRUE)
v21 <- sum(y2, na.rm = TRUE)

u11;u22

[1] 0.07692308

[1] 0.05263158

v11;v21

[1] 317.1

[1] 325.8

u11*v11;u22*v21

[1] 24.39231

[1] 17.14737

Aside from algebra, now I’ll show the equivalency between lm and aov that is the command used to perform an analysis of variance:

y <- mtcars$mpg
x1 <- mtcars$am
x2 <- ifelse(x1 == 1, 0, 1)

fit2 <- aov(y ~ x1 + x2 - 1)
fit2$coefficients

      x1       x2 
24.39231 17.14737

Case 2

Changing the design matrix to \(X = (\vec{1} \: \vec{x}_1)\) will lead to the estimate: \[ \hat{\vec{\beta}} = \begin{bmatrix}\bar{y}_2 \cr \bar{y}_1 - \bar{y}_2 \end{bmatrix} \] Fitting the model results in:

y <- mtcars$mpg
x1 <- mtcars$am

fit <- lm(y ~ x1)
fit$coefficients

(Intercept)          x1 
  17.147368    7.244939

So to see the relationship between the estimates and the group means I need additional steps:

x0 <- rep(1,length(y))
X <- cbind(x0,x1)
beta <- solve(t(X)%*%X) %*% (t(X)%*%y)
beta

        [,1]
x0 17.147368
x1  7.244939

I obtained the same estimates with the lm command so now I calculate the group means:

x2 <- ifelse(x1 == 1, 0, 1)

x1 <- ifelse(x1 == 0, NA, x1)
x2 <- ifelse(x2 == 0, NA, x2)

m1 <- mean(y*x1, na.rm = TRUE)
m2 <- mean(y*x2, na.rm = TRUE)

beta0 <- m2
beta1 <- m1 - m2

beta0;beta1

[1] 17.14737

[1] 7.244939

This means that the slope for the two groups is the same but the intercept is different, and therefore exists a positive effect of manual transmission on miles per gallon in average terms.

Again I’ll verify the equivalency between lm and aov in this particular case:

y <- mtcars$mpg
x1 <- mtcars$am
x2 <- ifelse(x1 == 1, 0, 1)

fit2 <- aov(y ~ x1)
fit2$coefficients

(Intercept)          x1 
  17.147368    7.244939

A simpler way to write the model is:

fit3 <- lm(mpg ~ am, data = mtcars)
summary(fit3)


Call:
lm(formula = mpg ~ am, data = mtcars)

Residuals:
    Min      1Q  Median      3Q     Max 
-9.3923 -3.0923 -0.2974  3.2439  9.5077 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   17.147      1.125  15.247 1.13e-15 ***
am             7.245      1.764   4.106 0.000285 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 4.902 on 30 degrees of freedom
Multiple R-squared:  0.3598,    Adjusted R-squared:  0.3385 
F-statistic: 16.86 on 1 and 30 DF,  p-value: 0.000285

I can calculate the residuals by hand:

mean_mpg <- mean(mtcars$mpg)
fitted_mpg <- fit3$coefficients[1] + fit3$coefficients[2]*mtcars$am
observed_mpg <- mtcars$mpg

TSS <- sum((observed_mpg - mean_mpg)^2) 
ESS <- sum((fitted_mpg - mean_mpg)^2)
RSS <- sum((observed_mpg - fitted_mpg)^2)

TSS;ESS;RSS

[1] 1126.047

[1] 405.1506

[1] 720.8966

Here its verified that \(TSS = ESS + RSS\) but aside from that I can extract information from aov:

summary(fit2)

            Df Sum Sq Mean Sq F value   Pr(>F)    
x1           1  405.2   405.2   16.86 0.000285 ***
Residuals   30  720.9    24.0                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

And check that, as expected, \(ESS\) is the variance explained by x1.

I also can run ANOVA over lm with:

anova(fit3)

Analysis of Variance Table

Response: mpg
          Df Sum Sq Mean Sq F value   Pr(>F)    
am         1 405.15  405.15   16.86 0.000285 ***
Residuals 30 720.90   24.03                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The table provides information on the effect of am over mpg. In this case the null hypothesis is rejected because of the large F-value and the associated p-values.

Considering a 0.05 significance threshold I can say, with 95% of confidence, that the regression slope is statistically different from zero or that there is a difference in group means between automatic and manual transmission.