# Chapter 5 Linear Models

## 5.1 Problem Setup

Example 5.1 (Bottle Cap Production) Consider a randomized experiment designed to study the effects of temperature and pressure on the diameter of manufactured a bottle cap.
Example 5.2 (Rental Prices) Consider the prediction of rental prices given an appartment’s attributes.

Both examples require some statistical model, but they are very different. The first is a causal inference problem: we want to design an intervention so that we need to recover the causal effect of temperature and pressure. The second is a prediction problem, a.k.a. a forecasting problem, in which we don’t care about the causal effects, we just want good predictions.

In this chapter we discuss the causal problem in Example 5.1. This means that when we assume a model, we assume it is the actual data generating process, i.e., we assume the sampling distribution is well specified. The second type of problems is discussed in the Supervised Learning Chapter 9.

Here are some more examples of the types of problems we are discussing.

Example 5.3 (Plant Growth) Consider the treatment of various plants with various fertilizers to study the fertilizer’s effect on growth.
Example 5.4 (Return to Education) Consider the study of return to education by analyzing the incomes of individuals with different education years.
Example 5.5 (Drug Effect) Consider the study of the effect of a new drug for hemophilia, by analyzing the level of blood coagulation after the administration of various amounts of the new drug.

Let’s present the linear model. We assume that a response6 variable is the sum of effects of some factors7. Denoting the response variable by $$y$$, the factors by $$x=(x_1,\dots,x_p)$$, and the effects by $$\beta:=(\beta_1,\dots,\beta_p)$$ the linear model assumption implies that the expected response is the sum of the factors effects:

\begin{align} E[y]=x_1 \beta_1 + \dots + x_p \beta_p = \sum_{j=1}^p x_j \beta_j = x'\beta . \tag{5.1} \end{align} Clearly, there may be other factors that affect the the caps’ diameters. We thus introduce an error term8, denoted by $$\varepsilon$$, to capture the effects of all unmodeled factors and measurement error9. The implied generative process of a sample of $$i=1,\dots,n$$ observations it thus \begin{align} y_i = x_i'\beta + \varepsilon_i = \sum_j x_{i,j} \beta_j + \varepsilon_i , i=1,\dots,n . \tag{5.2} \end{align} or in matrix notation \begin{align} y = X \beta + \varepsilon . \tag{5.3} \end{align}

Let’s demonstrate Eq.(5.2). In our cap example (5.1), assuming that pressure and temperature have two levels each (say, high and low), we would write $$x_{i,1}=1$$ if the pressure of the $$i$$’th measurement was set to high, and $$x_{i,1}=-1$$ if the pressure was set to low. Similarly, we would write $$x_{i,2}=1$$, and $$x_{i,2}=-1$$, if the temperature was set to high, or low, respectively. The coding with $$\{-1,1\}$$ is known as effect coding. If you prefer coding with $$\{0,1\}$$, this is known as dummy coding. The choice of coding has no real effect on the results, provided that you remember what coding you used when interpreting $$\hat \beta$$.

Remark. In Galton’s classical regression problem, where we try to seek the relation between the heights of sons and fathers then $$p=1$$, $$y_i$$ is the height of the $$i$$’th father, and $$x_i$$ the height of the $$i$$’th son.

There are many reasons linear models are very popular:

1. Before the computer age, these were pretty much the only models that could actually be computed10. The whole Analysis of Variance (ANOVA) literature is an instance of linear models, that relies on sums of squares, which do not require a computer to work with.

2. For purposes of prediction, where the actual data generating process is not of primary importance, they are popular because they simply work. Why is that? They are simple so that they do not require a lot of data to be computed. Put differently, they may be biased, but their variance is small enough to make them more accurate than other models.

3. For non continuous predictors, any functional relation can be cast as a linear model.

4. For the purpose of screening, where we only want to show the existence of an effect, and are less interested in the magnitude of that effect, a linear model is enough.

5. If the true generative relation is not linear, but smooth enough, then the linear function is a good approximation via Taylor’s theorem.

There are still two matters we have to attend: (i) How the estimate $$\beta$$? (ii) How to perform inference?

In the simplest linear models the estimation of $$\beta$$ is done using the method of least squares. A linear model with least squares estimation is known as Ordinary Least Squares (OLS). The OLS problem:

\begin{align} \hat \beta:= argmin_\beta \{ \sum_i (y_i-x_i'\beta)^2 \}, \tag{5.4} \end{align} and in matrix notation \begin{align} \hat \beta:= argmin_\beta \{ \Vert y-X\beta \Vert^2_2 \}. \tag{5.5} \end{align}
Remark. Personally, I prefer the matrix notation because it is suggestive of the geometry of the problem. The reader is referred to Friedman, Hastie, and Tibshirani (2001), Section 3.2, for more on the geometry of OLS.

Different software suits, and even different R packages, solve Eq.(5.4) in different ways so that we skip the details of how exactly it is solved. These are discussed in Chapters 16 and 17.

The last matter we need to attend is how to do inference on $$\hat \beta$$. For that, we will need some assumptions on $$\varepsilon$$. A typical set of assumptions is the following:

1. Independence: we assume $$\varepsilon_i$$ are independent of everything else. Think of them as the measurement error of an instrument: it is independent of the measured value and of previous measurements.
2. Centered: we assume that $$E[\varepsilon]=0$$, meaning there is no systematic error, sometimes it called The “Linearity assumption”.
3. Normality: we will typically assume that $$\varepsilon \sim \mathcal{N}(0,\sigma^2)$$, but we will later see that this is not really required.

We emphasize that these assumptions are only needed for inference on $$\hat \beta$$ and not for the estimation itself, which is done by the purely algorithmic framework of OLS.

Given the above assumptions, we can apply some probability theory and linear algebra to get the distribution of the estimation error: \begin{align} \hat \beta - \beta \sim \mathcal{N}(0, (X'X)^{-1} \sigma^2). \tag{5.6} \end{align}

The reason I am not too strict about the normality assumption above, is that Eq.(5.6) is approximately correct even if $$\varepsilon$$ is not normal, provided that there are many more observations than factors ($$n \gg p$$).

## 5.2 OLS Estimation in R

We are now ready to estimate some linear models with R. We will use the whiteside data from the MASS package, recording the outside temperature and gas consumption, before and after an apartment’s insulation.

library(MASS) # load the package
library(data.table) # for some data manipulations
data(whiteside) # load the data
head(whiteside) # inspect the data
##    Insul Temp Gas
## 1 Before -0.8 7.2
## 2 Before -0.7 6.9
## 3 Before  0.4 6.4
## 4 Before  2.5 6.0
## 5 Before  2.9 5.8
## 6 Before  3.2 5.8

We do the OLS estimation on the pre-insulation data with lm function, possibly the most important function in R.

whiteside <- data.table(whiteside)
lm.1 <- lm(Gas~Temp, data=whiteside[Insul=='Before']) # OLS estimation 

Things to note:

• We used the tilde syntax Gas~Temp, reading “gas as linear function of temperature”.
• The data argument tells R where to look for the variables Gas and Temp. We used Insul=='Before' to subset observations before the insulation.
• The result is assigned to the object lm.1.

Like any other language, spoken or programable, there are many ways to say the same thing. Some more elegant than others…

lm.1 <- lm(y=Gas, x=Temp, data=whiteside[whiteside$Insul=='Before',]) lm.1 <- lm(y=whiteside[whiteside$Insul=='Before',]$Gas,x=whiteside[whiteside$Insul=='Before',]$Temp) lm.1 <- whiteside[whiteside$Insul=='Before',] %>% lm(Gas~Temp, data=.)

The output is an object of class lm.

class(lm.1)
## [1] "lm"

Objects of class lm are very complicated. They store a lot of information which may be used for inference, plotting, etc. The str function, short for “structure”, shows us the various elements of the object.

str(lm.1)
## List of 12
##  $coefficients : Named num [1:2] 6.854 -0.393 ## ..- attr(*, "names")= chr [1:2] "(Intercept)" "Temp" ##$ residuals    : Named num [1:26] 0.0316 -0.2291 -0.2965 0.1293 0.0866 ...
##   ..- attr(*, "names")= chr [1:26] "1" "2" "3" "4" ...
##  $effects : Named num [1:26] -24.2203 -5.6485 -0.2541 0.1463 0.0988 ... ## ..- attr(*, "names")= chr [1:26] "(Intercept)" "Temp" "" "" ... ##$ rank         : int 2
##  $fitted.values: Named num [1:26] 7.17 7.13 6.7 5.87 5.71 ... ## ..- attr(*, "names")= chr [1:26] "1" "2" "3" "4" ... ##$ assign       : int [1:2] 0 1
##  $qr :List of 5 ## ..$ qr   : num [1:26, 1:2] -5.099 0.196 0.196 0.196 0.196 ...
##   .. ..- attr(*, "dimnames")=List of 2
##   .. .. ..$: chr [1:26] "1" "2" "3" "4" ... ## .. .. ..$ : chr [1:2] "(Intercept)" "Temp"
##   .. ..- attr(*, "assign")= int [1:2] 0 1
##   ..$qraux: num [1:2] 1.2 1.35 ## ..$ pivot: int [1:2] 1 2
##   ..$tol : num 1e-07 ## ..$ rank : int 2
##   ..- attr(*, "class")= chr "qr"
##  $df.residual : int 24 ##$ xlevels      : Named list()
##  $call : language lm(formula = Gas ~ Temp, data = whiteside[Insul == "Before"]) ##$ terms        :Classes 'terms', 'formula'  language Gas ~ Temp
##   .. ..- attr(*, "variables")= language list(Gas, Temp)
##   .. ..- attr(*, "factors")= int [1:2, 1] 0 1
##   .. .. ..- attr(*, "dimnames")=List of 2
##   .. .. .. ..$: chr [1:2] "Gas" "Temp" ## .. .. .. ..$ : chr "Temp"
##   .. ..- attr(*, "term.labels")= chr "Temp"
##   .. ..- attr(*, "order")= int 1
##   .. ..- attr(*, "intercept")= int 1
##   .. ..- attr(*, "response")= int 1
##   .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
##   .. ..- attr(*, "predvars")= language list(Gas, Temp)
##   .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
##   .. .. ..- attr(*, "names")= chr [1:2] "Gas" "Temp"
##  $model :'data.frame': 26 obs. of 2 variables: ## ..$ Gas : num [1:26] 7.2 6.9 6.4 6 5.8 5.8 5.6 4.7 5.8 5.2 ...
##   ..$Temp: num [1:26] -0.8 -0.7 0.4 2.5 2.9 3.2 3.6 3.9 4.2 4.3 ... ## ..- attr(*, "terms")=Classes 'terms', 'formula' language Gas ~ Temp ## .. .. ..- attr(*, "variables")= language list(Gas, Temp) ## .. .. ..- attr(*, "factors")= int [1:2, 1] 0 1 ## .. .. .. ..- attr(*, "dimnames")=List of 2 ## .. .. .. .. ..$ : chr [1:2] "Gas" "Temp"
##   .. .. .. .. ..$: chr "Temp" ## .. .. ..- attr(*, "term.labels")= chr "Temp" ## .. .. ..- attr(*, "order")= int 1 ## .. .. ..- attr(*, "intercept")= int 1 ## .. .. ..- attr(*, "response")= int 1 ## .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> ## .. .. ..- attr(*, "predvars")= language list(Gas, Temp) ## .. .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric" ## .. .. .. ..- attr(*, "names")= chr [1:2] "Gas" "Temp" ## - attr(*, "class")= chr "lm" In RStudio it is particularly easy to extract objects. Just write your.object$ and press tab after the  for autocompletion. If we only want $$\hat \beta$$, it can also be extracted with the coef function. coef(lm.1) ## (Intercept) Temp ## 6.8538277 -0.3932388 Things to note: • R automatically adds an (Intercept) term. This means we estimate $$Gas=\beta_0 + \beta_1 Temp + \varepsilon$$ and not $$Gas=\beta_1 Temp + \varepsilon$$. This makes sense because we are interested in the contribution of the temperature to the variability of the gas consumption about its mean, and not about zero. • The effect of temperature, i.e., $$\hat \beta_1$$, is -0.39. The negative sign means that the higher the temperature, the less gas is consumed. The magnitude of the coefficient means that for a unit increase in the outside temperature, the gas consumption decreases by 0.39 units. We can use the predict function to make predictions, but we emphasize that if the purpose of the model is to make predictions, and not interpret coefficients, better skip to the Supervised Learning Chapter 9. plot(predict(lm.1)~whiteside[Insul=='Before',Gas]) abline(0,1, lty=2) The model seems to fit the data nicely. A common measure of the goodness of fit is the coefficient of determination, more commonly known as the $$R^2$$. Definition 5.1 (R2) The coefficient of determination, denoted $$R^2$$, is defined as \begin{align} R^2:= 1-\frac{\sum_i (y_i - \hat y_i)^2}{\sum_i (y_i - \bar y)^2}, \end{align} where $$\hat y_i$$ is the model’s prediction, $$\hat y_i = x_i \hat \beta$$. It can be easily computed R2 <- function(y, y.hat){ numerator <- (y-y.hat)^2 %>% sum denominator <- (y-mean(y))^2 %>% sum 1-numerator/denominator } R2(y=whiteside[Insul=='Before',Gas], y.hat=predict(lm.1)) ## [1] 0.9438081 This is a nice result implying that about $$94\%$$ of the variability in gas consumption can be attributed to changes in the outside temperature. Obviously, R does provide the means to compute something as basic as $$R^2$$, but I will let you find it for yourselves. ## 5.3 Inference To perform inference on $$\hat \beta$$, in order to test hypotheses and construct confidence intervals, we need to quantify the uncertainly in the reported $$\hat \beta$$. This is exactly what Eq.(5.6) gives us. Luckily, we don’t need to manipulate multivariate distributions manually, and everything we need is already implemented. The most important function is summary which gives us an overview of the model’s fit. We emphasize that that fitting a model with lm is an assumption free algorithmic step. Inference using summary is not assumption free, and requires the set of assumptions leading to Eq.(5.6). summary(lm.1) ## ## Call: ## lm(formula = Gas ~ Temp, data = whiteside[Insul == "Before"]) ## ## Residuals: ## Min 1Q Median 3Q Max ## -0.62020 -0.19947 0.06068 0.16770 0.59778 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 6.85383 0.11842 57.88 <2e-16 *** ## Temp -0.39324 0.01959 -20.08 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.2813 on 24 degrees of freedom ## Multiple R-squared: 0.9438, Adjusted R-squared: 0.9415 ## F-statistic: 403.1 on 1 and 24 DF, p-value: < 2.2e-16 Things to note: • The estimated $$\hat \beta$$ is reported in the Coefficients’ table, which has point estimates, standard errors, t-statistics, and the p-values of a two-sided hypothesis test for each coefficient $$H_{0,j}:\beta_j=0, j=1,\dots,p$$. • The $$R^2$$ is reported at the bottom. The “Adjusted R-squared” is a variation that compensates for the model’s complexity. • The original call to lm is saved in the Call section. • Some summary statistics of the residuals ($$y_i-\hat y_i$$) in the Residuals section. • The “residuals standard error”11 is $$\sqrt{(n-p)^{-1} \sum_i (y_i-\hat y_i)^2}$$. The denominator of this expression is the degrees of freedom, $$n-p$$, which can be thought of as the hardness of the problem. As the name suggests, summary is merely a summary. The full summary(lm.1) object is a monstrous object. Its various elements can be queried using str(sumary(lm.1)). Can we check the assumptions required for inference? Some. Let’s start with the linearity assumption. If we were wrong, and the data is not arranged about a linear line, the residuals will have some shape. We thus plot the residuals as a function of the predictor to diagnose shape. plot(residuals(lm.1)~whiteside[Insul=='Before',Temp]) abline(0,0, lty=2) I can’t say I see any shape. Let’s fit a wrong model, just to see what “shape” means. lm.1.1 <- lm(Gas~I(Temp^2), data=whiteside[Insul=='Before',]) plot(residuals(lm.1.1)~whiteside[Insul=='Before',Temp]); abline(0,0, lty=2) Things to note: • We used I(Temp)^2 to specify the model $$Gas=\beta_0 + \beta_1 Temp^2+ \varepsilon$$. • The residuals have a “belly”. Because they are not a cloud around the linear trend, and we have the wrong model. To the next assumption. We assumed $$\varepsilon_i$$ are independent of everything else. The residuals, $$y_i-\hat y_i$$ can be thought of a sample of $$\varepsilon_i$$. When diagnosing the linearity assumption, we already saw their distribution does not vary with the $$x$$’s, Temp in our case. They may be correlated with themselves; a positive departure from the model, may be followed by a series of positive departures etc. Diagnosing these auto-correlations is a real art, which is not part of our course. The last assumption we required is normality. As previously stated, if $$n \gg p$$, this assumption can be relaxed. If $$n$$ is in the order of $$p$$, we need to verify this assumption. My favorite tool for this task is the qqplot. A qqplot compares the quantiles of the sample with the respective quantiles of the assumed distribution. If quantiles align along a line, the assumed distribution if OK. If quantiles depart from a line, then the assumed distribution does not fit the sample. qqnorm(resid(lm.1)) Things to note: • The qqnorm function plots a qqplot against a normal distribution. For non-normal distributions try qqplot. • resid(lm.1) extracts the residuals from the linear model, i.e., the vector of $$y_i-x_i'\hat \beta$$. Judging from the figure, the normality assumption is quite plausible. Let’s try the same on a non-normal sample, namely a uniformly distributed sample, to see how that would look. qqnorm(runif(100)) ### 5.3.1 Testing a Hypothesis on a Single Coefficient The first inferential test we consider is a hypothesis test on a single coefficient. In our gas example, we may want to test that the temperature has no effect on the gas consumption. The answer for that is given immediately by summary(lm.1) summary.lm1 <- summary(lm.1) coefs.lm1 <- summary.lm1coefficients
coefs.lm1
##               Estimate Std. Error   t value     Pr(>|t|)
## (Intercept)  6.8538277 0.11842341  57.87561 2.717533e-27
## Temp        -0.3932388 0.01958601 -20.07754 1.640469e-16

We see that the p-value for $$H_{0,1}:\hat \beta_1=0$$ against a two sided alternative is effectively 0, so that $$\beta_1$$ is unlikely to be $$0$$.

### 5.3.2 Constructing a Confidence Interval on a Single Coefficient

Since the summary function gives us the standard errors of $$\hat \beta$$, we can immediately compute $$\hat \beta_j \pm 2 \sqrt{Var[\hat \beta_j]}$$ to get ourselves a (roughly) $$95\%$$ confidence interval. In our example the interval is

coefs.lm1[2,1] + c(-2,2) * coefs.lm1[2,2]
## [1] -0.4324108 -0.3540668

### 5.3.3 Multiple Regression

Remark. Multiple regression is not to be confused with multivariate regression discussed in Chapter 8.

The swiss dataset encodes the fertility at each of Switzerland’s 47 French speaking provinces, along other socio-economic indicators. Let’s see if these are statistically related:

head(swiss)
##              Fertility Agriculture Examination Education Catholic
## Courtelary        80.2        17.0          15        12     9.96
## Delemont          83.1        45.1           6         9    84.84
## Franches-Mnt      92.5        39.7           5         5    93.40
## Moutier           85.8        36.5          12         7    33.77
## Neuveville        76.9        43.5          17        15     5.16
## Porrentruy        76.1        35.3           9         7    90.57
##              Infant.Mortality
## Courtelary               22.2
## Delemont                 22.2
## Franches-Mnt             20.2
## Moutier                  20.3
## Neuveville               20.6
## Porrentruy               26.6
lm.5 <- lm(data=swiss, Fertility~Agriculture+Examination+Education+Education+Catholic+Infant.Mortality)
summary(lm.5)
##
## Call:
## lm(formula = Fertility ~ Agriculture + Examination + Education +
##     Education + Catholic + Infant.Mortality, data = swiss)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -15.2743  -5.2617   0.5032   4.1198  15.3213
##
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)
## (Intercept)      66.91518   10.70604   6.250 1.91e-07 ***
## Agriculture      -0.17211    0.07030  -2.448  0.01873 *
## Examination      -0.25801    0.25388  -1.016  0.31546
## Education        -0.87094    0.18303  -4.758 2.43e-05 ***
## Catholic          0.10412    0.03526   2.953  0.00519 **
## Infant.Mortality  1.07705    0.38172   2.822  0.00734 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.165 on 41 degrees of freedom
## Multiple R-squared:  0.7067, Adjusted R-squared:  0.671
## F-statistic: 19.76 on 5 and 41 DF,  p-value: 5.594e-10

Things to note:

• The ~ syntax allows to specify various predictors separated by the + operator.
• The summary of the model now reports the estimated effect, i.e., the regression coefficient, of each of the variables.

Clearly, naming each variable explicitely is a tedios task if there are many. The use of Fertility~. in the next example reads: “Fertility as a function of all other variables in the swiss data.frame”.

lm.5 <- lm(data=swiss, Fertility~.)
summary(lm.5)
##
## Call:
## lm(formula = Fertility ~ ., data = swiss)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -15.2743  -5.2617   0.5032   4.1198  15.3213
##
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)
## (Intercept)      66.91518   10.70604   6.250 1.91e-07 ***
## Agriculture      -0.17211    0.07030  -2.448  0.01873 *
## Examination      -0.25801    0.25388  -1.016  0.31546
## Education        -0.87094    0.18303  -4.758 2.43e-05 ***
## Catholic          0.10412    0.03526   2.953  0.00519 **
## Infant.Mortality  1.07705    0.38172   2.822  0.00734 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.165 on 41 degrees of freedom
## Multiple R-squared:  0.7067, Adjusted R-squared:  0.671
## F-statistic: 19.76 on 5 and 41 DF,  p-value: 5.594e-10

### 5.3.4 ANOVA (*)

Our next example12 contains a hypothetical sample of $$60$$ participants who are divided into three stress reduction treatment groups (mental, physical, and medical) and two gender groups (male and female). The stress reduction values are represented on a scale that ranges from 1 to 10. The values represent how effective the treatment programs were at reducing participant’s stress levels, with larger effects indicating higher effectiveness.

twoWay <- read.csv('data/dataset_anova_twoWay_comparisons.csv')
head(twoWay)
##   Treatment   Age StressReduction
## 1    mental young              10
## 2    mental young               9
## 3    mental young               8
## 4    mental   mid               7
## 5    mental   mid               6
## 6    mental   mid               5

How many observations per group?

table(twoWay$Treatment, twoWay$Age)
##
##            mid old young
##   medical    3   3     3
##   mental     3   3     3
##   physical   3   3     3

Since we have two factorial predictors, this multiple regression is nothing but a two way ANOVA. Let’s fit the model and inspect it.

lm.2 <- lm(StressReduction~.,data=twoWay)
summary(lm.2)
##
## Call:
## lm(formula = StressReduction ~ ., data = twoWay)
##
## Residuals:
##    Min     1Q Median     3Q    Max
##     -1     -1      0      1      1
##
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)
## (Intercept)         4.0000     0.3892  10.276 7.34e-10 ***
## Treatmentmental     2.0000     0.4264   4.690 0.000112 ***
## Treatmentphysical   1.0000     0.4264   2.345 0.028444 *
## Ageold             -3.0000     0.4264  -7.036 4.65e-07 ***
## Ageyoung            3.0000     0.4264   7.036 4.65e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9045 on 22 degrees of freedom
## Multiple R-squared:  0.9091, Adjusted R-squared:  0.8926
## F-statistic:    55 on 4 and 22 DF,  p-value: 3.855e-11

Things to note:

• The StressReduction~. syntax is read as “Stress reduction as a function of everything else”.

• All the (main) effects and the intercept seem to be significant.

• The data has 2 factors, but the coefficients table has 4 predictors. This is because lm noticed that Treatment and Age are factors. Each level of each factor is thus encoded as a different (dummy) variable. The numerical values of the factors are meaningless. Instead, R has constructed a dummy variable for each level of each factor. The names of the effect are a concatenation of the factor’s name, and its level. You can inspect these dummy variables with the model.matrix command.

model.matrix(lm.2) %>% lattice::levelplot()

If you don’t want the default dummy coding, look at ?contrasts.

If you are more familiar with the ANOVA literature, or that you don’t want the effects of each level separately, but rather, the effect of all the levels of each factor, use the anova command.

anova(lm.2)
## Analysis of Variance Table
##
## Response: StressReduction
##           Df Sum Sq Mean Sq F value    Pr(>F)
## Treatment  2     18   9.000      11 0.0004883 ***
## Age        2    162  81.000      99     1e-11 ***
## Residuals 22     18   0.818
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Things to note:

• The ANOVA table, unlike the summary function, tests if any of the levels of a factor has an effect, and not one level at a time.
• The significance of each factor is computed using an F-test.
• The degrees of freedom, encoding the number of levels of a factor, is given in the Df column.
• The StressReduction seems to vary for different ages and treatments, since both factors are significant.

If you are extremely more comfortable with the ANOVA literature, you could have replaced the lm command with the aov command all along.

lm.2.2 <- aov(StressReduction~.,data=twoWay)
class(lm.2.2)
## [1] "aov" "lm"
summary(lm.2.2)
##             Df Sum Sq Mean Sq F value   Pr(>F)
## Treatment    2     18    9.00      11 0.000488 ***
## Age          2    162   81.00      99    1e-11 ***
## Residuals   22     18    0.82
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Things to note:

• The lm function has been replaced with an aov function.
• The output of aov is an aov class object, which extends the lm class.
• The summary of an aov does not like the summary of an lm object, but rather, like an ANOVA table.

As in any two-way ANOVA, we may want to ask if different age groups respond differently to different treatments. In the statistical parlance, this is called an interaction, or more precisely, an interaction of order 2.

lm.3 <- lm(StressReduction~Treatment+Age+Treatment:Age-1,data=twoWay)

The syntax StressReduction~Treatment+Age+Treatment:Age-1 tells R to include main effects of Treatment, Age, and their interactions. Here are other ways to specify the same model.

lm.3 <- lm(StressReduction ~ Treatment * Age - 1,data=twoWay)
lm.3 <- lm(StressReduction~(.)^2 - 1,data=twoWay)

The syntax Treatment * Age means “mains effects with second order interactions”. The syntax (.)^2 means “everything with second order interactions”

Let’s inspect the model

summary(lm.3)
##
## Call:
## lm(formula = StressReduction ~ Treatment + Age + Treatment:Age -
##     1, data = twoWay)
##
## Residuals:
##    Min     1Q Median     3Q    Max
##     -1     -1      0      1      1
##
## Coefficients:
##                              Estimate Std. Error t value Pr(>|t|)
## Treatmentmedical            4.000e+00  5.774e-01   6.928 1.78e-06 ***
## Treatmentmental             6.000e+00  5.774e-01  10.392 4.92e-09 ***
## Treatmentphysical           5.000e+00  5.774e-01   8.660 7.78e-08 ***
## Ageold                     -3.000e+00  8.165e-01  -3.674  0.00174 **
## Ageyoung                    3.000e+00  8.165e-01   3.674  0.00174 **
## Treatmentmental:Ageold      4.246e-16  1.155e+00   0.000  1.00000
## Treatmentphysical:Ageold    1.034e-15  1.155e+00   0.000  1.00000
## Treatmentmental:Ageyoung   -3.126e-16  1.155e+00   0.000  1.00000
## Treatmentphysical:Ageyoung  5.128e-16  1.155e+00   0.000  1.00000
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1 on 18 degrees of freedom
## Multiple R-squared:  0.9794, Adjusted R-squared:  0.9691
## F-statistic:    95 on 9 and 18 DF,  p-value: 2.556e-13

Things to note:

• There are still $$5$$ main effects, but also $$4$$ interactions. This is because when allowing a different average response for every $$Treatment*Age$$ combination, we are effectively estimating $$3*3=9$$ cell means, even if they are not parametrized as cell means, but rather as main effect and interactions.
• The interactions do not seem to be significant.
• The assumptions required for inference are clearly not met in this example, which is there just to demonstrate R’s capabilities.

Asking if all the interactions are significant, is asking if the different age groups have the same response to different treatments. Can we answer that based on the various interactions? We might, but it is possible that no single interaction is significant, while the combination is. To test for all the interactions together, we can simply check if the model without interactions is (significantly) better than a model with interactions. I.e., compare lm.2 to lm.3. This is done with the anova command.

anova(lm.2,lm.3, test='F')
## Analysis of Variance Table
##
## Model 1: StressReduction ~ Treatment + Age
## Model 2: StressReduction ~ Treatment + Age + Treatment:Age - 1
##   Res.Df RSS Df   Sum of Sq F Pr(>F)
## 1     22  18
## 2     18  18  4 -3.5527e-15

We see that lm.3 is not (significantly) better than lm.2, so that we can conclude that there are no interactions: different ages have the same response to different treatments.

### 5.3.5 Testing a Hypothesis on a Single Contrast (*)

Returning to the model without interactions, lm.2.

coef(summary(lm.2))
##                   Estimate Std. Error   t value     Pr(>|t|)
## (Intercept)              4  0.3892495 10.276186 7.336391e-10
## Treatmentmental          2  0.4264014  4.690416 1.117774e-04
## Treatmentphysical        1  0.4264014  2.345208 2.844400e-02
## Ageold                  -3  0.4264014 -7.035624 4.647299e-07
## Ageyoung                 3  0.4264014  7.035624 4.647299e-07

We see that the effect of the various treatments is rather similar. It is possible that all treatments actually have the same effect. Comparing the effects of factor levels is called a contrast. Let’s test if the medical treatment, has in fact, the same effect as the physical treatment.

library(multcomp)
my.contrast <- matrix(c(-1,0,1,0,0), nrow =  1)
lm.4 <- glht(lm.2, linfct=my.contrast)
summary(lm.4)
##
##   Simultaneous Tests for General Linear Hypotheses
##
## Fit: lm(formula = StressReduction ~ ., data = twoWay)
##
## Linear Hypotheses:
##        Estimate Std. Error t value Pr(>|t|)
## 1 == 0  -3.0000     0.7177   -4.18 0.000389 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## (Adjusted p values reported -- single-step method)

Things to note:

• A contrast is a linear function of the coefficients. In our example $$H_0:\beta_1-\beta_3=0$$, which justifies the construction of my.contrast.
• We used the glht function (generalized linear hypothesis test) from the package multcompt.
• The contrast is significant, i.e., the effect of a medical treatment, is different than that of a physical treatment.

## 5.4 Bibliographic Notes

Like any other topic in this book, you can consult Venables and Ripley (2013) for more on linear models. For the theory of linear models, I like Greene (2003).

## 5.5 Practice Yourself

1. Inspect women’s heights and weights with ?women.
1. What is the change in weight per unit change in height? Use the lm function.
2. Is the relation of height on weight significant? Use summary.
3. Plot the residuals of the linear model with plot and resid.
4. Plot the predictions of the model using abline.
5. Inspect the normality of residuals using qqnorm.
6. Inspect the design matrix using model.matrix.
2. Write a function that takes an lm class object, and returns the confidence interval on the first coefficient. Apply it on the height and weight data.
3. Use the ANOVA function to test the significance of the effect of height.

4. Read about the “mtcars” dataset using ? mtcars.Inspect the dependency of the fuel consumption (mpg) in the weight (wt) and the 1/4 mile time (qsec).
1. Make a pairs scatter plot with plot(mtcars[,c("mpg","wt","qsec")]) Does the connection look linear?
2. Fit a multiple linear regression with lm. Call it model1.
3. Try to add the transmission (am) as independent variable. Let R know this is a categorical variable with factor(am). Call it model2.
4. Compare the “Adjusted R-squared” measure of the two models (we can’t use the regular R2 to compare two models with a different number of variables).
5. Do the coefficients significant?
6. Inspect the normality of residuals and the linearity assumptions.
7. Now Inspect the hypothesis that the effect of weight is different between the transmission types with adding interaction to the model wt*factor(am)`.
8. According to this model, what is the addition of one unit of weight in a manual transmission to the fuel consumption (-2.973-4.141=-7.11)?

### References

Friedman, Jerome, Trevor Hastie, and Robert Tibshirani. 2001. The Elements of Statistical Learning. Vol. 1. Springer series in statistics Springer, Berlin.

Venables, William N, and Brian D Ripley. 2013. Modern Applied Statistics with S-Plus. Springer Science & Business Media.

Greene, William H. 2003. Econometric Analysis. Pearson Education India.

1. The “response” is also know as the “dependent” variable in the statistical literature, or the “labels” in the machine learning literature.

2. The “factors” are also known as the “independent variable”, or “the design”, in the statistical literature, and the “features”, or “attributes” in the machine learning literature.

3. The “error term” is also known as the “noise”, or the “common causes of variability”.

4. You may philosophize if the measurement error is a mere instance of unmodeled factors or not, but this has no real implication for our purposes.

5. By “computed” we mean what statisticians call “fitted”, or “estimated”, and computer scientists call “learned”.

6. Sometimes known as the Root Mean Squared Error (RMSE).

7. The example is taken from http://rtutorialseries.blogspot.co.il/2011/02/r-tutorial-series-two-way-anova-with.html